MANISH KUMAR
MATHEMATICS
CIRCLES CHORD PROPERTIES OF A CIRCLE
CIRCLE A circle is the locus of a point which moves in a place in such a way that its distance from a fixed point remains constant. The fixed point is called the centre and the constant distance is called the radius of the circle. The given figure consists of a circle with centre O and radius equal to r units.
TERMS AND FACTS RELATED TO CIRCLES.
Radius : A line segment joining the centre and a point on the circle is called its radius, generally denoted by r In the adjoining figure, OA, OB and OC are the radii of a circle.
Circumference : The perimeter of a circle is called its circumference. Circumference = 2r
Position of a Point With Respect To a Circle Let us consider a circle with centre O and radius r. (i) inside the circle, if OP < r: (ii) (iii)
on the circle, if OP = r. outside the circle, if OP > r.
In the adjoining figure of a circle with centre O and radius r.
(i) (ii)
The points A, O, B lie inside the circle. The points P, Q, R lie on the circle;
(iii)
The points X, Y, Z lie outside the circle.
Interior and Exterior of a Circle. The region consisting of all those points which lie inside a circle, is called the interior of the circle. The region consisting of all those points which lie outside a circle, is called the exterior of the circle.
Circular Region or Circular Disc The region consisting of all those points which are either on the circle or lie inside the side, is called the circular region.
Chord : A line segment joining any two points on a circle is called a chord of the circle. In the adjoining figure, PQ, RS and AB are the chords of a circle with centre O.
MANISH KUMAR
MATHEMATICS
Diameter : A chord of the circle passing through the centre of a circle is called its diameter. In the figure. AOB is a diameter of a circler with centre O. Diameter = 2 × Radius Properties : (i) Diameter is the largest chord of a circle. (ii) All diameters of a circle are equal in length.
Secant : A line which intersects a circle in two distinct points in called a secant of the circle. In the adjoining figure, the line cuts the circle in two points C and D. So, is a secant of the circle.
Tangent : A line that intersects the circle is exactly one point is called a tangent of the circle. The point at which the tangent intersects the circle is called its point of contact. In the adjoining figure, SPT is a tangent at the point P of the circle with centre O. Clearly, P is the point of contact of the tangent with the circle. Facts About Tangents : (i)
No tangent can be drawn to a circle through a point inside the circle:
(ii)
One and only one tangent can be drawn to a circle at a point on the circle.
(iii)
Two tangents can be drawn to a circle from a point outside it. In the adjoining figure, PT1 and PT2 are the tangents to the circle from point P.
Touching Circles : Two circles are said to touch each other if and only if they have one and only one point in common. Two circles may touch externally (Fig. (i) o internally [Fig (ii)].
The common point is called the point of contact, and the line joining their centre is called the line of centre. A line touching the two circles is called common tangent. Thus, in the above figure, P is the point of contact, AB is the line of centers are PT is a common tangent.
Direct Common Tangents : A common tangent to two circles is called a direct common tangent if both the circle lie on the same side of it. In the adjoining figure, AB and CD are two direct common tangents.
MANISH KUMAR
MATHEMATICS
Transverse Common Tangents : A common tangent to two circles is called a transverse common tangent if the circles lie on its opposite sides. In the adjoining figure, PQ and RS are two transverse common tangents.
Arc : A continuous piece of a circle is called on arc of the circle. Let P and Q be any two points on a circle with centre O. Then, clearly the whole circle has been divided into two pieces. namely arc PAQ and arc QBP, to be denoted by PAQ and QBP respectively. We any denote them by PQ and QP respectively.
Minor and Major Arc: An arc less than one-half of the whole arc of a circle is called a minor arc, and an arc greater than one-half of the whole are of a circle is called a major arc of the circle. Thus, in the above figure, PQ is a minor arc, while QP is a major arc.
Central Angle : An angle subtended by an arc at the centre of a circle is called its central angle. In the given figure, central angle of PQ = POQ.
Degree Measure of An Arc : Let PQ be an arc of a circle with centre O. If POQ = 0 , we say that the degree measure of PQ is 0 and we write, m(PQ) 0 . If m(PQ)= 0 , then m(PQ) = (360 - 0 )0. Degree measure of a circle is 3600
Congruent Arcs : Two arcs AB and CD are said to be congruent, if they have same degree measure.
AB CD m( AB) m(CD) AOB = COD.
Semi-Circle : A diameter divides a circle into two equal arcs. Each of these two arcs is called a semi-circle. The degree measure of a semi-circle is 1800. In the given figure of a circle with centre O, ABC as well as ADC is semi-circle.
MANISH KUMAR
MATHEMATICS
Congruent Circles : Two circles of equal radii are said to be congruent.
Concentric Circles : Circles having same centre but different radii are called concentric circles.
Concyclic Points : The points, which lie on the circumference of the same circle, are called concyclic points. In the adjoining figure, points A, B, C and D lie on the same circle and hence, they are concyclic.
Segment : A segment is a part of circular region bounded by an arc and a chord, including the arc and the chord. The segment containing the minor arc is called a minor segment, while the other one is a major segment. The centre of the circle lies in the major segment. Alternate Segments of a Circle : The minor and major segments of a circle are called alternate segments of each other.
Sector of a Circle : The part of the plane region enclosed by an arc of a circle and its two bounding radii is called a sector of the circle. Thus, the region OABO is the sector of a circle with centre O.
Quadrant : One-fourth of a circular disc if called a quadrant.
Cyclic Quadrilateral : If all the four vertices of a quadrilateral lie on a circle, then such a quadrilateral is called a cyclic quadrilateral.
If four points lie on a circle, they are said to be concyclic.
MANISH KUMAR We also say the quad. ABCD is inscribed in a circle with centre O.
MATHEMATICS
MANISH KUMAR Theorem 1.
MATHEMATICS
The straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is
perpendicular to the chord. Given : AB is a chord, other than the diameter of a circle with centre O and OL bisects AB. To prove : OL AB. Construction : Join OA and OB. Proof :
STATEMENT 1.
REASON
In OLA and OLB, we have
Radii of the same circle
(i) OA = OB
Given, OL bisects AB
(ii) AL = BL
Common
(iii) OL = OL
SSS-Axiom of congruence
OLA OLB
Corresponding part of congruent s are congruent.
OLA = OLB
2.
OLA + OLB = 1800 ....(II)
3.
OLA = OLB = 900
....(I)
ALB is a straight line From (I) and (II).
Hence, OL AB Theorem 2. (Converse of Theorem 1): The perpendicular to a chord from the centre of a circle bisects the chord. Given : AB is a chord of a circle with centre O and OL AB. To prove : LA = LB. Construction : Join OA and OB. Proof :
STATEMENT
REASON
In OLA and OLB, we have (i)
OA = OB
Radii of the same circle
(ii)
OLA = OLB
Each equal to 900, since OL AB
(iii)
OL = OL
Common
OLA OLB
RHS axiom of congruency
LA = LB
Hence, La = LB
c.p.c.t.
MANISH KUMAR
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Theorem 3. One and only one circle can be drawn, passing through three non-collinear points. Given : Three non-collinear points A, B, C. To prove : One and only one circle can be drawn, passing through A, B and C. Construction : Join AB and BC. Draw the perpendicular bisector of AB and BC, meeting at a point O. Proof :
STATEMENT
REASON
1.
O lines on the perpendicular bisector of AB
OA = OB
2.
O lies on the perpendicular bisector of BC
From A and B.
OB = OC
Each point on perpendicular bisector of BC is equidistant
3.
OA = OB = OC
O is equidistant from A, B and C
Any circle drawn with centre O and
Each point on perpendicular bisector of AB is equidistant
...(i)
...(ii)
From B and C
radius OA will pass through B and C also. 4.
O is the only point equidistant from A, B and C
From (i) and (ii)
Perpendicular bisector of AB and BC cut each other at point O only.
Hence, one and only circle can be drawn to pass through three non-collinear points, A B and C.
Theorem 4.
Equal chords of circle are equidistant from the centre.
Given : A circle with centre O in which chord AB = chord CD; OL AB and OM CD. To prove : OL = OM. Proof :
STATEMENT 1. 2. 3.
AL
1 AB 2
…..(i)
1 CD 2 Now, AB = CD CM
…..(ii)
REASON Perpendicular from centre bisects the chord. Perpendicular from centre bisects the chord. Given Halves of equals are equal. From (i) and (ii).
1 1 AB = CD 2 2
AL = CM
4.
In OLA and OMC, we have
(i)
OA = OC
(ii)
AL = CM
(iii)
OLA = OMC
OLA OMC
....(iii) Radii of the same circle. From (iii). Each equal to 900, as OL AB and OM CD. RHS-axiom of congruency of s. c.p.c.t.
MANISH KUMAR
MATHEMATICS
OL = OM
Hence, the chords AB and CD are equidistant from the centre O. Theorem 5. (Converse of Theorem 4) : Chords of a circle that are equidistant from the centre of the circle, are equal. Given : AB and CD are two chords of a circle with centre O. OL AB, OM CD and OL = OM. To prove : AB = CD Construction : Join OA and OC. Proof :
STATEMENT
REASON
In OLA and OMC we have (i)
OL = OM
Given,
(ii)
OA = OC
Radii of the same circle.
(iii)
OLA = OMC
Each equal to 900, as OL AB and OM CD.
OLA = OMC
RHS-axiom of congruency of s.
AL = CM
c.pc.t.
1 1 AB CD 2 2
AB = CD
Perpendicular from centre bisect the chord Doubles of equal are equal.
Hence, chord AB = chord CD
Ex.1
A chord of length 30 cm is drawn in a circle of radius 17 cm. Find its distance from the centre o the circle.
Sol.
Let AB be a chord of a circle with centre O and radius = 17 cm such that AB = 30 cm. From O, draw OL AB. Join OA. Since perpendicular to the chord from the centre bisects the chord, we have AL =
1 1 AB 30 cm 15 cm. 2 2
OA = Radius of the circle = 17 cm. From right-angles OLA, we have
OL OA 2 AL2
MANISH KUMAR (17 )2 (15)2 289 225 64 8cm. Hence, the distance of the chord from the centre is 8 cm.
MATHEMATICS
MANISH KUMAR Ex.2
MATHEMATICS
In the figure, the diameter CD of a circle with centre O is perpendicular to the chord AB. If AB = 12 cm and CE = 3 cm, find the radius of the circle.
Sol.
Join OA. Since perpendicular from the centre to a chord, bisects the chord, we have AE = EB = 6 cm. Let r be the radius of the circle. The, OA = OC = r cm. OE (OC - CE) = (r - 3) cm. From right angles OEA, we have OA2 = AE2 + OE2
[By Pythagoras Theorem]
r2 = (6)2 + (r - 3)2
r2 = 36 + r2 - 6r + 9
6r = 45
r = 7.5.
[ OA = r cm, AE = 6 cm & OE = (r - 3) cm]
Hence, the radius of the circle is 7.5 cm. Ex.3
AB and CD are two parallel chords of a circle of length 24 cm and 10 cm respectively and lie on the same side of its centre O. If the distance between the chords is 7 cm, find the radius of the circle.
Sol.
Draw OL AB and OM CD. Since AB || CD, it follows that O, L, M are collinear. Now, OL AB and OM CD
AL
and CM =
1 1 AB 24 cm 12 cm 2 2
1 1 CD 10 cm 5 cm. 2 2
Also, distance between the chords = LM = 7 cm. Join OA and OC. Let OL = x cm. Then, OM = (x + 7) cm. Let the radius of the circle be r cm. Then, OA = OC = r cm. Now, from right-angles OLA and OMC, we have OA2 = OL2 + AL2 and OC2 = OM2 + CM2
r2 = x2 + (12)2 .... (i) and r2 = (x + 7)2 + (5)2
x2 + (12)2 = (x + 7)2 + 52
x2 + 144 = x2 + 14x + 74
14x = 70
x=5
.....(ii)
[From (i) and (ii)]
Substituting = 5 in (i), we get : r2 = x2 + (12)2 = (5)2 + (12)2 = (25 + 144) = 169
MANISH KUMAR
r=
MATHEMATICS
169 = 13 cm.
Hence, the radius of the circle is 13 cm. Ex.4
AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle.
Sol.
Let O be the centre of the circle and let its radius be r cm. Draw OL AB and OM CD Then AL =
1 1 AB = 5 cm and CD = CD = 12 cm. 2 2
Since AB || CD, it follows that the points O, L, M are collinear and therefore, LM = 17 cm. Let OL = x xm. Then, OM = (17 - x) cm. Join OA and OC. Then, OA = OC = r cm. Now, from right-angles OLA and OMC, we have OA2 = OL2 + AL2 and OC2 = OM2 + CM2 [By Pythagoras Theorem]
r2 = x2 + 52 ....(i) and r2 = (17 - x)2 + (12)2 .....(ii)
x2 + 52 = (17 - x)2 + (12)2
x2 + 25 = x2 - 34x + 433
34x = 408
x= 12.
[From (i) and (ii)]
Substituting x = 12 in (i), we get : r2 = (12)2 + 52 = (144 + 25) = 169
r=
169 = 13 c,
Hence, the radius of the circle is 13 cm. Ex.5
In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of BAC.
Sol.
Given : AB and AC are equal chords of circle with centre O and O has been joined with A. To prove : BAO = CAO. Construction : Join OB and OC.
Proof :
REASON
STATEMENT 1.
In OAB and OAC, we have
(i)
AB = AC
Given
(ii)
OB = OC
Radii of the same circle.
(iii)
OA = OA
Common.
OAB OAC
SSS-axiom of congruency.
MANISH KUMAR
BAO = CAO
MATHEMATICS c.p.c.t.
Hence, the centre of the circle lies on the bisector of BAC.
MANISH KUMAR Ex.6
MATHEMATICS
In the adjoining figure, AB and AC are two equal chords of circle of radius 5 cm. IF AB = AC = 6 cm, find the length of chord BC.
Sol.
Let O be the centre of the circle, Join OB. Let AD be the bisector of BAC, meeting BC and D. In s BAD and CAD, we have AB = AC [Given]
BAD = CAD
[By contraction]
AD = AD
[Common]
BAD CAD
[SAS-axiom of congruency]
BDA = CDA [c.p.c.t.]
But, BDA = + CDA = 1800
[ BDC is a straight line]
BDA = CDA = 900
This shows that AD in the perpendicular bisector of BC and so, it when produced passes through O.
OB = OA = 5 cm.
Let OD = x cm and BD = y cm. The, AD = (OA - OD) = (5 - x)cm. From right angles OBD and ADB, we get: OB2 = OD2 + BD2 and AB2 = AD2 + BC2
52 = x2 + y2 and 62 = (5 - x)2 + y2
y2 = 25 - x2
.....(i) and y2 = 36 - (5 - x)2 .....(ii)
From (i) and (ii), we get: 25 = x2 = 36 - (5 - x)2
10x = 14 i.e. x = 1.4.
Substituting x = 1.4 in (i) we get : y2 = 25 - (1.4)2 = 23.04 y =
23.04 = 4.8 cm.
BD = 4.8 cm.
BC = 2 × BC = (2 × 4.8)cm = 9.6 cm
Hence, the length of chord BC = 9.6 cm.
MANISH KUMAR
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Ex.7
If two equal chords of a circle intersect, then prove that their segments are equal.
Sol.
Given : Two equal chords AB and CD of a circle with centre O, intersect at a point P. To prove : BP = CP and AP = DP. Construction : Draw OL AB, OM CD. Join OP. Proof :
STATEMENT 1.
AL = BL =
1 AB 2 1 CD 2
2.
DM = CD =
3.
AL = DM
...(i)
BL = CM
...(ii)
REASON Perpendicular from centre bisects the chord. Perpendicular from centre bisects the chord. From (i) and (ii).
Equal chords are equidistant from the centre,
4.
In right s OLP and OMP, we have:
Each equal to 900
(i)
OL = OM
RHS-axiom of congruence.
(ii)
OLP = OMP
c.p.c.t
(iii)
OP = OP
From (i) and (iii), adding the corresponding sides.
PLP OMP
LP = MP
5.
AL + LP = CM + MP
AP = DP
6.
BL - LP = CM - MP
BP = CP
....(iii)
From (ii) and (iii)
Hence, BP = CP and AP = DP
ANGLE PROPERTIES OF A CIRCLE Theorem-6.
The angle subtended by an arc of a circle at the centre is double the angle subtended by its an
any point on the remaining part of the circle. Given : A circle with centre O and an arc AB subtends AOB at the centre and ACB at any point C on the remaining part of the circle. To prove : AOB = 2ACB. Construction : Join CO and produce it to some point D.
MANISH KUMAR
MATHEMATICS STATEMENT
1.
2.
In AOC, we have OA = OC
Radii of the same circle.
Angle opposite to equal sides of a are equal.
OAC = OCA ....(i)
AOD = OAC + OCA
Ext. angle of a = Sum of its into opp. s.
= OCA + OCA
Using (i)
= 2 OCA 3.
....(ii)
Similarly, we have Adding the corresponding sides of (ii) and (iii).
BOD = 2OCB 4.
REASON
....(iii)
In figure (i), we have :
Adding the corresponding sides of (ii) and (iii).
AOD + BOD = 2OCA + 2OCB = 2(OCA + OCB) = 2ACB
Subtracting the corresponding sides of (iii) and (ii).
AOB = 2ACB. In Figure (iii), we have :
AOD + BOD = 2OCA + 2OCB. =2ACB In figure (ii), we have.
BOD - AOD = 2OCB - 2OCA = 2(OCB - OCA) = 2ACB
AOB = 2ACB.
Hence, AOB = 2ACB. Remark : In the above theorem, fig (i) refers to minor arc while fig. (iii) refers to major arc. Theorem-7 : Angles in the same segment of a circle are equal. Given : A circle with centre O and two angles ACB and ADB in the same segment of the circle. To prove : ACN = ADE/ Construction : Join OA and OB.
(i)
(ii )
MANISH KUMAR
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Proof :
STATEMENT
REASON
In fig (i) : 1.
Arc AB subtends AOB at the centre and
ACB at a point C of the remaining part Angle at the centre is double the angle at any point on
of the circle.
AOB - 2ACB 2.
...(i)
remaining part of the circle.
Arc AB subtends AOB at the centre and
ADB at a point D on the remaining part of the circle.
Same as above
AOB = 2ADB 3.
....(ii)
From (i) and (ii)
2ACB = 2ADB
ACB = ADB 4.
Similarly, in fig (ii)
ACB = ADB
1 reflex AOB 2
ACB = ADB.
Hence, the angels in the same segment of a circle are equal. Theorem-8. The angle in a semi-circle is a right angle. Given : A semi-circle ACB of a circle with centre O. To prove : ACB = 900
STATEMENT 1.
REASON
Arc AB subtends AOB at the centre
Angle at the centre is double the angle at any point on
and ACB at a point C on the remaining
remaining part of the circle.
part of the circle. AOB is a straight line.
AOB =2ACB
From (i) and (ii)
1 ACB = AOB 2
...(i)
2.
AOB = 1800
....(ii)
3.
ACB = 1800 = 900
1 2
MANISH KUMAR
MATHEMATICS
Hence, the angle in a semi circle is a right angle. Theorem-9.
(Converse of Theorem 8) : if an arc of a circle subtends a right angle at any point on the
remaining part of the circle, then the arc is a semi-circle. Given : A circle with centre O and an arc AB subtending ACB at a point C on the remaining part of the circle such that
ACB 900. To prove : Arc AB is semi-circle. Construction : Join OA and OB Proof :
STATEMENT 1.
REASON
Arc AB subtends AOB at the centre
Angle at the centre is double the angle at a point of the
and ACB at a point C on the
remaining part of the circle.
remaining part of the circle.
AOB = 2ACB
...(i)
Given From (i) and (ii)
2.
ACB = 900
3.
AOB = (2 × 90 ) = 180
AOB is a straight line
AOB is a diameter
Arc AB is a semi-circle.
Chord AB passes through the centre O. 0
0
Hence, arc AB is a semi-circle Theorem-10. The opposite angles of a quadrilateral inscribed in a circle are supplementary. OR The sum of the opposite angles of a cyclic quadrilateral is 1800 Given : A quadrilateral ABCD inscribed in a circle with centre O. To prove : ADC + ABC = 1800 and BAD + BCD = 1800 Construction : Join OA and OC. Proof :
MANISH KUMAR
MATHEMATICS
STATEMENT Arc ABC subtends AOC at the centre and ADC at a point D on the remaining part of the circle. AOC = 2ADC 1 ADC = AOC ...(i) 2 Major arc CDA subtends reflex AOC at the centre and ABC at a point B on the remaining part of the circle. reflex AOC = 2AOC 1 ABC = reflex AOC .....(ii) 2 From (i) and (ii), we get ADC + ABC 1 1 = AOC + reflex AOC 2 2 1 = (AOC + reflex AOC) 2
1.
2.
3.
1 = 3600 = 1800 2
REASON
Angle at the centre is double the angle at any point on remaining part of the circle.
Same as above
(AOC + reflex AOC) =sum of the angles around a point O = 3600
ADC + ABC = 1800 Similarly BAD + BCD = 1800
4.
Hence, the opposite angles of a cyclic quadrilateral are supplementary. Theorem-11. (Converse of Theorem 10) : If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. Given : A quadrilateral ABCD in which B + D = 1800 To prove : ABCD is a cyclic quadrilateral. Construction : if possible, let ABCD be not a cyclic quadrilateral . Draw a circle passing through three non-collinear points A, B, C. Suppose this circle meets
(i)
CD or CD produced at D’, as shown in Fig. (i) and
(ii )
Fig. (ii) respectively. Join D’A. Proof :
STATEMENT 1. 2. 3.
B + D = 180 B + D = 1800 B + D = B + D D = D’
4.
But, this is not possible.
0
REASON Given. ABCD is a cyclic quadrilateral and so its opposite s are supplementary. From (i) and (ii) An exterior angle of a is never equal to its into opp.
MANISH KUMAR
MATHEMATICS
Our supposition is wrong.
angle.
Hence, ABCD is a cyclic quadrilateral. Theorem-12. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Given : A cyclic quadrilateral whose side AB is produced to a point E. To prove : CBE = ADC Proof :
STATEMENT 1. 2. 3.
REASON
ABC + ADC = 180 ABC + CBE = 1800 ABC + ADC = ABC + CBE ADC = CBE
ABCD is a cyclic quadrilateral and so the sum of its opp. s is 1800 ABE is a straight line. From (i) and (ii). ABC is common to both sides.
0
Hence, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Ex.8
In the given circle with diameter AB, find the value of x.
Sol.
ADB = 900
[Angle in a semi-circle]
ABD = ACD = 30
[s in the same segment]
0
In ADB, we have :
BAD + ADB + ABD = 1800 (Sum of the s of a is 1800)
x + 900 + 300 = 1800
x = (1800 - 1200) = 600
Ex.9
Hence, x = 600 If O is the centre of the circle, find the value of x in each of the following figure, giving reasons.
Sol.
We have:
(i)
(ii )
(i) BDC, = BAC = 350
DBC + BCD + BDC = 180
0
(iii )
[s in the same segment]. [Sum of the s of a is 1800]
700 + x0 + 350 = 1800 x0 = (1800 - 1050) = 750 Hence x = 750. (ii)
BCD = 900.
In BCD, we have
[Angle is semi-circle]
(iv )
MANISH KUMAR
MATHEMATICS
BCD + CDB + DBC = 1800
[Sum of the s of a is 1800]
900 + CDB + 550 = 1800
CDB = (1800 - 1450) = 350
x = BAC = CDB = 350
[s in the same segment]
0
Hence, x = 35 . (iii)
Reflex AOC = (360 - 110)0 = 2500 Major arc CA subtends reflex AOC at the centre and ABC at a point B on the remaining part of the circle.
ABC =
1 1 reflex AOC = × 2500 = 1250 2 2
Hence, x = 1250. (iv)
Since ABD is a straight line, we have
ABC + CBD = 1800
ABC + 700 = 1800
ABC = (1800 - 70)0 = 1100.
Reflex AOC = 2ABC = (2 × 1100) = 2200 [Major arc CA subtends reflex AOC at the centre and ABC at a point B on remaining part of the circle.]
x = 2200 Ex.10 In the adjoining figure, AB is a diameter of a circle with centre O and CD || BA. If BAC = 200, find the value of (i) BOC Sol.
(i)
(ii)
(ii) DOC
(iii) DAC
(IV) ADC.
Arc BC subtends BOC at the centre and BAC at the circumference.
BOC = 2BAC = (2 × 200) = 400
[ Angle at the centre is double the angle at circumference]
OCD = BOC = 400
[Alt. into s as CD || BA]
Now, OC = OD
ODC = OCD = 40
[Radii of the same circle] 0
In OCD, we have
DOC + OCD = 40 In OCD, we have
DOC + OCD + ODC = 1800
DOC + 400 + 400 = 1800
DOC = (1800 - 800) = 1000.
[Sum of the s of a is 1800]
MANISH KUMAR
MATHEMATICS 1 1 DOC = 1000 = 500 2 2
(iii)
DAC =
(iv)
ACD = CAB = 200
[Angle at the centre is double the at circumference] [Alt. Int s, as CD || BA\
In ACD, we have
ADC + ACD + DAC = 1800
ADC + 200 + 500 = 1800
ADC = (1800 - 700) = 1100
[Sum of the s of a is 1800]
Hence, BOC = 400, DOC = 1000, DAC = 500 and ACD = 1100. Ex.11 Two circles with centers O and O’ intersect at points A and B. AC and AD are their diameters respectively. Prove that the points C, B and Dare collinear. Sol.
Given: Two circle with centre O and O’ intersect at points A and B; AC is a diameter of circle with centre of O and AD is a diameter of circle with centre O’. To Prove :
The points C, B, D are collinear.
Construction : Join AB, CB and BD. Proof :
STATEMENT
REASON
1.
CBA = 900
Angle in a semi-circle
2.
DBA = 900
Angle in a semi-circle.
3.
CBA + DBA = 1800
Adding 1 and 2
CBD is a straight line C, B, D are collinear
Hence, the points C, B and D are collinear
Ex.12 Prove that every cyclic parallelogram is a rectangle. Sol.
Given : ABCD is a cyclic parallelogram. To prove : ABCD is a rectangle. Proof :
STATEMENT
REASON
MANISH KUMAR
MATHEMATICS
1.
ABC = ADC
Opposite s of a ||gm are equal.
2.
ABC + ADC = 1800
Sum of opposite s of a cyclic quad. is 180.
3.
ABC = ADC = 900
From (i) and (ii).
ABCD is a rectangle
A || gm one of whose s is 900 is a rectangle.
Hence, every cyclic parallelogram is a rectangle.
Ex.13 Prove that an isosceles trapezium is always cyclic. Sol.
Given :
A trapezium ABCD in which AB || DC and AD = BC.
To prove :
ABCD is a cyclic trapezium.
Construction : Draw DE AB and CF AB. Proof :
MANISH KUMAR
MATHEMATICS
STATEMENT 1.
2.
3. 4.
REASON
In DEA and CFB, we have (i) AD = BC (ii) DEA = CFB = 900 (iii) DE = CF DEA CFB A = B ....(i) and ADE = BCF ...(ii) ADE = BCF ADE + 900 = BCF + 900 ADE + CDE = BCF + DCF D = C ...(iii) A = B + and C = D A + B + C + D = 3600 2 (B + D) = 3600 B + D = 1800 Sum of a pair of opp. s of quad. ABCD is 1800 ABCD is a cyclic trapezium,
Given DE AB and CF AB Distance between parallel lines remains constant. RHS axiom of congruency. c.p.c.t. c.p.c.t. From (ii)
ADE + CDE = D, BCF + DCF = C. From (i) and (iii) Sum of the s of quadrilateral is 3600 Using (3)
Hence, ABCD is a cyclic trapezium. Ex.14 Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral ABCD is also cyclic. Sol.
Given : A cyclic quadrilateral ABCD in which AP, BP, CR and DR are the bisector of A, B, C and D respectively , forming a quadrilateral PQRS. To prove : PQRS is a cyclic quadrilateral Proof :
STATEMENT 1.
APB + PA B + PBA = 180
1 1 A + B = 1800 (i) 2 2 CRD + RCD + RDC = 1800 1 1 CRD + C + D = 1800 ....(ii) 2 2 APB + CRD + 1 (A + B + C + D) = 3600 2 APB + CRD + × 3600 = 3600 APB + CRD = 1800 Sum of a pair of opposite s of quad, PQRS is 1800 PQRS is a cyclic quadrilateral.
2.
3.
REASON 0
APB +
Sum of the s of PAB is 1800 1 1 PAB = A and PBA = B. 2 2 Sum of the s of RCD is 1800 1 1 RCD = C and RDC = D. 2 2 Adding (i) and (ii) A + B + C + D = 3600
MANISH KUMAR
MATHEMATICS
Hence, PQRS is a cyclic quadrilateral.
ARC PROPERTIES OF A CIRCLE Theorem-13.
In equal (or in the same circle), if two arcs subtend equal angles at the centre, they are equal.
Given : Two equal circles C1 and C2 with O and O. as their centre respectively. AB subtends OB and CD subtends CO’D such that AOB = CO’D To prove AB = CD Proof :
STATEMENT 1.
REASON
Place circle C1 on circle C2 such that
OA = O’C (Radii of equal circles).
O falls on O’ and OA falls along O’C.
AOB’ = CO’D (Given)
2.
Then, A falls on C and OB falls along O’D.
OB = O’D (Radii of equal circles)
3.
Clearly, B falls on D.
A falls on C, B falls on D and AB falls along CD, as
AB completely coincides with CD.
circles are equal.
Hence, AB=CD Theorem-14. In equal circles (or in the same circle), if two arcs are equal, they subtend equal angles at the centre. Given : Two equal circle C1 and C2 with O and O’ as their respective centre such that AB = CD To prove: AOB - CO’D Proof :
STATEMENT 1.
REASON
Place circle C1 on circle C1 such that A falls on C, AO falls along CO’ and
2.
AB = CD
AO = CO’ (Radii of equal circles.)
Then, O falls on O’ and B falls on D.
and AB = CD (Given)
OB falls on O’D 3.
Sector AOB completely coincides with sector CO’D.
A falls on C, O falls on O’ and B falls on D.
MANISH KUMAR
AOB = CO’D.
Hence AOB = CO’D.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Theorem-15. In equal circles (or in the same circle), if two chords are equal, they cut off equal arcs. Given : Two equal circle C1 and C2 with centers O And, chord AB = chord CD To prove : AB CD Proof :
STATEMENT
REASON
Case - I When AB and CD are Minor Arcs 1.
2.
In OAB and O’CD, we have (i)
OA = O’C
Radii of equal circles
(ii)
OB = O’D
Radii of equal circles.
(iii)
AB = CD
Given,
OAB O’CD
SSS-axiom of congruency.
AOB = CO’D
c.p.c.t.
AB = CD
...(i)
Case -II When AB and CD are Major Arcs
In equal circles, two arcs subtending equal s at the centre, are equal
In this case, BA and DC are Minor Arcs.
AB = CD BA = DC
BA = DC
AB = CD
Chord AB = chord BA, chord CD = chord DC Result being true for Minor Arcs Equal arcs subtracted from equal circles give equal arcs.
Case-III. When AB and CD are diameters
Si mi-circles of equal circles are equal.
In this case, AB and CD are semi-circles.
AB = CD
Hence, chord AB = chord CD AB = CD
Theorem-16. In equal circles (or in the same circle), if two arcs are equal, then their chords are equal. Given : Two equal circles C1 and C2 with centers O and, O’ respectively and AB = CD To prove : Chord AB = Chord CD. Construction : Join OA, OB, O’ C and O’D.
MANISH KUMAR Proof :
MATHEMATICS
MANISH KUMAR
MATHEMATICS
STATEMENT
REASON
Case-I, When AB and CD are Minor Arcs 1.
AB = CD AOB = CO’D
2.
In OAB and O’CD, we have
Equal arcs of equal circles subtend equal angles at the
(i)
OA = O’C
centre.
(ii)
OB = O’D
(iii)
AOB = CO’D
...(i)
Radii of equal circles. Radii of equal circles. From (i)
OAB O’CD
Chord AB = chord CD
....(ii)
SAS-axiom of congruency. c.p.c.t.
Case-II, When AB and CD are Major Arcs In this case BA and DC are Minor Arcs. Now, AB = CD
BA = DC
BA = DC
B = CD
Chord BA = chord AB, chord DC = chord CD.
Case-III. When AB and CD are semi-circles Diameters of equal circles are equal.
In this case, AB and CD are diameters.
AB = CD
Hence, in all the case AB = CD chord AB = chord CD. Ex.15 In the given diagram, O is the centre of the circle and chord AB = chord BC. (i) What is the relation between arc AB and arc BC ? Sol.
(ii)
What is the relation between AOB and BOC ?
(i)
Since equal chords is a circle cut equal arcs, so chord AB = chord BC arc AB = arc BC.
(ii)
Equal arcs in a circle make equal angles at the centre.
chord AB = chord BC arc AB = arc BC AOB = BOC.
Ex.16 In the adjoining figure A, D, B, C are four points on the circumference of a circle with centre O. Arc AB = 2 Arc BC and AOB = 1080, find : (i) ACB Sol.
(ii) CAB
(iii) ADB
Let BOC = x0. Them AOB = 2x0 Now 2x0 = 108 x = 54.
BOC = 540 and AOB = 1080 (i)
ACB =
1 1 AOB = × 1080 = 540 2 2
[ Angle at the centre is double the at a point on the
circumference] (ii)
CAB =
1 1 1 CPB = × BOC = × 540 = 270 2 2 2
MANISH KUMAR
MATHEMATICS
1 1 1 reflex AOB = (3600 - 1080 (3600 - 1080) = × 2520 = 1260 2 2 2 Ex.17 The adjoining figure shows a pentagon inscribed in a circle with centre O. Given AB = BC = CD and (iii)
ADB =
ABC = 1300 Find (i) AEB (ii) AED Sol.
(i)
(iii) COD.
ABCE is a cyclic quadrilateral.
ABC + AEC = 1800
1300 + AEC = 1800 AEC = (1800 - 1300) = 500 AEB + BEC = 500 2AEB = 500
[ BEC = AEB, since equal chords subtend equal angles at a point on the
circumference]
AEB = 250 (ii)
CD = AB
CED = AEB = 250
[Equal chords subtend equal s at a point on the circumference]
CED = 250 AED = AEC + CED = (500 + 250) = 750 (iii)
COD = 2CED 0
[ Angle at the centre is double the at a point on the circumference]
0
= (2 × 25 ) = 50 .
COD = 500 Ex.18 In the given figure, O is the centre of the circle. Chord AB is parallel to chord CD and CB is a diameter. Prove that : arc AC = arc BD. Sol.
Given : Chord AB || chord CD and COB is a diameter. To prove : Arc AC = Arc BD. Construction : Join OA and OD. Proof :
STATEMENT 1.
AOC =2ABC
2.
ABC =
Angle at the centre is double the angle at a point
1 AOC 2
...(i)
on the circumference.
BOD = 2BCD BCD =
3.
REASON
1 BOD 2
But ABC = BCD
Same as above ...(ii) Alternate s, as AB || CD From (i) and (ii)
MANISH KUMAR
1 1 AOC BOD 2 2
AOC = BOD
arc AC = arc BD
MATHEMATICS
In a circle, the arcs subtending equal s at the centre are equal
Ex.19 If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal. Sol.
Given : A cyclic quadrilateral ABCD in which AB = DC. To Prove : Diagonal AC = Diagonal BD. Proof :
STATEMENT
REASON
1.
BAC = BDC ...(i)
s in the same segment of a circle.
2.
CAD = ADB ....(ii)
Equal arcs CD and AB subtend equal s at the
3.
BAC + CAD = BDC + ADB
circumference
BAD = ADC
Adding (i) and (ii)
BD = AC
Hence, diagonal AC = diagonal BD.
Equal s on the same circle cut off equal chords.
MANISH KUMAR
EXERCISE
MATHEMATICS
CIRCLE
SUBEJCTIVE TYPE QUESTIONS. (A)
CHORD PROPERTIES OF A CIRCLE
1.
A chord of length 16 cm is drawn is a circle of radius 10 cm. Calculate the distance of the chord from the centre of the circle.
2.
A circle of radius 2.5 cm has a chord of length 4.8 cm. Find the distance of the chord from the centre of the circle.
3.
The radius of a circle is 40 cm and the length of perpendicular drawn from its centre to chord is 24 cm. Find the length of the chord.
4.
A chord of length 48 cm is drawn at a distance of 7 cm from the centre of a circle. Calculate the radius of the circle.
5.
A chord of length 16 cm is at a distance of 15 cm from the centre of the circle. Find the length of the chord of the same which is at a distance of 8 cm from the centre.
6.
Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of circle of radius 17 cm. Find the distance between the chords.
7.
Two parallel chords of lengths 80 cm and 18 cm drawn on the same side of the centre of circle of radius 41 cm. Find the distance between the chords.
8.
Two parallel chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre of a circle. If AB = 1.4 cm and CD = 4 cm, find the radius of the circle.
9.
AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. If they are 1 cm apart and lie on the same side of the centre of the circle, find the radius of the circle.
10.
PQR is an isosceles triangle inscribed in a circle. If PQ = PR = 25 cm and QR = 14 cm, calculate the radius of the circle to the nearest cm.
11.
An isosceles ABC is inscribed in a circle. IF AB = AC = 12 5 cm and BC = 24 cm, find the radius of the circle.
12.
An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.
13.
If a line intersects two concentric circles at the points, A, B, C and D, as shown in the figure, prove that AB = CD.
14.
The radii of two concentric circles are 17 cm and 10 cm. A line segment PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, find the length PQ.
15.
Two circles of radii 17 cm and 25 cm intersect each other at two points A and B. If the length of common chord AB of the circle be 30 cm, find the distance between the centers of the circles.
16.
In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD. prove that AB = CD.
MANISH KUMAR 17.
MATHEMATICS
The adjoining figure shows a circle with centre O in which a diameter AB bisects the chord PQ at point R. If PR = RQ = 8 cm and RB = 4 cm, find the radius of the circle.
18.
In the adjoining figure, AB is a chord of a circle with centre O and BC is a diameter. If OD AB, show that CA = 2 OD and CA || OD.
19.
In the adjoining figure, P is a point of intersection of two circles with centre C and D. If the straight line APB is parallel to CD, prove that AB = 2 CD.
20.
If a diameter of a circle bisects each of the two chords of a circle, then prove that the chords are parallel.
21.
If two chords of a circle are equally inclined to the diameter through their point of intersection, prove that the chords are equal.
22.
Show that equal chords of a circle subtend equal angles at the centre of the circle.
23.
In the given figure, equal chords AB and CD of a circle with centre O, cut at right angles at P. If L and M are midpoints of AB and CD respectively, prove that OLPM is a square.
24.
Prove that the perpendicular bisector of a chord of a circle always passes through the centre.
25.
AB and CD are two parallel chords of a circle and line is the perpendicular bisector of AB. Show that is the perpendicular bisector of CD also.
26.
Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.
27.
Prove that a diameter of a circle, which bisects a chord of the circle also bisects the angels subtended by the chord at the center of the circle.
28.
In the given figure, L and M are mid-points of two equal chords AB and CD of a circle with centre O. Prove that (i) OLM = OML
(ii)
ALM = CML
MANISH KUMAR
MATHEMATICS
MANISH KUMAR 29.
MATHEMATICS
In the given figure, AB and AC are equal chords of a circle with centre O and OP AB, OQ AC. Prove that PB = QC.
30.
In an equilateral triangle, prove that the centroid and the circum centre of the triangle coincide.
(B)
ANGLE PROPERTIES OF A CIRCLE
1.
In the given figure, O is the centre of the circle OAB = 300 and OCB = 400. Calculate AOC.
2.
In the given figure, O is the centre of the circle and AOC = 1300. Find ABC.
3.
In the given figure, O is the centre of the circle and AOB = 1100. Calculate (i) ACO (ii) CAO.
4.
In the given figure, AB || D and BAD = 1000. Calculate : (i) BCD (ii) ADC (iii) ABC.
5.
In the given figure, ACB = 520 and BDC = 430 Calculate (i) ADB (ii) BAC (iii) ABC.
6.
In the given figure, O is the centre of the circle. If AOB = 1400 and OAC = 500, find (i) ABC (ii) BCO (iii) OAB
MANISH KUMAR (iv) BCA.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
7.
In the given figure, BAD = 700, ABD = 560 and ADC = 720. Calculate (i) DBC (ii) BCD (iii) BCA
8.
In the given figure, O is the centre of the circle. If ADC = 1400, find BAC.
9.
In the given figure, O is the centre of the circle and ABC is equilateral. Find (i) BDC (ii) BEC.
10.
In the given figure, O is the centre of the circle and AOC = 1600. Prove that 3y - 2x = 1400
11.
In the given figure, O is the centre of the circle. If CBD = 250 and APB = 1200, find ADB.
12.
(i)
In the given figure, AOB is a diameter of the circle O and AOC = 1000, find BDC.
(ii)
In the given figure, O is the centre of the circle; AOD = 400 and BDC = 1000. Find OCB.
MANISH KUMAR
MATHEMATICS
MANISH KUMAR
MATHEMATICS
13.
In the figure, AB is parallel to DC, BCE = 800 and BAC = 250. Find : (i) CAD (ii) CBD (iii) ADC.
14.
In the given figure, O is the centre of the circle and OBC = 500 Calculate (i) ADC (ii) AOC.
15.
In the given figure, ABCD is a cyclic quadrilateral in which CAD = 250, ABC = 500 and ACB = 350. Calculate (i) CBD
16.
(ii) DAB
(iii) ADB
In the adjoining figure, BAD = 650, ABD = 700 and BDC = 450. Find (i) BCD (ii) ADB Hence, show that AC is a diameter.
17.
In the given figure, AB is a diameter of a circle with centre O and chord ED is parallel to AB and EAB = 650 Calculate (i) EBA
18.
(ii) BED
(iii) BCD
In the given figure, ABCD is a cyclic quadrilateral whose side CD has been produced to E. If BA = BC and BAC = 460, find ADE.
MANISH KUMAR 19.
In the given figure, O is the centre of a circle and ABE is a straight line. If CBE = 550, find : (i) ADC
20.
MATHEMATICS (ii) ABC
(iii) the value of x.
In the given figure AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that AEB is isosceles.
21.
In the given figure, chords AB and CD of a circle are produced to meet at O. Prove that ODB and OAC are similar. If BO = 3 cm, DO = 6 cm and CD = 2 cm, find AB.
22.
In the given figure, O is the centre of the circle, If AOD = 1400 and CAB = 500, calculate : (i) EDB
23.
(ii) EBD
In the given figure, AB is diameter of a circle with centre O. If ADE and CBE are straight lines, meeting at E such that BAD = 350 and BED = 250, find : (i) DCB
24.
(ii) DBC
(iii) BDC
In the given figure, find whether the points A, B, C, D are concyclic, when (i) x = 70 (ii) x = 80.
MANISH KUMAR
MATHEMATICS
MANISH KUMAR 25.
MATHEMATICS
In the given figure, the straight lines AB and CD pass through the centre O of the circle. if AOD = 750 and
OCE = 400, find (i) CDE (ii) OBE.
26.
In the given figure, the two circles intersect at P and Q. If A = 800 and D = 840 calculate : (i) QBC
(ii) BCP
27.
In the adjoining figure, AB = AC = CD, ADC = 350. Calculate : (i) ABC (ii) BEC
28.
In the adjoining figure, two circles intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. If PAC and PBD are straight lines and APB = 750, find (i) AOB
29.
(ii) ACB
(iii) ADB.
The exterior angles B and C in ABC are bisected to meet at a point P. Prove that BPC = 900
A . Is ABPC a 2
cyclic quadrilateral ? 30.
In the given figure, is the incentre of ABC. AT produced meets the circum circle of ABC at D: ABC = 550 and
ACB = 650. Calculate : (i) BCD
(ii) CBD
(iii) DCI
(iv) BIC
MANISH KUMAR
MATHEMATICS
MANISH KUMAR
MATHEMATICS
(C)
ARC Properties of Circles
1.
In the given figure, arc AC and arc BD are two equal arcs of a circle. Prove that chord AB and chord CD are parallel.
2.
Prove that the angle subtended at the centre of a circle, is bisected by the radius through the mid-pint of the arc.
3.
In the given figure, P is the mid-point of arc APB and M is the midpoint of chord AB of a circle with centre O. Prove that : (i)
PM AB
(ii)
PM produced will pass through the centre O;
(iii)
PM produced will bisect the major arc AB.
4.
Prove that in a cyclic trapezium, the non-parallel sides are equal.
5.
P is a point on a circle with centre O. if P is equidistant from the two radii OA and OB, prove that arc AP = arc BP.
6.
In the given figure, two chords AC and BD of a circle intersect at E. If arc AB = arc CD, prove that: BE = EC and AE = ED.
7.
In the given figure, two chords AB and CD of a circle intersect at a point P. If AB = CD, prove that : arc AD = are CB.
8.
If two sides of a cyclic quadrilateral are parallel, prove that : (i) its other two sides are equal,
9.
(ii) its diagonals are equal.
In the given figure, AB BC and CD are equal chords of a circle with centre O and AD is a diameter. If DEF = 1100 find (i) AEF
(ii) FAB.
MANISH KUMAR 10.
MATHEMATICS
In the given figure, ABCDE is a pentagon inscribed in a circle. If AB = BC = CD, BCD = 1100 and BAE = 1200, find : (i) ABC (ii) CDE (iii) AED (iv) EAD
11.
In the given figure, ABC is an isosceles triangle inscribed in a circle with centre O. If AB = AC, prove that : AP = bisects BPC.
12.
In the given figure, AB is a side a regular 6-sided polygon and AC is a side a regular 8-sided polygon inscribed in a circle with centre O. Find : (i) AOB (ii) ACB ABC.
(D)
OBJECTIVE TYPE QUESTIONS :
1.
O is the centre of the circle, If chord AD = chord CD, then x = (A) 700 (B) 500 (C) 350 (D) 450
2.
O is the centre of the circle. If 1 2 , then (A) x > z (B)
x
(C) x + y = 2z (D) None of these 3.
O is the centre of the circle having radius 5 cm. OM on chord AB. If OM = 4 cm, then the length o the chord AB = (A) 6 cm (B) 5 cm (C) 8 cm
MANISH KUMAR (D) 10 cm
MATHEMATICS
MANISH KUMAR 4.
MATHEMATICS
O is the centre of the circle, AB is a chord of the circle. OM AB. If AB = 20 cm, OM = cm, then radius of the circle is (A) 15 cm (B) 12 cm (C) 10 cm (D) 11 cm
5.
O is the centre of the circle with radius 5 cm. Chords AB and CD are parallel. AB = 6 cm and CD = 8cm. If PQ is distance between AB and CD then PQ = (A) 10 cm (B) 8 cm (C) 7 cm (D) 7 2 cm
6.
O is the centre of the circle. AB and CD are two chords of the circle. OM AB and ON CD. If OM = ON = 3 cm and AM = BN = 4.5 cm, then CD = (A) 8 cm (B) 9 cm (C) 10 cm (D) None of these
7.
If BAD ADC , then (A) AB = CD (B) AB CD (C) AD = BC (D) AD BC
8.
If AB CD , then (A) 1 = 2 (B) 2 = 3 (C) 3 = 4 (D) None of these
9.
O is the centre of the circle having radius 5 cm. AB and AC are two chords such that AB = AC = 6 cm. If OA meets BC at M then OM = (A) 3.6 cm (B) 1.4 cm (C) 2 cm (D) 3 cm
10.
O is the centre of the circle. BC is a diameter of the circle. OD AB (chord). If OD = 4 cm, BD = 5 cm, then CD = (A) 13 cm (B)
71 cm
MANISH KUMAR (C)
89 cm
(D) None of these
MATHEMATICS
MANISH KUMAR
MATHEMATICS ANSWER KEY
(A)
CHORD PROPERTIES OF A CIRCLE 1. 6 cm,2. 07 cm
3. 64 cm
4. 25 cm
8. 2.5 cm
9. 5cm 10. 13 cm
5. 30 cm
11. 15 cm
6. 23 cm
12. 3 3
7. 31 cm
14. 9 cm
15. 28 cm
17. 10 cm (B)
ANGLE PROPERTIES OF A CIRCLE 1. AOC = 1400 2. ABC = 1150
4. (i) BCD = 800 (ii) ADC = 800 (ii) ABC = 1000
3. (i) 550 (ii) 550
5. (i) ADB = 520 (ii) BAC = 430 (iii) ABC = 850 6. (i) ABC = 400 (ii) BCO = 600 (iii) OAB = 200 (iv) BCA = 1100 7. (i) BDC = 1800 (ii) BCD = 600 (iii) BCA = 540 8. BAC = 500 9. (i) BDC = 600 (iii) BEC = 1200
10. (i) BAD = 62.50 (ii) BCD = 11750
11. ADB = 95012. BDC = 400 (ii) OCB = 600
13. (i) CAD = 550 (ii) CBD = 550 (iii) ADC - 1000
14. (i) ADC = 1300 (ii) AOC = 1000
15. CBD = 250 (ii) DAB = 700 (iii) ADB = 350
17. (i) EBA = 250 (ii) BED = 250 (iii) BCD = 1550
18. 880
19. (i) ADC = 550 (ii) ABC = 1250 (iii) x = 250
21. AB = 13 cm
22. (i) EDB = 500 (ii) EBD = 1100
25. (i) CDE = 500 (ii) OBE = 250
24. (i) Yes (ii) No.
27. (i) ABC = 400 (ii) BEC = 400
(D)
(iii) DBC = 300
26. (i) QBC = 1000 (ii) BCP = 960
28. (i) AOB = 1500 (ii) ACB = 300 (iii) ADB = 300
30. (i) BCD = 250 (ii) CBD = 350 (iii) DCI = 550 (iv) BIC = 1200
29. No (C)
23. (i) DCB = 350 (ii) DBC = 1150
ARC Properties of Circles 9. (i) AEF = 200 (ii) FAB = 1300
10. (i) ABC = 1100 (ii) CDE = 950 (iii) AED = 1050 (iv) EAD = 500
12. (i) AOB = 600 (ii) ACB = 300
(iii) ABC = 22030
Objective Type Questions : 1. C
2. C
3. A
4. B
5. C
6. B
7. A
8. B
9. B
10. C
MANISH KUMAR
MATHEMATICS
GEOMETRICAL CONSTRCUTIONS
INTRODUCTION In the chapter “Lines and Angles” and “Triangles” we have proved many theorem and properties by using diagrams in which angles and sides of triangles were drawn in approximate measurement. The diagrams were drawn to have the idea of the situations according to the given conditions. In this chapter, we shall construct some angles and triangles in precise measurement by using only two geometrical instruments. These two instruments’ are, a graduated ruler and a compass.
CONSTRCUTION OF PERPENDICULAR BISECTOR OF A LINE SEGMENT We want to construct the perpendicular bisector of the given line segment AB. Steps of Construction:
1 AB , draw, arcs on both sides of the line segment AB. 2
1.
Taking A and B as centers and radius
2.
The arcs are drawn in such as way that on both sides of AB, we get intersection points P and Q.
3.
Join PQ. PQ intersects AB at M. Here, PMQ is the required perpendicular bisector of AB. Justification. In figure, Join AP, AQ, BT and BQ. Here, AP = BP = BQ = AQ (Each = radius of the arc)
APBQ is a rhombus.
Diagonals AB and PQ are right bisectors of each other.
Hence, PQ is perpendicular bisector of the chord AB.
CONSTRUCTION OF BISECTOR OF A GIVEN ANGLE We want to construct the bisector of given angle ABC. Steps of Construction:
1.
Taking B as centre, we draw an arc of circle which meets BC at P and BA at Q.
2.
Now, taking P and Q as centre and radius
3.
Join BM.
1 PQ draw two arcs so that they intersect at a point M. 2
Here, they ray BM is the required bisector of ABC. Justification. In figure, Join PM and QM. In BPM and BQM, we have
BP = BQ
(Radius of the first arc)
PM = QM
(Radius of the second arc)
BM = BM
(Common)
BPM BQM
ABC.
By CPCT, we have
Construction of
600
angle.
MANISH KUMAR We will construct 600 angle at the initial point A of the given ray AB.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Step of Construction : 1. 2. 3.
Taking A centre and radius = r (say), we draw an arc of a circle. The arc intersects AB and P. Now, taking P as centre and same radius r, we again draw an arc of a circle which intersects the previous arc at Q. Join AQ and produce this as ray AC.
CAB = 600
Here,
Justification : Joint PQ. We know that AP = AQ = PQ = r.
PAQ is equilateral 0 Construction of 30 Angle
PAQ = 600
We will construct 300 and at the initial point A of the given ray AB.
Step of construction : 1. 2.
Taking A as centre and radius = r (say), we draw an arc of a circle. The arc intersect AB at P. Now, taking P as centre and same radius r, we again draw an arc of a circle which intersects the previous arc at Q.
3. 4. 5.
Join AQ and produce the ray AC. Here BAC = 600 Now, taking P and Q as centers and some radius r > r, we draw arcs which intersect at R. Join AR and produce the ray AD along AR.
6.
Here, AD is bisector of BAC = 600 Therefore, we have BAD = 300
Construction of 450 Angle We will construct 450 angle at the initial point A of the given ray AB.
1. 2. 3. 4.
Steps of construction : First of all, we construct BAC = 600 We find AD bisector of BAC. Here; BAD = DAC = 300 Now, we find AE bisector of CAD. Here, DAE = CAE = 150 BAE = BAD + DAE = 300 + 150 = 450 Construction of 900 Angle Steps of Construction :
1.
We construct CAB = CAD = 600
2.
Now, we find AE bisector of CAD.
3.
BAE = BAC + CAE = BAC +
1 1 CAD = 600 + × 600 2 2
MANISH KUMAR Hence, BAE = 900
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Ex.1 Sol.
Take a line segment AB = 6.3 cm. Find the right bisector (perpendicular bisector) of AB. Steps of Construction:
1.
We take AB = 6.3 cm.
2.
Taking A and B as centers and radius >
3.
we draw arcs on both sides of B. Two arcs intersect at P on one side and
4.
other two arcs intersect at Q on the second side of AB. We join PQ ; it meets AB at M.
Ex.2
Now PMQ is the required right bisector of AB. Construct an angle of 750 at the initial point of a given ray. Justify the construction.
Sol.
We have
1 AB. 2
BAC = 600
and
CAD = 600,
Then,
CAE =
1 × 600 = 300 2
Further, AF bisect CAE, Then
CAF = =
1 CAE 2 1 × 300 = 150 2
Hence, BAF = BAC + CAF = 600 + 150 = 750.
CONSTRCUTION OF TRAINGLES In this section, we shall construct some triangles with given data by using a graduated ruler and a compass.
To construct a triangle, given its base, a base angle and sum of other two sides. We have to construct ABC. When base BC, B and AB + AC is given. Here, AB + AC > BC.
Steps of Construction : 1.
Draw the given base BC.
2.
Construct the base angle CBX = B as given.
3.
Cut the line segment BD = AB + AC along BX.
4. 5. 6.
Join CD. Draw perpendicular bisector PQ of CD. PQ meets CD at L and BC at A. Join AC. Here, ABC is the required triangle.
Justification. In ABC, AL is perpendicular bisector of CD.
AD = AC.
Now, in ABC we have AB + AC = AB + AD = BC
Note : If sum of two sides is less than the given base, then triangle is not possible. To construct a triangle, given its base, a base angle and the difference of the other two sides.
MANISH KUMAR Case (a) : We shall construct ABC when base BC and B are given. Also, we are given AB - AC. Here, AB > AC.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Steps of Construction : 1.
Draw the given base BC.
2. 3. 4. 5. 6. 7.
Construct CBX = B as given. Along BX, cut BD = AB - AC, (Here, BD and BA are is same direction). Join CD. Construct the perpendicular bisector PQ of CD and PQ intersects CD at L. QP (produced) meets BX at A. Join AC. Here , AC = AD and hence the ABC is required triangle. Justification. In ACD. AL is perpendicular bisector of CD
AC = AD
Now, in ABC, we have AB - AC = AB - AD = BD
Case (b) : Let us construct ABC, when BC and B are given. Also, AC > AB, i.e., AB < AC and AC - AB is given
Steps of Construction : 1.
Draw the given base BC.
2. 3.
Construct CBX = B as given. Cut BD = AC - AB along XB (produced). Here, BD and BA are in opposite directions. Join CD. Construct the perpendicular bisector PQ of CD. QP meets BX at A and CD at L. Join AC. Here, AC = AD = AB + BD i.e., AC - AB = BD. Justification. In ACD, Al perpendicular bisector of CD.
4. 5. 6. 7.
AC = AD.
Now, in ABC, we have AC - AB = AD - AB = BD.
To construct a triangle, given its two base angles and the perimeter of the triangle. We shall construct ABC if its base angles B and C are given and its perimeter AB + BC + CA is given equal to p.
Steps of Construction: 1.
Cut GH = AB + BC + CA = p units.
2.
Construct XGH = B (as given) and YHG = C (as given)
3.
Draw bisectors of XGH and YHG and both intersect at A.
MANISH KUMAR 4. 5.
MATHEMATICS
Construct PQ and RS perpendicular bisectors of GA and HA respectively. PQ and RS intersect GH at B respectively, Also PQ intersects GA at L and RS intersects HA at M.
MANISH KUMAR 6.
MATHEMATICS
Join AB and AB. Here, ABC is the required triangle.
Justification BL is right bisector of GA.
BA = BG
and
BAG = BGA ...(ii) ABC = BAG + BGA = 2BGA (By ii)
and
=2×
...(i)
1 XGH = XGH 2
(Given angle B)
Similarly, ACB = YHG (Given angle C) Now, BA = BG; Similarly, CA = CH Thus, we have AB + BC + CA = BG + BC + CH = GH Ex.3 Sol. 1.
Construct ABC in which BC = 8 cm, B = 300 and AB + AC = 12 cm. Steps of Construction : Draw BC = 8 cm
2. 3. 4. 5. 6. 7.
Construct CBX = 300 Along BX, cut BD = 12 cm Join CD Draw PQ right bisector of CD. PQ intersects BD at A and CD at L. Join CA; CA = AD because AL is perpendicular bisector of CD. Now, ABC is the required triangle.
Ex.4 Sol.
Construct ABC such that B = 5.8 cm, BC + CA = 7 cm and B = 600 Steps of Construction :
1. 2. 3. 4. 5.
Draw AB = 5.8 cm and ABX = 600 Cut BD = 7 cm along BX Join AD and draw PQ, right bisector of AD. PQ intersects AD at L and BD at C. Join AC. We observe that in CAD, CL is perpendicular bisector of AD.
CA = CD BC + CA = BC + CD = BD = 7 cm.
Hence, ABC is the required triangle. Ex.5 Sol. 1. 2. 3. 4. 5. 6.
Construct ABC such that BC = 6 cm, B = 450 and AC - AB = 2 cm. Steps of Construction: Draw BC = 6 cm and CBX = 450 Produce XB to D so that BD = 2 cm i.e., BD = AC - AB Here, BD and BA are to be in opposite directions Join CD. Draw PQ, right bisector of CD. PQ intersects BX at A and CD at L. Join AC. ABC it the required triangle.
MANISH KUMAR
MATHEMATICS
Ex.6 Sol.
Construct ABC such that BC = 6 cm, B = 450 and AB - AC = 3 cm. Steps of Construction :
1. 2. 3. 4. 5. 6.
Draw BC = 6 cm and CBX = 450 On BX cut BD = 3 cm. Join CD. Draw PQ, right bisector of CD. PQ intersects BX at A and CD at L. Join AB. Here, ABC is the required triangle.
Ex.7 Sol. 1.
Construct ABC such that B = 600, C = 750 and AB + BC + CA = 13 cm. Steps of Construction : Draw GH = 13 cm.
2.
Draw HGX = 600 and GHY = 750
3. 4.
Bisector of HGX and GHY intersect at A. Draw PQ, right bisector of GA. PQ intersects GH at B and GA at L. Draw RS, right bisector of HA. RS intersects GH at C and HA at M. Join AB and AC. The ABC is the required triangle. Construct a right triangle whose base is 6 cm and sum of its hypotenuse and the other side is 10 cm. Steps of Construction : Draw base BC of the triangle equal to 6 cm.
5. 6. Ex.6 Sol. 1. 2. 3. 4. 5. 6.
Construct CBX = 900 Cut BD = 10 m along BX. Join CD and draw PQ, right bisector of CD. PQ intersects BD at A and CD at L. Join AC. We observe that is ACD, AL is right bisector of CD. Therefore AC = AD. Thus, in ABC, AC is hypotenuse and AB + AC = AB + AD = BC = 10 cm. i.e., AB + AC = 10 cm. Hence, the ABC is the required right triangle.
Ex.9 Sol. 1.
Construct ABC in which AC = 7 cm, C = 600 and AB + BC = 12 c m. Steps of Construction: Draw AC = 7 cm
2. 3. 4. 5. 6.
Draw ACX = 600 Cut CD = 12 cm along CX. Join AD and draw right bisector PQ and AD. PQ intersects AD at L and CD at B. Join AB. Now, ABC is the required triangle. Justification In ABC, BL is right bisector of AD. Therefore , AB = BD. We have AB + BC = BD + BC = CD = 12 cm
MANISH KUMAR i.e.
AB + BC = 12 cm as required.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
EXERCISE 1.
Take AB = 7.4 cm and find its perpendicular bisector.
2.
Construct the following angles at the initial point of a given ray : (i) 150 (ii) 1050 (iii) 7
1 2
(iv) 22
1 2
3.
Construct ABC such that BC = 5 cm, B = 300 and AB + AC = 8 cm.
4.
Construct ABC such that BC = 6 cm, BC + CA = 10 cm and B = 300
5.
Construct ABC such that BC = 8 cm, B = 600 and AB + AC = 14 cm.
6.
Construct ABC such that BC = 4 cm, AB + AC = 7 cm and B = 600.
7.
Construct ABC such that BC = 4.5 cm, AB + AC = 6.5 cm and B = 450
8.
Construct ABC such that BC = 5 cm, AB - AC = 3 cm and B = 300
9.
Construct ABC such that BC = 5 cm, AC - AB = 3 cm and B = 300.
10.
Construct ABC such that BC = 6 cm, AB - AC = 2.5 cm and B = 450.
11.
Construct ABC such that B = 600, C = 450 and AB + BC + CA = 11 cm.
12.
Construct PQR such that P = 900, Q = 600 and the perimeter of the triangle is 12 cm.
13.
Construct a right triangle whose base is 8 cm and the sum of its hypotenuse and the other side is 16 cm.
14.
Construct ABC in which AC = 5 cm, C = 600 and AB + BC = 9 cm.
15.
Construct PQR in which QR = 6.2 cm, R = 900 and PQ + PR = 8.6 cm.
16.
Construct ABC in which BC = 6.4 cm, C = 900 and AB + AC = 7.4 cm.
17.
Construct ABC in which BC = 4.5 cm, C = 450 and AB + AC = 7.4 cm.
18.
Construct ABC in which AC = 6.2 cm, C = 600 and AB - BC = 2.5 cm.
NCERT Questions. 1.
Construct an angle of 900 at the initial point of a given ray and justify the construction.
2. 3.
Construct an angle of 450 at the initial point of a given ray and justify the construction. Construct the angles of the following measurements :
4.
1 (iii) 150 2 Construct the following angles and verify by measuring them by a protractor :
5.
(i) 750 (ii) 1050 (iii) 1350 Construct an equilateral triangle, given its side and justify the construction.
6.
Construct a triangle ABC in which BC = 7 cm, B = 750 and AB + AC = 13 cm.
7.
Construct a triangle ABC in which BC = 8 cm, B = 450 and AB - AC = 3.5 cm.
8.
Construct a triangle PQR in which QR = 6 cm, Q = 600 and PR - PQ = 2 cm.
9.
Construct a triangle XYZ in which Y = 300, Z = 900 and XY + YZ + ZX = 11 cm.
(i) 300 (ii) 22
MANISH KUMAR 10.
MATHEMATICS
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
MANISH KUMAR
MATHEMATICS
STATISTICS
INTRODUCTION In various field, we need information in the form of numerical figure called data. These data may relate to the marks obtained by the pupils of a class in a certain examination; the weights, height, ages, etc., of pupils in a class; the monthly wedges earned by workers in a factory; the population of a town or the profits of a company during last few years, etc. Evaluation of such data helps analysis study the various growth patterns all formulate future targets or policies or derive certain inferences.
STATISTICS It is the science which deal with the collection, presentation, analysis and interpretation of numerical data. In singular form, statistics is taken as a subject. And, in plural form, statistics means data.
DATA The word data means a set of given facts in numerical figures. Fundamental Characteristics of DATA (i) Numerical facts alone form data. Qualitative characteristics, line honesty, poverty, etc., which be measured numerically do not form data. (ii) Data are aggregate of facts. A single observation does not form data. (iii) Data collected for a definite purpose may not be suitable for another purpose. Types of Data (i) Primary Data : The data collected by the investigator himself with definite plan in mind are known as primary data. (ii) Secondary Data : The data collected by someone, other than the investigator are known as secondary data.
1.
2.
VARIABLE A quantity which can take different values is called a variable. Ex. : Height, Age and Weight of pupils in a class are three variable. If we denote them by x, y and z respectively, then values of x give the heights of the pupils; the value of y give the ages of the pupils and the values of z give the weights of the pupils. Continuous and Discrete Variables Variables are of 2 types Continuous Variable : A variable which can take any numerical value within a certain range is called a continuous variable, Ex. (i) Wages of workers in a factory (ii) Heights of children in a class (ii) Weights of persons in a group etc. Discontinuous (or Discrete) Variable : A variable which cannot take all possible values between two given values, is called a discontinuous or discrete variable. Ex, (i) Number of members in a family (ii) Number of workers in a factory Such variable cannot take any value between 1 and 2, 2 and 3, etc.
IMPORTANT TERMS Range : The difference between the maximum and minimum values of a variable is called its range. Variate : A particular value of a variable is called variate. Presentation of Data : Putting the data in condensed form in the form of a table is known as presentation of data.
MANISH KUMAR
MATHEMATICS
Frequency : The number of times an observation occur is called its frequency. Frequency Distribution : The tabular arrangement of data showing the frequency of each observation is called its frequency distribution.
RAW OR UNGROUPED DATA The data obtained in original form are called raw data or ungrouped data. Ex. The marks obtained by 25 students in a class in a certain examination are given below :25, 8, 37, 16, 45, 40, 29, 12, 42, 40, 25, 14, 16, 16, 20, 10, 36, 33, 24, 25, 35, 11, 30, 45, 48. This is the raw data. Array : An arrangement of raw data in ascending or descending order of magnitude is called any array. Arranging the marks of 25 students in ascending order, we get the following array. 8, 10, 11, 12, 14, 16, 16, 16, 20, 24, 25, 25, 29, 30, 33, 35, 36, 37, 40, 40, 42, 45, 45, 48.
TO PREPARE A FREQUENCY DISTRIBUTION TABLE FOR RAW DATA USING TALLY MARKS We take each observation from the data, one at a time and indicate the frequency (the number of times the observation has occurred in the data) by small lines, called tally marks. For convenience, we write tally marks in bunches of five, the fifth one crossing the fourth diagonally. In the table so formed the sum of the frequencies is equal to the total number of observation in the given data. Ex. The sale of shoes of various sizes at a shop, a particular day is given below :7 8 5 4 9 8 5 7 6 8 9 6 7 9 8 7 9 9 6 5 8 9 4 5 5 8 9 6 The above data is clearly raw data. From this data, we may construct a frequency table, as given below :
Size 4
II
2
5
IIII
5
6
IIII
4
7
IIII
4
8
IIIII
6
9
IIIIII
7
Total
Frequency Table Tally Marks Frequency
28
GROUPED DATA To put the data in a more condensed form, we make groups of suitable size, mention the frequency of each group. Such a table is called a grouped frequency distribution table. Class-Interval : Each group into which the raw data is condensed, is called a class-interval. Each class is bounded by two figures, which are called class limits. The figure on the left side of a class is called its lower limit and that on its right is called its upper limit.
1.
Types of Grouped Frequency Distribution Exclusive Form (or Continuous Interval Form): A frequency distribution in which the upper limit of each class in excluded and lower limit is included, is called an exclusive form.
MANISH KUMAR
MATHEMATICS
Ex. Suppose the marks obtained by some students in an examination are given. We many consider the classes 0 - 10, 10 - 20 etc. In class 0 - 10, WE include 0 and exclude 10. In class 10 - 20, we include 10 and exclude 20.
2.
Ex.1
Sol.
Inclusive Form (or Discontinuous Interval Form) : A frequency distribution in which each upper limit as well as lower limit in included, is called an inclusive form. Thus, we have classes of the form 0 -10, 11 20, 21 - 30 etc. In 0 - 10, both 0 and 10 are included. Given below are the marks obtained by 40 students in an examination: 3, 25, 48, 23, 17, 13, 11, 9, 46, 41, 37, 45, 10, 19, 39, 36, 34, 5, 17, 21, 39, 33, 28, 25, 12, 3, 8, 17, 48, 34, 15, 19, 32, 32, 19, 21, 28, 32, 20, 23. Arrange the data in ascending order and present it as a grouped data in : (i) Discontinuous Interval form, taking class-intervals 1 - 10, 11 - 20, etc. (ii) Continuous Interval form, taking class-intervals 1 - 10, 10 - 20, etc. Arranging the marks in ascending order, we get : 3, 3, 5, 8, 9, 10, 11, 12, 13, 15, 17, 17, 17, 19, 19, 19, 20, 21, 21, 23, 23, 25, 25, 28, 28, 32, 32, 32, 33, 34, 34, 36, 37, 39, 39, 41, 45, 46, 48, 48. We many now classify them into groups as shown below :
(i)
Discontinuous Interval Form (or Inclusive Form) Marks (Class-intervals) 1-10 11-20
Tally Marks IIIII IIII
IIIII
Number of Students (frequency) 6 11
21-30
IIII III
8
31-40
IIII IIII
10
41-50
IIII
5
Total
40
Note that the class 1 - 10 means, marks obtained from 1 to 10, including both.
(ii)
Continuous Interval Form (or Exclusive Form) Marks (Class-intervals)
Tally Marks
Number of Students (frequency)
1-10
IIII
5
10-20
IIII
IIIII
11
20-30
IIII IIII
9
30-40
IIII IIII
10
40-50
IIII
5
Total
40
Here, the class 1 - 10 means, marks obtained from 1 to 9, i.e., excluding 10.
1.
IMPORTANT TERMS RELATED TO GROUPED DATA Class Boundaries Or True Upper And True Lower Limits : -
MANISH KUMAR
MATHEMATICS
(i)
In the exclusive form, the upper and lower limits of a class are respectively known as the true upper limit and true lower limit. (ii) In the inclusive form, the number midway between the upper limit of a class and lower limit of the subsequent class gives the true upper limit of the class and the true lower limit of the subsequent class :Thus, in the above table of inclusive form, we have :
10 11 true upper limit of class 1 - 10 is 10.5 = 10.5 , and, true limit of class 11 - 20 is 10.5. 2 20 21 Similarly, true upper limit of class 11 - 20 is 20.5 = 20.5 and, true lower limit of class 21 - 30 is 20.5. 2
2.
Class Size : The difference between the true upper limit and the true lower limit fo class is called its class size.
3.
True upper limit True lower limit Class Mark of A Class = 2 The difference between any two successive class marks given the class size.
Ex.2
The class marks of a frequency distribution are 7, 13, 19, 25, 31, 37, 43. Find the class-size and all the class-intervals.
Sol.
Class size = Difference between two successive class-marks = (12 - 7) = 6. Let the lower limit of the first class interval be a. Then, its upper limit = (a + 6)
a (a6) = 7 2a = 8 a = 4. 2
So, the first class-interval is 4 - 10. Let the lower limit of last class-interval be b. Then, its upper class limit = (b + 6)
b (b 6) 43 2b = 80 b = 40. 2
Some the last class-intervals is 40 - 46. Hence, the required class-intervals are 4 - 10, 10 - 16, 16 - 22, 22 - 28, 28 - 34, 34 - 40 and 40 - 46.
METHOD OF FORMING CLASSES OF A DATA.
1.
Determine the maximum and minimum values of the variate occurring in the data.
2.
decide upon the number of classes to be formed.
3.
Find the range, i.e. the difference between the maximum value and the minimum value. Divide the range by the number of classes to be formed to get the class - size.
4.
Be sure that there must be classes having minimum and maximum values occurring in the data.
5.
By counting, we obtain the frequency of each class.
Ex.3
The water tax bills (in rupees) of 30 houses in a locality are given below :147, 167, 136, 178, 175, 116, 155, 121, 115, 156, 176, 141, 189, 167, 177,
MANISH KUMAR
MATHEMATICS
208, 212, 143, 203, 210, 188, 175, 212, 118, 197, 145, 134, 133, 196, 185.
Construct a frequency-distribution table with class-size 10. Sol.
Minimum observation = 115, Maximum observation = 212. Range = (Maximum observation) - (Minimum observation) = (212 - 115) = 97. Class size = 10. Since 97 ÷ 10 = 9.7, we should have 0 classes, each of size 10. These classes are : 115 - 125, 125 - 135, 135 - 145, 145 - 155, 155- 165, 165 -175, 175 - 185, 185 - 195, 195 - 205 and 205 - 215.
The frequency distribution table many be presented as shown below :
Bill (in Rs)
Tally Marks
Frequency
115-125
IIII
4
125-135
II
2
135-145
III
3
145-155
II
2
155-165
II
2
165-175
II
2
175-185
IIII
5
185-195
IIII
4
195-205
III
3
205-215
III
3
Total
Ex.4
30
RULE TO CONVERT DISCONTINUOUS (OR INCLUSIVE) FORM TO CONTINUOUS (OR EXCLUSIVE) FORM In a discontinuous interval or inclusive form, we have: 1 Adjustment factor = [(lower limit of one class - Upper limit of precious lass)] 2 1 Thus, if the classes are 1 - 10, 11 - 20, etc. then adjustment factor = (11 - 10) = 0.5. 2 To convert data given in discontinuous form to the continuous form, we subtract the adjustment factor from each lower limit and add the adjustment factor to each upper limit to get true limits. Convert the following frequency distribution form discontinuous to continuous form : Marks Frequency (Class-intervals) 1-10 7 11-20
5
21-30
9
MANISH KUMAR
Sol.
MATHEMATICS 31-40
11
41-50
6
1 (11 - 10) = 0.5. Subtract 0.5 form each lower limit and add 0.5 to each upper limit. 2 Then, the required table in continuous form may be prepared as under : Marks Marks Frequency (before adjustment) (after adjustment 1-10 0.5-10.5 7 Adjustment factor =
11-20
10.5-20.5
5
21-30
20.5-30.5
9
31-40
30.5-40.5
11
41-50
40.5-50.5
6
Total
38
CUMULATIVE FREQUENCY OF A CLASS INTERVAL. The sum of the frequencies of all the previous classes and that particular class, is called the cumulative frequency of the class.
Cumulative Frequency Table
Ex.5
Sol.
A table which shows the cumulative frequencies over various classes is called a cumulative frequency distribution table. Following are the ages (in years) of 360 patients, getting medical treatment in a hospital. Age (in years)
10-20
20-30
30-40
40-50
50-60
60-70
Number of Patients
90
50
60
80
50
30
Construct the cumulative frequency table for the above data. The cumulative frequency table for the above data is given below. Class Interval
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
90
50
60
80
50
30
Cumulative Frequency
90
140
200
280
330
360
This table may be presented in ‘less that form’ as under.
Age (in years)
Number of Patients
Less than 20
90
Less than 30
140
Less than 40
200
Less than 50
280
Less than 60
330
MANISH KUMAR
MATHEMATICS Less than 70
Ex.6
360
The monthly wages (in rupees) of 28 laborers working in a factory, are given below :220, 268, 258, 242, 210, 267, 272, 242, 311, 290, 300, 320, 319, 304, 302, 292, 254, 278, 318, 306, 210, 240, 280, 316, 306, 215, 256, 358. Form a cumulative frequency table with class intervals of length 20.
Sol.
We may form the table as under : Class Interval
Tally Marks
Frequency
Cumulative Frequency
210-230
IIII
4
4
230-250
III
3
7
250-270
IIII
5
12
270-290
III
3
15
290-310
IIII II
7
22
310-330
IIIII
6
28
MANISH KUMAR GRAPHICAL REPRESENTATION OF STATISTICAL DATA
MATHEMATICS
The tabular representation of data is a ideal way of presenting them is a systematic manner. When these numerical figures are represented pictorially or graphically, they become more noticeable and easily intelligible, leaving a more lasting effect on the mind of the observer. With the help of these pictures or graphs, data can be compared easily. There are various types of graphs. In this chapter, we shall be dealing with the following graphs : 1. Bar Graphs 2. Histogram
3. Frequency Polygon
BAR GRAPH (OR COLUMN GRAPH OR BAR CHART) A bar graph is a pictorial representation of numerical data in the form of rectangles (or bars) of equal width and varying heights. These rectangles are drawn either vertically or horizontally. The height of a bar represents the frequency of the corresponding observation. The gap between two bars is kept the same.
Ex.7
The following table shows the number of students participating in various games in a school. Cricket
Tennis
Football
Badminton
27
9
18
12
Draw a bar graph to represent the above data. Sol.
Take the games along x-axis and the number of students along Y-axis.
Along y-axis, take the sale 1 cm = 6 students. The bar-graph may, thus, be drawn as shown alongside. Ex.8
Given below are data showing number of students of a school using different modes of travel to school. Mode
School Bus
Walking
Bicycle
Other Vehicles
No. of Boys
100
160
240
40
No. of Girls
180
60
120
20
MANISH KUMAR Draw a bar graph to represent the above data.
MATHEMATICS
MANISH KUMAR Sol.
MATHEMATICS
Take the mode along x-axis and the number of students along y-axis Scale : Along y-axis, take 1 cm = 40 students. The bars of equal width and proportionate heights with same gap between the two consecutive bars, may be drawn as shown below. Shading for boys and girls may be done as under :
HISTOGRAM A histogram is a graphical representation of frequency distribution in an exclusive form in the form of rectangles with class intervals as bases and the corresponding frequencies as heights, there being no gap between any two successive rectangles.
METHOD OF DRAWING A HISTOGRAM Step 1: If the given frequency distribution is in inclusive form, convert it into an exclusive form. Step 2: Taking suitable scales, mark the class-intervals along x-axis and frequencies along y-axis. Note that the scales chosen for both the axes need not be the same. Step 3: Construct rectangles with class-intervals as bases and the corresponding frequencies as heights.
Ex.9
Sol.
Draw a histogram to represent the following data : Class interval
30-36
36-42
42-48
48-54
54-60
Frequency
15
25
20
30
10
Draw rectangles with bases 30 - 36, 36 - 42, 42 - 48, 48 - 54 and 54 - 60 and heights 15 , 25, 20, 30 and 10 respectively.
MANISH KUMAR Note : Since the scale on x-axis starts at 30, we make a kink (
MATHEMATICS ) in the beginning.
MANISH KUMAR
MATHEMATICS
Ex.10 Draw a histogram for the following data :
Sol.
Class interval
1-10
11-20
21-30
31-40
Frequency
5
15
10
15
The given table is in inclusive-form. So, we first convert it into an exclusive form as given below. Class interval
0.5-10.5
10.5-20.5
20.5-30.5
30.5-40.5
Frequency
5
15
10
15
Now, we may draw the histogram, as shown below. Note : Since the scale on x-axis start at 0.5, a kink is shown near the origin.
FREQUENCY POLYGON Let x1, x2,....xn be the class marks (i.e., mid points) of the given frequency distribution and let f 1, f2, .... fn be the corresponding frequencies. We plot the points (x1, f1), (x2, f2)..., (xn, fn) on a graph paper and join these points by line segments. WE complete the diagram in the form of a polygon by taking two more classes (called imagined classes), one at the beginning and the other at the end, each with frequency zero. This polygon is known as the frequency polygon of the given frequency distribution.
Ex.11 The following table shows the number of diabetic patients at various age groups. Age (in years)
0.5-10.5
10.5-20.5
20.5-30.5
30.5-40.5
Number of diabetic patients
5
15
10
15
Represent the above data by a frequency polygon. Sol.
Take two more intervals one at the beginning and the other at the end, each with frequency 0. Thus, we have class-intervals 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70 and 70 - 80 with corresponding frequencies as 0, 3, 6, 14, 8, 5, 2, 0 respectively. Thus, we have :
MANISH KUMAR
MATHEMATICS
Class mark
5
15
25
35
45
55
65
75
Frequency
0
3
6
14
8
5
2
0
MANISH KUMAR
MATHEMATICS
Now, plot the points (5, 0), (15, 3), (25, 6) (35, 14) (45, 8) (55, 5), (65, 2), (75, 0) on a graph paper and join them successively to get the required graph.
Ex.12 Draw the frequency polygon representing the following frequency distribution.
Sol.
Class interval
30-34
35-39
40-44
45-49
50-54
55-59
Frequency
12
16
20
8
10
4
Though the given frequency table is in inclusive form, yet we find that class marks in case of inclusive and exclusive form are the same. We take the imagined classes 25 -29 at the beginning and 60 -64 at the end, each with frequency zero. Thus, we have : Class interval
25-29
30-34
35-39
40-44
45-49
50-54
55-59
60-64
Class Mark
27
32
37
42
47
52
47
62
Frequency
0
12
16
20
8
10
4
0
Now plot the points (27, 0), (32, 12), (37, 16), (42, 20), (47, 8), (52, 10), (57, 4) and (62, 0) and join them successively to obtain the required frequency polygon, as shown below :
MANISH KUMAR HISTOGRAM AND FREQUENCY POLYGON ON THE SAME GRAPH
MATHEMATICS
When a histogram and a frequency polygon are to be drawn on the same graph, we first draw the histogram with the given data. We then join the mid-points of the tops of adjacent rectangles by line segments to obtain the frequency polygon. Ex.13 The following table gives the number of doctors working in government hospitals in a city in various age groups. Draw a histogram and frequency polygon for the given data. Age (in years)
20-25
25-30
30-35
35-40
40-45
Number of doctors
40
60
50
20
10
Sol.
Step-1: Draw rectangles with bases 20-25, 25-30, 30-35, 35-40 are 40-45 and heights 40, 60, 50, 20 and 10 respectively. Since the scale on x-axis start at 20, we make a kink in the beginning. Thus, we obtain the required histogram. Step-2: Mark the mid-point of the top of each rectangle of the histogram Step-3: Mark the mid-points of class-intervals 15 - 20 and 45 - 50 on x-axis. Step-4: Join the consecutive mid-points by line segments to obtain the required frequency polygon.
ARTHMETIC MEAN The average of number in arithmetic is known as the Arithmetic Mean or simply the mean of these number in statistics.
Mean
Sum of observations Number of observation
MEAN OF UNGROUPED DATA The means of n observations x1, x2,..... xn is given by
Mean, x
( x1 x 2 x 3...... xn ) x1 n n
where the symbol called sigma stands for the summation of the terms. Ex.14 The heights of 6 boys in a group are 142 cm, 154 cm, 146 cm, 145 cm, 151 cm and 150 cm. Find the mean height per boy. Sum of the heights of the boys (142 154 146 145 151 150) Sol. Mean height : cm Number of boys 6
MANISH KUMAR
MATHEMATICS
888 cm 148cm 6 Hence, the mean height is 148 cm.
MANISH KUMAR
MATHEMATICS
Ex.15 Find the mean of the first five multiples of 7. Sol. The first five multiples of 7 are 7, 14, 21, 28 and 35.
Required mean
(7 14 21 28 35) 105 21. 5 5
Ex.16 If the mean of 7, 9, 11, 13 x , 21 is 13, find the value of x. Sol.
Mean of the given numbers =
(7 9 11 13 x 21) (61 x ) 6 6
But, mean = 13 (given
61 x = 13 61 + x = 78 x = 17. Hence, x = 17. 6
Ex.17 If the mean of five observations x, x + 2,x + 6 and x + 8 is 11, find the value of x. Sol.
Mean of the given observations
x ( x 2) ( x 4) ( x 6) ( x 8) 5 x 20 5 5 But, mean = 11 (given)
5x 20 = 11 5x + 20 = 55 5x = 35 x = 7 5 Hence, x = 7
MEAN FOR AN UNGROUPED FREQUENCY DISTRIBUTION 1. Direct Method Let n observations consist of values x1, x2, ....xn of a variable x, occurring with frequencies f1, f2,.... fn respectively. Then, the mean of these observations is given by :
Mean, x
( f 1 x1 f 2 x 2 .... f n x n ) f i xi ( f 1 f 2 ..... f n ) f i
Ex.18 The ages of 40 students in a class are given below :
Sol.
Age (in years)
12
13
14
15
16
17
Number of students
6
8
5
7
9
5
Find the means age of the class. We prepare the table as given below : Age (in years) Number of students 1;, xi
Age (in years) (xi)
Number of students
fixi
12
6
72
13
8
104
14
5
70
15
7
105
16
9
144
17
5
85
fi =40
fixi 580
MANISH KUMAR
MATHEMATICS
Mean age =
fixi 580 14.5 years. fi 40
Ex.19 If the mean of the following frequency distribution is 28.25, find the value of p/ xi fi So.
15 8
20 7
25 p
30 14
35 15
40 6
We prepare the table as under :
xi
fi
fixi
15 20 25 30 35 40
8 7 P 14 15 6
120 140 25p 420 525 240
fi (50 p)
fi x i 1445 25p
fix i 1445 25p fi 50 p But, mean = 28.25 1445 25p = 28.25 1445 + 25p = 1412.5 + 28.25 p 50 p 32.5 3.25p = 32.5 p = = 10 Hence p = 10. 3.25 Ex.20 The mean of the following frequency distribution is 16.6. Mean =
xi fi Sol.
8
12
15
18
20
25
30
Total
12
16
p
24
16
q
4
100
Find the missing frequencies p and q. We prepare the table given below :
xi
fi
fi xi
8 12 15 18 20 25 30
12 16 p 24 16 q 4
9+6 192 15p 432 320 25q 120
fi 72 p q
fi xi 1160 15p 25q
Here, fi = 72 + p + q. But, fi = 100 (given) 72 + p + q = 100 p + q = 28.
MANISH KUMAR
MATHEMATICS
fix i 1160 15p 25q 1160 15(p q) 10q 1160 15 28 10q 1580 10q fi 72 p q 72 (p q) 72 28 100 But mean = 16.6 (given) 1580 10q = 16.6 1580 + 10q = 1660 100 10q = 80 q = 8. p + q = 28 p = 28 - q = 28 - 8 = 20 Hence, p = 20 and q = 8. Also, mean =
MEDIAN OF UNGROUPED DATA Median : After arranging the given data in a ascending or a descending order of magnitude, the value of the middle-most observation is called the media of the data. Method for Finding the Median of An Ungrouped Data Arrange the given data in an increasing or decreasing order of magnitude. Let the total number of observations be n. (i)
n 1 In n is odd, then median = value of th the observation. 2
(ii)
If n is even, then median
1 n n th observations 1 th observation. 2 2 2
Ex.21 The marks of 13 students (out of 50) in an examination are : 39, 21. 23, 17, 32, 1, 18, 26, 30, 24, 27, 36, 9. Sol.
Arranging the marks in an ascending o order, we have : 9, 17, 18, 21, 23, 24, 26, 27, 30, 32, 36, 39, 41. Here, n = 13, which is odd.
Median marks = Value of
1 (13 + 1) th term = Value of 7th term = 26. 2
Ex.22 The weights (in kg) of 10 children are : 40, 52, 34, 47, 31, 35, 48, 41, 44, 38 Find the median weight. Sol.
Arranging the weights in ascending order, we have : Here, n = 10, which is even.
Median weight =
=
1 1 1 1 10 {40 + 41} th term 1th term {5th term + 6th term} = 2 2 2 2 2
81 kg = 40.5 kg 2
Hence, median weight = 40.5 kg
MANISH KUMAR
MATHEMATICS
EXERCISE SUBBJECTIVE TYPE QUESTIONS (A)
PRESENTATION OF DATA
1. 2. 3. 4. 5. 6.
Discuss the meaning and scope of statistics. What are different types of statistical data ? Which one is more reliable ? What are the steps taken to prepare a frequency table ? What do you mean by frequency distribution ? State its advantages. What are the characteristics and limitations of statistics ? Explain the following terms: (i) Variate (ii) Array (iii) Class interval (iv) Frequency (v) Cumulative frequency (vi) Class limits (vii) True class limits (viii) Class mark(ix) Class size. Array the data and form a frequency table for the following variables : 450, 453, 458, 459, 451, 460, 452, 453, 449, 460 ,451, 454, 457, 459 , 450, 459, 451, 454, 451, 459, 452, 455, 456, 458, 452, 460, 459, 455, 456, 457, Form an array and prepare a frequency tale for the following values of annual gold output (in lakhs kg) for 20 different years: 94, 95, 96, 93, 87, 89, 93, 89, 87, 88, 95, 96, 92, 87, 88, 90, 92, 91, 90, 94 The marks obtained by 30 students, out of 10 are : 2, 1, 6, 8, 4, 3, 0, 3, 3, 9, 5, 8, 1, 6, 7, 5, 0, 4, 4, 7, 3, 4, 2, 3, 3, 2, 7, 2, 4, 5 Array the data and form the frequency distribution. The following data gives the weight (in grams) of 0 oranges picked from a basket: 106, 107, 76, 109, 187, 95, 125, 92, 70, 139, 128, 100, 88, 84, 99, 113, 204, 141, 136, 123, 90, 115, 115, 110, 97, 90, 107, 75, 80, 118, 82 Construct a grouped frequency distribution taking classes of equal width 20 in such a way that the mid value of first class is 70. From the frequency table, find the number of oranges whose weights are (i) less than 100 gms (ii) more than 180 gms. Time taken in seconds by 25 students to solve a questions is given below : 20, 16, 20, 26, 27, 28, 30, 33, 37, 50, 40, 42, 4, 38, 43, 46, 46, 48, 49, 53, 58, 59, 60, 64, 52 By taking class interval of 10 seconds, make a frequency table. Construct a frequency distribution table for the following data of marks obtained by 25 students in a test in Mathematics in a school. Take 20 - 30 (30 not included) as one of the classes : 9, 25, 17, 12, 28, 20, 7, 31, 14, 43, 11, 19, 23, 37, 6, 24, 48, 10, 32, 17, 40, 31, 18, 24, 29. Now find the following: (i) the number of students getting less than 40 marks. (ii) the number of students getting 30 or more marks. Find the class mark of the class 11 - 19 Find the class mark of the class 8.5 - 15.5 If the class marks of a distribution of 32, 39, 46, 3 find the width of each class. If the class marks of a frequency distribution are 20, 25, 30, 35, 40 and 45, find the classes are their width. If the class marks of a frequency distribution are 19.5, 26.5, 33.5, 40.5, 47.5, 54.5, 61.5 then find the size of the class sand the classes. The class marks of a distribution are 11, 14, 17, 20, 23, 26, and 29. Find the width of the class and the true class limits.
7.
8.
9.
10.
11.
12.
13. 14. 15. 16. 17. 18.
MANISH KUMAR 19.
MATHEMATICS
The class marks of a distribution are 2.03, 2.23, 2.43, 2.63, 2.83, 3.03 and 3.23 Determine the class size and the class boundaries.
MANISH KUMAR
MATHEMATICS
20.
Find the true class limits 6 - 10, 11- 15, 16- 20, 21 - 25, 26 - 30, 31 - 35.
21.
Heights of seven persons in cm are 120, 125, 142, 134, 150, 155 and 128. Calculate the range.
22.
The weights in grams of 30 mangoes are : 70, 105, 108, 106, 76, 116, 110, 108, 188, 86, 87, 90, 110, 84, 75, 131, 81, 107, 90, 186, 100, 130, 138, 137, 140, 80, 84, 89, 110, 139 From the grouped frequency table by taking suitable classes. Also form the cumulative frequency table.
23.
The distances (in km) covered by 24 cars in 2 hours are given below : 125, 140, 128, 108, 96, 149, 136, 112, 84, 123, 130, 120, 103, 89, 65, 103, 145, 97, 102, 87, 67, 78, 98, 126 Represent the data as a cumulative frequency table using 60 as the lower limit of the first group and all the classes of class size 15.
24.
In a study of diabetic patients, the following data are obtained :
Age (in year)
No. of diabetic patients
10-20 20-30 30-40 40-50 50-60 60-70
3 6 14 9 5 2
Construct a cumulative frequency table for the above data. 25.
The ages of workers in a factory are as follows :
Age (in year)
No. of workers
21-23 23-25 25-27 27-29 29-31 31-33 33-35
3 4 5 6 5 4 3
Construct a cumulative frequency table for the above data. 26.
From the frequency distribution table given below, form (i) less than series and (ii) more than series.
MANISH KUMAR
MATHEMATICS
Daily Expenses
No. of persons
0-20 20-40 40-60 60-80 80-100 80-100
13 12 20 13 13 23
MANISH KUMAR s27.
28.
29.
30.
MATHEMATICS
Change the following cumulative frequency table into simple frequency table:
Class
Cumulative frequency
0-7 7-14 14-21 21-28 28-35 35-42 42-49
5 14 25 42 55 61 65
Form a frequency table from the following table :
Marks
No. of Students
Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80
15 35 60 84 96 127 198 250
Make a frequency table from the following :
Age (in years)
No. of persons
More than 100 More than 90 More than 80 More than 70 More than 60 More than 50 More than 40 More than 30 More than 20 More than 10 More than 0
0 7 18 26 41 63 75 98 117 139 150
The following data represents the cumulative frequency distribution of daily earnings of 100 laborers in a factory :
Marks
No. of Students
Less than 62 Less than 74 Less than 86 Less than 98 Less than 110
15 22 35 60 100
MANISH KUMAR
MATHEMATICS
Contruct of frequency table from the above data.
(B)
GRAPHICAL PRESENTATION OF DATA
1.
From the figures given below, write which of the following are histograms ?
2.
Subjects offered by the number of students is a school is given below : Subject No. of students
3.
Mathematics 50
Hindi 40
English 70
G.Sc. 60
Soc. St. 60
Daily sales (in Rs.) of Sushil Enterprises from Marth of August are given below Month Sale (in Rs.)
March 2000
April 2500
May 4500
June 3000
July
August
4000
3500
Draw a bar graph to depict the above data. 4.
Following table gives the birth rate per thousand of different countries over a certain period : Country Birth rate
India Germany U.K. 33 16 20 Represent the above data by a bar graph.
5.
15
Draw a bar graph to represent the following figures relating to manufacture o sewing machines : Year No. of sewing
6.
Sweden
China 40
1999 1000
2000 1500
2001 2000
2002 3000
2003
2004
2005
2500
2000
3500
The expenditure of M/s APC under different heads (in thousands of rupees) is as follows :
Head
Expenditure
Salary Travelling allowance Machines Rent Interest Miscellaneous
450 100 250 150 70 200
Draw a bar graph for the above data. 7.
The following table shows the number of books which are meant for only reference in the library :
MANISH KUMAR Subject Number
MATHEMATICS Mathmetics 40
Science 50
Languages 30
History 35
Commers 20
Draw a bar graph for the above data. 8.
The expenditure (in RS) of a family under different heads is shown in the bar graph. Observe the graph at answer the following questions :
(i)
What is the expenditure on rent ? (ii) On which item is the expenditure maximum and how much is the expenditure ? (iii) What is the expenditure of Education and Clothing together ? (vi) On which items is the expenditure minimum ? Draw a histogram for the following distribution : Class 80-100 100-120 120-140 140-160 Frequency 20 30 20 40
9.
10.
The ages of workers in a company are as follows : Age (in years) No. of workers
11.
20-25 20
100-120 30
120-140 20
140-160 40
160-180 90
Draw the histogram for the above data. The following is the distribution of monthly salaries of 925 employees of a company :
Salary (in Rs.)
No. of employees
1000-2000 2000-3000 3000-4000 4000-5000 5000-6000 6000-7000 7000-8000 Total
100 150 175 250 125 75 50 925
Represent the data by histogram.
MANISH KUMAR 12.
The daily profits (in rupees) of 100 shops are distributed as follows :
Profit per shop (in Rs.) No. of shapes 13.
MATHEMATICS 0-50 12
50-150 45
150-200 20
200-250 17
Draw the histogram for the above data. Weekly pocket expenses of 50 students of a school are given below : 10-20 20-40 40-50 Weekly pocket expences (in Rs.) 3 10 7 No. of students Draw a histogram for the above data
250-300 3
50-70 12
70-10 18
MANISH KUMAR 14.
MATHEMATICS
Draw a histogram of the marks obtained by the students of class IX in a class test :
Marks obtained No. of students 15.
15-30 24
30-40 12
40-60 20
60-65 8
65-80 27
0-10 3
10-20 7
20-30 12
30-40 8
40-50 6
50-60 4
Construct a frequency polygon from the following data :
Marks Frequency 17.
5-15 10
Draw a frequency polygon for the following distribution :
Marks No, of students 16.
0-5 4
0-50 5
5-10 11
10-15 4
15-20 9
20-25 10
25-303
30-35 2
35-40 5
Marks obtained by two groups of students of class IX in a school are given below :
Marks 30-40 40-50 50-60 60-70 70-80 80-90 90-100
No of students Gropus A Group B 10 12 25 14 15 5 1
16 7 9 20 22 2 3
Draw a frequency polygon for each of the groups on the same axes. 18.
Followings is the distribution of ages (in years) of two groups of teachers in a senior secondary school :
Marks 55-60 50-55 45-50 40-45 35-40 30-35 25-30 20-25
No of students Gropus A Group B 1 5 14 4 12 8 10 13
2 4 9 11 7 8 6 3
Represents the above data by means of a frequency polygon for each group of the same axes. 19.
Following is the distribution of ages (in years) of 32 teachers in a primary school.
Age (in years)
No. of Teachers
25-31 31-37 37-43 43-49 49-55
10 13 5 3 1
MANISH KUMAR
MATHEMATICS
Represent the above data by means of a histogram and a frequency polygon. 20.
21.
Draw a histogram and frequency polygon for the following data :
Wages (in Rupees)
No. of labourers
75-80 80-85 85-90 90-95 95-100 100-105
12 17 21 18 10 6
Histogram for the marks obtained by 80 students in a test in given in Fig. Answer the following questions :
(i)
How many students obtained marks between 20 and 30 ?
(ii)
How many students obtained marks less than 20 ?
(iii)
How many students obtained marks not less than 40 ?
22.
Histogram for the daily earning of 32 drug stores is given in Fig. Construct the corresponding grouped frequency table. Answer the following questions :
(i)
How many stores had daily earning between Rs. 1500 and Rs. 2500 ?
(ii)
How many stores had daily earnings between Rs. 3000 and Rs. 3500 ?
(iii)
How many stores had daily earning less than Rs. 2000 ?
(C)
MEASURES OF CETNRE TENDENCY
1.
Find the arithmetic mean of 17, 20, 18, 15, 24 and 23.
2.
In a cricket test match the scores of ten players are : 85, 32, 0, 54, 29, 101, 73, 64, 29 and 36 find the means of the runs.
3.
In a zonal athletic long jump meet the distances jumped by 10 athletes are : 205 cm, 200 cm, 275 cm, 260 cm, 259 cm, 199 cm, 252 cm, 239 cm, 288 cm and 281 cm. Find the arithmetic mean of the jumps.
4.
Find the mean of first ten odd natural numbers.
5.
Calculate the mean for the weekly pocket expenses of students given below : Pocket Expenses (in Frequency
6.
45 7
40 4
59 10
Find the mean of x, x + 2, x+ 4, x + 6 and x + 8.
71
58
47
65
6
3
8
1
MANISH KUMAR
MATHEMATICS
7.
If the arithmetic mean of 6, 8, 5, 7, x and 4 is 7, find the value of x.
8.
The mean of 15 observation is 20. If 8 is added to each observation, find the new mean.
9.
The mean of 27 observation is 35. If 5 subtracted from each observation, what will be the new mean ?
10.
The mean of 53 observation is 18. If each observation is multiplied by 3, what will be the new mean ?
MANISH KUMAR 11.
MATHEMATICS
The traffic police recorded the speed (in km/h) of 10 motorists as 47, 53, 49, 60, 39, 42, 55, 57, 52, 48. Later on an error in the recording instrument was found. Find the correct average speed of the motorist if the instrument
12.
recorded the speed 5 km/h less in each case. The mean of 25 observation is 16. If 4 is added to each of the first 10 observations, find the mean of the new set
13.
of 25 observations/ The mean height of 15 students is 154 cm. If is discovered later on that while calculating the mean, the item 175
14.
cm was read as 145 cm. Find the correct mean height. Mean of 25 observations was found to be 78.4. But later on its was discovered that 96 was misread as 69. Find
15.
the correct means. The mean marks of 25 students of X A is Board examination is 67 and that of 30 students of X B is 75. Find the
16.
mean of marks of all the 55 students correct to one decimal place. The average weight of 33 students of a class is 52 kg. The average weight of 18. of them is 48 kg. Find the
17.
average weight of the remaining students. The mean monthly salary of 12 employees of a firm is Rs 1450. If one more person joins the firm who gets Rs 1645 per month, what will be the mean monthly salary of 13 employees ?
20.
For the numbers 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9, verify that (x - x ) = 0. The average of 15 results is 50. If the average of first 8 results is 48 and that of the last 8 is 53, find the eighth result. Select the correct alternative for each of the following (20 - 21) : If each variate of a data is increased by 5, then arithmetic mean
21.
(i) remains the same (ii) is 5 times the original mean (iii) is increased by 5 Mean of a set of observations is the value which
22.
(i) occurs most frequently (ii) divides observations into two equal parts (iii) is the sum of the observations (iv) is a representative of the whole group. Find the median of each of the following data :
23.
(i) 7, 10, 5, 20, 18, 25, 17 (ii) 3, 5, 9, 2, 8, 7, 6, 7, 4 Find the median of each of the following data :
24.
(i) 5, 8, 16, 12, 11, 15, 10, 13, 6, 18, 20 (ii) 133, 73, 89, 108, 94, 140, 94, 86, 75, 135, 146 Find the median of each of the following data :
25.
(i) 12, 17, 3, 14, 5, 8, 7, 15 (ii) 25, 24, 26, 30, 27, 35, 24, 29 (iii) 25, 29, 35, 32, 46, 27, 24, 31, 23, 40 (iv) 25, 34, 31, 23, 26, 35, 29, 20, 32 The heights (in cm) of 12 girls are given below :L
18. 19.
(iv) is decreased by 5
(iii) 37, 42, 31, 46, 25, 27, 30, 32, 41
149, 153, 138, 148, 151, 160, 147, 143, 146, 142, 139, 152 27.
Calculate the median height. The marks obtained by 10 students in a test are : 12, 22, 32, 41, 26, 30, 14, 11, 18, 35
28.
Find the median marks. The runs scored by some players of a cricket team in a one day match are given below :
29.
83, 40, 36, 0, 69 ,105, 73, 21, 8 Find the median runs. The median of the following observations, arranged in ascending order is 32. Find the value of a.
30.
12, 14, 15, 27, a + 2, a + 3, 35, 36, 40, 49 Find the median of the following : 38, 46, 64, 87, 41, 58, 77, 35, 90, 25, 92, 33, 91.
MANISH KUMAR If 58 is replaced by 97 and 35 and 23, find the new median.
MATHEMATICS
MANISH KUMAR 31.
MATHEMATICS
Find the median of the following : 61, 58, 67, 60, 55, 72, 50 ,52, 64, 69, 70, 63. If 64 is replaced by 54 and 69 is replaced by 59, find the new median. Select the correct alternative for each of the following (32 - 33):
32.
The daily earning (in rupees) of 10 workers in a factory are 75, 90, 70, 50, 70, 50, 75, 90, 70, 72. The median
33.
wage is (i) Rs 70 (ii) Rs 71.5 (iii) Rs 72.5 (iv) Rs 75 Median of the observations 5, 7, 14, 12, 15, 17, 17, 5, 14, 7, 5 is
34. 35.
(i) 14 (ii) 12 (iii) 11.8(iv) 7 Calculate the mode for the following data : 20. 18, 20, 16, 15, 23, 16, 17, 15, 16 A cricket player scored the following runs in 11 one-day international matches :
36.
65, 30, 7, 60, 65, 65, 30, 28, 30, 15, 30 Find the model runs. Satish got the following marks in his weekly class test our of 50 : 48, 32, 36, 42, 38, 35, 39, 49, 34, 14, 32, 37, 31 What should he expect in the next weekly test out of the following ?
37.
(i) between 40 - 50 (ii) between 30 - 40 Calculate the mode for the following sizes of shoed sold out by a shopkeeper on Monday :
38.
6, 7, 6, 5, 10, 11, 10, 12, 6, 13, 13, 5, 11, 11, 10, 9, 8, 11, 8, 9, 11, 13 Using Empirical formula, calculate mode for the following date : 17, 16, 25, 23, 22, 23, 28, 25, 25, 23
39.
Find the mode for the following: 36, 32, 31, 31, 29, 31, 30, 32, 40, 46, 31, 32, 35, 37, 40.
40.
If one observation 31 is replaced by 32, will the modal value change ? If yes, find it. For what value of x, the mode for the following data is 9 ?
41.
8, 9, 6, 8, 9, 8, 7, 5, 8, 9, 9, x Select the correct alternative for each of the following (41 - 42): Mode of a set of observation is the value which (i) occurs most frequently
(ii) divides the observations into two equal parts
42.
(iii) is the mean of the middle two observations (iv) is the sum of the observations. Mode of the observations 5, 7, 2, 8, 5, 5, 8, 2, 5, 8, 7, 6 is
(D) 1.
(i) 7 (ii) 6 (iii) 6.5 (iv) 5 NCERD Questions. Give five examples of data that you can collect from your day-to-day life.
2. 3.
Classify the data in Q.1 above as primary or secondary data, The blood groups of 30 students of Class VIII are recorded as follows : A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
4.
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O, Represent third data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students ? The distance (in km) of 40 engineers from their residence to their place of work were found as follows : 5 19
3 10
10 12
20 17
25 18
11 11
13 32
7 17
12 16
31 2
7
9
7
8
3
5
12
15
18
3
12
14
2
9
6
15
15
7
6
12
MANISH KUMAR
MATHEMATICS
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation ?
MANISH KUMAR 5.
6.
7.
8.
9.
10.
11.
MATHEMATICS
The relative humidity (in %) of a certain city for a month of 30 days was as follows : 98.1 98.6 9.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1 89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3 96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89 (i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88 etc. (ii) Which month or season do you think this data is about ? (iii) What is the range of this data ? The heights of 50 students, measures to the nearest centimeters, have been found to be as follows : 161 150 154 165 168 161 154 162 150 151 162 164 171 165 158 154 156 172 160 170 153 159 161 170 162 165 166 168 165 164 154 152 153 156 158 160 160 161 173 166 161 159 162 167 168 159 158 153 154 159 (i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, et. (ii) What can you conclude about their heights from the table ? A study was conducted to find out the concentration of sulphur dioxide in the air in part per million (ppm) of a certain city. The data obtained for 30 days is as follows : 0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20 0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18 0.11 0.07 0.05 0.07 0.01 0.04 (i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08 and so on. (ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million ? Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows : 0 1 2 2 1 2 3 1 3 0 1 3 1 1 2 2 0 1 2 1 3 0 0 1 1 2 3 2 2 0 Prepare a frequency distribution table for the data given above. The value of upto 50 decimal places is given below : 3.1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8462643383279502884 19716939937510 (i) Make a frequency distribution of the digits from 0 to 9 after the decimal point. (ii) What are the most and the least frequently occurring digits ? Thirty children were asked about the number of hours they watched TV programs in the previous week. The results were found as follows : 1 6 2 3 5 12 5 8 4 8 10 3 4 12 2 8 15 1 17 6 3 2 8 5 9 6 8 7 14 12 (i) Make a grouped frequency distribution table for this data, raking class widths 5 and one of the class intervals as 5 - 10. (ii) How many children watched television for 15 or more hours a week ? A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows : 2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5 3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7 2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8 3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4 4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
MANISH KUMAR
MATHEMATICS
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5. 12.
A survey conducted by an organization for the cause of illness and death among the women between the ag 15 44 (in years) worldwide, found the following figures (in %): S.NO. Causes Female fatality rate (%)
13.
1. 2.
Reproductive health conditions Neuropsychiatric conditions
31.8 25.4
3. 4.
Injuries Cardiovascular conditions
12.4 4.3
5. 6.
Respiratory conditions Other causes
4.1 22.0
(i)
Represent the information given above graphically.
(ii) (iii)
Which condition is the major cause of women’s ill health and death worldwide ? Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii)
above being the major cause. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of India society is given below. Section
Number of girls per thousand boys
Scheduled Caste (SC) Scheduled Tribe (ST) Non SC/ST
940 970 920
Backward districts Non-backward districts
950 920
Rural 930 Urban 910 (i) Represent the information above by a bar graph. 14.
(ii) In the classroom discuss what conclusion can be arrived at from the graph. Given below are the seats won by different political parties in the polling outcome of a state assembly elections Political Party Seats Won
15.
A 75
B 55
C 37
D 29
E 10
F 37
(i) Draw a bar graph to represent the polling results. (ii) Which political party won the maximum number of seats ? The lengths of 40 leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table :
Length (in mm)
Number of leaves
118-126 127-135 136-144 145-153 154-162 163-171 172-180
3 5 9 12 5 4 2
MANISH KUMAR
MATHEMATICS
(i) (ii)
Draw a histogram to represent the given data. Is there any other suitable graphical representation for the same data ?
(iii)
Is it correct to conclude that the maximum number of leaves are 153 mm long ? Why ?
MANISH KUMAR 6.
17.
MATHEMATICS
The following table gives the life times of 400 neon lamps :
Life time (in hours
Number of lamps
300-400 400-500 500-600 600-700 700-80 800-900 900-1000
14 56 60 86 74 62 48
(i)
Represent the given information with the help of a histogram.
(ii)
How many lamps have a life time of more than 700 hours ?
The following table gives the distribution of students of two sections according to the marks obtained by them :
Section A Marks Frequencey 0-10 10-20 20-30 30-40 40-50
Section B Marks Frequency
3 9 17 12 90
0-10 10-20 20-30 30-40 40-50
5 19 15 10 1
Represents the marks of the students of both the sections of the same graph by two frequency polygons. From the two polygons compare the performance of the two sections. 18.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below.
Number of balls
Team A
Team B
1-6 7-12 13-18 19-24 25-30 31-36 37-42 43-48 49-54 55-60
2 1 8 9 4 5 6 10 6 2
5 6 2 10 5 6 3 4 8 10
Represent the data of both the teams of the same graph by frequency polygons. [Hint : First make the class intervals continuous.]
MANISH KUMAR 19.
20.
MATHEMATICS
A random survey of the number of children of various age groups playing in a park was found as follows :
Age (in years)
Number of children
1-2 2-3 3-5 5-7 7-10 10-15 15-17
5 3 6 12 9 10 4
Draw a histogram to represent the data above. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows : Number of letter 1-4 4-6 6-8 8-12 12-20
(i) 21.
22.
23. 24. 25.
26.
Number of surnames 6 30 44 16 4
Draw a histogram to depict the given information. (ii) Write the class interval in which the maximum number of surnames lie. The following number of goals were scored by a team in a series of 10 matches : 2, 3, 4, 5, 0, 1, 3, 3, 4, 3 Find the mean, median and mode of these scores. In a mathematics, test given to 15 students, the following marks (our of 100) are recorded : 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60 Find the mean, median and mode of this data. The following observations have been arranged in ascending order. If the median of the data is 63. find the value of x : 29, 32 , 48, 50, x, x + 2, 72, 78, 84, 95 Find the mode of 14, 25, 14, 28, 17, 18, 14, 23, 22, 14, 18. Find the means salary of 60 workers of a factory form the following table :
Salary (in Rs.)
Number of workers
3000 4000 5000 6000 7000 8000 9000 10000 Total
16 12 10 8 6 4 3 1 60
Given one example of a situation in which (i) the mean is an appropriate measure of central tendency.
MANISH KUMAR (ii)
MATHEMATICS
the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
MANISH KUMAR
MATHEMATICS ANSWER KEY
STATISTICS Marks 0 1 2 3 4 Value 5of variable 6449 7450 8 451 9 453 O F D A T A ( P g .
7.
No. of Students 2 2 4 6 5 Frequency 3 2 1 3 2 2 44 1 3
453
2
454
2
455
1
456
2
457
2
458
2
459
5
460
3
(A)
8.
87, 87, 87, 88, 88, 89, 89, 90, 90, 91, 92, 92, 93, 93, 9, 94, 95, 95, 96, 96.
9.
0, 0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9
PRE SEN TATI ON
555 1 N o. 365-367/Ex14.1) 449, 450, 450, 451, 451, 451, 451, 452, 452, 452, 453, 453, 454, 454, 455, 456, 456, 457, 457, 458, 458, Annual gold output Frequency 459, (in lakhs kg) 459, 87 3 459, 88 2 459, 460, 89 2 460, 90 2 460, 91 1 555 92 2 93 94 95 96
EXERCISE
10. Weight (in gms) 60-80 80-10 100-120 120-140 140-160 160-180 180-200 200-220 (i) 13 (ii) 2
2 2 2 2
No. of Oranges 3 10 9 5 1 0 1 1
11.
12.
Second
No. of Students
15-25 25-35 35-45
3 5 5
45-55 55-65
8 4
MANISH KUMAR
MATHEMATICS Marks obtained
No. of students
0-10 10-20 20-30
3 8 7
30-40 40-50
4 3
21-23
13. 16.
(i) 22 (ii) 7 15 14. 12 15. 7 17.5-22.5, 22.-27.5, 27.5-32.5, 32.5-37.5. 37.5-42.5,
17.
3
23-25
7
25-27
12
42.5-47.5, width of the class is 5 16-23, 23-30, 30-37, 37-44, 44-51, 51-58, 58-68,
27-29
18
29-31
23
31-33
27
18.
size of the class is 7. Width of the class is3, true class limit are 9.5-12.5,
33-35
30
19.
12.5-15.5, 15.-18.5, 18.5-21.5, 21.5-24.5, 24.5-27.5, 27.5-30.5 Class size is 0.20. Class boundaries are 1.93-2.13,
20.
2.13-2.33, 2.33-2.53, 2.53-2.73, 2.73-2.93, 2.933.13, 3.13-3.33. True class limits are 5.5-10.5,10.5-15.5, 15.5-20.5,
21. 22.
26.
20.5-25.5, 25.5-30.5, 30.5-35.5. 35 cm.
There will be deferent answers for different size of the class. 23. Distance covered (in km)
No. of cars (Cumulative frequency)
60-75
2
(i) Less than frequency Distribution Distribution Table Daily Expenses Number of persons (in Rs) Less than 20
13
Less than 40
25
Less than 60
45
Less than 80
58
Less than 100
81
Weight (in grams)
No. of mangoes
70-90 90-110
10 8
Cumulative frequency 10 18
4 6 0
22 28 28
2
75-90
6
110-130 130-150 150-170
90-105
12
170-190
105-120
14
120-135
20
135-150
24
Age (in years)
No. of patients (Cumulative requency)
Less than 120
(ii) More than Frequency Distribution Table Daily Expenses Number of persons (in Rs)
24.
10-20
3
20-30
9
30-40
23
40-50
32
50-60
37
60-70
39
25. Age (in years )
No. of workers (Cumulative frequency )
30 100
More than 120
0
More than 100
19
More than 80
42
More than 60
55
More than 40
75
More than 20
87
More than 0
100
Class
Frequency
0-7
5
7-14
9
27.
MANISH KUMAR
MATHEMATICS
14-21
11
21-28
17
28-35
13
35-42
6
42-49
4
Marks
No. of students
0-10
15
10-20
20
20-30
25
30-40
24
40-50
12
50-60
31
60-70
71
70-80
52
Age (in years)
No. of persons
0-10
11
10-20
22
20-30
19
30-40
23
40-50
12
50-60
22
60-70
15
28.
29.
70-80
8
80-90
11
90-100
7
Earnings (in Rs)
No. of labourers
50-62 62-74 74-86 86-98 98-110
15 7 13 25 40
30.
(B) 1. 8.
GRAPHICAL PRESENTATION OF DATE (a) and (d) (i) Rs. 1500 (ii) Food, Rs 2250 (iii) rs. 1750 (iv ) Education
21.
(i) 24
Marks
No. of students
0-10 10-20 20-30 30-40 40-50 50-60
12 16 24 16 4 8
(ii) 28
(iii) 12
28 Daily earnings (in repees) 1500-2000 2000-2500
No. of stores 14 9
2500-3000 3000-3500
(i) 23
3 6
(ii) 6
(iii) 14
MANISH KUMAR
MATHEMATICS 3.
(C)
MEASURES OF CENTRAL TENDENCEY
1.
19.5
2.
50.3 runs
3.
239.8 cm
4.
10
5.
54
6.
x+4
7.
12
8.
28
9.
30
10.
54
11.
55.2 km/hr
12.
17.6
13.
156 cm
14.
79.48
15.
71.4 marks
16.
56.8 kg
17.
Rs. 1465
19.
58
20.
(iii)
21.
(iv)
22.
(i) 17.
(ii) 6.
23.
(i) 12.
(ii) 94
24.
(i) 10.
(ii) 26.5 (iii) 30. (iv) 27.5
25.
43 kg
26.
Distances (in km) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 Total
28.
40 runs
29.
29.5
30.
58, 64
31.
62, 59.5
32.
(iii)
33.
(ii)
34.
16
35.
30 runs
36.
Between 3.-40
37.
11
38.
23.6
39.
31, Yes, 32
40.
9
41.
(i)
42.
(iv)
(D)
NCERT Questions
1.
Five examples of data we can gather from our day to day life are : (i) Number of students in our class (ii) Number of fans in our school. (iii) Electricity bills of our house for last two years (iv) Election results obtained from television or newspapers. (v) Literacy rate figures obtained from Educational Survey. Of course, remember that there can be many more different answer . Primary data : (i), (ii) and (iii) Secondary data : (iv) and (v)
B 6
O 12
AB 3
Total 30
Tally Marks |||| |||| |||| |||| |||| |||| |||| | | ||
Frequency 5 11 11 9 1 1 2 40
5 (i)
147.5 cm
24 Marks
A 9
4.
(iii) 31
27.
2.
Blood group No. of students
Relative humidity (in%)
Frequency
84-86 86-88 88-90 90-92 92-94 94-96 96-98 98-100
1 1 2 2 7 6 7 4
Total
30
6. (i)
Heights (in cm)
Frequency
150-155 155-160 160-165 165-170 170-175
12 9 14 10 5
Total
50
7. (ii)
(ii)
Concentration of sulphur dioxide (in ppm)
Frequency
0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.24
4 9 9 2 4 2
Total
30
The concentration of sulphur dioxide was more than 0.11 ppm for 8 days.
8. Number of heads Frequency
0 6
1 10
2 9
3 5
Total 30
MANISH KUMAR
MATHEMATICS
9. (i) Digits
Frequencncy 0
2
1
5
2
5
3
8
-4
4
5
5
6
4
7
4
8
5
9
8
Total
50
19.
(ii) The most frequently occurring digits are 3 and 9. The least occurring is 0.
Age (in years)
Freqnecy
Width
Length of the rectanble
1-2
5
1
2-3
3
1
3-5
6
2
5-7
12
2
7-10
9
3
10-15
10
5
15-17
4
2
5 1 5 1 3 1 3 1 6 1 3 2 12 1 6 2 9 1 3 3 10 5 2 5 4 1 2 2
10. (i) Number of hours Frequency
050 10
510 13
1015 5
1520 2
Total 30
(ii) 2 children. 11.
Now, you can draw the histogram using these lengths. Life of batteries (in years)
Frequency
2.0-2.5
2
2.5-3.0
6
3.0-3.5
14
3.5-4.0
11
4.0-4.5
4
4.5-5.0
3
Total
40
20.
12.
(i) Reproductive health conditions
14.
(ii) Party A
15.
(ii) Frequency polygon (iii) No
16.
(ii) 184
(i) Number of letters 1-4
Frequency
6
Width of interval 3
4-6
30
2
6-8
44
2
8-12
16
4
12-20
4
8
Now, draw the histogram. (ii) 6-8
Length of the rectangle
6 2 4 3 30 2 30 2 44 2 44 2 16 8 4 4 2 1 8
MANISH KUMAR 21.
Mean = 2.8; Median = 3; Mode = 3
22.
Mean = 54.8; Median = 52; Mode = 52
MATHEMATICS 23.
x = 62
24.
14
25.
Mean salary of
60 workers is Rs. 5083.33
PROBABILITY
INTRODUCTION In our daily life, we say that there is a probability that something’s will happen when it is likely to happen. We7often use phrases such as : “There is a probability that it will rain today”. “There is a probability that it will be a hot day tomorrow”. “There is a probability that he may be right”. etc. The statements : “It will rain today”. “It will be hot day tomorrow”. “He may be right”. are the events (or situations) about which have an element of uncertainty in our mind, that the event may happen or may not happen. The happening of an event rules out the possibility of the non-happening of the event and vive-versa. The element of uncertainty about the happening and on-happening of an event can be measured mathematically. The branch of Mathematics, which deals with it, is called Theory of Probability.
HISTORY OF PROBABILITY THEORY In old times, dice games was a common mode of gambling. In 1654, a French gambler Chevalier De-Mere, approached the well known 17th century French philosopher and mathematician Blaise Pascal (1623-1662) regarding some dice problems. Pascal discussed this problem with another French mathematician, Pierre DeFermat (1601-1665). Both these mathematician solved the problem independently. This work of Pascal and Fermat laid the foundation of the probability theory. The first book on the subject was published in 1663. The title of the book was “Book on Games of Change”. Significant contributions in the field were also made by mathematician J. Bernoulli (1654-1705), P. Laplace (1749-1827), .A. Markov (1856-1922) and A.N. Kolmogorov (in the 20th century).
DIFFERENT APPROACHES There are following three approaches to theory of probability. (i) Experimental approach or Empirical approach or Observed frequency approach. (ii) Classical approach. (iii) Axiomatic approach. Here, in this chapter, we shall study only the Empirical probability and not the other two approaches. The remaining two kinds of approaches are to be left for the time being and we will study these in the next classes.
RANDON EXPERIMENT In theory of probability, we deal with situations which are outcomes of a random experiment. This experiment is of a different nature is comparison to the experiment made in physical sciences. Here, an experiment repeated under same conditions, may not give the same result (or outcome). For example, we toss a coin and we get head but there is not guarantee that we must have the same result on the second toss. In this case, it may be tail. When we perform a random experiment, it is called a trial. For example, if we toss a cont (fair coin) once, the possible outcomes (i.e., results) are H and T. Here, H means head and T means tail. Now, we toss a pair of two similar fair coins, i.e., one trial are.
MANISH KUMAR
MATHEMATICS
(i) Tail on both the coins, i.e., no head. (ii) Tail on one coin and head on the other coin, i.e., exactly one head and one tail. (iii) Head on both the coin’s i.e., we get two heads. The above outcomes can be expressed in numerical numbers as 0, 1, 2, So, there are three possible outcomes of one trial of tossing two coins simultaneously and these outcomes are written as 0, 1, 2. Here, 0 means that no head occurs ; 1 means that exactly on head occurs ; and 2 means that exactly two heads occur. There is no possibility of any other value or outcome of this random experiment. Similarly, if we toss three similar (fair) coins simultaneously, then the possible outcomes will be 0, 1, 2 and 3.
MANISH KUMAR TOSSING OF A DIES AND ITS POSSIBLE OUTCOMES
MATHEMATICS
A die is fair and balanced cube with its six faccs marked with bots form one to six as below :
If a die is tossed , we find one face out of the six faces at the top. We count the number of dots on this face. The outcome is ‘n’ if the number of dots on the top face are n ; n = 1, 2, 3, 4, 5, 6. Thus, in this random experiment, there are six possible outcomes. The possible outcomes are : 1. i.e., we get 1 on the top face ; 2. i.e., we get 2 on the top face ; 3. i.e., we get 3 on the top face ;
6.
i.e. we get 6 on the top face.
EVENT ASSOCIATED WITH A RANDOM EXPERIEMENT An even associated with a random experiment is a collection of some outcomes out of the possible outcomes of the random experiment. Events are generally denoted by the capital letters A, B, E, F etc. For example, a pair of coins are tossed simultaneously and the possible outcomes are, 0 number of heads, 1 head and 2 heads. If E be the event of getting at least one head, then E = {1, 2}. Here we say that the outcomes 1 and 2 of the random experiment favours the happening of the event E but the outcome 0 does not favour the happening of the event E or in other words, the outcome 0 favours the happening of the event not E’. A teacher given four coins (fair) of same type to a student and asks him throw these four coins on the floor. The teacher makes enquiry about the number of heads from the other student. The correct answer given by that students is as below : Either 0 number of heads (all four are tails) or 1 head (other three are tails) or 2 heads (other two are tails) or 3 heads (one is tail) or 4 heads (no tail) Now, the teacher say that an event E is to get more than 1 heads. The third student is asked by the teacher to name the outcomes which favour the happening of the event E. The third students reply is an under : The favourable outcomes are 2, 3 and 4. We can write E = {2, 3, 4}. Two students play a game by throwing a die. Students X says that he will win if he gets an even number on the die and he will lose the game to the second student Y if the outcome is more than 4. Now, the questions arises, who is more beneficial, X o Y ? Let us examine the truth. The possible outcomes are 1,2, 3, 4, 5 and 6. Here, X wins the game if the outcomes are 2, 4 and 6, i.e., the three outcomes favours X. But if the outcomes are of values more than 4, then X loses the game and Y wins the game. Such outcomes are 5 and 6. Thus, only two outcomes favours Y. Clearly X is more beneficial than Y.
MANISH KUMAR
MATHEMATICS
In the above activity, event E is to get an even number and the favourable outcomes are 2, 4 and 6. Now, we denote the event of getting a number greater than 4 as F. So, the favourable outcomes for the, happening of the even F are 5 and 6.
DATA OF MORE THAN ONE TRIALS A single trial as a random experiment gives exactly one outcome at a time. On this basis, we cannot claim that the occurrence of all outcomes have equal changes. We will confirm, through practical activity, that the chances for the happening of all outcomes is equally likely. Practical Activity Take a fair coin, toss it 50 times and record the outcome after every toss in the form of value H (for head) and T (for tail). Here, the number of trials are 50 and the data has two distinct values as h and T. Now, prepare a frequency distribution table as below :
Outcomes
Number of times i.e.m frequency
H
f1
T
f2
Total Number of trials
=f1 + f2 = 50
Observe the value of the fractions (i)
f1 f and 2 . You will notice the following : 50 50
f1 f 0.5 (approximately) and 2 0.5 (approximately) 50 50
i.e., the possibility of the occurrences of head and tail is equally likely for each.
f f 1and 0 2 1 50 50
(iii)
0
(iii)
f1 f 2 1. 50 50
Practical Activity Toss a pair of coins simultaneously 50 times and record the outcomes as 0, 1 and 2 for each toss or trial. AS usual, 0 means no head and all tails ; 1 means one head and one tail; and 2 means two heads and no tail. Now, prepare a frequency distribution table as below :
Outcomes as number of heads
Number of times i.e., frequency
0 1
f1 f2
2
f3
Total number of tosses or trials
= f1 + f2 + f3 = 50
Observe the value of the fractions
f1 f2 f and 3 . You will notice the following , 50 50 50
MANISH KUMAR (i)
MATHEMATICS
All the fractions are approximately equal to
1 i.e., 0.3.... equally likely changes for the happening of 0. 1 3
and 2. (ii) 0
(iii)
f1 f f 1, 0 2 3 1 0 50 50 50
f1 f 2 f 3 1 50 50 50
EMPIRICAL PROBABILITY OF AN EVENT In practical activity 1, the value
f1 f and 2 are the empirical probabilities for th happening of the outcomes H 50 50
and T respectively. In practical activity 2, the values
f1 f2 f and 3 are the empirical probabilities for the happening of the , 50 50 50
outcomes 0, 1 and 2 respectively. In this activity, if we say that the even E happens when we get atleast one head and then prepare frequency distribution table for two events which are E and not E as below :
Event
Frequen cy
E (happens for (1 and 2) not-E (happens for 0)
(f2 + f3) f1
Total number of trials
50
Here, the values of the fractions
( f2 f3 ) f and 1 are empirical probabilities of the events E and not-E 50 50
respectively. For example, if we toss a pair of coins 20 times and we get 0,7 times ; 1, 7 times ; and 2, 6 times, we have the table for the events E and not-E as below :
Event
Frequen cy
E Not-E
7+6=13 7
Total number of trials
20
Probability of the event =
13 7 and probability of the even not E 20 20
Also, we find the Probability of E + Probability of not E
13 7 1 20 20
P (E) denotes the empirical probability for the happening of an event E. In this chapter, we shall use word probability in place of empirical probability, We define.
(Numbr of trial in which the event E happens) P(E) (The total number of trial made REMARK 1. REMARK 2.
P(E) + P (not-E) = 1 0 P (E) 1.
REMARK 3. Sure Event.
Sum of the probabilities of all the outcomes of a random experiment is 1.
MANISH KUMAR Ex.1
MATHEMATICS
When all the outcomes of random experiment favour an event, the event is called an impossible event and its empirical probability is 0. There are 40 students in a class and their results is presented as below :
Result (Pas/Fail)
Pas 30
Number of Students
Fail 10
If a student chosen at random out of the class, find the probability that the student has passed the examination. Sol.
Total number of chances = 40 Chances or trials which favour a student to pass = 30 The probability of the required event, i.e., the student has passed the examination
Ex.2
30 0.75 40
A coin is tossed 150 times and the outcomes are recorded. The frequency distribution of the outcomes H (i.e., head) and T (i.e., tail) is given below :
Outcome
H
T
Frequency
85
65
Find the value of P (H), i.e., probability of getting a head in a single trial. Sol.
Total number of trials = 150 Changes or trials which favour the outcome H = 85.
P(H) Ex.3
85 = 0.567 (approx) 150
A die is tossed 120 times and the outcomes are recorded as below : Outcome
1
Even Number less than 6
Odd number greater than 1
6
Frequency
20
35
30
15
Find the probability in a trial of getting (i) The number 1
(ii) The number 6
(ii) The even number les than 6. Sol.
(iv) The odd number greater than 1.
Total number of trials = 120 (i)
Changes which favour the outcome 1 are 20. So, P (getting 1)
(ii)
P (getting 6)
20 0.125 (approx) 120
15 = 0.125 120
MANISH KUMAR (iii)
MATHEMATICS
Now, getting an even number less than 6 implies that the outcomes 2 or 4. The total number of chances of getting 2 or 4, i.e., an even number less than 6 = 35
P (even number less than 6) = (iv)
35 7 = 0.292 (approx) 120 24
Getting an odd number greater than 1, i.e., 3 or 5. The changes favouring an odd number greater than 1 = 30. SO. P (an odd number greater than 1)
30 1 = 0.25 120 4
MANISH KUMAR Ex.4
MATHEMATICS
Two similar coins were tossed simultaneously 1000 times and the frequency distribution of heads obtained on each toss is as below :
No. of heads Frequency
Sol.
2
200
500
300
(i)
Total number of chances favouring the event of getting one head = 500. So, P (one head)
(ii)
P (two heads) =
(iii)
P (at least one head) = P (1 head or 2 heads) =
(iv)
P (less than two heads) = P (0 head or 1 head) =
(v)
P (three heads) =
500 =0.5 1000
300 = 0.3 1000 500 300 = 0.8 1000 200 500 = 0.7 1000
0 (No chance favour the occurrence of three heads) = 0 1000
200 500 300 =1 ( all changes are favorable) 1000 A dies is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the data below : P (not more than two heads )
1 180
Out come Frequency
So.
1
Find the probabilities of the following (i) Probability of getting one head. (ii) Probability of getting two heads. (iii) Probability of getting at least one head (iv) Probability of getting less than two heads (v) Probability of getting three heads. (vi) Probability of getting not more than two heads. Total number of trials = 1000
(vi)
Ex.5
0
2 150
3 158
4
5
6
145
176
191
Find the probability of getting each outcome. Also, show that the sum of the probabilities of all outcomes is 1. We denote the event of getting 1 by E1 and similarly, E2 , E3, E4 E4 and E6 for getting 2, 3, 4, 5 and 6 respectively. From the table, we have
P(E1 )
{Frequency of getting 1} 180 0.180 6 1000 f1 Total number of Trials i1
Similarly,
P( E 4 )
P( E 2 )
150 0.150 1000
145 0.145 1000
P(E3 )
P( E 5 )
P( E 6 )
191 0.191 1000
176 0.176 1000
150 0.158 1000
MANISH KUMAR Now,
Ex.6
MATHEMATICS
P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6)
= 0.180 + 0.150 + 0.158 + 0.145 + 0.176 + 0.191 = 1.000, i.e. 1. Hence, the sum of the probabilities of the outcomes is 1. There are 500 packets in a large box and each packet contains 4 electric devices in it. On testing at the time of packing, it was noted that there are some faulty pieces in the packets. The data is as below : No. of faulty Number of packets
0 300
1 100
2 50
3 30
4 20
Total number of packets 500
If one packet is drawn from the box, what is the probability that all the four devise in the packet are without any fault ? Sol.
When the packet has all the four devices without fault, it means the number of faulty devices in the packet is 0. Number of chances which are favourable to 0 are 300 as given in the table above. Thus, the probability of packet containing all the four devices without any fault
300 3 0.6 500 5
Ex.7
A factory manufacturing car batteries made a survey in the field about the life of these batteries. The data obtained is as under : Life time (in Freuency or the
Less than 24 40
24 to 36 220
36 to 48 540
more than 200
Total number of 1000
If you put a battery of this company in your car, what is the probability that (i) the battery will last for more than 36 months ? (ii) the battery will last for less than 48 months ? (iii) the battery will last for 36 to 48 months ? Sol.
Total frequency or the total number of trials made = 1000 (i)
Te total number of batteries which last for more than 36 months = 540 + 200 = 740 Now, P(battery will last for more than 36 months) =
(ii)
The total number of batteries which last for less than 48 months = 40 + 220 + 540 = 800 So, P (battery will last for less than 48 months) =
(iii)
800 = 0.80 1000
The total number of batteries which last for 36 to 48 months = 540 So, P(battery will last for 36 to 48 months) =
Ex.8
740 0.74 1000
540 = 0.54 1000
400 students of class X of a school appeared in a test of 100 marks in the subject of social studies and the data about the marks secured is as below : Marks secured Number of
0-25 50
26-50 220
51-75 100
Above 75 30
Total number of 400
MANISH KUMAR
MATHEMATICS
If the result card of a student he picked up at random what is the probability that the student has secured more than 50 marks. Sol.
Total member of students, i.e. the total frequency = 400 The total number of students who secured more than 50 marks = 100 + 30 = 130
130 1.3 = 0.325 400 4
Probability that the marks secured are more that 50 = Ex.9
100 plants each, were planted in 100 schools during Van Mahotsava. After one month, th number of plants that survived were recorded as in data below : Number of plants survied
Less than 25
26-50
51-60
61-70
More than 70
Total number of Scools
15
20
30
30
5
100
Number of Schools = frequency
When a school is selected at random for inspection, what is the probability that :
Sol.
(i)
More than 25 plants survived in the school ?
(ii) (iii)
Less than 61 plants survived in the school ? 61 to 70 plants survived in the school ?
Total frequency or the total number of schools in which plants were planted = 100 (i)
Number of schools in which more than 25 plants survived = 20 + 30 + 30 + 5 = 85 P (more than 25 plants survived in the school)
(ii)
85 = 0.85 100
Number of school in which less than 61 plants survived = 15 + 20 + 30 = 65 P (Less than 61 plants survived)
(iii)
P (61 to 70 plants survived) =
65 = 0.65 100
30 = 0.30 100
Ex.10 Fifty seeds were selected at random from each of 5 bags A, B, C, D, E of seeds, and were kept under standardized conditions equally favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follow : Bag
A
B
C
D
E
Number of seeds germinated
40
48
42
39
41
What is the probability of germination of (i) More that 40 seeds in a bag ?
Sol.
(ii) (iii) (i)
49 seeds in a bag ? more than 35 seeds in a bag ? Number of bags in which more than 40 seeds out of 50 seeds germinated = 3
(These are B, C and E) Total number of bags = 5
MANISH KUMAR
MATHEMATICS
So, P (more than 40 seeds in a bag germinated (ii)
Number of bags in which 49 seeds germinated = 0 So, P (49 seeds germinated in bag)
(iii)
3 = 0.60 5
0 =0 5
Number of bags in which more than 35 seeds out 50 seeds germinated = 5 Total number of bags = 5 So, P (more than 35 seeds in a bag germinated)
5 =1 5
Ex.11 An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained are given in the following table : Age group of drivers
Number of accidents in one year
(in years)
0
1
2
3
More than 3
18-29
440
160
110
61
35
30-50
505
125
60
22
18
Above 50
360
45
35
15
9
Find the probability of the following events for a driver selected at random from the city :
Sol.
(i)
being 18-29 years of age and having exactly 3 accidents in one year.
(ii)
being 30-50 years of age and having one or more accidents in a year.
(iii)
having no accident in one year.
(i)
The number of drivers in the age group 18-29 having exactly 3 accident = 61 Total number of trials = 2000 So, P (driver in age group 18-29 having exactly 3 accident in one) =
(ii)
61 = 0.0305 2000
The number of drivers in the age group 30-50 and having one or more than one accident in one year = 125 + 60 + 22 + 18 = 225 P (driver in angel group 30-50 having one or more accidents in one year) =
(iii)
225 = 0.1125 2000
The number of drivers having no accident in one year = 440 + 505+ 360 = 1305 So, P (driver having no accident) =
1305 = 0.6525 2000
MANISH KUMAR
MATHEMATICS
EXERICE 1.
Answer the following in one word, one sentence or as per the exact requirement : (i)
IF P (E) = 0.2 find P (not = E)
(ii)
“Probability of an event can not be greater than I”. Is the statement true or false ?
(iii)
“P (E) > P (not - E)” Is the statement true or false ?
(jv)
What is the probability of a sure even ?
(v)
If a coin is tossed 40 times and 19 times head comes and 21 times tail comes, write the probability of getting a head in a trial out of these 40 trials of the experiment.
2.
There are 50 students in a class and their results in as below : Result (Pass/Fail)
Pass
Fail
No. of students
35
15
If a student chosen at random out of the class (i.e., without any bias), find the probability that the student is not failing (i.e., the student passed the examination). 3.
A coin is tossed 400 times and the data of outcomes is a below : Outcome (H/T)
H
T
Frequency
280
120
Find (i) P (H), i.e., probability of getting head and (ii) P (T), i.e., probability of getting tail. (iii) the value of P (H) + P(T). 4.
A die having six faces is tossed 80 times and the data is as below : Outcome
1
2
3
4
5
6
Frequency
10
20
10
28
8
4
Find (i) P (1) (ii) P (4) (ii) P (6) (iv) P (5). 5.
Three similar coins were tossed simultaneously for 100 times and the data recorded is an given below : No. of heads
0
1
2
3
Total number of tosses
No. of tosses = fequency
22
30
28
20
100
(i)
Find the probability of two heads and one tail in a toss.
(ii)
Find the probability three heads and no tail is no toss.
(iii)
Find the probability atleast one head is a toss.
(iv)
Show that the sum of the probabilities of all the possible outcomes in a toss is equal to 1.
(v)
Find the probability of getting 4 heads in a toss.
MANISH KUMAR
7.
MATHEMATICS
(vi)
Find the probability of getting not more than 3 heads in a toss.
(vii)
Find the probability of getting more than one head in a toss.
(viii)
Find the probability of getting less than three heads in a toss.
A die is thrown 200 times and the outcomes 1, 2, 3, 4, 5, 6 have frequencies as below : Outcome
1
2
3-Jan
4
5
6
Frequency
40
38
43
29
28
22
Find the probabilities of the following events in a toss (trial): (i) (iii) (v)
getting 6 (ii) getting 1 getting 3 (iv) getting an even number getting and odd number (vi) getting a number more than 3
(vii) getting number less than 3 (viii) getting a number less than 5 (ix) getting a number more than 6 (x) getting a number less than one. (xi) getting a number more than 1 and less than 6. 7.
One page of a telephone directly, there were 200 telephone numbers. The frequency distribution of their unit place digit (for example in the number 25828573, the unit place digit is 3) is given is table below : Digit
0
1
2
3
4
5
6
7
8
9
Frequency
22
26
22
22
20
10
14
28
16
20
Without looking at the page, the pencil is placed on one of these number, i.e., the number is chosen at random. What is the probability that the digit in its unit place is (i) 6 (ii) less than 3 (iii) more than 7. 8.
A type manufacturing company kept a record of the distance covered before a type needed to be replaced. The table show the result of 1000 cases : Distance
Less thn
4000 to
9000 to
(in km)
4000
9000
14000
Frequency
20
210
325
More than 14000 445
If you buy a tyre of this company what is the probability that :
9.
(i) (ii)
it will need to be replaced before it has covered 4000 km? it will last more that 9000 km ?
(iii)
it will need to be replaced after it had covered somewhere between 4000 km and 14000 km ?
Twenty bags of sugar, each marked 10 kg, actually give the following data : Weight of a bag (in kg)
9.5-9.8
9.8-9.9
9.9-10.0
10.0-10.1
Frequency
1
2
5
12
The lower limits of the classes are inclusive and the upper limits are exclusive.
MANISH KUMAR
MATHEMATICS
What is the probability that the bag selected at random (without any reference) weight 10 kg or more ? 10.
The record of weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times.
11.
(i) What is the probability that on a given day it was correct ? (ii) What is the probability that it was not correct on a given day. A survey was conducted by can manufacturing company in a metropolitan city on 1000 persons having monthly income from Rs. 30,001 to Rs. 50,000. The data about the number of persons in various categories is an under Monthly income
Number of Cars 2 More than 2
(in rupees)
1
30,001-40,000
400
50
25
40,001-50,000
100
300
125
Find the probability that a person selected at random
1.
(i) in the income slab 40,001-50,000 have more than 2 cars (ii) in the income slab 30,001 to 50,000 have to cars. NCERT Questions: In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she
2.
did not hit a boundary. 1500 families with 2 children were selected randomly, and the following data were recorded. No. of girls in a No. of families
2 475
1 814
0 211
Compute the probability of a family, choses at random having (i) 2 girls (ii) 1 girls (ii) No girl 3.
Also check whether the sum of these probabilities is 1. Refer to example 5, Section 14.4, Chapter 14 of NCERT. Find the probability that a student of the class was born in August. The statement of the data in the example is states as below : In a particular section of class IX, 40 students were asked about the months of their birth and the following graph was prepared to represent the data :
MANISH KUMAR
MATHEMATICS Month of Birht
4.
Three coins are tossed simultaneously 200 limes with following frequencies of different outcomes : Outcome Frequency
3 heads 23
2 heads 72
1 heads 77
No head 28
If the three coins are simultaneously loosed again, compute the probability of 2 heads coming up.
MANISH KUMAR 5.
MATHEMATICS
An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below : Monthly Income
Vehicles per family 0 1 2 Above 2 Less than 7000 10 160 25 0 Suppose a family is chosen. Find the probability that the family chosen is
6.
(i)
earning Rs. 10000 - 13000 per month and owning exactly 2 vehicles.
(ii)
earning Rs 16000 or more per month and owning exactly 1 vehicle.
(iii)
earning less that Rs. 7000 per month and does not own any vehicle.
(iv)
earning Rs. 13000 - 16000 per month and owning more than 2 vehicles,
(v)
owning not more than 1 vehicle.
Refer to chapter 14 (NCERT) the table below :-
Marks (our of 100) No. of students
7.
0-20 7
20-30 10
30-40 10
40-50 20
50-60 20
60-70 15
70-above 8
(i)
Find the probability that a student obtained less than 20% in the mathematics test.
(ii)
Find the probability that a student obtained marks 60 or above.
Total 90
To know the opinion of the student about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table. Opinion No. of Students
Like 135
Dislike 65
Find the probability a student chosen at random (i) likes statistics, 8.
(ii) does not like it.
Refer to Q.2, Exercise 14.2 (NCERT). What is the empirical probability that an engineer lives. (i) less than 7 km from her place of work (ii) more than or equal to 7 km from her place of work. (iii) within
9.
1 km from her place of work ? 2
Activity : Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
MANISH KUMAR 10.
MATHEMATICS
Activity : - Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by by/her is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
11.
Eleven bags of wheat flour, each marked 5 kg. actually contained the following weight of flour (in kg ): 4.97
5.05
5.08
5.03
5.00
5.06
5.08
4.98
5.04
5.07
5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour. 12.
In Q. 5, Exercise 14.2 (NCERT), you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of certain city for 30 days. Using this table, find the probability, of the concentration of sulphur dioxide in the interval 0.1 - 0.16 on any of these days.
13.
In Q. 1, Exercise 14.2 (NCERT), you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. use this table to determine the probability that a student of this class, selected at random, has blood group AB.
PROBABILITY
ANSWERS KEY
1.
(i) 0.8 (ii) True (iii) False (iv) 1 (v) 1940
2. 07
4.
(i 0.125 (ii) 0.35 (iii) 0.05 (iv) 0.1
5. (i) 0.28 (ii) 0.20 (iii) 0.78 (v) 0 (vi) 1 (vii) 0.48 (iii) 0.80
6.
(i) 0.11 (ii) 0.20 (iii) 0.215 (iv) 0.445 (v) 0.555 (vi) 0.395 (vii) 0.75 (iii) 0 (ix) 0 (x) 0.69
7.
(i) 0.07 (ii) 0.35 (iii) 0.18
8. (i) 0.02 (ii) 0.77 (iii) 0.535
10. (i) 0.7 (ii) 0.3
11. (i) 0.125 (ii) 0.05
EXERCISE
3. (i) 0.7 (ii) 0.3 (iii) 1
9. 0.6
NCERT Questions: 1.
24 4 , i.e., 30 5
2. (i)
19 407 211 (ii) (iii) 60 750 1500
5. (i)
29 579 1 1 1031 (ii) (iii) (iv) (v) 2400 240 240 240 1200
8. (i)
9 31 (ii) (iii) 0 40 40
11.
7 11
3. 6. (i)
12.
1 15
2 20
7 23 (ii) 90 90 13.
1 10
4.
9 25
7. (i)
27 13 (ii) 40 40
MANISH KUMAR
MATHEMATICS
Importatnt Notes
