Class 6 Imo Wkbsol e Book

International Mathematics Olympiad

WORK BOOK

6

INSTANT

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Chapter - 1: KNOWING OUR NUMBERS 1.

(B) : In the given number 605, 116, 218 ; 5 is at Million’s place. Therefore place value of 5 = 5 × 10 6 = 5 million.

2.

(C)

3.

(B) : For the formation of greatest 4 digit number using different digit we have 4 places. Thousand’s place, Hundred’s place, Ten’s place, One’s place 9 ® The largest digit at thousand’s place 8 ® The second largest digit at hundred’s place 6 ® Given that 6 is at ten’s place 7 ® Third largest digit at unit’s place \ Number is 9867.

4.

(A) : Mustard oil in a bottle = 920 ml Mustard oil in 25 bottles = 25 × 920 ml = 23000 ml 23000 1 l = 23 l l , then 23000 ml = 1000 1000 (B) : Difference of 500 and 200 = 500 – 200 = 300 Since C = 100, then CCC = 300.

Since 1 ml =

5. 6.

(B) : Roman numeral starts from I.

7.

(D) : Expanded form of 37034 = 3 × 10000 + 7 × 1000 + 0 × 100 + 3 × 10 + 4 × 1 = 30000 + 7000 + 0 + 30 + 4.

8.

(D) : 10 = 10 + 9 = X + IX = XIX.

9.

(D) : Capacity is not measured in centimetres as it is measuring unit of length.

10. (C) : 498 = (500 – 100) + (100 – 10) + 8 = CDXCVIII. 11. (B) : To write a number in which the smaller digit placed before the greater digit, we subtract the value of the small digit from that of the greater digit. 12. (C) : XX = 10 + 10 = 20, XXXVI = 10 + 10 + 10 + 6 = 36, V = 5 Ascending order = V, XX, XXXVI. 13. (D) : CDI + V + C + X = ((500 – 100) + 1) + 5 + 100 + 10 = (400 + 1) + 5 + 100 + 10 = 516. 14. (C) : Underlined digit is at the hundred’s place.

Class 6

15. (A) : Using the place value chart, we may write

Ten Crores Crores 3

1

Ten Lakhs Ten Thousands Hundreds Tens Lakhs Thousands 5

5

3

2

8

Ones

0

0

Hence the standard numeral = 31,55,32,800. 16. (B) : Greatest 7 digit number = 9999999 Smallest 5 digit number = 10000 Difference between them = 9999999 – 10000 = 9989999. 17. (B) : Tens place comes on immediate right of the hundreds place. 18. (C) : Roman Numbers

I

V

X

L

C

D

M

HinduArabic Numerals

1

5

10

50

100

500

1000

Since C is lying between two roman numerals greater than itself therefore C is subtracted from digit at right MCDXVIII = 1000 + (500 – 100) + 10 + 8 = 1000 + 400 + 18 = 1418. 19. (B) : Million period in international number system is given below :

Hundred Ten Millions Millions Millions 8

10

7

10

Ten Crores Crores

6

10

Ten Lakhs

20. (D) : Largest number using 4, 0, 3, 7 = 7430 Smallest number using 4, 0, 3, 7 = 3047. 21. (C) : The face value of a digit in a numeral is its own value, at whatever place it may be. Therefore face value of 3 in 31005660 is 3. 22. (C) : Beginning from the right, a comma is put after the first three digits and then after two digits. 23. (C) : Sugar distributed among 12 persons = 55 kg 200 g = (55000 + 200)g = 55200 g \ Sugar received by 1 person = 55200 ÷ 12 = 4600 g = 4 kg 600 g. 24. (B) : Place value and face value are always equal at one’s place.

IMO Work Book Solutions

25. (A) : In international system 315233729 can be written as Three hundred fifteen million, two hundred thirty three thousand seven hundred twenty nine. 3 1 5 2 3 3 7 2 9 Units Tens Hundreds Thousands Ten Thousands Hundred Thousands Millions Ten Millions Hundred Millions

26. (B) 27. (B) : Since X is lying between C and L therefore X is subtracted from L \ CCCXLVII = 100 + 100 + 100 + (50 – 10) + 7 = 300 + 47 = 347. 28. (B) : Since it is given that digit cannot be repeated therefore three digit number are 350, 305, 550, 503 i.e. 4 numbers. 29. (A) : Number of screws produced in a day = 2825 Number of screws produced in 30 days = 2825 × 30 = 84750 Number of screws 5 dealers got = 84750 Number of screws each dealer got = 84750 ¸ 5 = 16950 . 30. (A) : Number name for 900800700 is ninety crores eight lakhs and seven hundred. 9 0 0 8 0 0 7 0 0 Units Tens Hundreds Thousands Ten Thousands * Lakhs Ten lakhs Crores Ten crores

vvv

Class 6

Chapter - 2 : NUMBER SYSTEM 1.

(C) : 2 is the only even prime number.

2.

(D) : Definition

Example

(A) Closure property : A set is closed under an operation when the result of the operation on any two members in the set is also in that set.

Let a = 1, b = 2, c = 4 (A) a ÷ b = 1 ÷ 2 = 0.5 (not a whole number) \ Closure property is not satisfied under division (B) Commutative property : An operation is (B) a ÷ b = 1 ÷ 2 = 0.5 commutative if you can change the order b ÷ a = 2 ÷ 1 = 2 without affecting the result. \ Commutative property is not satisfied under division. 1 (C) Associative property : If a, b, c are three (C) (a ÷ b) ÷ c = (1 ÷ 2) ÷ 4 = 8 whole numbers then it should satisfy a ÷ (b ÷ c) = 1 ÷ (2 ÷ 4) = 2 (a * b) * c = a * (b * c) under the operation * \ (a ÷ b) ÷ c ¹ a ÷ (b ÷ c) \ Associative property is not satisfied under division. 3. 4. 5.

(B) : As explained in Q. 2, Associative property is used between three numbers. (C) : Remainder r can be zero or less than divisor. (B) : Whole number : Starts with 0 and goes to infinity. Natural number : Starts with 1 and goes to infinity. ‘0’ is the least whole number and there in no largest definite whole number.

6.

(B) : As explained in Q. 3, the given equation satisfies the associative property.

7.

(B) : a and b are whole numbers and a > b and c ¹ 0, c > 0 so a × c > b × c. For example : a = 2 and b = 1 as a > b Let c = 3 ¹ 0 and 3 > 0, then a × c = 6 > 1 × 3 = 3 = b × c.

8.

(C) : Face value of 5 in 165,234 = 5 Face value of 9 in 842,928 = 9 Difference between them = 9 – 5 = 4.

9.

(D) : Let a, b be two whole numbers (A) Closure property : a, b are two whole numbers, but a – b may not be a whole number. (B) Commutative property : a, b are two whole numbers but a – b ¹ b – a (C) Associative property : a, b and c are three whole numbers but (a – b) – c ¹ a – (b – c).


10. (B) : Number of red balls in basket = 222 \ 222 ÷ 6 = 37 groups of 6 balls are replaced by 12 white balls 37 times. \ Number of white balls = 37 × 12 = 444 11. (A) : Multiplicative Identity : The result of multiplying a number by 1 is equal to the number itself, then 1 is the identity element with respect to multiplication. It is also called the multiplicative identity. 12. (D) : 82 × 1 = 82. 13. (A) : Commutative law : If a and b be two whole numbers then, a – b = b – a. It is possible when a = b. 14. (D) 15. (C) : If a × b = 0 then either a = 0 or b = 0. 16. (D) : In option (D) it should be reciprocal of 0 which is not defined. 17. (A) : On the number line negative integers are on the left of 0 and positive integers are on the right of 0. Hence positive integers > zero > negative integer. 18. (D) : Let a and b be two integers. Let a = –6, a × b = –48 Þ –6 × b = –48 Þ b = 8 19. (A) : (–7 + (–2)) = (–7 –2) = –9 Integer 2 more than –9 = –9 + 2 = –7. 20. (A) : Positive sign preceding the bracket does not change the sign of term when bracket is removed. 21. (B) : L.H.S. = (–25) – (–42) – (–27) = – 25 + 42 + 27 = 44 R.H.S. = (–42) – (–25) + (–22) = – 42 + 25 – 22 = – 39 \ (–25) – (–42) – (–27) > (–42) – (–25) + (–22). 22. (B) : Square of any –ve integer is positive. 23. (B) : Negative sign preceding the bracket changes the sign of terms when bracket is removed. 24. (A) : 3 + 4 – 2 changes to 3 × 4 ¸ 2 = 3 × 2 = 6. 25. (C) : We first remove the ( ), then { } and lastly [ ]. N A

26. (D) : W

3 km E 8 – 3 = 5 km

8 km

\

Final position is 5 km towards south.

B S

27. (B) : 0 is greater than negative integer and less than positive integer. If a > b, then –a < –b Hence if 99 > 37, then –99 < –37 and –37 < 0 < 2 Þ –99 < –37 < 0 < 2

Class 6

28. (B) : Let –a be negative number. Then (–a) × (–a) × (–a) = a 2 × (–a) = – a 3 (Negative) (–a) × (–a) × (–a) × (–a) × (–a) = a 2 × a 2 × (–a) = –a 5 (Negative) And so on. Hence, we will get a negative number. 29. (D) : Let a and b be two integers. According to question a = b + 4 and b = –16 \ a = –16 + 4 = –12 30. (A) : Natural numbers (N) : N = {1, 2, 3, 4,........} Whole numbers (W) = If “0” is included in the natural numbers it forms the set of whole numbers. W = (0, 1, 2, 3, 4, .........) Integers (I) : The natural numbers 1, 2, 3, ...... together with “0” and their negatives –1, –2, – 3,... are called integers. Hence N Ì W Ì I. vvv


Chapter - 3 : PLAYING WITH NUMBERS 1.

(A) : Coprime number : Two numbers are said to be coprime numbers if highest common factor between them is 1. Prime number : A number which is divisible by 1 and itself only is called a prime numbers. \ (9, 8) are coprime numbers.

2.

(B) :

Pairs

Common factors

7 and 63 36 and 25 35 and 21 63 and 81

1, 7 1 1, 7 1, 3, 9

3.

(D) : A natural number is divisible by 11 if and only if the difference between the sums of alternate digits is divisible by 11. Difference between sums of alternate digits in 653 * 47 = (6 + 3 + 4) – (5 + * + 7) = (13) – (12 + *) If * = 1, then difference = 13 – 13 = 0 and 0 is divisible by 11. Hence 653147 is divisible by 11, if * = 1.

4.

(D) : 70060800 + 109003 = 70169803 (A) 70169803 is not divisible by 7. (B) A natural number is divisible by 3 if and only if the sum of its digits is divisible by 3. \7 + 0 + 1 + 6 + 9 + 8 + 0 + 3 = 34 is not divisible by 3. (C) A natural number is divisible by 9 if and only if the sum of its digits is divisible by 9. \7 + 0 + 1 + 6 + 9 + 8 + 0 + 3 = 34 is not divisible by 9. (D) Difference between sum of alternate digit in 70169803 = (7 + 1 + 9 + 0) – (0 + 6 + 8 + 3) = 17 – 17 = 0. 0 is divisible by 11, hence 70169803 is divisible by 11.

5.

(B) : Product of two numbers = Their L.C.M. × H.C.F. Þ a × b = L.C.M. (a, b) × H.C.F. (a, b)

6.

28 ´ 336 = 84. 112 (B) : Twin Primes : The prime numbers differing by 2 are called Twin Primes.

7.

(A) : We have (20 – 14) = 6, (25 – 19) = 6, (35 – 29) = 6 and (40 – 34 ) = 6

Þ 112b = 28 × 336 Þ b =

2 20, 25, 35, 40 2 10, 25, 35, 20 5

5, 25, 35, 10 1, 5, 7, 2

Class 6

L.C.M. = 2 × 2 × 5 × 5 × 7 × 2 = 1400 \ Required Number = (L.C.M. of 20, 25, 35, 40) – 6 = 1400 – 6 = 1394. 8.

(B) : Odd numbers : A number which is not divisible by 2 is called an odd number.

9.

(A) : Maximum capacity of a container which can measure the Kerosene out of both the tankers when used an exact number of times = H.C.F. (850, 680) = 170 litres.

10. (C) : Observing the given pattern the value of sum can be find by multiplying number of terms to number of terms. Hence 1 + 3 ...... + 19 = 10 × 10 = 100. 11. (A) 12. (A) : L.C.M. of 2, 3, 4, 5, 6 and 7 = 2 × 2 × 3 × 5 × 7 = 420.

2 2, 3, 4, 5, 6, 7 3 1, 3, 2, 5, 3, 7 1, 1, 2, 5, 1, 7 When 10000 is divided by 420 we get 23 as quotient and 40 as remainder. So, two numbers nearest to 10000 divisible by 2, 3, 4, 5, 6 and 7 are 420 × 23 and 420 × 24 = 9660 and 10080. 13. (B) : Prime numbers have two factors while composite numbers have more than two factors. 14. (C) : The side of square tiles needed to pave a plot 225 m by 30 m = H.C.F. of 225 and 30 = 5 m. 15. (D) : Let x divided by 7 gives quotient y & remainder 2 Þ x = 7y + 2 Now x divided by 6 gives quotient z & remainder 3 Þ x = 6z + 3 Since number is same, so 7y + 2 = 6z + 3 1 + 6z 7 1 + 6z The least value of z for which is a whole number is z = 1. 7 \ x = 6 × 1 + 3 = 9.

Þ

7y – 6z = 1 Þ

y =

16. (C) : L.C.M. of two coprime numbers is their product because their H.C.F. is 1. 17. (D) : Product of two numbers = H.C.F. × L.C.M. Þ H.C.F. = Product of numbers ÷ their L.C.M. 18. (C) : The greatest number of fruits possible in each heap = H.C.F. (153, 119) = 17. 19. (C) : Prime numbers between 1 to 50 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. 20. (D) : 2 × 45 = 90, 5 × 18 = 90 and 9 × 10 = 90. Also all these pairs are coprime.


21. (B) : Smallest prime number = 2 1 Reciprocal of this number = . 2

22. (A) 23. (C) 24. (C) : 3,116,365 is divisible by 5 and 7 both. 25. (C) : 16 – 9 = 7 is not divisible by any of the given number 48 – 9 = 39 is not divisible by any of the given number 57 – 9 = 48 is divisible by each of the given number 39 – 9 = 30 is not divisible by any of the given number. or L.C.M. of 12, 16, 24 and 48 = 2 × 2 × 2 × 2 × 3 = 48 2 12, 16, 24, 48 2 6, 8, 12, 24 2 3, 4, 6, 12 2 3, 2, 3, 6 3 3, 1, 3, 3 1, 1, 1, 1

\ Required number = 48 + 9 = 57 26. (C) : 36 = 6 2 and 81 = 9 2 27. (C) : a × 8 = 5 × 8 = 40 28. (D) : 1265 has 5 at units place, so it is not divisible by 2 and 10. Sum of digits of 1265 = 1 + 2 + 6 + 5 = 14 which is not divisible by 3, so 1265 is not divisible by 3. Difference of the sums of alternate number in 1265 = (1 + 6) – (2 + 5) = 7 – 7 = 0. \ 1265 is divisible by 11. 29. (A) : Prime factorisation of 13915 = 5 × 11 × 11 × 23. 5 13915 11 2783 11 253 23 23 1 30. (A) : L.C.M. × H.C.F. = Product of two numbers (a, b) \ 16 × 192 = 64 × b Þ

b=

16 ´ 192 = 48 . 64

vvv

Class 6

Chapter - 4 : BASIC GEOMETRICAL IDEAS 1.

(A) : Semicircle : Each diameter of the circle divides the circle into two equal parts, each equal part is called semicircle.

2.

(B) : Parallel lines : Two or more straight lines are said to be parallel to each other if they are in the same plane and do not meet when produced on either side.

3.

(A) : Line : A line has a length only, but neither breadth nor thickness. It extends indefinitely in both directions.

4.

(C) : Concurrent lines : If three or more lines passes through the same point, they are called concurrent lines and their point of intersection is called concurrent point (or) point of concurrence.

5.

(C) :

6. 7.

A

O

B

(D) : Triangle : A figure which has three sides, 3 vertices and three angles is called a triangle. uuur uuur (C) : Ray PQ means P is starting point and ray QP means Q is starting point. uuur uuur \ PQ ¹ QP P

Q

8.

(B)

9.

(B) : Only one line segment can be drawn through two given points in a plane.

10. (D) : Angles formed = Ð1, Ð2, Ð3, Ð4, Ð5, Ð6, ...... 6 2

1 3 4 5

O

11. (C) :

A

Major segment B

Minor segment


12. (C) : Three points are used to draw a triangle. 13. (*) : * = None of the options is correct. Angle in the major segment is the reflex angle.

O

A

Reflex angle Acute angle B

14. (C) : Measure of AB = 4.6 cm , Measure of CD = 3.8 cm Then measure of AB - CD = 4.6 - 3.8 = 0.8 cm . 15. (C) : Point has position on a piece of paper but it has no length, breadth and thickness. 16. (B) : A plane is a flat surface extending indefinitely in all directions. 17. (B) : If two rays are drawn in different directions from a common initial point, they are said to be two arms of angle. suur 18. (B) : A line is symbolically written as PQ . P

Q

19. (B)

Corners of a room They are points.

Railway track

Never meet

Slides of triangle Intersect each other

Surface of ball Curved surface

Class 6

20. (A) : Diameter

Diameter A

Chord

B

21. (A) : Diameter = 2 × Radius Let initial radius = x Þ Diameter = 2x New radius = x + 3x = 4x Þ New diameter = 4x × 2 = 8x New diameter = 8x = 4(2x) = 4 (old diameter). 22. (C) : Circumference : The length of the boundary of the interior of a circle is its circumference. 23. (B) : Line Segment : A line segment is limited by two end points. It is a part of straight line between two points P and Q. P

Q

24. (D) 25. (B) : Collinear points : Three or more points in a plane are said to be collinear if they lie on the same line. A

B

C

26. (B) : Two arcs are made by one chord. Major arc A

B

Minor arc 27. (D) : Rectangle : A quadrilateral having four sides, four vertices and four angles of each 90° is said to be rectangle.

28. (C) : Scale is used to measure radius and compass to draw circle of corresponding radius. 29. (A)

30. (C) vvv


Chapter - 5 : UNDERSTANDING ELEMENTARY SHAPES

1.

(B) :

2.

(A)

3.

(C) : In D ABE, ÐABE + ÐBAE + ÐAEB = 180° (Angle sum property of triangle) A Þ 30° + x + 90° = 180°

1 1 straight angle 4

x x

Þ

x = 60°

Þ

ÐBAC = x + x = 60° + 60° = 120°

30° B E C D (Angle sum property of triangle)

In D ABC, ÐBAC + ÐABC + ÐACB = 180° Þ ÐACB = 180° – 30° – 120° = 30° Now ÐACB + ÐACD = 180° (Linear pair) Þ ÐACD = 180° – 30° = 150° 4.

5.

(D) :

P Q R Q is mid point of PR Þ PQ = QR R is mid point of QS Þ QR = RS From (i) and (ii), PQ = RS

S

...(i) ...(ii)

(B) : Number of triangles (1, 2, 4, 234, 135) = 5. 1 4

2 3 5

6.

(D)

Class 6

7.

(B) :

Figure

Face

1

6

2 5 8.

(C) : According to angles sum property of triangle sum of all the three interior angles of a triangle is equal to 180°.

9.

(A) : We represent perpendicular symbolically as “^”.

P

S

10. (D) : POR and SOQ are two straight lines, then ÐPOQ = ÐSOR = 110° (Vertically opposite angles).

O

11. (C) : By the angle sum property of quadrilateral. ÐA + ÐB + ÐC + ÐD = 180° ÐA + ÐC = 180° (Given) Þ ÐB + ÐD = 180°

110°

Q

R

12. (C) 13. (A) 14. (A) C

15. (B) : ÐAOB + ÐAOC = 134° (Given) Þ

ÐAOB = 134° – 90° = 44° A

O

B

16. (A) : O is the vertex and OA is common arm of ÐAOC and ÐCOB which are adjacent angles. 17. (B) : ÐACD + ÐACB = 180° (linear pair) Þ ÐACB = 180° – 115° = 65° In D ABC, ÐABC + ÐBAC + ÐAÐB = 180° (Angle sum property of D) Þ ÐABC = 180° – 65° – 40° = 75°

A 40° y° B

115° C D

18. (B) : 360° is a complete angle. 19. (D)


20. (B) : ÐABC + ÐACB + ÐBAC = 180° (Angle sum property of triangle)

A

Þ 90° + 35° + ÐBAC = 180° Þ ÐBAC = 180° – 90° – 35° = 180° – 125° = 55° 21. (B) : Obtuse angle measures greater than 90° but less than 180°.

35°

90° B

C

22. (C) : Each face of a cube is of square shape. 23. (C) : ABC is a triangle, if AB + BC > AC, AB + AC > BC, BC + AC > AB. A

B

C

24. (B) 25. (D) : Properties of parallelogram (i) The opposite sides of a parallelogram are equal and parallel. (ii) The opposite angles of a parallelogram are equal. (iii) The diagonals of a parallelogram bisect each other.

D

O

B

A

26. (B) : Wheel of bicycle is circular. Number of right angles is one turn = 4 Number of right angles is 4 turns = 16 Number of right angles in half turn = 2 Therefore total number of right angles = 16 + 2 = 18 27. (A) : ÐABC = 90°

C

A E

{given}

Also ÐABC = ÐABE + ÐEBD + ÐDBC

12° ?

Þ 12° + ÐDBE + 27° = 90° Þ ÐDBE = 90° – 39° = 51°

B

28. (B) : ÐABC = ÐDBF (Given) Since ABD is straight line. Þ ÐABC + ÐCBD = 180° (Linear pair) Þ ÐABC = 180° – 33° = 147° Þ ÐDBF = 147° Also ÐDBF + ÐDBC = 147° + 33° = 180° Þ FBC is a straight line. Þ ÐABF = ÐCBD = 33° (Vertically opposite angles)

Class 6

D 27° C

F

A

E B 33° C

D

rd 29. (A) : III row nd I row st I row

Number of cubes in I st row = 3 Number of cubes in II nd row = 8 Number of cubes in III rd row = 3 \ Total number of cubes in the solid = 14

E

D

30. (B) : ABDE is a rhombus and ÐA = 48° \

?

ÐBDE = 48° {opposite angles of rhombus}

48° A

B

C

vvv


Chapter - 6 : FRACTIONS 1.

(C) : Number of days in month of September 2008 = 30 1 13 litres One day consumption = 3 litres = 4 4 390 1 13 litres = 97 litres litres = 30 days consumption = 30 ´ 4 2 4

2.

(B) : Number of flowers in the garden = 12 Number of red roses = 12 – (3 + 4) = 5 Fraction of red roses =

5 12 2 2 2 2 3 2 4 4 6 8 = ´ , ´ , ´ = , , 3 3 2 3 3 3 4 6 9 12

3.

(B) : Equivalent fractions of

4

(A) : Number of parts of watermelon = 18 Left part of watermelon = 18 – (7 + 4) = 7 Fraction of watermelon remained =

5.

7 18

(B) : Figure

Shaded Fraction 1 [out of 4 parts, 1 is shaded] 4 1 [out of 3 parts, 1 is shaded] 3 1 [out of 4 parts, 1 is shaded] 4 1 [out of 6 parts, 1 is shaded] 6

6.

(D) :

1 1 2 1 1 2 + = and + = 3 3 3 4 4 4

2 2´4 8 2 2´3 6 = = = and = 3 3 ´ 4 12 4 4 ´ 3 12 8 6 2 2 > So Þ > 12 12 3 4

Class 6

7.

(D) :

1 2 x 13 3 x 10 Þ + + = + = 7 7 7 7 7 7 7

Subtracting Þ 8.

...(i)

3 on both sides of (i), we get 7

x 10 3 7 = - = Þ x = 7 7 7 7 7

(D) : L.C.M. of denominators (2, 9, 18, 27) = 54 Make the same denominators of the each of fraction 1 27 27 ´ = (Multiplying numerator and Denominator by 27) 2 27 54 4 6 24 (Multiplying numerator and Denominator by 6) ´ = 9 6 54 5 3 15 (Multiplying numerator and Denominator by 3) ´ = 18 3 54 7 2 14 ´ = (Multiplying numerator and Denominator by 2) 27 2 54 27 24 15 14 1 4 5 7 Now, i.e. > > > > > > 54 54 54 54 2 9 18 27

9.

(C) : (A) and (B) are false as fractions with equivalent numerator and denominator are called like fractions. (D) is also false as fraction with numerator less than denominator is called proper fraction.

10. (D) : According to given question

Þ

1 2 2 4 5 5 5 1 + ? + = + + = + + 15 15 15 15 15 15 15 15 3 11 11 3 8 +?= Þ ? = = 15 15 15 15 15

(On subtracting

3 from the both sides) 15

11. (A) : In figure (I) 5 squares are shaded out of 9 and in figure (II) 6 squares are shaded out of 9. Hence the given fractions will be

5 6 5 6 11 and \ Sum = + = 9 9 9 9 9

+ Fig. I

Fig. II

12. (C) : Number of children in the group = 7 (3 girls, 4 boys) Fraction of girls among 7 children =

3 . 7


13. (C) : Figures

Shaded fraction 4 1 = (4 parts are shaded out of 8) 8 2 1 (1 part is shaded out of 2) 2 1 (1 part is shaded out of 4) 4 3 1 = (3 parts are shaded out of 6) 2 6

14. (C) : Number of total marbles in bag = 3 + 1 + 6 + 2 = 12 Number of red marbles in bag = 3 Fraction of red marbles in bag =

3 12

1 2 5 < < [Numerators are in ascending order and denominators are same] 8 8 8 1 4 3 3 1 2 4 2 1 < > ; > < ; > > 2 2 2 4 4 4 16 16 16

15. (A) :

16. (D) 17. (D) : 3

é 3 ì 1 æ 1 1 ö üù 1 - ê1 + í2 - ç1 - ÷ ýú 12 ë 4 î 2 è 2 3 ø þû

37 é 7 ì 5 -ê +í 12 ë 4 î 2 37 é 7 ì 5 = +í 12 êë 4 î 2 37 é 7 ì 5 +í = 12 êë 4 î 2

=

=

æ 3 1 ö üù çè - ÷ø ýú 2 3 þû 3 ´ 3 - 1 ´ 2 üù ýú 6 þû 7 üù ý 6 þúû

37 é 7 3 ´ 5 - 7 ù + ú 12 êë 4 6 û

37 é 7 8 ù 37 é 3 ´ 7 + 8 ´ 2 ù + = ú 12 êë 4 6 úû 12 êë 12 û 37 37 37 - 3 7 0 = = = = 0 12 12 12 12

=

18. (B) : Let total number of pages in book = x Piyush read = Pages left = 80

Class 6

3 3 x of x = 5 5

(Expanding the mixed fractions) (L.C.M. (2, 3) = 6)

(L.C.M. (2, 6) = 6) (L.C.M. (4, 6) = 12)

3 x 80 3 x + 80 ´ 5 3 x + 400 + Þ x = = 5 1 5 5 Þ 5x = 3x + 400 (Multiply by 5 on both sides) Þ 2x = 400 (Subtract 3x on both sides) Þ x = 200 (Divide by 2 on both sides)

Hence x =

19. (D) : Shaded fraction shown in Model I =

9 12

Model I

Model II 3 3 2 6 Shaded fraction shown in Model II = = ´ = 6 6 2 12 9 6 > And 12 12

20. (D) : 6 more equal sized pieces would be needed to make a whole for the given figure. 1 2 3

6 5 4

21. (D) :

Figure

Shaded Fraction

Unshaded fraction

A

1 4

3 4

Unshaded fraction of Figure A = Shaded fraction of figure B Total parts in figure B = 16 3 x x x 12 3 4 = ´ Þ = Let the shaded part of figure B is x then, = Þ Þ x = 12 4 16 16 16 16 4 4 22. (B) :

So,

Figure

Equal parts

Shaded parts

36

18

12

x

Fractional number 18 1 = 36 2 x 12

x 1 x 1 6 x 6 = Þ = ´ Þ = Þ x = 6 12 2 12 2 6 12 12


23. (C) : Alphabets K E S H A V

Number of straight lines = 3 = 4 S = 0 = 3 = 3 = 2

Þ Fraction of alphabets made of 3 straight lines =

3 . 6

24. (D) : Only alphabet S is made of semicircles Alphabets S

Number of semicircle = 2

Þ Fraction of alphabet made of semicircles =

1 . 6

25. (A) : 1 metre = 100 cm, Given length = 10 cm 10 1 = \ Required fraction = 100 10 26. (A) 27. (B) :

Figure

Shaded fraction

6 ; 6 parts are shaded out of 15 parts 15

2 ; 2 parts are shaded out of 5 parts 5

Hence

6 6¸3 2 = = 15 15 ¸ 3 5

3 8 3 ö æ 28. (C) : Reena eats ç1 + ÷ chocolate bar = 1 = è ø 5 5 5 5 x y = = 7 35 49 5 x = 7 35

29. (A) : Þ

Class 6

[Given] [Take first two fractions]

5 ´ 35 7 Þ x = 5 × 5 = 25

Þ

x=

Now take

5 y = 7 49

5 ´ 49 7 Þ y = 5 × 7 = 35

Þ

[Multiplying 35 both sides]

y =

[Multiplying 49 both sides]

30. (A) : Quantity of milk bought = 7 Milk consumed = 5 Milk left =

1 litres 2

=

15 litres 2

3 23 litres = litres 4 4

15 23 15 ´ 2 - 23 30 - 23 7 3 = = = = 1 litres 2 4 4 4 4 4

vvv


Chapter - 7 : DECIMALS 1.

(B) : Weight of boy = 56.74 kg Weight of father = 1.5 times heavier than son = 1.5 × 56.74 kg = 85.11 kg

2.

(B) : 0.111 =

1 9 Multiply by 7 both sides, we get (0.111) ´ 7 =

1 7 ´ 7 Þ 0.777 = 9 9

3.

(D) : 0.005 < 0.055 < 0.550 < 5.500 Hence 0.005 < 0.055 < 0.55 < 5.5

4.

(C) : (A) 63% = (B) 0.66 =

(Given)

63 60 6 > = = 0.6 Þ 63% > 0.6 100 100 10

66 5 5 > Þ 0.66 > 100 9 9

(C)

455 11 > 1 25

(D)

1 100 17 102 1 = , 0.17 = = Þ < 0.17 6 600 100 600 6 1 3 = 2 + 0.01 + 3 × 0.001 = 2 + 0.01 + 0.003 = 2.013 + 100 1000

5.

(C) : 2 +

6.

(C) :

7.

(C) : Ant went up a tree = 4 m 26 cm Ant came down tree = 2 m 30 cm Height of ant from the ground = 4 m 26 cm – 2 m 30 cm [Q 1 m = 100 cm] = 426 cm – 230 cm = 196 cm = 1 m 96 cm

8 7 1 1 +0+ =8´ + 0 + 7 ´ = 8 × 0.1 + 0 + 7 × 0.001 10 1000 10 1000 = 0.8 + 0 + 0.001 = 0.800 + 0.007 = 0.807

Class 6

8.

(A) :

Transport used

Distance travelled

Bus

5 km 52 m

Car

2 km 265 m

Walk

1 km 30 m

Distance (km) 52 = 5.052 1000 265 2+ = 2.265 1000 30 1+ = 1.030 1000 5+

Total distance travelled = (5.052 + 2.265 + 1.030) km = 8.347 km

9.

(A) : Length of pole = 8 m 57 cm

66 ù é = ê4 + m = (4 + 0.66) m = 4.66 m 100 úû ë Length of pole inside the pole = depth of pond Þ Depth of pond = (8.57 – 4.66) m = 3.91 m

Pond

8 m 57 cm

57 ù é = ê8 + m = (8 + 0.57) m = 8.57 m 100 úû ë Length of pole outside the water = 4 m 66 cm

4 m 66 cm

1 é ù ê 1 km = 1000 or 1 m = 1000 km ú ë û

10. (C) : 0.001 < 0.011 < 0.101 < 0.111 Hence 0.001 is the smallest possible decimal fraction upto three decimal places. 11. (B) : 3 Hundredths =

3 1 = 3´ = 3 ´ 0.01 = 0.03 100 100

12. (C) : 3.92 × 0.1 × 0.0 × 6.3 = 0 [If 0 is multiply by any number gives result 0] 13. (B) : Clearly 12.1280 < 12.1290 Þ 12.1280 < 12.129 14. (B) :

1 = 0.3 = 0.333... = 0.33 3

2 = 0.6 = 0.6666... = 0.67 3 3 3 = 0.75 , = 0.6 4 5 0.1 0.01 1 1 + = 0.1 ´ + 0.01 ´ 0.01 0.1 0.01 0.1 1 100 1 10 1 = ´ + ´ = 10 + = 10 + 0.1 = 10.1 10 1 100 1 10

15. (A) :

16. (A) : Let one of number = x, other number = 11.56 Sum = 31.021 Now x + 11.56 = 31.021 Þ x = 31.021 – 11.56 = 19.461 [Subtracting 11.56 from both sides]


17. (C) : Cost of 1 litre milk = ` 7.50 \ Cost of 30.5 litres milk = ` 7.50 × 30.5 = ` 228.75 18. (B) : Vulgar fraction : The numerator and denominator are both integers 23 100 19. (B) : Copper : Zinc = 9 : 5 Let weight of copper = 9x and weight of zinc = 5x given that 5x = 9.5 gm Þ x = 1.9 gm \ Weight of copper = 9 × 1.9 = 17.1 gm

Hence 0.23 =

20. (A) : Scientific notation of 0.000076 = 7.6 × 10 –5 21. (B) : 9 tens + 50 ones + 3 tenths + 7 hundredths = 9 ´ 10 + 5 ´ = 90 + 5 + 0.3 + 0.07 = 95.37 =

1 1 + 7 ´ 10 100

9537 100

3 43 43 ´ 125 5375 22. (A) : Required decimal of 5 = = = = 5.375 8 8 8 ´ 125 1000

23. (B) : Standard form of 2 + (8 × 0.1) + (6 × 0.01) + (4 × 0.001) = 2 + 0.8 + 0.06 + 0.004 = 2.864 24. (C) : Given question in the equation is 16 = 6 + 8 m, where m is money at each table. Þ 16 – 6 = 8 m (Subtract 6 on both sides) Þ 8 m = 10 Þ m = 10 ¸ 8 = 1.25 (Divide by 8 both sides) Hence money spent on each table is ` 1.25 25. (B) :

Fruits

Weight

Apples

4 kg 90 g

Grapes

2 kg 60 g

Mangoes

5 kg 300 g

Weight in kg 90 = 4.09 kg 1000 60 2+ = 2.06 kg 1000 300 5+ = 5.3 kg 1000 4+

Total weight of all fruits = (4.09 + 2.06 + 5.30) kg = 11.45 kg = 11.450 kg 26. (D) : Number of equal parts = 10

Shaded parts = 4

Class 6

4 10 Decimal fraction of shaded part = 0.4

Fraction of shaded parts =

27. (C) : Initial amount of money = ` 7.45 Money spent on toffees = ` 5.30 Balance amount left = ` [7.45 – 5.30] = ` 2.15 28. (A) :

Figure

Shaded part (Fraction)

Decimal form

3 10

0.3

1 4

0.25

1 4

0.25

10 30

0.333....

1 ö æ 29. (C) : 2.06 = 2 + 0.06 = ( 2 ´ 1) + çè 6 ´ ÷ 100 ø

30. (A) :

Hours st

1 hour 2 nd hour 3 rd hour Total distance travelled

Distance travelled 64 km 324 m = 64.324 km 58 km 56 m = 58.056 km 62 km 8 m = 62.008 km 184.388 km = 184 km 388 m vvv


Chapter - 8 : DATA HANDLING 1.

(C) : Since 1 = 1 student Hence 7 + 5 = 12 students scored 95 or more in test.

2.

(C) : Total friends = 20, students whose favourite colour is red = 6 æ 6 ö ´ 100 ÷ % = 30 %. Required percentage = ç è 20 ø

3.

(B) : Fans sold on Saturday = 4 × 3 = 12

4.

(D) :

Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday

Number of fans 6 × 3 = 18 8 × 3 = 24 4 × 3 = 12 5 × 3 = 15 7 × 3 = 21 4 × 3 = 12 9 × 3 = 27

Clearly maximum fans were sold on Sunday. 5.

(C) : Number of fans sold on Friday = 7 × 3 = 21 Cost of one fan = ` 900 Cost of 21 fans = ` (21 × 900) = ` 18900

6.

(A) : Total students = 60 + 100 + 120 + 20 = 300 Number of student selected cricket = 60 æ 60 ö ´ 100 ÷ % = 20% Required percentage = çè ø 300

7.

(D) :

Saturday

Number of cars washed

1 2 3 4

3 × 10 = 30 2 × 10 = 20 2 × 10 = 20 5 × 10 = 50

Total number of car washed = 30 + 20 + 20 + 50 = 120 Amount charged for 1 car = ` 5.00 Amount charged for 120 cars = ` 5.00 × 120 = ` 600.00

Class 6

8.

(C) : Month August has maximum rainfall.

9.

(D) : Month March and November have no rain.

10. (A) : Horizontal line in the graph is called Xaxis. Y

X

11. (A) : Vertical line in the graph is called yaxis, shown above. 12. (A) : In tally 5 is represented as

.

13. (D) : People liking red colour refrigerator = 5 × 10 + 5 = 50 + 5 = 55 14. (A) : People liking blue colour refrigerator = 4 × 10 = 40 People liking green colour refrigerator = 3 × 10 = 30 \ Total people = 40 + 30 = 70 15 19 :

Subject English Hindi Maths Science SST

Marks scored 15 10 30 20 25

minimum maximum

15. (C) 16. (A) : Maximum marks of 5 subjects = 250 Total marks obtained = 15 + 10 + 30 + 20 + 25 = 100 æ 100 ö ´ 100 ÷ % = 40 % Required percentage of marks = ç è 250 ø

17. (A) : Let the marks obtained to get 60% = x æ x ö x ´ 100 ÷ % = 60 % Þ ´ 100 = 60 Þ x = (60 × 250) ¸ 100 Hence ç è 250 ø 250 Þ x = 150 Hence he need 150 – 100 = 50 more marks to raise percentage upto 60%.

18. (A) : He failed in 1 subject (Hindi). 19. (A) : In Hindi he scored minimum marks. 20. (B) : 7

= 4200 Þ

= 4200 ¸ 7 = 600.


21. (B) : 600 represents = 1 3600 represents = 3600 ÷ 600 = 6 22. (B) : 7J = 4200 Þ J = 4200 ÷ 7 = 600 23. (A) : 600 represents = 1J 2400 represents = 2400 ÷ 600 = 4 J 24. (D) : 5 balloons represents = 1] 1 60 balloons represents = ´ 60 ] = 12 ] 5 25. (B) 26. (A) 27. (A) :

Day Monday Tuesday Wednesday Thursday Friday Saturday

Number of scooter sold 5 × 6 = 30 4 × 6 = 24 3 × 6 = 18 5 × 6 = 30 7 × 6 = 42 2 × 6 = 12

28. (B) 29. (B) 30. (B) vvv

Class 6

maximum minimum

Chapter - 9 : MENSURATION 1.

(C) : Distance covered by a farmer around a rectangular field = perimeter of rectangular field = 2 [length + breadth] = 2 × [120 + 80] m = 400 m

2.

(B)

3.

(C) : Diameter of circle = side of square Hence side of square = 2 × 4 = 8 cm Perimeter of the square = 4(side) = 4 × 8 cm = 32 cm

4.

(B) : Length covered by 1 student = 1.75 m Length covered by 80 students = 80 × 1.75 m = 140 m

5.

(B) : Measuring units or SI unit of area is square units.

6.

(D) : Area of rectangle = length × breadth (A) (8 × 2) m 2 = 16 m 2 (B) (16 × 1) m 2 = 16 m 2 1 ö 2 æ (C) çè 32 ´ ÷ø m = 16 m 2 2 (D) (8.5 × 2) = 17 m 2

7.

(A) : Perimeter of square = Perimeter of rectangle 4 × (Sides) = 2(length + breadth) Þ 4 × 40 m = 2 (50 m + breadth) [1 dac = 10 m] Þ 160 m = 100 m + 2 breadth Þ Breadth = (160 – 100) m ¸ 2 = 30 m Area of rectangle = length × breadth = (50 × 30) m 2 = 1500 m 2

8.

(D) : LM = 2 m D

16 m L M

MN = 16 m E F Area of rectangle LMNO = Length × Breadth Q 2 2 16 m = (16 × 2) m = 32 m P H G Similarly area of rectangle PQRS = 32 m 2 2 2 Area of square EFGH = 2 × 2 m = 4 m O N A Area of shaded region = Area (LMNO + PQRS) – Area (EKGH) 7 m 2 m = [(32 + 32) – (4)] m 2 = [64 – 4] m 2 = 60 m 2 9.

(C) : Length of park = 50 m Breadth of park = 250 dm = 25 m Distance covered in 1 round = 2(length + breadth)

Park

C R S 7 m B

250 dm

50 m


= 2(50 + 25) m = 2 × 75 m = 150 m Distance covered in 5 rounds = 150 × 5 m = 750 m 10. (D) : 1 mm =

1 1 cm 2 cm \ 1 mm 2 = 10 100

11. (A) : Let the breadth of rectangle = x m 6 Then length of rectangle = x m 5 Perimeter of rectangle = 2(length + breadth) 6 x ö æ x 6 x + = 132 ¸ 2 Þ 132 = 2 çè x + ÷ Þ 5 ø 1 5 5 x + 6 x Þ = 66 Þ 11x = 66 × 5 = 330 Þ x = 330 ¸ 11 Þ x = 30 5 6 6 Hence length = x = ´ 30 = 36 m 5 5 Breadth = x = 30 m \ Area = Length × Breadth = (36 × 30) m 2 = 1080 m 2

12. (C) 13. (A) : Let the length of side of one square is 1 unit Figure

Perimeter

1 1 1

1

1

1

8 units

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

10 units

10 units

1 1 1 1

1 1

1 1 1

1

10 units

1 1

14. (C) : Let side of original square = x units Area of original square = x 2 units 2 Perimeter of original square = 4 x units

Class 6

Smallest

Now according of question, perimeter would be tripled, Þ Perimeter of new square = 3 (4 x) = 4(3 x) units Þ Side of new square = 3 x units Þ Area of new square = 3x × 3x = 9x 2 Þ Area of new square = 9 (Area of original square) 15. (B) : Area of sheet of paper= length × breadth = 384 × 172 cm 2 = 66048 cm 2 Area of piece of paper to each envelope = 16 × 12 cm 2 = 192 cm 2 Number of envelopes =

Area of paper 66048 = = 344 Area of each envelope 192

16. (D) : Length of rectangular field = 260 m Breadth of rectangular field = 130 m Area of rectangular field = length × breadth = 260 × 130 = 33800 m 2 Also 10000 m 2 = 1 hectare \ 33800 m 2 = 33800 ÷ 10000 = 3.38 hectares. 17. (C) : Area of square = Perimeter of square Þ 4 × side = side × side Þ Side = 4 units 18. (B) : Area of rectangular veranda = (60 × 6) m 2 = 360 m 2 10m 9 m 90 m 2 ´ = = 0.9 m 2 10 10 100 Area of rectangular veranda 360 Number of stones = = = 400 Area of one paving stone 0.9

Area of one paving stone =

19. (C) : Area of square shaped park = Side × Side = (28 × 28) m 2 = 784 m 2

28 m

2 m 3 m

28 m Area of rectangular pond = length × breadth = (3 × 2) m 2 = 6 m 2 Area of park excluding the pond = Area of park – Area of pond = (784 – 6) m 2 = 778 m 2 .

20. (D) : Perimeter of the figure is A O N

P

B C

M L K J

D E H

F G

I


AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LM + MN + NO + OP + PA = 3.8 + 4.2 + 6.3 + 4.6 + 7.2 + 4.3 + 8.2 + 1.7 + 3.5 + 1.4 + 4.5 + 3.6 + 4.2 + 3.8 + 4.4 + 2.6 = 68.3 cm. 8 A B 21. (D) : ABEF and EFDC are squares of side 8 cm each form a rectangle of length 8 cm and breadth 16 cm. E F Dimensions of rectangle formed = 8 cm × 16 cm Perimeter of the formed figure = 2(length + breadth) = 2 (8 + 16) cm = 2(24) cm = 48 cm. 22. (D) : Distance covered in one round = Perimeter of square = 4 × side = 4 × 85 = 340 m Distance covered in 5 rounds = 340 × 5 m = 1700 m

D

24. (B) : Length of the original rectangle = x Breadth of original rectangle = y Perimeter of original rectangle = 2(x + y) Now according of question, Length of new rectangle = 2x Breadth of new rectangle = 2y Perimeter of new rectangle = 2(2x + 2y) = 2 × 2 (x + y) = 2[2 (x + y)] = 2 [Perimeter of original rectangle]. 25. (C) : Let breadth of rectangle = x cm Then length of rectangle = 2(x) cm Area of rectangle = length × breadth = 288 = (2x × x) Þ 2x 2 = 288 Þ x 2 = 288 ¸ 2 = 144 = 12 × 12 Þ x = 12 cm Þ Breadth = 12 cm and Length = 2 × 12 = 24 cm. 26. (A) : Area of rectangular plot = 15.4 × 6.5 m 2 = 100.10 sq.m D C 6.5 m

A B 15.4 m Perimeter of rectangular plot = 2(15.4 + 6.5) m = 43.8 m.

Class 6

4 cm

B

Þ 4 + 8 + BC + 5 = 23 Þ BC = 6 cm.

27. (B) : Total cost of flooring the room = ` 510 Rate of flooring = ` 8.50 per sq. m. \ Area of room = 510 ÷ 8.50 = 60 sq. m. Length of room = 8 m \ Breadth = 60 ÷ 8 = 7.5 m

C

5 cm

Perimeter of trapezium ABCD = AB + CD + BC + AD = 23

Rectangular plot

8

A

23. (A) : Since the given figure is in the form of trapezium \

D

? 8 cm

C

28. (A) : Area of sheet of paper = (324 × 172) cm 2 = 55728 cm 2 Area of piece of paper = (18 × 12) cm 2 = 216 cm 2 Number of envelopes =

Area of sheet 55728 = = 258 Area of piece of paper for one envelope 216

29. (B) : Area of rectangular floor = (12.5 × 8) m 2 = 100 m 2

8 m

8 m

8 m 12.5 m Area of square carpet = (8 × 8) m 2 = 64 m 2 Area of square not carpeted = (100 – 64) = 36 sq.m 30. (B) : Figure 1 m

10 m

Area

I

Area of (I + II) = 10 × 1 + 8 × 2 = 26 m 2 8 m 2 m

II

Area of (I + II + III) = 12 × 2 + 7 × 1.5 + 9 × 1 = 24 + 10.5 + 9 = 43.5 m 2

12 m 2 m

2 m

I II

7 m 1 m

III 1.5 m 7.5 m 12 m I

7.5 m

1.5 m

II

III

10 m

2 m

Area of (I + II + III – II) = Area of ( I + III) = 12 × 1.5 + 7.5 × 2 = 18 + 15 = 33 m 2 .

vvv


Chapter -10 : ALGEBRA 1.

(B) : Distance travelled by Rakhi in (km) Means Distance Foot

4 x

Cycle

2 y

Bus

9

Required algebraic expression = (4x + 2y + 9) km 2.

(B) : Let the number = x According to question, 1. Half the number x i . e x ´ 2. Added 10 in it i.e.,

1 x = 2 2

x + 10 2

\ Required algebraic expression is 3.

(A) : Money I had

x + 10 = 15 2

= ` 200

Money given to Anwar = ` x x 2 x Left money with me = ` 2 Algebraic expression to explain above is

Money given to Vidhu = `

æx x xö 2 x + x + x \ 200 = çè + + ÷ø Þ 200 = 1 2 2 2

Þ 4x = 200 × 2 Þ 4x = 400

[Multiply by 2 both sides] Þ

Amount gave to Vidhu = ` 4.

x = 400 ¸ 4 Þ x = 100 x = ` 50 2

(B) : Let the ages of the friends be 2x, x sum of their age is 51 i.e. 2x + x = 51 Þ 3x = 51 Þ x = 51 ¸ 3 [Divide by 3 both sides]

Class 6

Þ x = 17 Age of two friends are 34 years, 17 years. 5.

(B) : Let the length of the room = l Thrice the length of room = 3l Now according to question, 3l = 340

6.

(D) : Let the number = x Six times the number = 6 × x = 6x According to question, 6x = 48 Þ x = 48 ¸ 6 Þ x = 8 Hence the number = 8

[Divide by 6 both sides]

7.

(B) : Since according to the given expression y is subtracted from II Þ y < 11 Hence, y is less than 11.

8.

(D) : 6 (2a – 1) + 8 = 14 Þ 6(2a – 1) = 6 (Subtract 8 from both sides) Þ 2a – 1 = 1 (Divide by 6 both sides) Þ 2a = 2 (Add 1 both sides) a = 2 ¸ 2 = 1 (Divide by 2 both sides)

9.

(B) : Let the number = x 1 x = 2 2 x Add 18 in half of the number = + 18 2 According to the question

Half of the number = x ´

x x + 18 = 46 Þ = 28 2 2 Þ x = 56

(Subtract 18 from both sides) (Multiply by 2 both sides)

10. (B) 11. (C) : ÐABD + ÐDBC = 180° (Linear pair) Þ x° + 3x°= 180° Þ 4x° = 180° Þ x° = 45° (Divide by 4 both sides) A Hence, ÐABD = x° = 45° & ÐDBC = 3x° = 3 × 45° = 135° x 1 + = 4 4 2 x 1 = 4 4 2

12. (C) : Þ

D x°

3x ° B

C

(Given) (Subtract

1 from both sides) 2


x 4 ´ 2 - 1 = 4 2 x 8 - 1 7 = = Þ 4 2 2 Þ x = 7 × 2 = 14

Þ

(Taking L.C.M. (2, 1) = 2) Þ x=

7 ´ 4 2

(Multiplied by 4 both sides)

13. (A) : Number of packs of trading cards = x (Given) Number of cards in each pack = 10 Total number of cards Meera has = 10 × x = 10 x She gave away card = 7 According of question, she left with 10x – 7 cards 14. (C) : 6a = 30 Þ a = 30 ¸ 6 Þ a = 5 15

(Given) (Divide by 6 both sides)

2 1 1 p - 2 = 3 (Given) 3 2 2 2 5 7 p- = [Change numbers from mixed fraction to improper fraction] 3 2 2 5 é ù 2 7 5 Add both sides ú p = + ê 2 3 2 2 ë û 2 7 + 5 12 p= = = 6 3 2 2 2p = 18 (Multiply by 3 both sides) p = 9 (Divided by 2 both sides)

(C) : Þ Þ Þ Þ Þ

16. (B) : Linear equation : Power of variable is 1. 17. (A) : 16 (x + 7) = 144 Þ x + 7 = 9 Þ x = 9 – 7 = 2

(Given) (Divide by 16 both sides) (Subtract 7 from both sides)

18. (B) : y – 8 = 6 y = 6 + 8 = 14

(Given) (Add 8 from both sides)

3 x + 8 = 17 4 3 x Þ = 17 – 8 = 9 4 Þ 3x = 36 Þ x = 12

19. (C) :

(Given) (Subtract 8 from both sides) (Multiplied by 4 both sides) (Divide by 3 both sides)

20. (D) : Perimeter of equilateral triangular garden A

Class 6

B

C

= AB + BC + CA = 3(AB) [Since the triangle is equilateral, therefore all sides are equal] Þ 66 = 3(AB) Þ AB = 22 (Divide by 3 both side). 21. (C) : Product of x and a = xa Product of b and y = yb So, the given expression will be yb – xa = by – ax. a a a + 2 a 3 a + = 6 Þ = 6 Þ = 6 8 4 8 8 Multiplying both side by 8, we get

22. (C) : Given

3 a ´ 8 = 6 ´ 8 or 3a = 48 8 Now dividing both side by 3, we get 3 a 48 = Þ a = 16. 3 3

23. (B) : Suppose total jumps that the grasshopper made be x. Distance covered in one jump = 91.4 cm Total distance covered in x jumps is 91.4 × x. So 91.4 × x = 731.2 Þ

31.4 x 731.2 = Þ x = 8. [Dividing both sides by 91.4] 31.4 31.4

24. (A) : We have a = b Þ a + x = b + x

[Adding x on both sides]

25. (C) : x multiplied by itself = x × x = x 2 . 26. (B) : We have a = b Þ ax = bx [Multiplying by x both sides] 27. (A) : We have (x 2 – y 2 + 2xy + 1) – (x 2 + y 2 + 4xy – 5) = x 2 – y 2 + 2xy + 1 – x 2 – y 2 – 4xy + 5 (By opening brackets, = 0 – 2y 2 – 2xy + 6 (Adding like terms) = –2y 2 – 2xy + 6. 28. (B) : 2a – [3b – {a – (2c – 3b) + 4c – 3 (a – b – 2c)}] = 2a – [3b – {a – (2c – 3b) + 4c – 3a + 3b + 6c}] = 2a – [3b – {a – (2c – 3b) – 3a + 3b + 10 c}] = 2a – [3b – {a – 2c + 3b – 3a + 3b + 10c}] = 2a – [3b – {–2a + 6b + 8c}] = 2a – [3b + 2a – 6b – 8c] = 2a – [2a – 3b – 8c] = 2a – 2a + 3b + 8c = 3b + 8c


29. (D) : 5x – [4y – {7x – (3z – 2y) + 4z – 3(x + 3y – 2z)}] = 5x – [4y – {7x – (3z – 2y) + 4z – 3x – 9y + 6z}] = 5x – [4y – {7x – (3z – 2y) – 3x – 9y + 10z}] = 5x – [4y – {7x – 3z + 2y – 3x – 9y + 10z}] = 5x – [4y – {4x – 7y + 7z}] = 5x – [4y – 4x + 7y – 7z] = 5x – [–4x + 11y – 7z] = 5x + 4x – 11y + 7z = 9x – 11y + 7z 30. (A) : Suppose the number be x. Since Raju multiplies x by 17 to get 17x. Also 4 added to product to get 17x + 4. The result obtained = 225. So 17x + 4 = 225 Transposing 4 to R.H.S., we get 17x = 225 – 4 or 17x = 221. Dividing by 17 both side, we get 17 x 221 = or x = 13 17 17 \ That number = 13. vvv

Class 6

Chapter - 11 : RATIO AND PROPORTION 1.

(B) : The middle terms in proportionality of four numbers is called means.

2.

(C) : Total children = 1200 Selected children = 900 \ nonselected children = 1200 – 900 = 300 \ Ratio of nonselected to total children =

3.

300 1 = = 1 : 4 . 1200 4

(B) : Distance travelled by bus = 126 km Time taken by bus = 3 h Distance 126 = = 42 km / h Time 3 Distanced travelled by train = 315 km Time taken by train = 5 h

\ Speed of bus =

315 = 63 km / h 5 42 km / h 2 = = 2 : 3 . Ratio of their speed = 63 km / h 3

\ Speed of train =

4.

(D) : In 1 hour diesel required = 2.875 litres \ In 24 hours diesel required will be = 2.875 × 24 litres = 69 litres.

5.

(A) : Suppose share of C = x \ B’s share = 2x and A’s share = 4x Sum of each share is 3010, so we get Þ 4x + 2x + x = 3010 {Adding like terms} Þ 7x = 3010 {Dividing both side by 7} Þ x = 430 \ A’s share 4 × 430 = ` 1720 B’s share 2 × 430 = ` 860 C’s share 1 × 430 = ` 430

6.

(A) : We have

15 x 10 = = 18 6 y

Dividing numerator and denominator of 15 5 = 18 6

15 by 3 we get 18

so x = 5


Now multiplying numerator and denominator of

5 by 2, we get 6

15 5 10 = = so y = 12 18 6 12 So x = 5, y = 12

7.

(A) : The proportion is p : q :: r : s Þ ps = qr {In proportion, product of extremes values is equal to product of middle mean values}

8.

(B) : In D ABC, we are given 3ÐA = 4ÐB = 6ÐC = p (suppose) p p p , ÐB = & C = 3 4 6 Now L.C.M. of 3, 4, 6 is 12

so ÐA =

p p p ´ 12 : ´ 12 : ´ 12 = 4p : 3p : 2p 3 4 6 Also sum of angles of D is 180° So, 4p + 3p + 2p = 180° or 9p = 180° or p = 180 ¸ 9 {Dividing both sides by 9} or p = 20° \ ÐA = 4 × 20° = 80° ÐB = 3 × 20° = 60° ÐC = 2 × 20° = 40°.

So ÐA : ÐB : ÐC =

9.

(A) : Ratio of A, B and C =

1 1 1 : : 2 3 5

L.C.M of 2,3,5 is 30 So A : B : C = 15 : 10 : 6 Suppose A’s share = 15x B’s share = 10x C’s share = 6x Sum of their shares is 4340. \ 15x + 10x + 6x = 4340 or 31x = 4340 or x = 4340 ¸ 31 = 140 \ A’s share = 15 × 140 = ` 2100 B’s share = 10 × 140 = ` 1400 C’s share = 6 × 140 = ` 840 10. (C) : or

Class 6

persons : 10 persons = 7000 paise : ` 100

W persons ` 70 W 70 = or = 10 persons ` 100 10 100

{100 paise = ` 1}

Dividing numerator and denominator of 70 7 . = So 100 10

70 by 10 we get 100

contains 7.

11. (B) : Q12 men can do a work in 18 days \ 1 men can do this work in = 18 × 12 days \ 8 men will do the work in

18 ´ 12 days = 27 days 8

12. (C) : Ratio of shaded portion =

3 4

\ percentage of shaded portion =

3 ´ 100 = 75% 4

13. (D) : Weight of 1 litre of Kerosene oil = 0.819 kg and 10 litres of petrol weight = 7.02 kg \ 1 litre of petrol weight =

7.02 kg = 0.702 kg 10

\ Ratio of weight of kerosene oil and petrol in one litre =

0.819 7 = . 0.702 6

14. (C) : Ratio is a method in which division is used. 15. (B) : If antecedent and consequent of the ratio in interchanged, we will get different ratio. 16. (B) : Cost of one toffee = 50 paise Cost of one chocolate = ` 10 = 10 × 100 paise = 10000 paise \ Ratio =

{Q ` 1 = 100 paise}

50 1 = = 1 : 20 . 1000 20

Father's age Mother's age = Son's age Daughter's age and Mother’s age = 48 years Son’s age = 24 years Daughter’s age = 18 years

17. (C) : Given,

Father's age 48 8 Father's age 8 = = or = 24 18 3 24 3 8 Multiplying numerator and denominator of by 8, we get 3 Father's age 8 64 = = 24 3 24 So father’s age = 64 years.

So,


18. (B) :

Correct answer 5 = Incorrect answer 2

Multiplying numerator and denominator of

5 by 8 we get 2

Correct answer 5 40 = = Incorrect answer 2 16 \ Correct answer = 40. 19. (B) : In the word MATHEMATICS Number of consonants = 7 Number of vowels = 4 \ Ratio = 7 : 4. 20. (A) : Ratio of one rupee, 50 paise & 25 paise coins = 5 : 6 : 8 Suppose no. of one rupee coin = 5x no. of 50 paise coin = 6x and no. of 25 paise coin = 8x Value of 5x, one rupee coins = ` 5x 6 x = ` 3x 2 8 x Value of 8x, 25 paise coins = = ` 2x 4 Total values of coins = 5x + 3x + 2x = 10x We have, 10x = 420 Dividing both side by 10, we get x = 42 So, total coins = 5x + 6x + 8x = 19x = 19 × 42 = 798.

Value of 6x, 50 paise coins =

21. (A) : We are give 4(first part) = 5(2nd part) = 7(3rd part) = x \ First part =

x 4

Second part =

x 5

x 7 L.C.M of 4, 5, 6 is 140 First part = 35x, Second part = 28x, Third part = 20x Sum of three part = 1162 or 35x + 28x + 20x = 1162 or 83x = 1162 or x = 1162 ¸ 83 = 14 {Dividing both side by 83} \ 1st part = 35 × 14 = 490, 2nd part = 35 × 14 = 392, 3rd part = 20 × 14 = 280

Third part =

Class 6

1 1 of ` 9.30 = ´ 9.30 = ` 3.10 = 310 paise and 0.6 of ` 1.55 3 3 = 0.6 × 1.55 = ` 0.93 = 93 paise

22. (B) :

\ Ratio =

310 10 = = 10 : 3 93 3

23. (C) : We will do it by ‘hit & trial’ method. When 1 is subtracted from each no., we get 53, 70, 53 74 74 and 98. Since is not equal to . So they are not proportional. 70 98 When 2 is subtracted from each number we get 52, 69, 73, & 97. 52 73 is not equal to . So they are not proportional. 69 97 When 3 is subtracted from each number, we get 51, 68, 72 and 96

and

51 3 72 3 51 72 = & = so = 68 4 96 4 68 96 \ The nos. are in proportion. When 6 is subtracted the no. obtained will also not be in proportional.

and

24. (A) : Ratio of salary of A, B and C = 2 : 3 : 5 Let A’s salary = 2x B’s salary = 3x C’s salary = 5x Q C’s salary is ` 1200 more than A’s salary So 5x – 2x = 1200 or 3x = 1200 Dividing both sides by 3, we get x = 400 \ B’s monthly salary = 3 × 400 = ` 1200. Now B’s annual salary = 12 × 1200 = ` 14400 25. (C) : Ratio of prices of scooter and television = 3 : 2 Let price of scooter = 3x and price of television = 2x Q Scooter costs ` 6000 more than television So, 3x = 2x + 6000 Transposing 2x to L.H.S. we get 3x – 2x = 6000 or x = 6000 \ Cost of television = 2 × 6000 = ` 12000. 26. (A) : Ratio of shares of Kavita and Reena = 4 : 3 Suppose Kavita’s share = 4x and Reena’s share = 3x \ Total amount = 4x + 3x = 7x As Reena’s share = ` 2400 \ 3x = ` 2400 Dividing both sides by 3, we get


3 x ` 2400 = or x = ` 800 3 3 \ Total amount = 7 × 800 = ` 5600.

27. (C) : Ratio of incomes of A and B = 3 : 2 Let A’s income = 3x and B’s income = 2x Ratio of their expenditure = 5 : 3 Let A’s expenditure = 5y, B’s = 3y A’s saving = 3x – 5y B’s saving = 2x – 3y Since each saves ` 1000 so 3x – 5y = 1000 = 2x – 3y We have, 3x – 5y = 2x – 3y Transposing 2x to L.H.S. and – 5y to R.H.S., we get 3x – 2x = –3y + 5y or x = 2y Also 3x – 5y = 1000 Putting value of x as 2y, we get 3(2y) – 5y = 1000 or 6y – 5y = 1000 or y = 1000 \ x = 2 × y = 2000 \ A’s salary = 3 × 2000 = ` 6000. 28. (C) : We have a : b = c : d a c = b d ma nc = or mb nd Interchanging antecedent and consequents, we get or

ma mb = nc nd Using components, we have or

ma + nc mb + nd = nc nd Again interchanging antecedent and consequents, we get

ma + nc nc = mb + nd nd ma + nc c = or mb + nd d or

Class 6

ma + nc a = mb + nd b

ì a c ü íQ = ý î b dþ

29. (C) : A : B = 4 : 5 and B : C = 2 : 3

or or

4 B 5 2 C B= 3 A=

As A’s money = ` 800 4 B = 800 5 or 4B = 4000 or B = 1000

{Multiplying both sides by 5} {Dividing both sides by 4}

2 C = 1000 3 or 2C = 3000

{Multiplying by 3 both sides}

or

Also

3000 2 \ C = ` 1500

or C =

{Dividing both sides by 2}

30. (B) : Given A 's share B 's share C 's share 2 = = = B 's share C 's share D 's share 3 A : B = 2 : 3 B : C = 2 : 3 C : D = 2 : 3 \ A : B : C = 2 : 3 2 : 3 = 4 : 6 : 9 A : B : C : D = 4 : 6 : 9 2 : 3 = 8 : 12 : 18 : 27 Let A’s share = 8x B’s share = 12x C’s share = 18x D’s share = 27x Sum of their shares = 1300 So, 8x + 12x + 18x + 27x = 1300 or 65x = 1300 or x = 20 {Dividing both side by 65} \ A’s share = 8 × 20 = ` 160 vvv


Chapter - 12 : SYMMETRY AND PRACTICAL GEOMETRY 1.

(D)

4 lines of symmetry

2.

(D)

Many lines of symmetry

3.

(C)

4.

(C) : After bisecting 60° we get 30° and then bisecting it again we get 15° angle.

5.

(C)

7.

(A) : Angle between 120° & 180° is 60°. After bisecting it we get 30°. So total angle = 120° + 30° = 150°.

8.

(C) :

9.

1 1 PQ = ´ 8.2 = 4.1 cm 2 2 (B) : Q YU bisects ÐXYZ

6.

(B)

8.2 cm O P Since O bisects PQ so

Q

PO =

\

Class 6

1 ÐXYZ 2 1 75 ° = ´ 75° = = 37.5 ° 2 2

Z

ÐUYZ =

U Y

X

10. (D) : Clearly MN ^ NP P

M N 11. (A) : A circle has infinite lines of symmetry so O has the same.

12. (A) : Line AB has no end point. A

13. (D) :

H

B

E

B

All has horizontal line of symmetry. .

14. (A) 15. (A) : Scalene triangle has no line of symmetry. 16. (B) : Between 60° & 120° we have angle 60°. After bisecting it we get an angle measuring 30°. So net angle = 60° + 30° = 90° 17. (B) : AB has two end points viz A and B. A

B

18. (B)

19. (A) : Since AB = AC. It is symmetric about AD only.

A

E

F

B

D

C

20. (A) A B

D

C

vvv


Chapter - 13 : ANALYTICAL AND LOGICAL REASONING 1.

(A) : Here

is alternatively followed by brush & Cup.

2.

(D)

3.

(B) : In each case there is gap of two letter’s between two consecutive letters.

4.

(D) : x is square root of x 2 which is fifth term in the pattern.

5.

(A) : Wed Sat Tues Fri

6.

(B)

C

D

A B

7.

(A) : Alphabates C, D, E and F and their mirror images are in each figure.

8.

(A) : M O N D A Y 1 2 3 4 5 6

B E L T

\

T O M B A Y 9 2 1 0 5 6

0 7 8 9

9.

(B) : Oven makes somethings hot Airconditioner makes cool. Refrigerator makes cold. Icecream and snow are rejected as they are cold itself. 10. (D) B A E N C

S W D

11. (B) : By this figure we have 5 & 6 and 2 & 4 opposite to each other so 1 & 3 will be opposite to each other. 12. (D) : LANDFALL, NIGHTFALL, WATERFALL, PITFALL

Class 6

13. (C) : Z Y X W V U T S R Q P O N M L K J I H G F E D C B A th

th

12 from left 7 to the right

14. (D) 15. (B) : Minimum 13 lines are required. 16. (C) : There are 22 triangles. 17. (C) : There are 7 squares. 18. (C) : 9 quadrilaterals E

F G A

D

H B C ABCD, AFED, ABDE, ACDF, AHDG, AGDB, AHDE, AGDC, AHDF

19. (D) : 6 Pentagons are BCDGA, EFAHD, ACDEF, BDEFA, DFABC, EABCD 20. (B) : 34 2 = 1156 \

16 2 = 256 7 2 = 49

7 1156

16

256

34

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IMO_Class-6

2011- 5th SOF IMO 1. (B)

21. (B) : Here p = – 9 and q = 3 \ p + q = – 9 + 3 = – 6.

2. (D) : 8 × 2 + 2 = 16 + 2 = 18 18 × 2 + 1 = 36 + 1 = 37 \ 5 × 2 + 2 = 10 + 2 = 12 12 × 2 + 1 = 24 + 1 = 25

22. (B) : 1 million = 1000000 = 10,00,000 = Ten hundred thousand.

3. (D) : All except (D) can be found by certain rotation. 4. (D) :

23. (C) 24. (C) : Prime factors of 63 = 3 and 7 \ Sum = 3 + 7 = 10 25. (B) : Right angles in one complete turn = 4. \ Right angles in 3 complete turns = 4 × 3 = 12 152 2 litres = litres 5 5 113 2 litres Added volume of oil = 37 litres = 3 3 505 1 Pumped out volume of oil = 56 litres = litres. 9 9

26. (A) : Volume of oil in tank = 30 5. (D) 6. (C) 7. (B) –2

8. (A) : 1

4 ×4

–2 2

8

–2 6

×4

24 ×4

22

–2 88

\ Volume of remaining oil in tank = 86

=

152 × 9 + 113 × 15 − 505 × 5 1368 + 1695 − 2525 = 45 45

=

538 43 litres = 11 litres 45 45

×4

9. (A) 10. (D) : H < M < G, H < S ≠ M, E ≠ G So, we cannot find the answer with given data. 11. (B) :

152 113 505 + − 5 3 9

27. (C) : H.C.F. of 56 and 77 = 7 So k – 8 = 7 ⇒ k = 15. 28. (B) : A triangular prism has 9 edges.

12. (D)

29. (C) : XCCMXCIX = 1089

13. (D) : In each option except (D), there are 5 circles and two stars.

30. (A) : 21397 + 42505 = 63902, when rounded to nearest thousand it becomes 64000.

14. (C) : P R I C E 14. (x) : 3 2 4 5 1

31. (A) : Only sum of the digits of number 9076185 is 36 which is divisible by 9 so given number is divisible by 9.

15. (C)

32. (B) : – 12 – (– 48) – [(– 13) + (– 8) – (– 3) + 4] = – 12 – (– 48) – [(– 13) – 8 + 3 + 4] = – 12 + 48 + 13 + 8 – 3 – 4 = 48 + 13 + 8 – 12 – 3 – 4 = 69 – 19 = 50.

16. (B) :

17. (C) : 1 5 7 8 9 and X T Z A L

2 3 4 6 \ 2 3 5 4 9 N P S U NPTSL

2 3 7 9 8 33. (B) : , , , , 3 5 9 11 9 L.C.M. of 3, 5, 9 and 11 = 3 × 3 × 5 × 11 = 495.

18. (A) : Outer figure is inside the inner most figure and second outer figure is inverted. 19. (C) : (4 × 6) ÷ 3 = 24 ÷ 3 =8 (6 × 10) ÷ 5 = 60 ÷ 5 = 12 (4 × 8) ÷ 2 = 32 ÷ 2 = 16. 20. (B) : 16 ÷ 64 – 8 × 4 + 2 changes to 16 + 64 ÷ 8 – 4 × 2 = 16 + 8 – 4 × 2 = 16 + 8 – 8 = 24 – 8 = 16.

Now 2 = 2 × 165 = 330 , 3 = 3 × 99 = 297 , 3 3 × 165 495 5 5 × 99 495 7 7 × 55 385 9 9 × 45 405 , = = , = = 9 9 × 55 495 11 11 × 45 495

IMO_Class-6 42. (A) : Let the number be x Original multiplication = 25x Multiplication by 52 = 52x \ According to question, 52x – 25x = 324 ⇒ 27x = 324 ⇒ x = 324 ÷ 27 = 12.

8 8 × 55 440 = = 9 9 × 55 495 \

295 330 385 405 440 < < < < 495 495 495 495 495

⇒

3 2 7 9 8 < < < < . 5 3 9 11 9

43. (C) : One-third of the tank holds 80 litres. \ Whole tank will hold 80 × 3 = 240 litres. \ Half of the tank holds 240 ÷ 2 = 120 litres.

34. (D) : 832.58 – 242.31 = 779.84 – x. Transposing x to L.H.S. and all constants to R.H.S., we get x = 779.84 – 832.58 + 242.31 ⇒ x = 189.57

44. (B) : Total students in class = 50 Number of girls = 15 Number of boys = 50 – 15 = 35 Number of girls chosen Match = 5 2 = 10 7 \ Total students chosen for Match = 5 + 10 = 15. Number of boys chosen Match = 35 ×

35. (A) : In January only 15 cars were sold. 36. (D) : Cars sold in last quarter of the year i.e. in Oct., Nov. and Dec. = 35 + 50 + 20 = 105. 37. (C) : Breadth of hall = 25 m Length of hall = 25 + 5 = 30 m Area of hall = 30 × 25 = 750 m2 38. (B) :

For horizontal line of symmetry only 2 squares must be shaded. 39. (D) : In DCDE, CE = DE = 6.2 cm So DCDE is isosceles triangle. 40. (C) : Line segments are AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. Total number = 10. 41. (D) : Times take to beep by four devices respectively are 1 30 mins, 1 hr, 1 hr and 1 hr 45 mins 2 i.e. 30 mins, 60 mins, 90 mins and 105 mins. Now,

45. (C) : Let number of boys = x. \ Number of girls = 5x So, total students = x + 5x = 6x. That means total students of the class should be multiple of 6. 35 is not a multiple of 6, so it can not be total students in the class. 46. (A) : Floor on which car is parked = 6th. Floor of Finance department = 6 + 17 = 23rd. Floor of Tax department = 23 – 9 = 14th. \ Lady is on 14th floor. 47. (B) : Number of boxes of sweets = 19 Sweets in each box = 228. \ Total sweets = 19 × 228 = 4332 Sweets given to friends = 519 \ Remaining sweets = 4332 – 519 = 3813. 1 48. (D) : Holiday work completed on Monday = 5 3 Holiday work completed on Tuesday = 4 \ Total holiday work completed in two days 1 3 4 + 15 19 = + = = 5 4 20 20 49. (D) : Quantity of wheat flour = 11.775 kg Share of Pia = 11.775 ÷ 3 = 3.925 kg Now, total chocolate cake baked = 5 \ Wheat flour needed in each cake = 3.925 ÷ 5 = 0.785 kg.

\ L.C.M. = 2 × 2 × 3 × 3 × 5 × 7 = 1260 So, they will beep together again after 1260 mins i.e. after 21 hrs. So, time = 12 : 00 noon + 21 hr = 9 : 00 a.m.

50. (D) : Temperature of City A = – 20°C Temperature of City B = 10°C Difference between temperature of two cities = 10°C – (– 20°C) = 10°C + 20°C = 30°C

vvv

Class 6 Imo Wkbsol e Book

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