CLASS 8 SET-4
HINTS AND SOLUTIONS 1.
x + 2
2 2 4 5
2 −2 ( x −1) = 5
x + 2
8 ⇒ 2 5
2 −2×( x −1) = 5
8 x +16
77° + x x + + 48° = 180° 125° + x x = = 180° x = 180° – 125° = 55° Ext∠ BOC = ∠OAB OAB + + ∠ ABO
−3
[∵ ( (a am)n = amn]
= 48° + 55° = 103° Now,, again in D BOC , Now ∠ BOC + + ∠OCB OCB + + ∠CBO CBO = = 180° [By angle sum property of a triangle] ⇒ 103° + 48° + y = 180° [ ∵ ∠ BAC = = ∠ ACB ACB = = 48°] ⇒ 151° + y y = = 180° ⇒ y y = = 180° – 151° = 29° Since, BA Since, BA || || CD CD and and AC AC is is a transversal line. \ ∠ BAC = = ∠ ACD ACD = = 48 48°° [By alternate interior angles] = 180° – 132° = 48° ⇒ ∠ BCA BCA = Also, ∠ BCA BCA + + ∠ ACD ACD + + ∠ DCE = = 180° [∵ BCE is is a straight line] = 180° ⇒ 48° + 48° + z = [ ∵ AB AB = = BC ⇒ ∠ BAC = = ∠ BCA BCA]] ⇒ 96° + z = = 180° ⇒ z = 180° – 96° = 84°
[∵ a m = an
⇒ m =
n]
(A) : The given numbers are 43, 91 and 183.
Now,, required number Now = H.C.F. [(91 – 43), (183 – 91) and (183 – 43)] = H.C.F (48, 92 and 140) 48
92 – 48 44
1 48 – 44 4
3.
1 44 – 44 0
11
4 140 – 12 20 – 20 0
35
Thus, x Thus, x = = 55°, y 55°, y = = 29° and z = = 84° 6.
(A) : We have, C
(B) : We have given,
A
Wages of 15 labourers for 6 days = ` 8100 8100 ` 8100 ⇒ Wages of 1 labourer for 1 day = = ` 90 90 15 × 6 Now,, wages of 21 labourers for 1 day = ` (90 Now (90 × 21)
F
8 cm
m 9 c
6 N
area (D ABC ) =
∵
(B) : According to given question,
1 2 1
26 cm
P
4 cm
G
H
× BC × × AM × 9 cm × 6 cm = 27 cm 2
BP = = BH – – PH PH = = BH – – EF
[∵ PH = = EF ]
= (26 – 8) cm = 18 cm
(A's 1 day work) : (B's 1 day work) = 2 : 1 1 and (A + B)'s 1 day work = 20 1 2 1 Now,, A's 1 day work = Now × = 20 3 30
area ( BPDC ) = BP × × BC = 18 cm × 9 cm = 162 cm2 Also, area ( PHFE ) = PH × × HF
Thus, A will alone nish the work in 30 days. Also, ∠ DOC = = ∠ AOB
6 cm
M c
2 Draw a perpendicular DP on on BH .
= ` 9450 9450
DOC = 77° (C) : We have ∠ DOC =
E
m
=
Thus, wages of 21 labourers for 5 days = ` (1890 (1890 × 5)
5.
D
B
= ` 1890
4.
…(ii) [By exterior angle property in D AOB AOB]] [By (ii)]
−3
6 ( x −1)
2 ⇒ 2 = 5 5 ⇒ 8 x x + + 16 = 6( x – 1) x + + 16 = 6 x x – – 6 ⇒ 8 x ⇒ 8 x x – – 6 x x = = – 6 – 16 ⇒ 2 x x = = –22 ⇒ x x = = –11 2.
⇒ ⇒ ⇒
(B) : We have,
…(i)
[ ∵ Vertically opposite angles]
⇒
∠ AOB = 77° [By (i)] Now,, In D AOB Now AOB,, ∠ AOB AOB + + ∠OBA OBA + + ∠ BAO BAO = = 180° [By angle sum property
= 8 cm × 6 cm = 48 cm 2 1 and area (DGFH ) = × FH × × GN 2 1 = × 6 cm × 4 cm = 12 cm 2 2 Thus, the required area of the given gure = (27 + 162 + 48 + 12) cm 2 = 249 cm 2
of a triangle]
Class-8 | Set-4 | Level-2
1
1
1
7.
(B) : We have,
1
2
1 4 (9) 6 (0.81) 3 (0.3) 3 27 −1
2
and number of girls below ten years of age = =
−1
1
=
2
−1
−1
9 3 (3) 2 (3)5 4 10 1
33
1
10 3
34
× 1
=
(3)
2×
1
× 33 × 3
(3) 10
2
4×
2×
−1
3
2 3
2
11. (C) : We have,
3
m n
=
33
1
× 1
10 3
1
× 33 × 3
34
8
(3) 3
CAB
4
2
×
10 3
=
33
= 8.
1
10
=
+ −
3 3
27 10
×
1 −1
(3) 2
1
×
−5
(b) A solid having 5 faces, 8 edges and 5 vertices is rectangular pyramid. (c) A solid having 3 faces, 2 edges and 0 vertices is cylinder. (d) A solid having 2 faces, 1 edge and 1 vertex is cone. 9.
→ (iv), (b) → (i), (c) → (ii), (d) → (iii)
(D) : We have, a2 +
Now, we have either
5 5 or
50
×3
×3
16 5 But,
15 0
50 ×3
1 5 0 is correct Thus, A = 5, B = 0 and C = 1 12. (C) : Total marks scored = 720
= 2.7
(B) : (a) A solid having 4 faces, 6 edges and 4 vertices is tetrahedron.
Thus, (a)
Since, the ones digit of B × 3 = B, So B = 0 or B = 5 Now, A × 3 = A ⇒ A = 5 or A = 0 But A = 0, it will make AB a 1-digit number, So, we reject A = 0
(3) 4
∵
3 3 4 3 2 4 1 4 2
33
4
a m m− n n =a a and a m × a n = a m+ n
+ + − − + +
10 3
10 3 (3) 3
1 1 8 3 4 1 5
×3
[∵ (a ) = a ]
−5
10 3 1
6 7 5 x
AB
mn
× (3) 2 × (3) 4
2
3x
14 Total number of students below ten years of age 2 x + 5 x 7 x x x 5 x = + = = = 14 14 2 7 14 \ Total number of students above ten years of age = 500. x Now, + 500 = x 2 x x ⇒ x – = 500 ⇒ = 500 ⇒ x = 1000 2 2 Thus, total number of students = 1000.
1 3 3 3 1 4 [(3)2 ]6 81 3 10 3 100 2
×
\
(0.9) 3 (3) 2 (243) 4 1
5
1
= 786 a2 Subtracting 2 from both sides, we get 1 a2 + – 2 = 786 – 2 a2 2 ⇒ a − 1 = 784 = (28) 2 [∵ a2 + b2 – 2ab = (a – b)2] a ⇒ a – 1 = 28 [ ∵ am = bm ⇒ a = b] a
10. (B) : Let the total number of students in the school be x. 3 So, number of girls = x 7 7 x − 3x 4 x 3 and number of boys = x – x = = 7 7 7 1 4 x x = Number of boys below ten years of age = × 4 7 7
90°
× 720 = 180 360° 70° and marks scored in Hindi = × 720 = 140 360° Thus, Rishi scored 180 – 140 = 40 more marks in Mathematics than the marks scored in Hindi.
Marks scored in Mathematics =
13. (C) : Step IV is incorrect.
The correct step is, with O as centre and radii =
1
(8 cm) 2 = 4 cm each, draw arcs, cutting OX at D and OY at B.
14. (D) : Let the original number be x. 30 x = 13 x Then, increased number = x + 100 10 When this increased number is decreased by 30%,
then nally decreased number = = Net decrease = x –
91 x
=
9x
13 x − 30 × 13x 10 100 10 91 x 100
100 100 Thus, the required net decrease percent =
=
9 x 100 ×100 % x 9 ×100 % 100
= 9%
2
Class-8 | Set-4 | Level-2
20. (C) : Let the digit at one's place be a,
15. (A) : We have C.P. = ` 21080, prot = 25% 125 \ S.P. = ` × 2080 100
= ` 2600
the digit at ten's place be b, …(i)
Let the marked price be ` x.
\
25 x 25 x x = ` = ` . 100 100 4 3 x x Discount = ` x − = ` …(ii) 4 4
Discount allowed = `
Now, S.P. = M.P. –
So, by (i) and (ii), we get 3 x = 2600 ⇒ x = 2600 × 4 = 3466.67 3 4 Thus, the marked price of the article is ` 3466.67 16. (C) : We have, 7 n = 2401 ⇒ 7n = 74 ⇒ n = 4 [ ∵ am = an ⇒ m = n] Now, 7 n – 3 + 8n – 5 = 7 4 – 3 + 8 × 4 – 5 = 7 1 + 32 – 5 = 34 17. (C) : Total number of votes cast = 600000 8 Number of invalid votes = × 600000 = 48000 100 Number of valid votes = 600000 – 48000 = 552000 Now, number of valid votes polled in favour of candidates 65 = × 552000 = 358800. 100 18. (C) : Let r 1 and r 2 be the radii and h1 and h2 be the heights of the two cylinders. h c We have given, h1 : h2 = c : d ⇒ 1 = …(i) h2 d 2 r 1 c a pr12 h1 a = ⇒ × = ∵ [By (i)] pr22 h2 b r 2 d b 2
⇒ ⇒
r 1 a d ad r = b × c = bc 2 2r 1 r 1 ad = . Also =
\
the number formed is
1000 d + 100 c + 10 b + a.
…(i)
After reversing the order of the digits, we get a new number i.e., 1000 a + 100 b + 10 c + d …(ii) According to question, 1000 d + 100 c + 10b + a – 1000 a – 100 b – 10 c – d [By (i) and (ii)] = 999 d + 90 c – 90 b – 999 a = 9 [111 d + 10 c – 10 b – 111 a], which is divisible by 9. 21. (C) : Let the dimensions of the rectangular block of the stone be 3 x, 2 x and x respectively. ∵
Volume of stone = 10368 dm 3
⇒ ⇒ ⇒ \
3 x × 2 x × x = 10368 dm 3
\
⇒ x3 = 1728 dm 3 ⇒ x = 12 dm
6 x3 = 10368 dm 3 x =
3
1728 dm The dimensions of the rectangular block of the stone are 36 dm, 24 dm and 12 dm. Total surface area = 2(l × b + b × h + h × l ) = 2(36 × 24 + 24 × 12 + 12 × 36) dm2 = 2(864 + 288 + 432) dm2 = 2(1584) dm2 = 3168 dm2
Thus, the cost of polishing = ` 3168 ×
= ` 63.36 100 2
22. (C) : We have, ab( x + y ) + xy (a2 + b2)
ad
ad : bc
19. (B) : Let the third number be x. According to question, second number be 2 x and rst number = 2(2 x) = 4 x 1 1 1 The reciprocal of the numbers are , and . 4 x 2 x x 1 1 1 + + 7 Then, 4 x 2 x x = 3 72 7 1+ 2 + 4 7 7 4 x ⇒ ⇒ 4 x = = 3 42 3 72 7 7 7 × 42 = ⇒ ⇒ = x 12 x 42 7 × 12 7 ⇒ x = 2 7 Thus, the rst number = 4 × = 14, 2 7 7 second number = 2 × = 7 and third number = . 2 2
Class-8 | Set-4 | Level-2
the digit at thousand's place be d ,
2
2r 2 bc r 2 bc Thus, the required ratio of the diameters of two cylinders be
the digit at hundred's place be c and
2
=
abx2 + aby 2 + xya 2 + xyb 2
=
(abx2 + xya 2) + (aby 2 + xyb 2)
=
ax(bx + ay) + by(ay + bx)
=
(bx + ay) (ax + by) …(i) 23. (D) : We have, ∠ DFE = ∠ AFG = b [ ∵ Vertically opposite angles] In D DEF , ∠ DEF = ∠ DFE [∵ Sides opposite to equal angles are equal] …(ii) [By (i)] ∠ DEF = b Also, ext ∠ FDC = ∠ DEF + ∠ DFE [∵ Exterior angle is equal to sum of its two opposite interior angles] = b + b = 2b …(iii) [By (i) and (ii)] Also ∠ FDC + ∠ BCD = 180° [∵ AD || BC and DC is a transversal line \ sum of the co-interior angles are supplementary] ⇒ ∠ FDC = 180° – ∠ BCD = 180° – c …(iv) From (iii) and (iv), we get 180° − c c 2b = 180° – c ⇒ b = = 90° – 2 2
3
24. (B) : Since, Time =
Distance
Speed According to question, time taken to cover 10 km at 15 km/hr = and time taken to cover 16 km at 12 km/hr =
10 15 hrs = 16 hrs = 12
2 3 4 3
hrs hrs
Now, total distance covered = (10 + 16) km = 26 km 2 4 6 and total time taken = + hrs = hrs = 2 hrs 3 3 3 26 total distance Thus, average speed = = km/hr 2 tatal time 25. (A) : We have
3 x + 8 = 272 x + 1
⇒ 3 x + 8 = 36 x + 3 3 ⇒ 6 x – x = 8 – 3 ⇒ 5 x = 5 ⇒ x = 1
3 x + 8 = (33)2 x + 1 x + 8 = 6 x +
2
1
1
x
Now, consider
=
=
[∵ (a m)n = amn]
1
289 17 ÷ 3 216 4 1296 1 x 2 17 ×17 x 17 ÷ 4 3 6 × 6 × 6 6 × 6 × 6 × 6 x
=
= =
= 13 km/hr
⇒ ⇒
1 + 2a × (a 2 + 4a − 5) 1 + a 1 − a 2 (a 2 + 10a + 25) 1 − a + 2a × ( a 2 + 5a − a − 5) 1 − a 2 (a 2 + 5a + 5a + 25) (1 + a ) [a (a + 5) − 1(a + 5)] × (1 − a 2 ) [a (a + 5) + 5(a + 5)] −1 (1 + a ) (a − 1)(a + 5)] −(1 − a) × = = a+5 (1 + a )(1 − a ) ( a + 5)(a + 5) (1 − a )( a + 5)
28. (D) : We have,
29. (D) : Number of bottles served to 5 children = 8 8 \ Number of bottles served to 1 children = . 5 8 Thus, number of bottles served to 40 children = × 40 = 64 5 30. (B) : Let P = x \ according to question, A = 2 x. 1 3 Time = 6 years = 6 years 4 12 x × R × 25 Now, 2 x = x + , where R be the rate 100 × 4
⇒ ⇒
\ according to question, A = 3 x. x × 16 × T Now, 3 x = x + , where T be the time ⇒ 1.
⇒
26. (D) : 90356294 is divisible by 2, but not by 4.
Also, a number with an even digit in the units place is always divisible by 2. 27. (C) : We have,
A = ` (32000 + 5044) = ` 37044 r Let rate = r % p.a. \ rate = % per quarter 4 and time = 9 months = 3 quarters, i.e., n = 3
⇒
a2 + b2 + 2ab – c2 = 4
⇒
1+
⇒
r
⇒
400
400
– 1 =
21 − 20 20
=
(b + c) – a = 9 2
…(ii)
2
(c + a) – b = 36 …(iii)
⇒ a =
b]
(a + b + c)2 = 49 (a + b + c) =
49
a + b + c = 7
32. (C) : Earning of 3 men in a day = ` 480 480 Earning of 1 man in a day = ` = ` 160 3 \ Earning of 7 men in a day = ` 160 × 7 = ` 1120
1 20
= 20% 20 Thus, the required rate percent is 20%
4
r =
20
[ ∵ am = bm
⇒ ⇒ ⇒
3
r ⇒ 21 r = 1 + = 1 + 8000 400 20 400
21
…(i)
2
Adding (i), (ii) and (iii) we get
⇒
=
2
a2 + b2 + c2 + 2ab + 2bc + 2ca = 49
3
20
1
c2 + a2 + 2ca – b2 = 36
n
400
= T
years 2 31. (A) : We have, (a + b)2 – c2 = 4
and
\ A = P 1 + r 400 3 1 + r ⇒ 37044 = 32000 400
21
4
25
b2 + c2 + 2bc – a2 = 9
\
=
2 × 25
Thus time = 12
P = ` 32000, C.I. = ` 5044
r
2 x =
100 x × 4 × T
⇒
∵
9261
16
R = 16%
Now if P = x
1 17 x 17 x 2 17 1 17 1 2 ÷ = ÷ = (1) 2 = 6 6 6 6
3
1
x = x × R ×
\ \
Earning of 4 women in a day = ` 480 480 Earning of 1 women in a day = ` = ` 120 4 Earning of 11 women in a day = ` 120 × 11 = ` 1320 Earning of 7 men and 11 women in a day = ` (1120 + 1320) = ` 2440
Class-8 | Set-4 | Level-2
33. (C) : Let the total money be ` x.
Amount of money, his wife gets = `
(B)
x
September rainfall exceeds April rainfall
2 x
Remaining amount of money = ` x −
2
2 × x = ` x 3 3 2
Amount of money, three sons gets = `
Remaining amount of money for daughters x x x = ` − = ` 6 2 3 1 x \ Amount of money, each daughter gets = ` × 4 6 x = ` 24 x According to question, = 20000 24 ⇒ x = 480000 \ Amount of money, each son gets
x 1 x = ` × = ` 9 3 3 480000 = ` 9 = ` 53333.33 34. (D) :
14 + 6 5
+
14 − 6 5
=
9 + 5 + 2 × 3× 5
=
(3) 2
=
(3 + 5 ) 2
=3+
+(
+
5 + 3 –
9 + 5 − 2 × 3× 5
(3 − 5 )2 5
=6 35. (B) : Reciprocal of
3
5
and reciprocal of
7
is
3
3 7 3 5 3 35 + 9 44 Now, sum of the reciprocals = + = = 3 7 21 21 44 21 \ Reciprocal of is 21 44 36. (A) : Let the sides of cubes be 3 x, 4 x and 5 x. According to question, Sum of the volumes of the cubes = volume of new cube ⇒ (3 x)3 + (4 x)3 + (5 x)3 = (side)3 ⇒ 27 x3 + 64 x3 + 125 x3 = (side)3 ⇒ 216 x3 = (side)3
⇒
Side =
5
is
3
216 × x 3 = 6 x Also, side of the smallest cube = 3 x 3 x 1 Thus, required ratio = = or 1 : 2 6 x 2 37. (C) : (A) November rainfall : in place 1 = 25 cm in place 2 = 50 cm in place 3 = 75 cm in place 4 = 75 cm in place 5 = 75 cm in place 6 = 25 cm
Class-8 | Set-4 | Level-2
38. (A) : Let the numbers are 3 x, 4 x and 5 x. According to question, (3 x)2 + (4 x)2 + (5 x)2 = 1250 ⇒ 9 x2 + 16 x2 + 25 x2 = 1250 ⇒ 50 x2 = 1250 ⇒ x2 = 25 ⇒ x = 5 So, the numbers are 3 × 5, 4 × 5 and 5 × 5 i.e., 15, 20 and 25 \ Sum of numbers = 15 + 20 + 25 = 60 39. (B) : Let the original price of the sugar be ` x per kg. x × 20 After 20% reduction, the price of the sugar = ` x − 100 4 x = ` 5 160 1 160 +2 = According to question, 4x x 2
⇒
5 ) 2 + 2 × 3 × 5 + (3)2 + ( 5 )2 − 2 × 3× 5
+
(C) place.
in place 1 = (200 – 175) cm = 25 cm in place 2 = (300 – 175) cm = 125 cm in place 3 = (250 – 200) cm = 50 cm in place 4 = (250 – 200) cm = 50 cm in place 5 = (250 – 200) cm = 50 cm in place 6 = (200 – 100) cm = 100 cm November rainfall is lower than April rainfall in each
160
5
+ =
160
×5 ⇒
320 + 5 x
5
=
200
x 2 4x 2 x x ⇒ 320 + 5 x = 400 ⇒ 5 x = 400 – 320 ⇒ 5 x = 80 ⇒ x = 16 \ Original price of sugar is ` 16 per kg 40. (D) : We have, x + y = 10 y x 3 x
⇒ ⇒ ⇒ ⇒
y
y
+
( x )2
x
=
+(
x y 10 xy
=
10 3
10 3
y )2
=
10 3
⇒
x + y xy
=
10 3
( ∵ x + y = 10)
xy = 3
⇒ xy =
32
⇒ xy =
9
41. (C) : Both man and train are running in the same direction.
\
Relative speed = (68 – 8) km/hr = 60 km/hr 5 = 60 × m/s 18 50 = m/s 3 Length of the train = 150 m 150 \ Required time = 50 secs
3
=
150 × 3
50 = 9 secs
secs
5
⇒ ⇒ ⇒ ⇒ \
42. (D) : We have, 2 x = 4 y = 8 z
Sum of 49 observations + 45 = 49 x + 45 Sum of 50 observations = 49 x + 45 50 x = 49 x + 45 x = 45 Mean of 50 observations = 45 46. (A) : Let the age of man and his wife be 4 x and 3 x respectively. According to question, 4 x + 4 9
⇒
x = 2 y = 4 z 1 1 ⇒ y = x and z = x 2 4 So, xyz = 1728 x x ⇒ x × × = 1728 2 4 ⇒ x3 = 1728 × 8
⇒ ⇒ Now,
x =
3
x = 24 1 1 2 x = =
+
1 2 x 3
⇒
13824
4y
+ =
+ 1
2x
⇒ ⇒ \
1 8 z
+ 3
1
=
1
2 x 2 × 24 16 43. (C) : Let the digit at units place be x. Then the digit at tens place = x – 2
\
The number = 10( x – 2) + x = 10 x – 20 + x = 11 x – 20
And, the number obtained by reversing the digits of the number = 10 x + x – 2 = 11 x – 2 According to question, 6 3(11 x – 20) + (11 x – 2) = 108 7 66 x − 12 ⇒ 33 x – 60 + = 108 7 ⇒ 231 x – 420 + 66 x – 12 = 756
⇒ ⇒
297 x = 756 + 432 1188 x = = 4 297
x = 8 Age of man = 4 × 8 = 32 Age of his wife = 3 × 8 = 24
(∵ 2 x = 4 y = 8 z )
2x
=
3 x + 4 7 28 x + 28 = 27 x + 36
Now, let they were married T years ago. 32 − T 5 So, = 24 − T 3 ⇒ 96 – 3T = 120 – 5T
⇒ ⇒
2T = 24 T = 12
Hence, they were married 12 years ago. 3 5 1 1 + 3 − 2 7 5 7 1 47. (C) : ÷ × + 1 × 3 2 + 15.6 31.2 2 7 10 5 5 + 5−2 2 −2 3 11 4.8 9.6 10 5 30 50 − + 7 5 7 6 3 2 16 32 ÷ + × + 50 10 8 7 10 5 27 5− − 2 + 3 11 48 96 20 − 15 60 + 50 7×3 5 7 6 ⇒ ÷ + × 6 + 32 15 − 8 7 10 5 27 − 22 100 + 10 96 11
Units digit = 4 and tens digit = 4 – 2 = 2 \ Required sum = 4 + 2 = 6
=
44. (D) : Let the amount of money Z has be ` T 3 50T Amount of money Y has = T + = T 2 100 3 And amount of money X has = 2 × T = 3T 2 According to question, 3 3T + T + T 2 = 110 3 ⇒ 6T + 3T + 2T = 110 6 ⇒ 11T = 660
=
⇒ \
T = 60 Amount of money X has = 3T = ` (3 × 60) = ` 180
45. (C) : Let the mean of 50 observations be x.
Also, mean of 49 observations = x
⇒ 6
= =
5 7 6 5 × 11 110 × 96 ÷ + × + 7 7 10 5 6 × 5 110 × 32 21 5 7 11 ÷ + + 3 = 21 ÷ 5 7 + 22 + 3 7 7 10 5 7 7 10 21
21 14 5 29 ÷ + 3 = × + 3 7 29 7 7 10 42 42 + 87 129 = +3= 21
29
29
29
Required percentage error =
146 − 129 29 29 × 100 129 29
=
17 129
×100 = 13.17%
48. (B) : Let the inner and outer radii be r 1 and r 2 respectively. Also, let the inner and outer perimeters be 22 x and 23 x respectively.
Sum of 49 observations = 49 x
Class-8 | Set-4 | Level-2
Given, r 2 = r 1 + 5; 2pr 1 = 22 x
Now, DFE is a straight line.
and 2pr 2 = 23 x
\ ⇒
⇒ ⇒ ⇒ ⇒ \
⇒ 2 p(r 1 + 5) = 23 x 2pr 1 + 10p = 23 x ⇒ 22 x + 10 p = 23 x x = 10p ⇒ 2 pr 1 = 22 × 10 p = 220p 220p r 1 = = 110 m and 2 pr 2 = 23 × 10 p = 230p 2p 230p r 2 = = 115 m 2p 2 2 2 2 Area of path = pr2 − pr1 = p( r2 − r1 ) = = = =
49. (C) : ∠ DGC =
22 7 22
7 22 7
× 225 × 5 m2
× 1125 m2 = 3535.71 m 2 (Vertically opposite angles)
∠ AGB = b Now in D AGB, by angle sum property, 65° + b + 50° = 180° b + 115° = 180° b = 180° – 115° b = 65°
Also, 334° + ∠ FCE = 360°
a = 180° – 42° = 138°
Number of arrows used to cut down the arrows thrown by x Bheeshm = 2 Number of arrows used to kill the rath driver = 6 Number of arrows used to knocked down the rath, ag and the bow = 3
⇒ ⇒ ⇒ ⇒
[Linear pair]
50. (B) : Let total number of arrows Arjun had be x
(115 + 110)(115 – 110)m 2
∠ AGB
a = 180°
So, a – b = 138° – 65° = 73°
× (1152 – 1102)m2
7 22
∠ EFC +
[Complete angle]
⇒
∠ FCE = 360° – 334° = 26° Now, in DCEF , by angle sum property, 26° + 112° + ∠ EFC = 180° ⇒ ∠ EFC = 180° – 138° ⇒ ∠ EFC = 42°
According to question, x 4 x + 1 + + 6 + 3 = x 2 x ⇒ 4 x + + 10 = x 2 x ⇒ 4 x = – 10 2 ⇒ 4 x = x − 20 2 ⇒ 8 x = x – 20 Squaring both sides, we get 64 x = x2 + 400 – 40 x ⇒ x2 – 104 x + 400 = 0 ⇒ x2 – 100 x – 4 x + 400 = 0 ⇒ x( x – 100) – 4( x – 100) = 0 ⇒ ( x – 100) ( x – 4) = 0 ⇒ x = 100 or x = 4 but x = 4 is not possible. \ x = 100 Hence, total number of arrows Arjun had was 100.
JJJ
Class-8 | Set-4 | Level-2
7
