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SSC MAINS (MATH) MOCK TEST - 4 (SOLUTION) 1. (D) Total no. of marbles kept in 50th box = sum of factors fo 50 = 1 + 2 + 5 + 10 + 25 + 50 = 93 2. (C) Let x be the required number and a be be its first part part so, (x (x – a a ) will be its second part Now, A.T.Q, 0.8a 0.8a = = 0.6 (x ( x – a ) + 3 or, 0.8a 0.8a + + 0.6a 0.6a = = 0.6x 0.6x + 3 or, 1.4a 1.4a = 0.6x 0.6x + 3
a
0.6x 3 1 .4
(i)
also 0.9a 0.9a + + 6 = 0.8 (x ( x – a ) or, 0.9a 0.9a + + 0.8a 0.8a = 0.8 x x - 6 or, 1.7a 1.7a = 0.8x 0.8x – 6
a
0.8 x 6 -------------- (ii) 1.7
from (i) & (ii) , 0.6x 3 0.8x 6 1.4 1.7
1.02x + 5.1 = 1.12x 1.12x – – 8.4 1.02x 0.1x = 13.5 0.1x
x
13 .5 = 135 0.1
A l t er n a t i v e m e t h od
4 5
a
3 5
(x a ) 3
3 4 3 5 5 a 5 a 3 --------(i)
al also
9 4 a 6 (x a ) 10 5
4 9 4 10 5 a 5 x 6 --------(ii) from (i) & (ii), we get,
3 9 8 4 4 3 x 3 x 6 5 10 5 5
135
50
x 10 5
x = 135 3. (A) Let the two digit number be 10y 10y + + x where where x > y 10x+ y 10y 10y x = 63 10x+ 9x 9y = y = 63 x y = 7 = 7, 8, 9 and y and y = = 0, 1 , 2 x = possible values of x are 7,8,9 4. (A) For every every n 4 ; n! will be divisible by 8 remainder will be zero [ because for n 4 , 8 will be a factor of n!] So, remainder of 1! + 2! + 3! + 4! + ....... ....... + 100! wil be equal to the remainder of 1! + 2! + 3! only 1! + 2! + 3! = 1 + 2 + 3 = 9 9 ;R=1 8 5. (D) H.C.F. H.C.F. = x and L.C.M. = y = y × y A × B = x × So, A³ + B³ = (A + B)³ - 3AB(A + B) = (x (x + + y )³ )³ – 3xy 3xy (x (x + + y ) = x ³ + y ³ + 3xy 3xy (x + x + y ) – 3xy ( xy (x + x + y ) = x ³ + y ³
and
1
1
1
1
6. (A) x 2 y 3 x 1/3 y 1/ 2 x ³ y ² 7. (C) Required average age of 2 persons 30 34 (8 3 ) 44 yrs. 2 8. (*) Required average average age just just before the birth of the youngest member
=
=
(10 20 ) (10 10 ) 10 1
=
100 = 11.11 yrs. 9
9.(C) 9.(C) Weight Weight of first first member member = x kg kg Weight of second member = (x ( x +2) -------------Weight of fifth member = (x ( x + 8) kg
51 51 28 42 x x or, 50 10 25 5
10. 10. (A) Average Average speed speed =
51 42 56 51 x or, 10 5 50 50
=
Total distance covered Total time taken (6 8 40)km (15 15 15) min
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================================================================================== 10x = 7.5 (x + y ) 54 km 54 km = = 2.5x = 7.5 y 45 45 min hr. x = 3y 60 x : y = 3 : 1 = 72 km/hr. 17.(D) Ratio of equivalent capitals of A, B and 11. (C) Age of the captain C for 1 month = (11 × 30 ) – {(5 × 29) + (5 × 27)} = (40500 × 6 + 45000 × 6 ): (45000 × 12) = (330 – 280) yrs. : (60000×6+45000×6) = 50 yrs. = (405+450): (450 × 2 ): (600 + 450) 12.(B) (2m + 4b ) × 10 = (4m + 5b ) × 6 = 855 : 900 : 1050 20m + 40b = 24m + 30b = 171 : 180 : 210 4m = 10b 2m = 5b = 57 : 60 : 70 So, 5b = 2 × 40 Sum the ratios = 57 + 60 + 70 = 187
b =
2 40 = 16 5
Required difference =
Required ratio = 40 : 16 = 5 : 2 13.(B) In case I. Let the number of shirts of brand B be x . Let the cost of a shirt of brand B be 1 Original cost = 4× 2 + x = (8 +x ) In case II.
=
70 57 = 56100 187 13 × 56100 = 187
`
3900
18. (A) Let the c ost price of 1 orange = Re. 1.
`
`
4x + 2x = (8 + x )×
140 7 = (8 + x ) 100 5
20 + 10x = 56 + 7x 10x 7x = 56 20 = 36 3x = 36 x = 12 Required ratio = 4 : 12 = 1 : 3 14. (C) Half the sum = 18 total sum = 36 Nos. will be 6, 12 and 18 So, ratio of squares = 6² : 12² : 18² = 36 : 144 : 324 15. (B) A.T.Q, A J 2 R 6 4 5
A : J : R = 6 : 4
:
5 2
= 12 : 8 : 5 Apoorva's share =
12 (12 8 5 )
9 100 = Rs. 7.5/litre 120 Now, let the ratio of milk and water in the mixture = x :y
Cost of mixture =
(10 x ) (0 y ) x y
= 7.5
3 4
and C.P. of 1 apple = Rs.
3 2
New prices: 1 orange = Re. 1.1 1 banana
3
3
1 apple
4 2
110
110
100 100
= Rs. 0.825 = Rs. 1.65
Original price of (4 bananas + 2 apples + 3 oranges) = Rs. (3 + 3 + 3) = Rs. 9 New price of (4 banana + 2 apples + 3 oranges) = Rs. (4 × 0.825 + 2 × 1.65 + 3 × 1.1) = Rs. (3.3 + 3.3 + 3.3) = 9.9 Percentage increase 9.9 9 100 = 10% 9 19. (A) Let the no. be x % change
=
2250
= Rs. 1080 16. (A) S.P. at 20% profit = Rs. 9/litre
C.P. of 1 banana = Rs.
=
original result - changed result original result
=
x 5 100% 5 x
=
25x x 100 % 25 x
=
24x 100% = 96% 25x
100%
5x
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================================================================================== 20. (C) 20% decrease in price ( S.P. > C.P.) = 20% gain 25% increase in consumption 24.(C) If the C.P. of wrist watch be x then (when expenditure is constant) C.P. of wall clock = (390 x ) increased amount of sugar = 25% of 20kg = 5 kg x 10 (390 x ) 15 So, = 51.50 Total amount now of sugar = (20 + 5) kg `
`
= 25 kg 21. (C) Votes got by Rahul Gandhi = (100-10)% of = 90% of
4 of total voters 5
4 of total voters 5
100
100
10x + 5850 15x = 5150 5x = 5850 5150 = 700
x =
700 = 5
`
140
C.P. of wall clock = 390 140= 250 Required difference = 250 140 = 110 25.(A) (40 20)% = Re1 `
`
=
9 4 of total voters 10 5
So,
18 = of total voters 25
120% =
4 = (100-20)% of 1 th of the total voters 5 1 th of total voters 5
=
4 1 of total voters 5 5
=
4 of total voters 25
= 210
120
= 210
120 87.5
100
100 (100 12.5)
= 288 27. (D) Let the C.P. = x So, S.P. in 1st case = 1.05x now, C.P. in 2nd case = 0.95x and S.P. in 2nd case = 1.05x – 2 Now, A.T.Q. 0.95x × 1.1 = 1.05 x - 2 or, 1.045x = 1.05x – 2 1.05x – 1.045x = 2 0.005x = 2 `
216
4 = 48 voters 18 So, total no. of votes polled = (216 + 48) votes = 264 votes 22. (C) 10% reduction in price 11.11% i ncrease in quantity (when expenditure is constant) A.T.Q, 11.11% of original amount of wheat = 50 gm Original amount (ie.100% ) of wheat = (50 × 9) gm = 450 gm 23. (A) C.P. of each book sold by publishe r =
`
26. (B) Required Marked price
= 216 voters -------- (i) Now, Votes got by Varun Gandhi
= 80% of
1 ×120= 6 20
70,000 = 2000 400
x
2 400 0.005
28.(C) Discount on
`
36000 =
Discount on first
`
3600 7 = 100
20,000 = =
Discount on next
`
10,000 =
`
2520
20000 8 100 `
1600
10, 000 5 100
= 500 Discount on remaining 6,000 = 2520 (1600 + 500) 420 `
`
`
`
= 43.75 S.P. of each book sold by publisher = (100-30)% of 75 = 52.5 `
Required percent =
420 100 = 7% 6000
29. (B) Net discount given by A
`
`
So, % gain =
52.5 43.75 100 % 43.75
5 25 % = 28.75% = 5 25 100
& Net discount given by B
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================================================================================== 16 12 % = 26.08% = 16 12 100
Annual payment =
A is giving more discount it is more profitable to purchase the fan from A.
=
30.(C) Single equivalen t discount = p q
pq % 100
31. (A) Required ratio = (15× 22) : (11 × 25) = 330 : 275 = 6:5 32. (D) Let Rs. x = quaterly payment given, r = 16% per annum So, rate of interest per quarter 16 = = 4% per quarter 4
Also, No. of quarters in 2 years = 4 × 2 = 8 quarters So, 8x + (7 + 6 + 4 + 3 + 2 + 1)× 4% of x = 2280
7 8 × 4% of x = 2280 or, 8x + 2 n (n 1) 1 2 3 ... n 2 or, 8x + 112% of x = 2280 or, 800% of x + 112% of x = 2280 912% of x = 2280
2280 100 = 912 Direct Method:Quarterly Payment
So, x =
33. (D)
=
100 2280 487 100 8 2
=
100 2280 2 = 250 1600 224
3310 1331 (1210 1100 1000 )
= 1331
R 34.(C) A = P 1 100
R 3 = 1 1 100
T
3
On squaring both sides, 9 = 1 1 35. (A)
R 100
6
Cash price of refrigerator 10 100 1003 = 1500 + 1020 11 121 1000 990 1331
(10200121) (10030011) 990000 = 1500 1331 1234200 1103300 990000 = 1500 1331 = 1500 +
250
3310 10 100 1000 11 121 1331
3327500 1331
= 1500 + 2500 = 4000 Alternative Method:Cash price of refrigerator 1020
1003
990
= 1500 1.1 (1.1)2 (1.1)3 = 1500
Let x be the required annual payment. So, also, r = 10% p.a. and t = 3 years 10%
1.1 1 So, (1 + 1.1 + 1.21)x = 3310 × (1.1)3 or, 3.31x = 3310 × 1.331
1020 1.1
1003 1.21
990 1.331
= 1500 + 2500 = 4000 36. (D) When B works normally then days taken by B to complete the w ork
20 12 days 20 12 3310 1.331 = 30 days x = 3.31 Now, If B does the work only half a day daily = Rs. 1331 B will take twice the total days to Direct Method:momplete the whole work alone Now No. of days taken by B = (30 × 2) days = 60 days So, Ph: 011-27607854, (M) 8860-333-333 4
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================================================================================== Now days taken by (A + B) toget her to do the 2 d t2 = 40 20 60 2d whole work = = 15 days 20 60 2 1 37.(A) Let time taken by B in completing the w ork t2 = 40 2 = x day Time taken by A (x 10) days t = 10 minutes 2
time taken by a pipe of diameter 2d for doing the same job = 10 minutes 40.(C) Let the capacity of the tank be x gallons. Quantity of water filled in the tank in 1 minute when all the pipes A, B and C are
1 1 1 x x 10 12
x 10 x
1
x x 10 12
24x 120 = x 2 10x x 2 34x + 120 = 0 x 2 30x 4x + 120 = 0 x (x 30) 4(x 30) = 0 (x 4) (x 30) = 0 x 30 because x 4 38. (C) Let days taken by A to do the wh ole work = x days So, days taken by B to do the whole work = (x – 5) days and, days taken by C to do t he whole work = (x – 9) days Now, ATQ, Days taken by (A + B) together to do the whole work = Days taken by C alone to do the whole work
x ( x 5 ) = x – 9 x ( x 5)
or, x 2 – 5x = (x – 9)(2x – 5) or, x 2 – 5x = 2x 2 – 5x – 18x + 45 or, x 2 – 18x + 45 = 0 or, x 2 – 15x – 3x + 45 = 0 or, x (x – 15) – 3(x – 15) = 0 or, (x – 3) (x – 15) = 0 x = 3 or 15 but x = 3 is not possible as x – 5 or x – 9 will become negative. So, x = 15 days 1
39.(B) Time cross sectional area of the pipe Time 1
4
Time
t 2
d2
1 d2
(d 1 )2
= (d ) t 1 2 2
So, t2 = t1
d 2 1
d2
t1 = 40 minutes, d 1 = d, d2 = 2d
opened simultaneously = According to the question,
x x x 3 20 24 15
6x 5 x 8x 3 120
3x = 3 ×120
x =
x x 3 20 24 x x x 3 20 24 15
3 120 120 gallons 3
41. (B) Let spee d of car = x km/hr. Here, Distance covered by the car in 27 minutes = Distance covered by the sound in (28 minutes 30 seconds – 27 minutes)
27 x km/hr × 60 hr. 18 1.5 km /hr × hr = 330 5 60
x = 330 ×
18 5
.5 60
60 27
= 66
Speed of car = 66 km/hr. 42. (D) In the race between Sonu and Monu. Distance travelled by Sonu and Monu in same time = 600 mtr. & (600 – 60)mtr = 600 mtr. & 540 mtr. In the same time, Ratio of distance travelled by Sonu & Monu = 10 : 9 Similarly, In the same time, Ratio of distance travelled by Monu & Bablu = 500 : (500 – 25) = 500 : 475 = 20 : 19 So, In the same time, Ratio of distance travelled by Sonu, Monu & Bablu
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================================================================================== = 10 × 20 : 9 × 20 : 9 × 19 a 2 + b 2 + c 2 + 2bc = 2a 2 = 200 : 180 : 171 a 2 + b 2 + c 2 = 2a 2 2bc When Sonu travels 200 m, Bablu will = 2 (a 2 bc ) travel 171 m 2 a 2 b 2 c 2 2 a bc So, When Sonu travels 400 m, Bablu will =2 travel 342 m a 2 bc a 2 bc In 400 m race between Sonu & Bablu, 48. (D) (a 1) 2 3 b 2 a Required No. of metres by which Sonu a = 3 : a 1 = b will win the race 3 1 : b b = 2 = 400 m – 342 m a + b = 3 + 2 = 5 = 58 m 43. (A) Let x = length of the faster train (in mtr.) 49.(B) OP = 2 x 10 So, 36 seconds = (40 20 )kmph
x = 36 second × 20 ×
5 18
m/sec
B 0,3 Q 2
= 200 mtr. 44. (A) Speed of boat in still water =
X
O
Sdown Sup
P (2.0)
X
2
x y = 0.5 (x + y ) 2
=
3 2
OQ =
45.(C) Expression = 4 + 44 + 444 +........... to n terms = 4 (1 + 11 + 111 +.... to n terms)
PQ = OP 2 OQ 2 =
2 2
3 2 2
=
4 (9 + 99 + 999 + ...... to n terms) 9
=
4
=
4 [(10 1)+(100 1) + (1000 1)+........ to n 9
=
5 16 9 25 = = = 2.5cm 2 4 4
terms] 4 = [(10 +102 + 103 +......... to n terms) n] 9
=
4 [10(1 + 10 +102 + .......... to n terms) n ] 9
=
40 10n 1 4 n 9 9 9
4 9
50.(A)
Y
X
O
10n 1 [ 1 +10 +10 + ........ to n terms = ] 9
A (12.0) B(0-,9)
X
2
40 4 = (10n 1) n 81 9
46. (B) 180 = 2 × 2 × 3 × 3 × 5 a 3b = abc a 2 = bc a 3b = abc= 180 = 12×180 ×1 13 ×180 c =1 47. (C) a + b + c = 0 b+c=a On squaring both sides, we get (b + c )2 = a 2 b 2 + c 2 = 2bc = a 2
Y Putting x = 0 in 9x - 12y = 108, we get, y = 9 Putting y = 0 in 9x 12y = 108, we get, x = 12 OA = 12, OB = 9
AB = OA 2 OB 2 =
122 92
=
144 81
= 225 = 15 units
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51.(A)
x
1 x =3 x
x
2
1 = x
2b
2a a b
Apply C & D, x 2b 3a b = x 2b a b Now,
3
On cubing both sides. 1 1 3 x = 3 3 3 x x
x 3 +
=
x 2a
x 2b
..... (2)
a 3b
3a b
=
(a b )
x +1 = 0 x 206 + x 200 + x 90 + x 84 + x 18 + x 12 + x 6 + 1
=
a 3b 3a b a b
= x 200 (x 6+1) + x 84 (x 6 +1) + x 12 (x 6 +1)+(x 6 +1) = 0
=
x 2b
1 x + 3 = 3 3 3 3 = 0 x
x 2b
a b
3 6
7
52. (A)
16 6 7 16 6 7
7
=
9 7 23 7 9 7 23 7 7
=
=
=
sin 1 sin 1 cos cos cos cos
=
1 sin 1 sin cos cos
=
(3 7 )2 (3 7 )2 7 (3 7 ) (3 7 )
x 2 3xy 2y 2
53. (B)
2
2
x 3xy 2y
= =
2( x y )2 xy 2
2( x y ) xy
=
=
7
=
2 7
2
2
( x y 2xy ) xy 2 (6)2 1 2
2 (4 2 ) 1
cos2
=
cos2 cos2
=
2 36 1 2 32 1
2 +1 2-1
2 +1
x + y =
2-1 2 +1
and x – y =
y =
and xy = 1
2-1
+
-
2-1 2 +1 2-1 2 +1
1 = x
= (5 × 1) + (4 × 1) =5+4 =9 B
x 2a
Q
=
4 2
tan
b a
A
h ---------- (i) a
tan(90º )
a 3b
P a b
here, h = height of tower AB
x 2b = 2a a b Apply C & D,
=
h 90°-
=6
54. (D)
x 2a
2-1 2 +1
sec2
= 5(sin2 cos2 ) 4(sin2 cos2 )
57. (A)
x =
5
= 5 sin2 4 sin2 4 cos2 5 cos 2
71 65
= 1
9 5 4 cos ² cosec ² 1 tan ²
2 2 = 9 sin 4 cos
( x 2 y 2 2xy ) xy
=
1 sin2
1 2
= 2
1 1 1 1 cos cot cos cot
56. (C) =
a b
55. (C)
(3)2 ( 7 )2 2 3 7 (3)2 ( 7 )2 2 3 7 7
2a 2b
h b
..... (1)
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================================================================================== or, cot
x
h b
60.(B) sin 17º = y cos 17º = 1 sin2 17º
b tan ---------- (ii) h
from (i) and (ii), h b a a h ab
=
=
y 2 x 2 y 2
58.(A) CD = h metre, AB = 2h metre
2h
y 2 x 2 y 2
x2 y2
1
y
sec 17º =
sin 73º = sin (90º 17º)= cos 17º sec 17º sin 73º
h
=
y y 2 x 2
y x 2 2
y 2 x 2
y
y ² y ² x ² OB = OD = From
From
= y y ² x ² = y y 2 x 2
x metre 2
61.(B)
OCD,
tan =
2 6
(i)
(ii)
Multiplying both equations,
2h 4h x x
meter
tan 3 =
1 = cot 2 tan2
tan 3 = tan (90º 2 ) 3 = 90º 2 5 = 90º
= 18º
2cos2 = 2×
2
= XZ2 YZ2
sec X = tan X =
59. (C) tan2 . tan3 = 1
2 6
(ii) Adding both the equations, 2 × Z = 14 XZ = 7 YZ = 7 - 2 = 5
x = 8h
2 2
5 1 = 2cos2 45º 1 2
1 1 =0 2
(i)
24 = (XZ YZ) (XZ+YZ) XZ + YZ = 12
2
x
Z
XZ YZ = 2
OAB,
2h 4h cot = x = x 2
h =
5
Y
XY2 + YZ2 = XZ2
AB tan (90º ) = BO
2
X 7
h 2h = x x 2
tan . cot =
x 2
7 2 6 5 2 6
sec X + tan X =
7 2 6
+
5 2 6
=
12 2 6
=
6
62. (A) Angles of triangle (a d )º , aº , (a +d )º a d + a +a +d = 180º 3a = 180º a = 60º
a d 60 60 1 a d 180 3
60 d 1 = 60 d 3
180 3d = 60 + d 4d = 120º d = 30º a d = 60º 30º = 30º
a = 60º a + d = 60º+ 30º = 90º
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================================================================================== So, Angles of triangle are 30º ,60º and 90º d AE d d cot tan DE tan DE DE 63. (A) B (Balloon) So, the height of the house, In the given figure, after leaving the AB = CD + DE point A, balloon = d (tan cot ) reach to point B vertically upward 1 sin cos d in 1.5 min = d cos sin cos sin Here, O the = d sec cosec 60º observer 65. (A) 200m O A So, BOA = 60º (Observer)
c
tan 60º = AB
AB
C
OA
= OA tan 60º = 200 ×
b
B
a
Let ABC is a and a, b & c are the lengths of BC, CA & AB respective ly.
3 m
So, speed of the ballon distance = time
sin
A : sin B : sin C = 1 : 1 : By sine formula:
AB = time to reach from A to B 200 3m = 1.5 = 3.87 m/sec. sec 60
a sin A
b sin B
street
B
d
C
Here, AB height of the house & CD height of the window So, ATQ, ADB = 90º Also, here, line AD makes as angle º with the vertical line DE. ADE º also, BDC 90 º In BCD , d d tan(90º ) = = or, cot CD CD CD BC
CD
d = cot d tan
2 x
c 2 : (a 2 + b 2) = ( 2x )2 : (x 2 x 2 ) = =
D (window)
90º –
2
Let a = x , b = x & c = ATQ,
º
90º
sinC
a : b : c = 1 : 1 :
E
House
c
a : b = sin A : sin B & b : c = sin B : si n C
64. (C) A
2
2x 2 : 2x 2 1:1
66.(B) In s ACD and ABC.
CDA = CAB = 90º C is common. ACD ABC C
D
A
B
ACD AC2 ACD BC2
10 92 40 BC2
BC2 = 4 × 92
BC = 2 × 9 = 18 cm
also, In ADE , Ph: 011-27607854, (M) 8860-333-333
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================================================================================== A
67.(B)
Y X
D
O
O
90º
B
A
Z
B O D =
152 122
=
225 144
=
81 = 9
OD =
132 122
=
169 144 = 25 = 5
OO' = 9 + 5 = 14 cm
68. (B)
A
XY 2 XZ 2 ZY2 172= (r+9)2 + (r+2)2 289 = r 2 + 18r + 18 + r 2 + 4r + 4 2r 2 + 22r + 85 289 = 0 2r 2 + 22r 204 = 0 r 2 + 11r 102 = 0 r 2 + 17r 6r 102 = 0 r ( r + 17) 6( r +17) = 0 ( r 6) ( r +17) = 0 r = 6 cm
70.(A) E
D
B
ADE = ABC AED = ACB ADE ABC
BDEC
1 ADE 1
BDEC
ADE
XY2 (r1 r2 )2
=
DE BC
Length of transverse trangent
C
XY = 145
71. (C)
+1=1+1
ABC AB2 =2= AD2 ADE
AB = AD
AB 1= 2 1 AD
BD = 2 1 AD
1 AD = BD 2 1
2
AD : BD = 1 :
8 = XY2 92 64 = XY2 - 81 XY2 = 64 + 81 = 145
1 2 1
69.(B) XZ = r + 9 & YZ = r + 2
AB
is diameter
ADB = 90º also DO AB at 'O', the centre of the circle.
BDO (by SAS cong. Rule) ADO AD = DB (by CPCT) DAB = ABD = 45º But ACD = ABD (angles in the same segment of a circle) = 45º 72. (A) CAD CBD (Angles iin the same segment of a circle) = 60º Now BAD = BAC CAD = 30º + 60º = 90º Now BAD BCD 180º ( ABCD is cyclic) 90º + BCD = 180º
BCD = 180º – 90º = 90º 73. (A)
Perimeter of the rope = 3× (
1 of circumference of a circle) + 3
3 × diameter of a circle) Ph: 011-27607854, (M) 8860-333-333
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================================================================================== =3×
1 × 2 + 3 × 2 3
= 2 + 6 74. (A) DC||AB (given) AOB ~ COD (by AA similarity) ar (AOB )
AB 2
ar (COD ) = DC 2
=
(3DC )2 DC 2
9DC 2
=
DC 2
=
9
1 1 AC BD ² AC ² 2 4
AE ² CF ²
4 3 r V e 3 = 3 = 64 : 1 V m 4 r 3 4
78. (B) Perimeter = 2(l + b ) P = 2(l + w )
1
75. (B) In the given figure, ABC is a right angle triangle, where B 90º AE, BD and CF are the 3 medians Now, AB = 12 cm, BC = 9 cm & AC = 15 cm So, here, BD
r r 2 Rm : radius of the moon = = unit 2 4
5 AC ² 4
also,
1 5 BD ² AE ² CF ² AC ² 4 4
P 2
– w = l
its area = l × b
P k = w w 2 2k = Pw – 2w 2 2w 2 – Pw + 2k = 0 79. (C) Volume of the i ce-cream in cylindrical container = r 2h =
6 6 AC ² 225 337.5 cm 4 4 76. (A) Let x unit be the side of the square. its diagonal = 2 x unit A1 : Area of the square = x 2 sq. unit A2 : Area of the e quilateral described on the diagonal of the square.
=
A2 A1
=
3 4 3 4
( 2 x )2 sq.unit
2x 2 sq. unit
3 x ² 3 2 = 2 x ²
A2 : A1 = 3 : 2 77. (A) dm : diameter of the moon de : diameter of the earth 1 d 4 e Let r unit be the radius of the earth.
Case I,
dm =
then, dm =
7
6 6 15 cm3
Let r cm be the radius of the cone, its height = 4r cm Volume of 1 cone with hemispherical top =
1 2 3 r ²h r 3 3
=
1 2 3 r ² 4r r 3 3
=
4 3 2 3 r r 3 3
=
6 3 r = 2 r 3 3
=
=
22
V. of 10 such cones = 10 × 2 r 3 cm³ ATQ, 22 6 6 15 = 10 2r 3 7 22 22 6 6 15 = 10 2 r 3 7 7
r 3 =
6 6 15 10 2
=
666 2 2 2
6 cm = 3 cm 2 80. (*) figure is not gettiva matched with the
r =
question asked.
1 r 2r = unit 4 2
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================================================================================== 81. (A) Area of rectangular field =
86.(C)
1000 m² 1 4
= 4000 m² breadth
Length =
= 50 m
Required exenditure =
B
C
D
BD = Diagonal = 16cm
4000 = 80 m 50
New length of field = (80+20) m = 100 m New area = 100 × 50 = 5000sq. m
A
`
1 5000 = 4
= `
1250
82. (C) Increase in water level Volume of sphere
1 × BD2 2
Area of square =
1 ×16×16 = 128 sq. cm. 2
87. (A) Let 'r ' be the radius of the semi-circle and 'x ' be the side of the inscribed square. S R
= Area of base of cylinder 4 r ² 4 4 14 = 3 = r 3.5 cm. 3 3 3 r ²
r x A In OQR,
Required water level r
x = x + 2
r 2
=
x 2
=
2
14 7 =7 = cm 3 3
83. (A) Curved surface of cylinder = 2 rh Case II 1 3
Radius = r : height = 6h Curved surface = 2 ×
1 r ×6h = (2 rh )×2 3
Increase will be twice. 84. (A) BO = 4 units: OC = 3 units & BOC = 90º
P O x Q 2 2
2
5 4 4 5
x 2 r 2
Area of the square =
y
A
Þ
=3 ×
22 ×21 × 21 = 4158 sq. cm 7
2r 2
2
2r sq. unit = 2 = 2r ² sq. unit
22
r = 3 21 21 21 = 21cm Total surface area = 3 r 2
2y
Area of the square = y 2
3 3 7 r = 19404
19404 3 7 = 9261 2 22
y =
B
2 3 = 19404 3 r
r 3 =
sq. unit
2r =
r
2
5
C r
85.(A)
4r 2
Again, let 'y ' be side of the square inscribed in the circle of same radius. D
BC = 42 32 = 5 units BC2 = 25 sq. units
B
Required ratio =
4r 2 5
: 2r 2
2 = 2r 2 : 1 5 = 2r 2[2 : 5] =2:5
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Centres at:
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BADARPUR BORDER
================================================================================== ar( BCE) =
88. (A) O
1 0 c m
c m 1 0
A
3.14 10 10 90º 360º
=
34 17 9 8
=
2 17 17 3 3 2 2 2
= 2 × 2× 3 × 17 = 204 cm2
B
Area of the minor segment = sector area OABO – area of OAB =
34(34 17)(34 25)(34 26) cm2
1 2
1
2
or,
10 10
314 50 = 78.5 – 50 = 28.5 cm2 = 4 Area of the major segment = area of circle – area of minor segment = 3.14 × 10 × 10 – 2 8.5 = 314 – 28.5 = 288.5 cm2 Required difference = 285.5 – 28.5 = 257 cm2
89. (C)
1 2
× BE × height = 204 × 17 × CM CM
= 204 204 2 = 24 cm 17
=
1
ar(Trap. ABCD) =
2
1 × 137 × 24 2
=
= 1644 sq. cm. 91.(C) Required number of persons = 450 +250 +150 + 75 + 50 +25 = 1000 92.(B) Required answer = 250+150 = 400 93.(C) Required ratio = 250 : 75 = 10 : 3 450 9 = 500 10
94.(B) Age group 15 - 20 Total area of the square field = (44 × 44)m2 = 1936m2 At the rate of Re. 1 per sq. mtr; the total cost would be Rs. 1936, but the total cost = Rs. 3536 Difference = Rs. 3536 – Rs. 1936 = Rs. 1600 Rs. 1600 would be the extra cost on the flower bed and as the extra cost on the flower bed is Rs. 1 per sq. mtr. Area of flower bed = 1600 sq. mtr. Side of flower bed
2
= 1600 m = 40 m
So, width of the gravel path
=
44 40 2
= 2 metre 90. (B)
D
m c 6 2
60 cm m c 6 2
C2
5 c m
S=
17 25 26 2
=
95.(D) Required percentage =
68 2
25 100 = 5% 500
96. (D) Expenditure on clothing & miscellaneous = (20 + 30)% of 25000 = 12500 `
97. (C) Total expenditure =
15000 (10 20)
100
= 50,000 `
98. (D)
360º = 100%
54 % 15 % 3 .6 and among the options the two items having difference of 15% (and hence 54º) are miscellanouns and Food
54º
99. (B) Required % age =
E F B ABCD is a trapezium. Draw CE || DA intersec ting AB at E. ABCE is a ||gm. DA = CE = 26 cm. In BCE A
× (60 + 77) × 24
15 10 100 % 15
1 100% 33.33% 3
90 % 25% 3.6 and among the options the two items making central angle of 90º, together, are travelling and eutertainment.
100. (D) 90º =
=34
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BADARPUR BORDER
==================================================================================
SSC MAINS (MATHS) MOCK TEST - 4 (ANSWER SHEET) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
(D) (C) (A) (A) (D) (A) (C) (*) (C) (A) (C) (B) (B) (C) (B) (A) (D) (A) (A) (C)
21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40.
(C) (C) (A) (C) (A) (B) (D) (C) (B) (C) (A) (D) (D) (C) (A) (D) (A) (C) (B) (C)
41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.
(B) (D) (A) (A) (C) (B) (C) (D) (B) (A) (A) (A) (B) (D) (C) (C) (A) (A) (C) (B)
61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80.
(B) (A) (A) (C) (C) (B) (B) (B) (B) (A) (C) (A) (A) (A) (B) (A) (A) (B) (B) (*)
81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100.
(A) (C) (A) (A) (A) (C) (A) (A) (C) (B) (C) (B) (C) (B) (D) (D) (C) (D) (B) (D)
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