RBI Assistant Prelims Test Series 5, RBI Assistant Prelims Mock Test

Area of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}² – ∏[4/2]²
= ∏[2.25² – 2²] = ∏(0.25)(4.25)  { (a² – b² = (a – b)(a + b) }
= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq m

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 * 2)/(15 * 14) = 1/35
Probability that both are yellow = ²C₂/¹⁵C₂ = (2 * 1)/(15 * 14) = 1/105
Probability that one blue and other is yellow = (³C₁ * ²C₁)/¹⁵C₂ = (2 * 3 * 2)/(15 * 14) = 2/35
Required probability = 1/35 + 1/105 + 2/35
= 3/35 + 1/105 = 1/35(3 + 1/3)
= 10/(3 * 35) = 2/21

Let Rajan’s present age be x years.
Then, his age at the time of marriage = (x – 8) years.
x = 6/5 (x – 8)
5x = 6x – 48 => x = 48
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = 30 years.
Rajan’s sister’s present age = (30 + 8) = 38 years.

S.I. = (956 – 800) = Rs. 156.
Rate = (100 * 156) / (800 * 3) = 6 1/2 %
Now rate = (6 1/2 + 4) = 10 1/2 %
New S.I. = (800 * 21/2 * 3/100) = Rs. 252
New amount = (800 + 252) = Rs. 1052.

HCF of 1189, 1073 = 29
1073/29 = 37

S = 1
M = x
DS = x + 1
US = x – 1
35/(x + 1) + 35/(x – 1) = 12
x = 6

Speed = [90 * 5/18] m/sec = 25 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 25 → x = 25 * 60 / 2 = 750

130
14
——-
361—— 144
100%——? => 400

Let the number be x. Then,
2/3 x = 25/216 * 1/x
x2 = 25/216 * 3/2 = 25/144
x = 5/12

Sum of 10 natural no. = 110/2 = 55
Average = 55/10 = 5.5

C = 1/2 – 1/4 – 1/5 = 1/20 => 20 days

Work done by a women in one day = 1/2 (work done by a man/day)
One women’s capacity = 1/2(one man’s capacity)
One man = 2 women.
12 men = 24 women.
12 men + 6 women = 30 women
30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join, then new work force = 30 women – 12 women + 6 women = 24 women.
Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days.

HCF of 408, 468, 516 = 12

27, 36, 72 and 45 are divisible by 9, but not 30.

The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.
Relative speed = 72 -36 = 36 kmph = 36 * 5/18 = 10 mps.
Time required = d/s = 500/10 = 50 sec.

S.I. = (15500 – 12500) = Rs. 3000\
Rate = (100 * 3000) / (12500 * 4) = 6%

(217)² + (183)² = (200 + 17)² + (200 – 17)²
= 2[(200)² + (17)²] = 2(40000 + 289) = 80578

Simran : Nanda = (50000 * 36) : (80000 * 30) = 3:4
Simran’s share = 24500 * 3/7 = Rs. 10500

(4 + 5 + 2) – (1 + 6 + 3) = 1, not divisible by 11.
(2 + 6 + 4) – (4 + 5 + 2) = 1, not divisible by 11.
(4 + 6 + 1) – (2 + 5 + 3) = 1, not divisible by 11.
(4 + 6 + 1) – (2 + 5 + 4) = 0.
So, 415624 is divisible by 11.

2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432

CP = Rs. 540, MP = 540 + 15% of 540 = Rs. 621
SP = Rs. 496.80, Discount = 621 – 496.80 = 124.20
Discount % = 124.2/621 * 100 = 20%

g% = 20 + 20 + (20*20)/100
= 44%

Interest received by A from B = 10% of half of Rs.50000 = 10% * 25000 = 2500.
Amount received by A per annum for being a working partner = 1500 * 12 = Rs.18000.
Let ‘P’ be the part of the remaining profit that A receives as his share. Total income of A = (2500 + 18000 + P)
Total income of B = only his share from the remaining profit = ‘P’, as A and B share the remaining profit equally.
Income of A = Twice the income of B
(2500 + 18000 + P) = 2(P)
P = 20500
Total profit = 2P + 18000
= 2*20500 + 18000 = 59000

Let sum = x. Then, S.I. = 0.125x = 1/8 x, R = 10%
Time = (100 * x) / (x * 8 * 10) = 5/4 = 1 1/4 years.

x = √8 – √7 = 1/(√8 + √7) and y = √6 – √5 = 1/(√6 + √5) and z = √10 – 3 = 1/(√10 + 3)
As (√8 + √7) > (√6 + √5), (√8 – √7) (√8 + √7) > (√6 + √5)
z

7770
7077
————-
693

8 = 2³, 27 = 3³, 64 = 4³, 125 = 5³ and 343 = 7³.
8, 27, 125 and 343 are cubes of prime numbers but not 64.

520 = 26 * 20 = 2 * 13 * 2² * 5 = 2³ * 13 * 5
Required smallest number = 2 * 13 * 5 = 130
130 is the smallest number which should be multiplied with 520 to make it a perfect square.

At the time of meeting, let the distance traveled by the second train be x km. Then, distance covered by the first train is (x + 100) km.
x/40 = (x + 100)/50
50x = 40x + 4000 => x = 400
So, distance between P and Q = (x + x + 100)km = 900 km.

G^-1F⁺³I, Q^-2O⁺³R, L^-1K⁺³N, Y^-1X⁺³A and T^-1S⁺³V.
Except QOR, all other groups follows similar pattern.

SP = CP + g
150 SP = 150 CP + 30 SP
120 SP = 150 CP
120 — 30 CP
100 — ? => 25%

2C + 3T = 1300 — (1)
3C + 3T = 1200 — (2)
Subtracting 2nd from 1st, we get
-C + T = 100 => T – C = 100

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.

Two men, three women and one child can be selected in ⁴C₂ * ⁶C₃ * ⁵C₁ ways
= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5
= 600 ways.

400*(90/100)*(95/100) = 342

2) 105 (52                             5) 52 (10               7) 10 (1
104                                       50                           7
——–                                 ——–                      – —–
1                                          2                           3
=> 65
105/70 = 35 (Remainder)

CP = SP + 1CP = SP – g
800 + x = 980 – 2x
3x = 180 => x = 60
CP = 800 + 60 = 860

Given Exp. = 666 * 1/6 * 1/3 = 37

lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10

Let the present ages of Anand and Bala be ‘a’ and ‘b’ respectively.
a – 10 = 1/3 (b – 10) — (1)
b = a + 12
Substituting b = a + 12 in first equation,
a – 10 = 1/3 (a + 2) => 3a – 30 = a + 2
=> 2a = 32 => a = 16.

4/12 + x/15 = 1
x = 10

2 1/3 + 4 1/6 – 2 2/3 – 1 1/2 = 2 ?/3
= 7/3 + 25/6 – 8/3 – 3/2 = (14 + 25 – 16 – 9)/6 = 7/3 = 2 1/3
? = 1

Relative speed of the thief and policeman = 11 – 10 = 1 km/hr.
Distance covered in 6 minutes = 1/60 * 6 = 1/10 km = 100 m.
Distance between the thief and policeman = 200 – 100 = 100 m.

LCM = 1400
1400 – 6 = 1394

Let the present ages of son and father be x and (60 – x) years respectively.
Then, (60 – x) – 6 = 5(x – 6)
6x = 84 => x = 14
Son’s age after 6 years = (x + 6) = 20 years.

Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec.

3/7 x – 40/100 x = 12
x = 35 * 12
35 * 12 * 60/100 = 252/2 = 126

1/18 + 1/30 = 8/90 = 4/45
45/4 = 11 ¼ *2 = 22 ½ days

LCM = 120
120) 1000 (8
960
——-
40
1000 + 120 – 40 = 1080