RBI Assistant Prelims Online Test Series, RBI Assistant Mock Test

A can do 1/10 of the work in a day.
B can do 1/12 of the work in a 1 day.
Both of them together can do (1/10 + 1/12) part of work in 1 day = (6 + 5)/60 = 11/60
They take 60/11 days to complete the work together.
Given that they already worked for 2 days.
The number of days required to complete remaining work => 60/11 – 2 = 38/11 = 3 (5/11) days.

800 — 200
100 — ? => 25%

Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23
Therefore there were 23 members in the group initially.

3x                  4x           5x
6                    9              4
18x + 36x + 20x = 1480
74x = 1480 => x = 20
5x = 100 Rs.

x/7 = y/9 => x:y = 7:9
7/16 * 96 = 42

Let the numbers be x, x + 2, x + 4, x + 6, x + 8 and x + 10.
Given (x + 2) + (x + 10) = 24
=> 2x + 12 = 24 => x = 6
The fourth number = x + 6 = 6 + 6 = 12.

2 * π * 2r * 3r = 12 πr²

Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy
= 4k = 4(9) = 36 kg.

(800*3*3)/100 = 72
920 + 72 = 992

They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.
The time required = d/s = 100/35 = 20/7 sec.

x*(85/100) = (480 – x)119/100
x = 280

(3500*1.5*3)/100 => 157.50

If the cost price is Rs.100, then to get a profit of 20%, the selling price should be Rs.120.
If 120kg are to be sold, and the dealer gives only 100kg, to get a profit of 20%.
How many grams he has to give instead of one kilogram(1000 gm).
120 gm —— 100 gm
1000 gm —— ?
(1000 * 100)/120 = 2500/3 = 833 1/3 grams.

106 * 106 – 94 * 94 = (106)² – (94)²
= (106 + 94)(106 – 94) = (200 * 12) = 2400

[(62.25/2.5) + (12.5/0.5)] / [(0.75 * 9) + (3.5 * 3.5)]
= [(6225/250) + (125/5)] / (6.75 + 12.25)
= (249/10 + 25) / 19 = 499/190 = 2 119/190

Let C.P. of the article be Rs. x.
Then, required ratio = 104% of x / 106% of x
= 104/106 = 52/53 = 52:53

Time taken by one tap to fill the tank = 3 hrs.
Part filled by the taps in 1 hour = 4 * 1/6 = 2/3
Remaining part = 1 – 1/2 = 1/2
2/3 : 1/2 :: 1 : x
x = 1/2 * 1 * 3/2 = 3/4 hrs. i.e., 45 min
So, total time taken = 3 hrs 45 min.

Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/39 = 50/3
3x + 900 = 1950 => x = 350 m.

30 = 3³ + 3, 630 = 5⁴ + 5, 10 = 2³ + 2, 520 = 8³ + 8 and 130 = 5³ + 5.
30, 10, 130 and 520 can be expressed as n³ + n but not 630.

a³ : a³/8 * 4/3 π => 6: π

I. (a – 4)(a + 2) = 0
=> a = 4, -2
II. b² = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.

3x + 5x + 7x = 45
x =3
3x = 9

100 —- 300 — 3
900 — 3
—-
6 years

Let the number of hours working per day initially be x. we have M₁ D₁ H₁= M₂ D₂ H₂
30 * 24 * x = 20 * d₂ * (4x)/3  => d₂ = (30 * 24 * 3)/(24 * 4) = 27 days.

(5 + x)/20 + 5/25 = 1 => x = 11 days

A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18

Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min.

Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6
Total profit = Rs. 52000
Share of B = 6/13 (52000) = Rs. 24000

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.

X=100 y=125
125——–25
100——–? => 20%

M + Tu + W + Th = 4 * 48 = 192
Tu + W + Th + F = 4 * 46 = 184
M = 42
Tu + W + Th = 192 -42 = 150
F = 184 – 150 = 34

{20 – [7 – (3 -2)] + 1/3 (4.2)} / 1.4
=> [20 – (7 – 1) + 1.4] / 1.4
=> (20 – 6 + 1.4)/1.4 = 10 + 1 = 11

A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70.

C = 1/5 – 1/6 = 1/30 => 30 days

One days’s work of A, B and C = 1/90 + 1/30 + 1/45
= (1 + 3 + 2)/90 = 1/15
A, B and C together can do the work in 15 days.

Let the ten’s and unit’s digit be x and 8/x respectively.
Then,
(10x + 8/x) + 18 = 10 * 8/x + x
9x² + 18x – 72 = 0
x² + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = 2
So, ten’s digit = 2 and unit’s digit = 4.
Hence, required number = 24.

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.

LCM = 1260
1260 + 7 = 1267

SP = 800
Profit = 25%
CP = (SP)*[100/(100+P)]
= 800 * [100/125]
= 640
Loss = 25% = 25% of 640 = Rs.160
SP = CP – Loss = 640 – 160 = Rs.480

1/15 + 1/25 = 8/75
75/8 = 9 3/8 days

Let x = 5k and y = 2k.
Then, (8x + 9y)/(8x + 2y)
= [(8 * 5k) + (9 * 2k)] / [(8 * 5k) + (2 * 2k)] = 58k/44k = 29/22
(8x + 9y):(8x + 2y) = 29:22

B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k – 5k = 3k.
Quantity of water in container C = 8k – 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k – 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.

Amount = [1600 * (1 + 5/(2 * 100)² + 1600 * (1 + 5/(2 * 100)]
= [1600 * 41/40(41/40 + 1)
= [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321.
C.I. = 3321 – 3200 = Rs. 121.

Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day
Let us say 70 men take x days to consume 150 kg.
Quantity consumed by each item in x days = (7x/30) kg
Quantity consumed by 70 men in x days = (7/30 x)(70) kg
= (7/30 x) * (70) = 960
x = 60 days.

Let C.P. be Rs. 100.
Then, S.P. = Rs. 123.50
Let marked price be Rs. x. Then, 95/100 x = 123.50
x = 12350/95 = Rs. 130
Now, S.P. = Rs. 130, C.P. = Rs. 100
Profit % = 30%.

Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters.

Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.

Investments of X, Y and Z respectively are Rs. 20000, Rs. 25000 and Rs. 30000
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x)
= 240 : 300 : 30x = 8 : 10 : x
The share of Z in the annual profit of Rs. 50000 is Rs. 14000.
=> [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7
=> 25x = 7x + (18 * 7) => x = 7 months.
Z joined the business after (12 – 7) months. i.e., 5 months.