RBI Assistant Mains Mock Test, RBI Assistant mains online test series 2

[10 + { 4 * (2/3 + 1/4 * √144/121 + 23) ÷ 12 + 5} – 3] = ? [10 + { 4 * (11/12 * 12/11 + 23) ÷ 12 + 5 } – 3] = ?
= [10 + {4 * 24 ÷ 12 + 5} – 3] = ?
= [10 + 13 – 3] = ? = > 20

Let the maximum mark in each test be M.
The candidate failed by 36 marks in the second test.
pass mark in the second test = 216 + 36 = 252
45/100 M = 252
Pass mark in the first test = 40/100 M = 40/45 * 252 = 224.

By the time A covers 1000 m, B covers (1000 – 50) = 950 m.
By the time B covers 1000 m, C covers (1000 – 100) = 900 m.
So, the ratio of speeds of A and C =
1000/950 * 1000/900 = 1000/855 So, by the time A covers 1000 m, C covers 855 m.
So in 1000 m race A beats C by 1000 – 855 = 145 m.

(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days.

[3/7 of {21 * (33.33% of (27 + 3) ÷ 5) – 7} – 8] = ?
= [3/7 of {21 * (1/3 * 30 ÷ 5) – 7} – 8] = ?
= 3/7 of [{21 * 2 – 7} – 8] = ? => {3/7 * 35 – 8} = ? => 7

Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such numbers = 14.
Required percentage = (14/70 * 100) = 20%

(5 – 7)² * (6 – 9)² / (2³)^-2 = (-2)² * (-3)² * 1/(2³)^-2
= 4 * 9 * 64 = 2304.

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, [(6x + 6) + 4] / [(5x + 6) + 4] = 11/10
10(6x + 10) = 11(5x + 10) => x = 2
Sagar’s present age = (5x + 6) = 16 years.

Let Mudit’s present age be ‘m’ years.
m + 18 = 3(m – 4)
=> 2m = 30 => m = 15 years.

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

7*5: 5*x = 7:10
x = 10

Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.

60*5:36*6:75*3
100: 72: 75
100/247 * 7410 = 3000

4.5          4.5
30%        70%
30% —– 4.5
70% ——? => 10.5 – 4.5 = 6 ml

25 = E/(1000 – E) * 100
1000 – E = 4E
1000 = 5E => E = 200
1000 – 200 = 800

Boxing, Chess, Wrestling and Squash are individual events, while Baseball is a team event.

Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117

1/2:1/3:1/4 = 6:4:3
3/13*1300 = 300

[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + [(2/5 * 3/5 – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + {(6 – 5 /25) * 625 + 7} ÷ 4 = ?
=[8 + {32} ÷ 4] = ? = 16

11 = 1 1³, 28 = 2 2³, 327 = 3 3³, 416 = 4 4² and 5125 = 5 5³
Except 416, other numbers follow similar pattern.

If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate.
Let the earlier rate of interest be x%.
Now it will be (x + 3)%
% increase = (x + 3) – x/x * 100 = 25
=> x = 12

S = (1 + 2 + 3)/2 = 3
=> No triangle exists

Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals

bisect at right angles in a rhombus.
BE² + AE² = AB²
25² = 15² + AE² AE = √(625 – 225) = √400 = 20,
AC = 20 + 20 = 40 cm.
Area of a rhombus = 1/2 * d₁d₂
= 1/2 * 40 * 30 = 600 sq.cm.

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.

Given numbers are 1.08, 0.36 and 0.90.
H.C.F of 108, 36 and 90 is 18.
H.C.F of a given numbers = 0.18

Except Japanian, all others are appropriate usage of citizenship.

2(l + 100) = 600 => l = 200 m

Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

4 πr² = 64 => r = 4

P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days.

LCM = 1400
1400 – 6 = 1394

S.I. for 3 years = (12005 – 9800) = Rs. 2205
S.I. for 5 years = Rs. 2205/3 * 5 = Rs. 3675.
Principal = (9800 – 3675) = Rs. 6125
Hence, rate = (100 * 3675) / (6125 * 5) = 12%

Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.

Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? → 28 m

(50 – ?/29)% of 4200 = 3√196

Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
(x – 7)/x = 7/9
2x = 63 => x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years.

Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000

100
75
——-
75 ——- 25
100 —— ? => 33 1/3%

(3x-9):(5x-9) = 9:17
x = 12 => 3x = 36

110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (110 * 560)/112 = Rs. 500

3h = 5c
6c = 10s
A = 21h*3 + 15c*2
= 63h + 30c
= 105c + 30c = 135c
B = 15c*6 + 40s*7 1/2
= 90c + 300s
= 90c + 180c = 270c
A:B = 135:270
27:52
A = 27/79 = 1/3

7770
7077
————-
693

72 kmph = 72 * 5/18 = 20 mps
D = Speed * time = 20 * 30 = 600 m.

30 — 3             DS = 10
? —- 1
18 —- 3            US = 6
? —- 1              M = ?
M = (10 + 6)/2 = 8

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

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.

74330 Largest
30347 Smallest
————-
43983

Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively.
Then, 4x – 3x = 1000 => x = 1000.
B’s share = Rs. 2x = 2 * 1000 = Rs. 2000.