RBI Assistant Mains MCQ Series 3, RBI Assistant Mains Online Test Series

The least number of trees that are required = LCM(4, 5, 6) = 60.

Six members can be selected from ten members in
¹⁰C₆ = ¹⁰C₄ ways.

100
125
—–
125 —– 25
100 —— ? => 20%

Let the five numbers be P, Q, R, S and T.
=> P + Q + R + S + T = 655.
(P + Q)/2 = 85 and R = 125
P + Q = 170 and R = 125
P + Q + R = 295
S + T = 655 – (P + Q + R) = 360
Average of the last two numbers = (S + T)/2 = 180.

1 to 11 = 11 * 50 = 550
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6^th = 294 + 312 – 550 = 56

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85.

LCM = 1440
1440 – 1 = 1439

2x² + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = -3 or x = 3/2.

Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126

W = 62%    L = 38%
62% – 38% = 24%
24% ——– 288
62% ——– ? => 744

Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years.

8 * 84 = 672 – 7 = 665

Let their speed be x km/hr and y km/he respectively.
Then, 120/(x + y) = 1 => x + y = 120 — (i)
Now, when they move in same direction:
(Distance traveled by P in 6 hrs) – (Distance traveled by Q in 6 hrs) = 120 km
6x – 6y = 120 => x – y = 20 — (ii)
Sloving (i) and (ii), we get x = 70, y = 50
P’s speed = 70 km/hr.

675 —- 225
100 —- ? => 33 1/3%

Let the cost price of table for Sreenivas be Rs. x and given that, cost price of table for Mahesh = Rs. 2178.
=> (90%) of (110%) of x = Rs. 2178.
=> (90/100)(110/100)x = 2178
=> x = (2178 * 100)/(9 * 11)
=> x = Rs. 2200

Let C.P. be Rs. x.
Then, 5% of x = 350 – 340 = 10
x/20 = 10 => x = 200

Given that the area of the square = 1225 sq.units
=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10 = 140 sq.units

50*2k + 20*3k + 10*4k = 40000
200k = 40000 => k = 200
50p coins = 2k = 2*200 = 400
20p coins = 3k = 3*200 = 600
10p coins = 4k = 4*200 = 800

19x + 19y + 17 = -19x + 19y – 21
38x = -38 => x = -1

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.

X – (10/100) X = X * ?
? = 90%

Time taken to meet for the first time at the starting point
= LCM { length of the track / speed of A , length of the track / speed of B}
= LCM { 600/ (36 * 5/18) , 600/ (54 * 5 /18) }
= LCM (60, 40) = 120 sec.

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

M = 6
S = 1.2
DS = 6 + 1.2 = 7.2
US = 6 – 1.2 = 4.8
x/7.2 + x/4.8 = 1
x = 2.88

2 * 22/7 * 14 * x = 1056 => x = 12

I. x² + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = -4, -5
II. y² + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = -3, -2
= x

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.

SP = CP + g
50 SP = 50 CP + 15 SP
35 SP = 50 CP
35 — 15 CP gain
100 — ? => 42 6/7%

Let ages of Ramu and Shyamu be x and y respectively.
x/y = 3/4 => x = 3/4 y
Also (x – 7) / (y – 7) = 2/3
=> 3x – 21 = 2y – 14
3x = 2y + 7
But x = 3/4 y
3 * 3/4 y = 2y + 7
9y = 8y + 28 => y = 28 years
Ratio of their ages five years hence
= (21 + 5) / (28 + 5) = 26/33.

The total CP = Rs. 16000 + Rs. 6250 = Rs. 22250 and SP = Rs. 31150
Profit(%) = (31150 – 22250)/22250 * 100 = 40%

Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr.

Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 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

25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000

Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let the ten’s and unit’s digits be 2x and x respectively.
Then, (10 * 2x + x) – (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) – (2x – x) = 2x = 8.

A+B+C = 675
A = 3B
3B+B+B-25 = 675
5B = 700
B = 140

4500 * x = 3375
= x = 3375/4500 = 3/4

Three consecutive numbers can be taken as (P – 1), P, (P + 1).
So, (P – 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P – 1) = 34 – 1 = 33.

Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42 = 91.5 cm

4000
200 —- I
200
10 —- II
—————
4410
2210
——–
2000
110 —- III
110
5.50 —- IV
———–
2425.50
2210
———–
4635.50
4000
———-
635.50

P = (P*30*R)/100
R = 3 1/3%

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.

Let the number of pens that A, B, C and D get be a, b, c and d respectively.
a:b = 2:1
a = c + 25
b:c = 2:3
a:b:c:d = 4:2:3:3
a, d get 4p, 3p pens
=> 4p – 3p = 25 => p = 25
=> D gets 3p = 3 * 25 = 75 pens.

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.

Number of students passed = 60% of 140 = 60/100 * 140 = 84
Number of students failed = 140 – 84 = 56.
Required percentage = 28/56 * 100 = 50%.

Let the cost of each pen and pencil be ‘p’ and ‘q’ respectively.
16p + 8q = 352 — (1)
4p + 4q = 96
8p + 8q = 192 — (2)
(1) – (2) => 8p = 160
=> p = 20

Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm²

S:E = 1:3
E:M = 3:5
————
S:E:M = 3:9:15
3/27 * 162 = 18

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