RBI Assistant Mock Test, RBI Assistant Prelims Online Test Series 2

Let the numbers be 37a and 37b.
Then, 37a * 37 b = 4107 => ab = 3
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 * 1, 37 * 3) i.e., (1, 111).
Greater number = 111.

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

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.

70 a = 40 b
a: b = 4: 7

525 => 5 5²,
39 => 3 3²,
24 => 2 2²,
636 => 6 6². The above pattern is not followed in 426.

Let the present ages of Anil and his son be 7x and 3x years respectively.
Let the present age of the son’s mother be m years.
(m + 6) / (3x + 6) = 2/1
m + 6 = 6x + 12 => m = 6x + 6
Ratio of the present ages of Anil and the son’s mother = 7x / (6x + 6).
This cannot be found uniquely.

S = 180/6 * 18/5 = 108 kmph

B = 1/16 – 1/24 = 1/48 => 48 days

Let the two positive numbers be 5x and 8x respectively.
8x – 5x = 15 => 3x = 15 => x = 5.
Smaller number = 5x = 25.

(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6

4/15 A = 2/5 B => A = (2/5 * 15/4)B
A =3/2 B => A/B = 3/2 => A:B = 3:2
B’s share = 1210 * 2/5 = Rs. 484.

100*(20/100) = 20 — 10
? — 1 => Rs.2
100 — 80
? — 2 => Rs.2.5

2/7 – 1/x = 1/4
x = 28

Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 18 — (1)
and x – y = 10 — (2)
From (1) & (2) 2x = 28 => x = 14, y = 4.

Let the length of the train be x meters and its speed be y m/sec.
They, x / y = 15 => y = x/15
x + 100 / 25 = x / 15
x = 150 m.

12 + 16 – 10 = 18
and 13/18 + 11/27 – 5/9 = (39 + 22 – 30)/(9 * 2 * 3) = 31/54 The answer is 18 (31/54)

16*8: x*4 = 5:2
x = 12.8 => Rs.12800

2 * 22/7 * 1.75 * x = 11000
x = 1000

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

64000* 9/8 * 9/8 = 81000

Part, Cart, Dart and Mart are rhyming words, but Trap does not sound similarly.

12.5 + 10 1/4 of 100 ÷ 10% of 10
= 12.5 + 41/4 of 100 ÷ 10/100 of 10
= 12.5 + 41(25) ÷ 1 = 12.5 + 1025 = 1037.5

The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle = 1/2 * 24 * 10 = 120 cm²

Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l² = 324 => l = 18.

HCF of 323, 437 = 19
323 * 437 = 19 * 19 * x
x = 391

Speed = 144 * 5/18 = 40 m/sec
Time taken = 100/40 = 2.5 sec.

Speed of train relative to jogger = 45 – 9 = 36 km/hr.
= 36 * 5/18 = 10 m/sec.
Distance to be covered = 240 + 120 = 360 m.
Time taken = 360/10 = 36 sec.

The discriminant of the quadratic equation is (-12)² – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.

100
5 —- I
5
0.25 — II
———–
10.25%

A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only.
A             :                 B
8000*18 : (8000*4) + (4000*14)
14400     : 88000
A:B = 18:11

RS = 5 ¾ – 4 ½ = 1 ¼
T = 3 ½ h.
D = 5/4 * 7/2 = 35/8 = 4 3/8 km

When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = (900 * 700)/800 = 6300/8 = 787.5 m.
In a race of 1000 m, A beats C by (1000 – 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = (600 * 212.5)/1000 = 127.5 m.

Average of 9 = x
9x = 8 * 3 + x * 2 x = 3.25
Total = 9 * 3.25 = 29.25

Let the SP of the refrigerator and the mobile phone be Rs. r and Rs. m respectively.
r = 15000(1 – 4/100) = 15000 – 600
m = 8000(1 + 10/100) = 8000 + 800
Total SP – Total CP = r + m – (15000 + 8000) = -600 + 800 = Rs. 200
As this is positive, an overall profit of Rs. 200 was made.

The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph.

Male = 45,000* 5/9 = 25,000
Female = 45,000* 4/9 = 20,000
Married Male = 25,000* 40/100 = 10,000
Married Female = 10,000
20,000 ———— 10,000
100 ————- ? => 50%

Except mirage, all others are sources of water.

(5/8 * 12/5 – 5/4)/(5/4) (3/2 -5/4)/(5/4) = 1/5

1/18 + 1/24 = 7/72
72/7 = 10 2/7
7/72 * 10 = 35/36 —– 20 hours
WR = 1 – 35/36 = 1/36
1 h —– 1/18
? —– 1/36 = 1/2 hours
=> 20 1/2 hours

Initial average age of the 10 persons be P.
Age of the new person Q.
Sum of the ages of the initial 10 persons = 10P
New average = (P-3) 10(P-3) = 10P – 42 + Q => Q = 12

Applying BODMAS rule
= 7 + 1/2 of [8 – 4 / 2 * 3 – 2 + (4 + 2) – 2 * 2]
= 7 + 1/2 of [8 – 2 * 3 – 2 + 6 – 4]
= 7 + 1/2 of (2) = 8

Let the cost price of an article be Rs. x.
(19% of x) – (12% of x) = 105
19x/100 – 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs. 1500

I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b

5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 days

Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x – 8) + (7x – 8) + (9x – 8) = 56
20x = 80 => x = 4
Their present ages are 16, 28 and 36 years respectively.

1/10 + 1/15 = 1/6 * 4 = 2/3
1 – 2/3 = 1/3
1/10 + 1/15 – 1/x = 1/3
x = 8

Clearly, the two will meet when they are 726 m apart.
To be (4.5 + 3.75) = 8.25 km apart, they take 1 hour.
To be 726 m apart, they take (100/825 * 726/1000) hrs = (242/2750 * 60) min = 5.28 min.

X * (105/100) * (105/100) * (105/100) = 138915
X = 138915/1.157625
X = 120000

160 = (180*4*R)/100
R = 5%

Let P’s age and Q’s age be 6x and 7x years respectively.
Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8