IPPB Prelims Mock Test Series 2, IPPB Pre Online Test Series 2

Let the present worth be Rs. x. Then,
S.I. = (132 – x)
(x * 5 * 2) / 100 = (132 – x)
10x = 13200 – 100x
110x = 13200 => x = 120.

1/8 + 1/24 = 1/6
6 * 2 = 12

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

Let the numbers be x, x + 1 and x + 2
Then,
x + (x + 1) + (x + 2) = 87
3x = 84
x = 28
Greatest number, (x + 2) = 30.

Distance covered by Amar = 18/4.8 (1.6km) = 3/8(1600) = 600 m

Let the two successive numbers be n, n + 1.
It is given that n(n + 1) = 4556.
=> n² + n – 4556 = 0
=> n² + 68n – 67n – 4556 = 0
=> n(n + 68) – 67(n + 68) = 0
=> n = 67 and (n + 1) = 68.

Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 140 = 500 m
Required time = 500 * 2/25 = 40 sec

Let the number be x. Then,
1/3 of 1/4 of x = 15
x = 15 * 12 = 180
So, required number = (3/10 * 180) = 54.

Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.

6A = 8B = 10 C
A:B:C = 1/6:1/8:1/10
= 20:15:12

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

Given Exp. = 800/64 * 1296/36
= 12.5 * 36 = 450

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

Let 4207 – x = 3007
Then, x = 4207 – 3007 = 1200

X * (90/100) * (80/100) * (75/100) = 1944
X * 0.9 * 0.8 * 0.75
X = 1944/0.54
X = 3600

Let the sum be Rs. x. Then,
(x * 25/2 * 1/100) – (x * 10 * 1)/100 = 1250
25x – 20x = 250000
x = 50000

For any perfect square the index of the power of each of its prime factors is even. in the given number the power of 5 and 17 have odd indices, while 3 has an even index.
The least number = (5)(17) = 85.

3W = 2M
15M —— 21 * 8 hours
21 W —— x * 6 hours
14 M —— x * 6
15 * 21 * 8 = 14 * x * 6
x = 30

D = 170
S = 72 * 5/18 = 20 mps
T = 170/20 = 8 ½ sec

Let A be the amount received at the end of the three years.
A = 4800[1 + 10/100][1 + 20/100][1 + 25/100]
A = (4800 * 11 * 6 * 5)/(10 * 5 * 4)
A = Rs.7920
So the interest = 7920 – 4800 = Rs.3120

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.

The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.

4/10 + 9/x = 1 => x = 15
1/10 + 1/15 = 1/6 => 6 days

Relative speed = 45 + 30 = 75 km/hr.
75 * 5/18 = 125/6 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 6/125 = 48 sec.

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

4⁵⁰ = (2²)⁵⁰ = 2¹⁰⁰ => 16²⁵ = (2⁴)²⁵ = 2¹⁰⁰
Hence, 4⁵⁰, 2¹⁰⁰ and 164²⁵ are all equal.

(x*8*3)/100 = ((2665 – x)*3*5)/100
24x/100 = 39975/100 – 15x/100
39x = 39975 => x = 1025
Second sum = 2665 – 1025 = 1640

26² – 10² = 24²
d₁ = 20         d₂ = 48
1/2 * 20 * 48 = 480

12 m/s = 12 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 12 * 18/5 * 15/4 km = 162 km.

The fractions considered are 8/15 9/13 6/11
To compare them we make the denominators the same. So the fractions are
(8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145
1144/2145, 1485/2145 and 1170/2145
so in descending order the fractions will be
1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15

P(1 + R/100)⁵ = 2P => (1 + R/100)⁵ = 2
Let P(1 + R/100)ⁿ = 8P
=> (1 + R/100)ⁿ = 8 = 2³ = {(1 + R/100)⁵}³
=> (1 + R/100)ⁿ = (1 + R/100)¹⁵ => n = 15 Required time = 15 years.

27% of 6000 + 45% of 4200 = ?
=> ? = 27/100 * 6000 + 45/100 * 4200
=> ? = 1620 + 1890 = 3510 = 3500

Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 => x = 10
Speed of each train = 10 m/sec.
= 10 * 18/5 =- 36 km/hr.

In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 500 resolutions.
= 500 * 2 * 22/7 * 22.4 = 70400 cm = 704 m

Let the monthly salaries of Ram and Shyam be Rs. r and Rs. s respectively.
10/100 r = 8/100 s
r = 4/5 s
Monthly salary of Abhinav = (1.92 lakhs)/12 = Rs. 0.16 lakhs
s = 2(0.16 lakhs) = 0.32 lakhs
r = 4/5(0.32 lakhs) = Rs. 25600

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

P — 10 —- 600
P — 5 —– 300
3P — 5 —– 900
——
=> 1200

Speed downstream = d/t = 85/(2 1/2) = 34 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (34 – 18)/2 = 8 kmph

3x * 2x = 3750 => x = 25
2(75 + 50) = 250 m
250 * 1/2 = Rs.125

Treat all boys as one unit. Now there are four students and they can be arranged in 4! ways. Again five boys can be arranged among themselves in 5! ways.
Required number of arrangements = 4! * 5! = 24 * 120 = 2880.

Three consecutive even numbers (2P – 2), 2P, (2P + 2).
(2P – 2) + 2P + (2P + 2) = 42
6P = 42 => P = 7.
The middle number is: 2P = 14.

Let the initial quantity of milk in vessel be T litres.
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T – y)/T]n * T
For the given problem, T = 90, y = 9 and n = 2.
Hence, quantity of milk finally in the vessel
= [(90 – 9)/90]2 (90) = 72.9 litres.

2x * x = 128 => x= 8

√3600 * 52 = ?
=> ? = 60 * 52 = 3120

(1500 * R₁ * 3)/100 – (1500 * R₂ * 3)/100 = 13.50 4500(R₁ – R₂) = 1350
R₁ – R₂ = 0.3%

Let the numbers be 3x and 5x.
Then, (3x – 9)/(5x – 9) = 12/23
9x = 99 => x = 11
The smallest number = 3 * 11 = 33.

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.

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.