IPPB Pre Online Test Series 5, IPPB Prelims Mock Test Series 5th

Sravan’s one day’s work = 1/25 – 1/75 = 2/75
Sravan worked for 8 days. So, his 8 days work = 8 * 2/75 = 16/75
Sravan completed 16/75th of total work.
So, his share is 16/75 * 225 = Rs. 48.

Speed=(60 * 5/18) m/sec = (50/3) m/sec Length of the train = (Speed x Time) = (50/3 * 9) m = 150 m.

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 216 – 1 = 215.

x + y = 80
x – 4y = 5
x = 65 y = 15

Let the earnings of P and Q be 9x and 10x respectively.
New ratio = [9x + 1/4 (9x)]/[10x – 1/4 (10x)]
=> 9*(1 + 1/4)/10*(1 – 1/4)
=> 9/10 * (5/4)/(3/4) = 3/2

20                 15           5
25                  20           8
500 + 300 + 40 = 840/40 = 21

Let the number be x. Then,
15x – x = 196
x = 14

A.M. of 75 numbers = 35
Sum of 75 numbers = 75 * 35 = 2625
Total increase = 75 * 5 = 375
Increased sum = 2625 + 375 = 3000
Increased average = 3000/75 = 40.

25 2/5 + 8 4/5 + 4/5 = 34 4/5 ≡ 35

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 = (7x/30)(70)kg
(7x/30)(70) = 980
x = (980 * 30)/490 => x = 60 days

(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2

108 * 107 * 96 ≡ 1109400

πr² = 30  h = 6
1/3 * 30 * 6 = 60

Let the numbers be 13a and 13b.
Then, 13a * 13b = 2028 => ab = 12.
Now, co-primes with product 12 are (1, 12) and (3, 4).
So, the required numbers are (13 * 1, 13 * 12) and (13 * 3, 13 * 4).
Clearly, there are 2 such pairs.

C.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.
S.P. of 50 kg wheat = 130% of Rs. 630 = 130/100 * 630 = Rs. 819.
S.P. per kg = 819/50 = Rs. 16.38 = 16.30.

(1.9 * 0.021) / (0.095 * 0.105) = (19 * 21 * 100) / (95 * 105) = 4.

2/7 – 1/x = 1/4
x = 28

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.

Let the least number to be subtracted be ‘x’, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.
(11 – x):(15 – x) = (21 – x):(30 -x)(21 – x)
From the options, when x = 3
=> 8 * 27 = 12 * 18 => then x = 3. => (11 – x)(30 – x) = (15 – x)(21 – x)

The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.

Work done by the waste pipe in 1 minute = 1/15 – (1/20 + 1/24) = – 1/40
Volume of 1/40 part = 3 gallons\
Volume of whole = 3 * 40 = 120 gallons.

Let the numbers be x and y.
Then, x + y = 20 and x – y = 8
x² – y² = (x + y)(x – y) = 20 * 8 = 160.

200 + 130 = 330
800 – 330 = 470
200 —- 5000
470 —- ? => 11750

(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.

Let the ages of the children be x, (x + 3), (x + 6), (x + 9) and (x +12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20 => x = 4.
Age of youngest child = x = 4 years.

For a number to be divisible by 4, the number formed by the last two digits of that number must be divisible by 4.
For a number to be divisible by 9, the sum of its digits must be divisible by 9.
Among the choices, choices (a) and (b) are divisible by 4 but of the first two choices, only choice (b) is divisible by 9.
Hence choice (b) is divisible by both 4 and 9.

9/12 = 3/4 * 900 = 675
=                           650
=                    ————-
=                            25
1/4 ——– 25
1 ——— ?  => Rs.100

I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50

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%
Learn more problems on : Percentages

There are three ladies and five gentlemen and a committee of 5 members to be formed.
Number of ways such that two ladies are always included in the committee = ⁶C₃ = (6 * 5 * 4)/6 = 20.

109 * 34 – 108 * 33
= 3706 – 3564 = 142

By applying BODMAS rule.
51/4 * (19/17 / 49/6) / 19/3
= 51/4 * ( 19/17 * 6/49) * 3/19 = 27/98

P —- 2P —- 8 years
2 1/4 P —- 10 years

100 —- 300 — 3
900 — 3
—-
6 years

I = (500*9*6)/100 = 270

1200 * 12: x * 9 = 2:3
x = 2400

Using question number 11 and 12, we get the probability as
1 – (1/36 + 5/9) = 5/12

Part filled by A + B + C in 1 hour = 1/5 + 1/10 + 1/30 = 1/3
All the three pipes together will fill the tank in 3 hours.

Let x be the investment of Ravi in 6% and y be in 7%
x(6)(n)/100 = y(7)(n)/100
=> x/y = 7/6
x : y = 7 : 6

Speed of the train relative to man = 125/10 = 25/2 m/sec.
= 25/2 * 18/5 = 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.

30% ———— 50
45% ———— 25
———————-
15% ————- 75
30% ————– ?
150 + 50 = 200 Marks

10 CP = 12 SP
12 — 2 CP loss
100 — ? => 16 2/3%

x * 2/3 * 3/5 * 1/4 = 24
x = 240
240 * 30/100 = 72

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

P = (P*3*R)/100
R = 33 1/3%

2/3 : 2/5 = 10:6 = 5:3

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

Let the income and the expenditure of the person be Rs. 5x and Rs. 4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs. 3600.

Let the school ages of Neelam and Shaan be 5x and 6x years respectively.
Then, (1/3 * 5x)/(1/2 * 6x) = 5/9
15 = 15
Thus, Shaan’s age cannot be determined.

100 —- 200 —- 4
400 —- 4
800 —- 4
——
12 years