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## IBPS SO Quantitative Aptitude Prelims Questions and Answers With Solutions

1. What should come in place of the question mark (?) in the following number series?

7, 8, 35, 160, 503, 1232, ?

2540
2563
2560
2550
2580

Solution:

Pattern of the series is:
7 + 1 3  = 8
8 + 3 3  = 35
35 + 5 3  = 160
160 + 7 3  = 503
503 + 9 3  = 1232
1232 + 11 3  = 2563

2. If the number of persons increased by 50% on Saturday from Friday in
parks A and B together, then what is the total number of persons visiting park
A and B on Tuesday and Saturday together?

88
67
48
79
61

Solution:

The number of persons visiting the parks A and B on Friday=14+16=30
The number of persons visiting the parks A and B on Saturday=30*150/100=45
The number of persons visiting the parks A and B on Tuesday= 12 + 22 = 34
So, total number of persons on Tuesday and Saturday = 45+34 = 79
Hence, option D is the correct answer.

3. What will come in place of question mark (?) in the given number series?

15, 57, 168, 417, 942, ?

1816
1904
2019
2146
2251

Solution:

The pattern of the series is as follows
× 2 + 27, × 2 + 54, × 2 + 81, × 2 + 108, × 2 + 135…
Hence the no. is 942×2+135=2019

4. A and B started a business B’s investment was 2.5 times that of A. The respective ratio between the time period for which A invested and that for which B invested was 3 : 1. If the total investment made by A and B together was 28000 rupees and the annual profit earned was 2500 rupees less than A’s investment, what was the difference between A’s share and B’s share in the annual profit ?

Rs. 200
Rs. 800
Rs. 500
Rs. 400
Rs. 650

Solution:

Ratio of investment of A and B = 1: 2.5
Let investment of B = 5p
Investment of A = 2p
Total investment = 7p
7p28000
p4000
A’s investment = 40002 = 8000
B’s investment = 40005 = 20000
Time period for A and B investment is in ratio = 3 : 1
So, profit will be in ratio = 6 : 5 between A and B
Then total profit earned = 6q + 5q = 11q
11q = 8000 – 2500
11q = 5500
q = 500
So, the required difference is 500

5. If on a particular day, 20% boys of class III and 25% boys of class IV are absent,
then what is the ratio of no. of chocolates that are un – distributed in these two classes
respectively. All girls are present in these two classes on that day.
41 : 55
45 : 31
31 : 45
55 : 41
None of these

Solution:

6. The ratio of present age of Ankur to present age of Sanjeev is 3 : 11 while the ratio of
present age of Sanjeev to present age of Reena is 5 : 4. If the average age after 7 years
of all three of them will become 45 years , then find the present age of Reena.

40 years
28 years
24 years
44 years
48 years

Solution:

Ratio of present ages of Ankur and Sanjeev = 3 : 11
Ratio of present ages of Sanjeev and Reena = 5 : 4
Equating the ratios, we get
Ratio of present ages of Ankur, Sanjeev and Reena = 15 : 55 : 44
Let Ankur’s age be 15x, Sanjeev’s age be 55x and Reena’s age be 44x.
Using the data provided in the question, we get
15x + 55x + 44x + 21 = 45 × 3
114x = 114
x = 11
Reena’s present age = 44x = 44 years

7. What should come in place of the question mark (?) in the following number series?

6, 13.5, 30, 64.5, 135, 277.5, ?

564
560
578
572
562

Solution:

Pattern of the series is:
6 × 2 + 1.5 = 13.5
13.5 × 2 + 3 = 30
30 × 2 + 4.5 = 64.5
64.5 × 2 + 6 = 135
135 × 2 + 7.5 = 277.5
277.5 × 2 + 9 = 564

8. A box contains S white, T/295 red and 11 blue erasers. If five erasers are taken at
random then the probability that all the five are blue color is:
11 C 5 / 45 C 5
11 C 5 / 48 C 5
37 C 5 / 48 C 5
25 C 5 / 37 C 5
None of these

Solution:

Total number of erasers in the box = 12 + 7375/295 + 11 = 12+25+11=48.
Let S be the sample space.
Then, n(S) = number of ways of taking 5 out of 11.
Now, the required probability = 11 C 5 / 48 C 5

9. Rohit Sharma scores an average of 48 runs in 20 matches. His average for
next 5 consecutive matches is found to be 72 and an error of 20 runs is
reported in the runs scored in the 18 th  match. Find average runs scored by him
in 25 matches.

52
51.4
53.6
Either A or B
Either A or C

Solution:

Total runs scored in 20 matches = 48 × 20 = 960
Total runs scored in next 5 matches = 72 × 5 = 360
Actual runs scored in first 20 matches = 960 ± 20 = either 940 or 980
Total runs scored in 25 matches = either (940 + 360) or (980 + 360) = 1300 or
1340
Average runs scored in 25 matches = 52 or 53.6
Hence, option E is correct.

10. Average marks of boys of class is 78 while average marks of girls of the
same class is 86. If average marks of all students is M and ratio of number of
boys to number of girls in the class is N : 5 then which of the following can be
the value of (M – N).

90
96
88
87
80

Solution:

78 < M < 86
0 < N
M – N < 86
Only option possible is option E

11. If the Investment in cricket in 2007 and 2008 were equal, what is the
difference between the profits earned in two years, if the income in 2007
Rs.29,700,00?

1.85 lakh
1.55 lakh
1.45 lakh
1.65 lakh
1.75 lakh

Solution:

1.65 lakh ( income in football in 2007 and 2008 are equal;
income = expenditure + profit;
income in 2007 =&gt; 100% + 80% = 180%;
income = 180% = 29,700,00;
expenditure in 2007 = 100%;
expenditure = 100 x 29,700,00/180 = 1,650,000;
difference in profit = 90 – 80 = 10%;
difference in profit = 10% of 1,650,000;
the difference in profit in the year 2007 and 2008 is 1.65 lakh;)

12. The tank is being filled with the help of two pipes, G and D. At the same
time, an outlet pipe K is being used which can take off 12 litres of water in a minute. All the three pipes. i.e . G, D and K are used together when the tank is completely filled, now the tank got emptied in 1 hour. Find the capacity of the tank.

24 litre
48 litre
55 litre
60 litre
Cannot be determined

Solution:

Let, the capacity of tank be x litres
According to the question,
x + (60/10)x + (60/12)x − 60 × 12 = 0
⇒ 12x = 720 ⇒ 60
So option (d) is the correct answer.

13. In the following question, two equations I and II are given. You have to solve both the equations and give your answer accordingly:

I. 3 × (x + 3) (y + 2) = 36
II. 6xy + 12x + 15y = 33
x >; y
x <; y
x ≥ y
x ≤ y
x = y or relation cannot be established between ‘x’ and ‘y’

Solution:

xy + 3y + 2x + 6 =12
xy + 3y + 2x = 6—- (i)
6xy + 12x+ 15y =33
2xy + 5y + 4x = 11 —- (ii)
Multiplying eq. (i) by 2 and subtracing from (ii):
y = 1
Now, puting y = 1in equation (i)
x + 3x + 2 = 6
4x = 4
x=1
So, x = y

14. In the following number series, a wrong number is given. Find out that wrong number.

1024, 768, 625, 432, 324, 243

625
768
243
432
None of these

Solution:

The pattern of the series:
1024 × 0.75 = 768
768 × 0.75 = 576
576 × 0.75 = 432
432 × 0.75 = 324
324 × 0.75 = 243
∴The wrong number is 625.

15. Find the wrong term in the given series.

8, 7, 4, 6, 12, 30, 90
4
7
6
12
30

Solution:

8 × 0.5 = 4
4 × 1 = 4
4 × 1.5 = 6
6 × 2 = 12
12 × 2.5 = 30
30 × 3 = 90
So, 7 should be replaced by 4.

16. The population of a city in the year 2016 is 8% more than the
population of the same city in the year 2015 . In July 2016, 12% of total
population migrated to a different city and in the month of December
75% of it returned again to the city. If the population in the city in 2015
was 3 lakh, then what was the population of the city at the beginning of

2017?
235710
342520
178870
314280
288450

Solution:

Population of city in 2016 = 108% of 300000 = 324000
Number of people migrated = 12% of 324000 = 38880
Number of people left in the city = 324000 – 38880 = 285120
Number of people returned to the city = 75% of 38880 = 29160
So, the population of the city in 2017 = 285120 + 29160 = 314280

17. If speed of boat in still water was 30 km/hr, then on which day does the boat travel for least number of hours?

Day1
Day2
Day3
Day4
Day5

Solution:

Clearly, boat has travelled least on Day1.

18. Find the wrong number in the following series.

130, 64, 32, 14.5, 6.25, 2.125

32
64
130
14.5
6.25

Solution:

Pattern of the series is:

130 ÷ 2 – 1 = 64
64 ÷ 2 – 1 = 31
31 ÷ 2 – 1 = 14.5
14.5 ÷ 2 – 1 = 6.25
6.25 ÷ 2 – 1 = 2.125

Clearly 32 is the wrong number.

19. What should come in place of the question mark (?) in the following number series?

2, 9, 46, 181, 544, 1085, ?

1092
1090
1086
1089
1093

Solution:

Pattern of the series is:

2 × 6 − 3 = 9
9 × 5 + 1 = 46
46 × 4 − 3 = 181
181 × 3 + 1 = 544
544 × 2 − 3 = 1085
1085 × 1 + 1 = 1086

20. If the number of persons increased by 50% on Saturday from Friday in
parks A and B together, then what is the total number of persons visiting park
A and B on Tuesday and Saturday together?

88
67
48
79
61

Solution:

The number of persons visiting the parks A and B on Friday=14+16=30
The number of persons visiting the parks A and B on Saturday=30*150/100=45
The number of persons visiting the parks A and B on Tuesday= 12 + 22 = 34
So, total number of persons on Tuesday and Saturday = 45+34 = 79

Hence, option D is the correct answer.

21. What is the difference between the average number of qualified candidates from institute Q, R and V together and the average number of qualified candidates from institute S, T, and U together?

1200
1600
1500
1250
1150

Solution:

Required difference = [18000(16+15+18)/100]/3 – [18000*(10+8+11)/100]3
= 3600/3 = 1200

22. What should come in place of the question mark ‘?’ in the following number series?

26, 27, 31, ?, 56, 81, 117

38
40
46
42

Solution:

Pattern of the series is:

26 + 1 = 27
27 + 4 = 31
31 + 9 = 40
40 + 16 = 56
56 + 25 = 81
81 + 36 = 117

Here is series is addition of squares of consecutive natural numbers.

23. What is the % of qualified candidates with respect to appeared candidates
from Q and R together?

22.85
15.75
17.85
20
28.25

Solution:

Required % = {(10+8)% of 18000/(12+18)% of 60500}*100 = 17.85%

24. What is the ratio of population of literate males in cities P and R together to the population of illiterate females from the same cities together?

11:23
26:33
33:28
17:43
None of these

Solution:

Population of literate males in cities P and R together = 732 + 60 = 792
Population of illiterate females in cities P and R together = 122 + 550 = 672
Required Ratio = 792/672 = 33:28

25. Find out the wrong term in the series

79, 50, 37, 30, 26.5, 24.75

79
50
37
30
26.5

Solution:

The pattern of the series is:

79 – 28 = 51
51 – 14 = 37
37 – 7 = 30
30 – 3.5 = 26.5
26.5 – 1.75 = 24.75

26. If the square root of the average ‘A’ of 11 numbers is increased by A, the
result is another value B which is the average of first 10 of the 11 numbers
used before. Then the 11 th  number (in terms of A is) is –

A + 10√A
A – 10√A
10A + √A
10A – √A
None of the above

Solution:

Let the sum of first ten numbers be x &amp; let the 11 th  no. be y
So, A = (x + y) / 11 …(1) Also, B = x / 10 …(2)
ATQ, B = A + √A
or, x/10 = A+ √A {using (2)}
or, x =10(A+ √A) putting in (1)
y = 11A – x = 11A – (10 A+ 10√A) = (A – 10√A)

27. What is the ratio of 20% of the total number of students passed in exam2
from MP and Bihar together to the 60% of the total number of students passed
in the same exam from UP and UK together?

20 : 331
60 : 123
10 : 23
76 : 333
None of these.

Solution:

Required ratio = 20% of (40% of 50000 +20% of 90000): 60% (30% of 60000 +
50% of 75000)
= (200 + 180) : 3(180 + 375)
= 380 : 1665
= 76 : 333

28. Ajay, Amit and Sameer initially have some amount of money such that if
Amit gives 20% of his share to Ajay, then share of both becomes equal and
when Sameer gives 10% of his share to Amit, then share of both becomes
equal. Initially, if Sameer has Rs.2400, then what is the ratio of share of Ajay to
share of Amit?

3 : 5
2 : 3
5 : 7
4 : 5
4 : 7

Solution:

Sameer = 2400
According to the data in the question, we get
Amit + 10% of 2400 = 2400 – 10% of 2400
Amit + 240 = 2400 – 240
Amit = 2400 – 480 = Rs. 1920
And,
Ajay + 20% of 1920 = 1920 – 20% of 1920
Ajay = Rs. 1152
Ratio of shares of Ajay and Amit = 1152 : 1920
Ajay: Amit = 3 : 5

29. What should come in place of question mark (?) in the following number series?

0, 4, 16, 84, ?, 5332, 58656

590
470
562
420
592

Solution:

The pattern of the series is:
0 × 1 + 4 = 4
4 × 3 + 4 = 16
16 × 5 + 4 = 84
84 × 7 + 4 = 592
592 × 9 + 4 = 5332
5332 × 11 + 4 = 58656

30. The present age of a father is 3 years more than three times the age of his
son. Three years hence, father&#39;s age will be 10 years more than twice the age
of the son. Find the present age of the father.

32
33
27
36
None of these

Solution:

Let the son’s present age be x years. Then, father’s present age = (3x + 3) years
(3x + 3 + 3) = 2 (x + 3) + 10
3x + 6 = 2x + 16 x = 10.

Hence, father’s present age = (3x + 3) = ((3 * 10) + 3) years = 33 years.

31. If the amount invested in football in 2004 is 15 lakhs and the income of
2004 is equal to the investment in 2005, what is the profit earned in 2005 in
football?

1,575,000
1,258,750
1,878,750
1,277,750
1,250,000

Solution:

expenditure in 2004 = 100% = 15 lakhs;
and profit in football in 2004 = 50%;
which means the income in football during the year 2004
=&gt; investment  + profit 2004;
=&gt; 100% + 50% = 150%;
income in football in 2004 = 15 lakhs + 7.5 lakhs = 22.5 lakhs
lets say investment in football in 2005 = 100 %
income in football in 2004 = investment in football in 2005
100 % = 22.5 lakhs (in 2005)
then, 70% = (22.5/100) × 70 = 15.75 lakhs

32. If in March, Company X and Y produced 15% and 25% of their sanitizers respectively as defective, then find the average of number of sanitizers that are produced non-defective by both the companies in the same month.

4250
4350
4355
4365
None of these

Solution:

No. of sanitizers produced by Company X as non-defective = (100 – 15)% of 4800 =
4080
No. of sanitizers produced by Company Y as non-defective = (100 – 25)% of 6200 =
4650

∴ Required Average = 4365

33. In the given series one number is wrong, find out the wrong number.

10, 46, 180, 630, 1890, 4725

46
10
180
630
1890

Solution:

Pattern of the series is:

2 × 5 = 10
10 × 4.5 = 45
45 × 4 = 180
180 × 3 .5 = 630
630 × 3 = 1890
1890 × 2.5 = 4725

34. In the following series one number is wrong, Find the wrong number in the series.

47, 62, 79, 98, 118, 142

142
118
98
62
47

Solution:

Pattern of the series is:

7 × 6 + 5 = 47
8 × 7 + 6 = 62
9 × 8 + 7 = 79
10 × 9 + 8 = 98
11 × 10 + 9 = 119
12 × 11 + 10 = 142

35. C has earned 25% of Profits on Sugar and 14% profits on Pulses. Find out
the total cost C has to bear for 5 kg of Sugar and Pulses each together.

2250
3250
2750
4380
3650

Solution:

Profit on Sugar per kg = 25% = 75
Cost Price of sugar per kg = (75/25) × 100 = Rs. 300
Profit on Pulses per kg = 14% = 35
Cost Price of pulse per kg = (35/14) × 100 = Rs. 250
Total cost price for 1 kg of sugar and 1 kg of pulses = 300 + 250 = Rs. 550
Cost, C has to bear for 5 kg of Sugar and Pulses = 550 × 5 = Rs. 2750

36. What should come in place of the question mark ‘?’; in the following number series?

14, 20, 31, 49, ?, 114

76
65
72
74
92

Solution:

Hence, option A is correct.

37. What value will come in place of the question mark (?) in the following question?

594 × 19 = 36 × 750 ÷ 4 + ? + 4500

38
35
37
36
45

Solution:

594 × 19 = 6750 + ? + 4500
594 × 19 = 11250 + ?
11286 = 11250 + ?
? = 11286 – 11250
? = 36
Hence, option D is correct.

38. In the following number series only one number is wrong. Find out the wrong number.

21, 22, 46, 141, 565, 2845

565
46
21
2845
141

Solution:

The pattern of the series is:
21 × 1 + 1 = 22
22 × 2 + 2 = 46
46 × 3 + 3 = 141
141 × 4 + 4 = 568
568 × 5 + 5 = 2845

39. What will come in place of the question marks (?) in the following Number
series?

32, 16, 16, 24, 48, 120, (?)

360
340
320
392
None of these

Solution:

The given series is : *.5, *1, *1.5, *2, *2.5
So the next term = 120*3 = 360.

40. Find the wrong term in the given series:

3, 9, 31, 128, 651, 3913

651
31
128
3913
None of these

Solution:

Hence, option C is correct.

41. There are 60 employees in a company. Now the number of employees got increased by 12. Due to this the expenses of the mess increased by 20 rupees per day while the average expenditure is decreased by 3 rupees. Find the original expenditure.

1180 rupees
1240 rupees
1080 rupees
1280 rupees
None of these

Solution:

Let initial expenditure is E per day. Now it is increased by 20 rupees
per day,
Initial students = 60 and now they are 72,

(E/60) – (E+20/72) = 3

On solving we will get E = 180 rupee

42. Find the average number of students who participated in singing event from school A, D and C.

42
48
53.7
45
39

Solution:

In the question total no. of students of each school is given and no. of students who did not participate in any of the event is given. Also, no. of students who participated in singing as a percentage of students who participated from that school is given. So, from the given data we can calculate no. of students who participated from each school and no. of students who participated in any of the event singing/dancing. The table given below shows the same:

43. What approximate value will come in place of the question mark (?) in the
following question? (You are not expected to calculate the exact value)
(297.07 + 317.94 – ?) ÷ 6.05 = 99.98

15
14
22
28
19

Solution:

(297 + 318 – ?) ÷ 6 = 100

615- ? = 600
? = 15

44. If in March, Company X and Y produced 15% and 25% of their sanitizers respectively as defective, then find the average of number of sanitizers that are produced non-defective by both the companies in the same month.

4250
4350
4355
4365
None of these

Solution:

No. of sanitizers produced by Company X as non-defective = (100 – 15)% of 4800 =
4080
No. of sanitizers produced by Company Y as non-defective = (100 – 25)% of 6200 =
4650

45. In the given series one number is wrong, find out the wrong number.

7, 15, 27, 43, 63, 87, 116

116
15
27
43
63

Solution:

Pattern of the series is:

7 + 4 × 2 = 15
15 + 4 × 3 = 27
27 + 4 × 4 = 43
43 + 4 × 5 = 63
63 + 4 × 6 = 87
87 + 4 × 7 = 115

46. What should come in place of question mark (?) in the following number series?

11, 16.2, 24, ?, 47.4, 63

30.4
26.8
34.4
38.8
42.6

Solution:

The pattern of the series:

11 + 2 × 2.6 = 16.2
16.2 + 3 × 2.6 = 24
24 + 4 × 2.6 = 34.4
34.4 + 5 × 2.6 = 47.4
47.4 + 6 × 2.6 = 63

47. What is the difference between (In thousand tons) the total quantity of
wheat and rice produced by country A in the year 2012 and the total quantity
of wheat and rice produced by country A in the year 2014?

890
940
660
825
980

Solution:

Required difference = (4570 + 2715) – (3620 + 2840) = 825 thousand tons.

48. The respective ratio of the sums invested for one and half year each in two
schemes ‘A’ and ‘B’ is 3 : 5. In scheme ‘A’ offering interest of 10% p.a.
compounded annually and in scheme ‘B’ offering interest of 20% compounded
semi-annually. If the total interest is Rs. 2756, then how much total amount he
has invested?

Rs.10,200
Rs. 9,800
Rs. 10,400
Rs. 9,600
None of these

Solution:

Let the amount invested in scheme ‘A’ be Rs.3x, then the amount invested in
scheme ‘B’ is Rs.5x
Interest in scheme ‘A’ + Interest in scheme ‘B’ = 3x × 1.1 × 1.05 – 3x + 5x × 1.1 ×
1.1 × 1.1 – 5x = 2.12x = 2756 ⇒ x = 1300
Total amount invested = 8x = Rs. 10400

49. What should come in place of question mark (?) in following questions?

6607.99 ÷ 13.99 – 2196.01 ÷ 18.01 = ?

325
350
275
425
375

Solution:

? = 6608 ÷14 – 2196 ÷ 18
⇒ ? = 472 – 122 = 350

50. Find the ratio of sales of product C in 2014 to the sales of product D in
2015. (Round off the decimal to three places)

875 : 654
1452 : 1343
962 : 744
796 : 733
None of these

Solution:

Sales of C In 2014 = 3.6 lakhs × (110/100) × (110/100) = 4.356 lakhs
Sales of D In 2015 = 3 lakhs × (109/100) × (110/100) × (112/100) = 4.029 lakhs
So ratio = 4.356 : 4.029 = 4356 : 4029 = 1452 : 1343
The answer is = 1452 : 1343

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