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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 => 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 & 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'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

=> investment + profit 2004;

=> 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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