# IBPS PO Prelims Quantitative Aptitude Question Paper 2020

**IBPS** conducts various banks exams like **IBPS Clerk, IBPS PO, IBPS SO** etc. to encourage the Recruitment of young talent in various **National Public Sector Banks**. These exams are conducted to fill** 4-5K vacancies every year**.

When we talk about the **IBPS PO exam**, it is conducted for the Recruitment of **“Probationary Officer”** in the various Public sector banks in India. IBPS PO Notification 2022 will be released in **August/September 2022**, It will be a golden opportunity for those, who want to serve in Public Sector Banks.

In this article, we have provided** “Memory-Based Quantitative Aptitude Paper of IBPS PO Prelims 2020”**, to give you an idea of the **current syllabus and pattern of IBPS PO and score good marks in the exam**.

For more practice read our other articles related to it.

**Directions (1-5): **

#### In the following questions, you have to find out the approximate value of ‘x’. (You are not expected to calculate the exact value.)

**1.Question**

399/39 × 899/41 = x (9/24)^{ 2}

a) 1550

b) 1620

c) 1680

d) 1700

e) 1750

**Answer :- B**

**Explanation :-**

399/39 × 899/41 = x (9/24)^{ 2}

=> 400/40 × 900 /40 × 576/80 = x

x = 1620

Thus the approximate value of x = 1620 ( option B)

**2.Question**

x + 13.99 % of 1299 = 67.99 % of 1401

a) 770

b) 850

c) 540

d) 700

e) 900

**Answer :- A**

**Explanation :-**

=> x = 67.99 % of 1401 – 13 . 99% of 1299

=> x = 68 × 14 – 14 × 13

=> x = 14 ( 68 – 13 )

x = 770

Thus, the approximate value of x = 770 ( option A).

**3.Question**

x + 250.17 – 77 % of 910 = 41 % of 601

a) 530

b) 555

c) 690

d) 810

e) 770

**Answer :-C**

**Explanation :-**

=> x = 41 % of 601 + 77 % of 910 – 250.17

=> x = 40 × 6 + 700 – 250

x = 690

Thus, the approximate value of x = 690 ( option C).

**4.Question**

61/29 x = 52001 ÷ 41

a) 490

b) 530

c) 710

d) 400

e) 600

**Answer :- E**

**Explanation :-**

61/29 x = 52001 ÷ 41

=> x =( 52001 × 29) /(41 × 61)

:. x = 600

Thus, the approximate value of x = 600 (option E)

**5.Question**

x + 3245.01 – 1122.99 = 5466.97 – 2309.99

a) 1035

b) 1095

c) 1065

d) 980

e) 1010

**Answer :- A**

**Explanation :-**

x + 3245.01 – 1122.99 = 5466.97 – 2309.99

=> x = 5467 – 2310 + 1123 – 3245

x = 1035

Thus, the approximate value of x = 1035 ( option A)

**Directions (6-10): **

#### There are two equations I and II are given below. You have to solve both the equations and give answer as,

#### (a) if x < y

#### (b) if x > y

#### (c)if x ≤ y

#### d) if x ≥ y

#### (e) if x = y or no relation can be established.

**6.Question**

I. 3×2 + 11x + 10 = 0

II. 14y2 + 13y + 3 = 0

**Answer :- A**

**Explanation :-**

3×2 + 11x + 10 = 0

=> 3×2 + 6x + 5x + 10 = 0

=> (3x + 5 ) (x + 2 ) = 0

x = – 5/3 , – 2

14y2 + 13y + 3 = 0

=> 14y2 + 7y + 6y + 3 = 0

=> (7y + 3) ( 2y + 1) = 0

y = – 3/7 , – 1/2

**x < y**

**7.Question**

I. 9×2 + 31x + 12 = 0

II. 9y2 + 50y + 25 = 0

**Answer :-**

**Explanation :- E**

9×2 + 31x + 12 = 0

=> 9×2 + 27x + 4x + 12 = 0

=> ( 9x + 4 ) ( x + 3 ) = 0

x = – 4/9 , – 3

9y2 + 50y + 25 = 0

=> 9y2 + 45y + 5y + 25 = 0

=> (9y + 5 ) ( y + 5 ) = 0

y = – 5/9 , – 5

Relationship can not be established .

**8.Question**

I. 10×2 + 40x = 50

II. y2 + 22y = – 85

**Answer :- D**

**Explanation :-**

10×2 + 50x – 10x – 50 = 0

=> (10x – 10) )( x + 5) = 0

x = 1 , – 5

y2 + 22y = – 85

=> y2 + 5y + 17y + 85 = 0

=> ( y + 5 ) ( y + 17 ) = 0

y = -5 , – 17

** x ≥ y**

**9.Question**

I. 21×2 – 5x = 4

II. 12y2 + 29y = -14

**Answer :- B**

**Explanation :-**

21×2 – 5x = 4

=> 21×2 + 7x – 12x – 4 = 0

=> ( 7x – 4 ) ( 3x + 1 ) = 0

x = 4/7 , – 1/3

12y2 + 29y = -14

=> 12y2 + 8y + 21y + 14 = 0

=> ( 4y + 7 ) ( 3y + 2 ) = 0

y = -7/4 , – 2/3

** x > y**

**10.Question**

I. 3×2 = 27

II. 7y2 – 47y = 14

**Answer :- E**

**Explanation :-**

3×2 = 27

=> x2 = 9

x = 3 , – 3

7y2 – 47y – 14 = 0

=> 7y2 – 49y + 2y – 14 = 0

=> ( 7y + 2 ) ( y – 7) = 0

y = 7 , – 2/7

The relationship can not be established.

**Direction ( 11 – 15): **

#### In the following table shows the total TV, Mobile and Laptop sold by a merchant in five months of a year. Read the data carefully and answer the following questions-

Total article sell = TV + Mobile + Laptop

**11. Question**

Total Mobile sold by the merchant in January and May together are what per cent less than total laptop sold by the Merchant in March and April together?

a) 45 %

b) 50 %

c) 55 %

d) 60 %

e) 70 %

**Answer :- B**

**Explanation :-**

Total mobile sold in January and May together

= 480/48 × 32 + 680/68 × 22

= 320 + 220

= 540

Total laptop sold in March and April together

= 840/35 ×20 + 720/24 × 20

= 480 + 600

= 1080

Required % = ( 1080 – 540 ) /1080 × 100 %

= 50 %

**Thus, the required less percentage = 50 %**

**12.Question**

Find the difference between the average number of Mobile sold in February and April together and the average number of TV sold in April and May together?

a) 512

b) 520

c) 524

d) 546

e) 555

**Answer :- C**

**Explanation :-**

Average number of mobile sold in February and April together

=( 640/40 × 48 + 720 /24 × 56 ) / 2

= (768 + 1680) / 2

= 1224

Average number of TV sold in April and May month together

= (720 + 680 ) / 2

= 700

Difference = 1224 – 700

= 524

**Thus, the required difference = 524**

**13.Question**

If total Mobile sold by merchant in July is 25% more than that of total mobile sold in April and total Laptop sold in July is 300% more than that of total Laptop sold in May, then find total number of Mobile and Laptop sold by merchant in July?

(a) 2200

(b) 2500

(c) 2700

(d) 1900

(e) 2100

**Answer :- B**

**Explanation :-**

Total mobile sold in July

= 720/24 × 56 × 125 /100

= 2100

Total Laptop sold in July

= 680 / 68 × 10 × 400 / 100

= 400

Total mobile + Laptop = ( 2100 + 400 ) = 2500

**Thus, the total mobile and laptop sold in July = 2500**

**14.Question**

Find the ratio between total article sold by the merchant in January to total article sold by merchant in April?

(a) 1 : 3

(b) 1 : 2

(c) 1 : 4

(d) 1 : 7

(e) 1 : 5

**Answer :- A**

**Explanation :-**

Ratio = ( 480 /48 × 100 ) : ( 720 /24 × 100 )

= 1000 : 3000

= 1 : 3

**Thus, the required ratio = 1 : 3**

**15.Question**

Total Laptop sold by merchant in March is what percent more than total Laptop sold by merchant in January and February together (you have to find approximate value)?

(a) 26%

(b) 18%

(c) 22%

(d) 16%

(e) 28%

**Answer :-C**

**Explanation :-**

Total laptop sold in March = 840/35 × 20

= 480

Total laptop sold in January and February together

= (480/48 × 20) + (640/40 × 12)

= 392

Required% = (480 – 392) /392 × 100 %

= 22.44 %

= 22 %

**Thus, the required more percentage approximately = 22 %**

**(Direction 16 – 17):-**

#### A toy seller gives a 20% discount on MRP on each toy and if a boy bought 45 toys at discounted price, he will also get 15 toys free. Even after selling 60 toys on the price of 45, toy seller earns 20%.

**16.Question**

Find ratio of MRP of 1 toy to Cost price [for toy seller] of 1 toy.

(a) 4 : 1

(b)10 : 1

(c) 5 : 1

(d) 3 : 1

(e) 2 : 1

**Answer :- E**

**Explanation :-**

Let the MRP of each toy is Rs. 100.

Therefore, Selling price of each toy is Rs. 80.

When customer paid Rs. 80 × 45 = Rs. 3600

the toy seller will give him(the boy) 60 toys i.e. at a price of Rs. 60/toy

Even on this SP of Rs. 60, the toy seller earns 20%

Therefore, CP of 1 toy = 60 × 100/120 = Rs. 50

Required ratio = 2 : 1

**Thus, the ratio between the MRP and CP of 1 toy = 2 : 1**

**17.Question**

A man paid price for 60 toys and a woman paid the price for 90 toys. Average price per toy paid by the man is how much % more than that of the woman.

a) 5.25 %

b) 6.66 %

c) 9 %

d) 11.30 %

e) 15 %

**Answer :- B**

**Explanation :-**

When the man will pay for 60 toys, he will get 60 toys at a price of 45 and then 15 more balls at 20% discount.

And When the woman will pay for 90 toys she will get 120 balls. i.e. 15 free for every 45 balls.

The man will pay 60×80 = Rs 480 for 75 toys =Rs.6.4 / toy

While the woman will pay Rs 90 ×80 = 720 for 120 toys

= Rs. 6 / toy

Required % = (6.4 – 6)/6 × 100 %

= 6.66 %

**Thus the required more percentage paid by man = 6.66 %**

**18.Question**

In a running train a passenger counts the electric poles on the track . The electric poles are at a distance of 90 m. What will be his count in 6 hrs if the speed of the train is 50 Km/h?

(a) 2600

(b) 2903

(c) 3001

(d) 5000

(e) None of these

**Answer :- C**

**Explanation :-**

Total distance covered by him = 45 × 6 = 270 km

No. of Telegraph poles =(270 × 1000)/90 = 3000

His total count will be (3000+ 1 )= 3001

**Thus the total count = 3001**

**19.Question**

The present age difference between two sisters is 10 years. After 5 years, elder sister’s age is twice of youngers age. What will be younger sister’s age after 27 years?

(a) 28 years

(b) 27 years

(c) 25 years

(d) 32 years

(e) None of these

**Answer :- D**

**Explanation :-**

Let, younger sister’s present age = x

Elder sister’s present age = (x +10) years

According to question,

(x + 10) + 5 = 2(x+ 5)

⇒ x = 5 years

**∴ younger sister’s age after 27 years = 32 years**

**20.Question**

Two small traders started a business by investing their sum in the ratio of 3 : 5 . After 10 months, first trader leaves the business. After one year, if total profit is Rs. 900, what is the profit of first trader ?

(a) 400

(b) 410

(c) 330

(d) 300

(e) None of these

**Answer :- D**

**Explanation :-**

Profit ratio of two trader = 3 × 10 : 5 × 12

= 1 : 2

Profit of first trader = 900 × 1/3 = 300

**Thus, the required profit of the first trader = Rs. 300**

**21.Question**

A fruit seller allows 4% discount and allows 2 apple free on purchase of 14 apple . He earns 40% profit on the transaction. By what percentage above the cost price he marked price of the Apple ?

(a) 55.50%

(b) 33.33%

(c) 66.66%

(d) 50%

(e) None of these

**Answer :- C**

**Explanation :-**

Let, cost price (CP) of one apple = 100x

14 × SP = 16 × CP + 40% of CP of 16 apple [SP → Selling price]

14 × SP = 1600x + 40/100 × 1600x

14 × SP = 2440x

SP = 160x

SP = 96% of MP [MP → marked price].

MP = 160x / 96 × 100

= 500x / 3

Required percentage

= (500x/3 – 100x ) / 100x × 100%

= 66.66 %

**Thus, required more percentage of the marked price above the cost price = 66.66 %**

**Directions ( 22 – 23):-**

#### Study the following information carefully and answer the questions given below. In a cricket tournament , probability that A team will win this match is 1/2 and the probability that B team will win this match is 1/3.

**22.Question**

What is the probability that both A and B team will win this match?

a) 1

b) 1/6

c) 2/3

d) 1/3

**Answer :- B**

**Explanation :-**

Probability that A team win the match = 1/2

Probability that B team will win the match = 1/3

Probability that both team will win the match = Probability that A team will win and probability that B team will win

= 1/2 x 1/3

= 1/6

**Thus, the probability that both A and B team will win the match =1/6 **

**23.Question**

What is the probability that only 1 team [either A team or B team] will win the match?

1. 1/2

2. 1/3

3. 1/7

4. 2/3

**Answer:- A**

**Explanation :-**

Probability that A team will win the match = 1/2

So, Probability that A team will loss the match = 1/2.

Probability that B team will win the match = 1/3

So, Probability that B team will loss the match = 2/3.

Probability that only one person will win is when A team wins and B team loss or A team loss and B team win.

= (1/2) x (2/3) + (1/2) x (1/3)

= 1/3 + 1/6

= 1/2

**Thus, the probability that only one team will win = 1/2**

**24.Question**

Three labors A, B, C can do a work in 15 days, if they working together. But B can work only half time due to his illness. The work can be finished in how many days?

(a) 16 days

(b) 17 days

(c) 18 days

(d) 20 days

(e) None of these

**Answer :- C**

**Explanation :-**

Let, efficiency of each labor is 1 unit/day

For three labors, it is 3 units/day

Hence in 15 days, Total work to be done = 45 units

If B works for half a day only

Then total work in a day by three worker = 1+ 1 + 0.5 = 2.5

**Hence, No. of days required to finish = 45/2.5 days = 18 days**

**25.Question**

A shopkeeper marks his goods at a certain price. He allows a discount on marked price is 8 % and still he gains 15 % . At what percent above the cost price does he marks in his goods?

(a) 28

(b) 35%

(c) 22

(d) 25%

(e) None of these

**Answer :- D**

**Explanation :-**

Let, Marked price (MP) = 100

Selling price ( SP) = (100 – 8) = 92

Cost price ( CP) = 92 × 100/115 = 80

MP – CP = (100 – 80) = 20

Required % = 20/80 × 100 %

= 25 %

**Thus, the required percentage above the cost price, he marks his good = 25 %**

**26.Question**

Rs. 40 is the difference between the compound interest and the simple interest on a certain sum for two years, when the interest is compounded yearly .The rate is 10 % . If the interest is compounded half-yearly, what will be the difference between the

CI and the SI ?

(a) Rs 40

(b) Rs 52.550

(c) Rs 62.025

(d) Rs 64.50

(e) Rs 72.0125

**Answer :-C**

**Explanation :-**

Difference = Principal ×( r/100) ^2

=> Principal = difference × ( 100/r) ^2

=> principal = (40 × 100 × 100)/(10× 10)

= 4000

When interest is compounded half yearly ,

T = 4

r = 10/2 = 5 %

P = 4000

Amount = 4000 ×(1+ 5/100) ^4

= 4862.025

Compound interest (CI) = Amount – Principal

= 862.025

Simple Interest ( SI) = 4000 × 10 × 2 /100

= 800

Difference = CI – SI =( 862.025 – 800 ) = 62.025

**Thus, the difference between the compound interest and simple interest =Rs.62.025**

**27.Question**

Curved surface area of a right circular cone is 550 sq. Metre and its radius 7 metre . What is the volume of the circular cone?

(a) 1232 m3

(b) 1550 m3

(c) 1600 m3

(d) 1750 m3

(e) None of these

**Answer :- A**

**Explanation :-**

Curved surface Area = πrl

πrl = 550

=> 22/7 × 7 × l = 550

=> l =25 metre

Now, l^2 = r^2 + h^2

=> 25^2 – 7^2 = h^2

=> h^2 = 576

=> h = 24 metre.

Volume = 1/3 πr^2h

= 1/3 ×22/7 × 7 × 7 × 24

= 1232 m3

**Thus, the volume of the circular Cone is = 1232 metre3**

**28.Question**

A fraction is divided by its reciprocal and then multiplied by itself, thus the fraction obtained is 729/625 . What is the original fraction?

a) 4/7

b) 7/4

c) 9/7

d) 9/5

e) none of these

**Answer :- D**

**Explanation :-**

Let, the original fraction is = x/y

According to question,

( x/y ÷ y/x) * x/y = 729/625

=> x/y * x/y * x/y = 729/625

=> x^3/y^3 = 729/625

=> x/y = 9/5

**Thus, the original fraction = 9/5**

**29.Question**

A person is working in a company . Savings of the person is 10% of his total income. After promotion his income is increased by 20% but his savings remain the same. Find the hike percentage in his expenditure .

(a) 23.65%

(b) 22.95%

(c) 22.22%

(d) 15.95%

(e) None of these

**Answer :- C**

**Explanation :-**

In first case,

Let, Income = 100

Savings = 10

Expenditure = (100 – 10) = 90

In second case,

Income = 120

Savings = 10 ( savings remain same)

Expenditure =( 120 – 10 )= 110

Hike % in expenditure = (110 – 90) /90 × 100 %

= 22.22 %

**Thus, the hike percentage of expenditure of the person = 22.22%**

**30.Question**

The rate of the current is 3 km/h and upstream speed of a motorboat is 5 km/h, then how much time taken to cover a distance 55 km in downstream ?

(a) 3 h

(b) 5 h

(c) 4 h

(d) 4.5 h

(e) None of these

**Answer :- B**

**Explanation :-**

Rate of current = 3 km/h

Upstream speed(U) = 5 km/h

Rate of current = 1/2 ( downstream speed – upstream speed)

=> 1/2 ( D – U) = 3

=> D – 5 = 6

=> D = 11 km/h

Downstream speed = 11 km/h

Time taken = 55/11 = 5 h

**Thus, the time taken to cover the distance in downstream = 5 hour**

**Direction (31-35): **

#### The following line graph gives the percentage of the number of passengers who qualified for passport and visa out of the total number of passengers who applied for the passport and visa over a period of seven years from 2010 to 2016 .

**31.Question**

The total number of Passengers qualified in 2015 and 2016 together was 33500 and the number of Passengers applied in 2015 was 26500. What was the number of Passengers in 2016?

(a) 20500

(b) 22000

(c) 23500

(d) 19000

(e) 24000

**Answer :- A**

**Explanation :-**

Number of Passengers qualified in 2015

= (80% of 26500)

= 21200.

Number of Passengers qualified in 2016

= (33500 – 21200)

= 12300.

Let the number of Passengers applied in 2016 be x.

Then, 60% of x = 12300

x= (12300 x 100)/60

x = 20500.

**Thus, the number of total passengers in 2016 = 20500**

**32.Question**

The difference between the percentage of Passengers qualified to applied was maximum in which of the following pairs of years?

(a) 2010 and 2011

(b) 2013 and 2014

(c) 2014 and 2015

(d) 2015 and 2016

(e) none

**Answer :- B**

**Explanation :-**

The differences between the percentages of passengers qualified to applied for the giving pairs of years are:

For 2010 and 2011 = 50 – 30 = 20.

For 2014 and 2015 = 80 – 80 = 0.

For 2010 and 2013 = 50 – 30 = 20.

For 2013 and 2014 = 80 – 50 = 30.

For 2015 and 2016 = 80 – 60 = 20.

**Thus, the maximum difference is between the years 2013 and 2014 .**

**33.Question**

If the number of passengers qualified in 2014 was 21200, what was the number of passengers applied in 2014?

(a) 30000

(b) 26500

(c) 28500

(d) 24000

(e) 25000

**Answer :- B**

**Explanation :-**

The number of candidates appeared in 2014 be x.

Then, 80% of x = 21200

=> x =(21200 x 100)/80

x = 26500 .

**Thus, the number of passengers applied in 2014 = 26500**

**34.Question**

If the total number of passengers applied in 2012 and 2013 together was 50000, then the total number of passengers qualified in these two years together was?

(a) 30000

(b) 31000

(c) 31500

(d) Data inadequate

(e) 32000

**Answer :- D**

**Explanation :-**

The total number of passengers qualified in 2012 and 2013 together, cannot be determined until we know at least, the number of candidates applied in any one of the two years 2012 or 2013 or the percentage of passengers qualified to applied in 2012 and 2013 together.

**Hence, the data is inadequate.**

**35.Question**

In which pair of years was the number of passengers qualified, the same?

(a) 2012 and 2013

(b) 2014 and 2016

(c) 2011 and 2015

(d) Data inadequate

(e) 2013 and 2014

**Answer 😀**

**Explanation :-**

The graph gives the data for the percentage of candidates qualified to applied .

We cannot compare the absolute values for any two years if the absolute values of number of passengers qualified or passengers applied is not given.

**Hence, the data is inadequate .**