# IBPS PO Prelims Quantitative Aptitude Question Paper 2021

**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 of 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 2021”**, 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 number series a number is missing. Find out the missing number in the given questions.

**1. Question**

11 18 32 60 116 ?

a) 232

b) 228

c) 224

d) 198

e) 218

**Answer : B**

**Explanation :**

11 + 7 = 18

18 + 14 = 32

32 + 28 = 60

60 + 56 =116

116 + 112 = 228

**Thus, the number is = 228**

**2. Question**

55 27 13 ? 2.5

a) 8

b) 6

c) 7.5

d) 6.5

e) 4.5

**Answer : B**

**Explanation :**

(55 – 1) /2 = 27

(27 – 1) /2 = 13

(13 – 1) /2 = 6

(6 – 1) /2 = 2.5

**Thus, the number is = 6**

**3. Question**

6 18 36 60 ?

a) 75

b) 80

c) 90

d) 105

e) 112

**Answer : C**

**Explanation :-**

{6 + (5×2) +2} =18

{18 + (5×3) +3} = 36

{36 + (5×4) + 4} = 60

{60 + (5×5) + 5} = 90

Thus, the number is = 90

**4. Question**

355 385 445 565 ? 1285

a) 805

b) 750

c) 900

d) 955

e) 675

**Answer : A**

**Explanation :-**

355 + 30 = 385

385 + 60 = 445

445 + 120 = 565

565 + 240 = 805

805 + 480 = 1285

**Thus, the number is = 805**

**5. Question**

275 295 335 355 395 ?

a) 435

b) 425

c) 405

d) 415

e) 450

**Answer :- D**

**Explanation :-**

275 + 20 = 295

295 + 40 = 335

335 + 20 = 355

355 + 40 = 395

395 + 20 = 415

**Thus, the number is = 415**

**Directions (6-10): **In the following question contains two equations as I and II . You have to solve both the equations and determine the relationship between them and give answer as,

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot establish.

**6. Question**

I. 3x^{2 }+ 11x + 10 = 0

II. 3y^{2} + 13y + 14 = 0

**Answer :- B**

**Explanation :-**

3x^{2} + 11x + 10 = 0

3x^{2} + 6x + 5x + 10 = 0

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

x = – 5/3 , -2

3y^{2} + 13y + 14 = 0

3y^{2 }+ 6x + 7x + 14 = 0

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

y = -7/3 , – 2

**x ≥ y**

Thus, the value of x is either greater than or equal to y.

**7. Question**

I. 25x^{2} – 30x + 9 = 0

II. 49y^{2} – 84y + 36 = 0

**Answer :- C**

**Explanation :-**

25x^{2} – 30x + 9 = 0

(5x – 3)^{2 }= 0

5x – 3 = 0

x = 3/5

49y^{2} – 84y + 36 = 0

(7y – 6) ^{2} = 0

7y – 6 = 0

y = 6/7

**x < y**

Thus, the value of x is less than y.

**8. Question**

I. 4x + 4/x + 17 = 0

II. 16y^{2} -104y + 169 = 0

**Answer: C**

**Explanation:**

4x + 4/x + 17 = 0

4x^{2} + 4 + 17x = 0 (Both sides multiplied by x)

4x^{2 }+ 16x + x + 4 = 0

(4x + 1) ( x + 4) = 0

x = -1/4 , -4

16y^{2} – 104y +169 = 0

(4y – 13 )^{2 }= 0

4y – 13 = 0

y = 13/4

**x < y**

Thus, the value of x is less than y.

**9. Question**

I. 9x^{2} – 72x + 144 = 0

II. y^{2} + 16y + 16 = 16y + 32

**Answer : B**

**Explanation :**

9x^{2 }– 72x + 144 = 0

(3x – 12 ) ^{2 }= 0

3x – 12 = 0

x = 4

y^{2 }+ 16y + 16 = 16y + 32

y^{2} – 16 = 0

(y +4 ) (y – 4) = 0

y = 4 , -4

**x ≥ y**

Thus, the value of x is either greater than or equal to y.

**10. Question**

I. x^{2} + x – 42 = 0

II. y^{2} + 7y + 12 = 0

**Answer: E**

**Explanation:**

x^{2} + x – 42 = 0

x^{2 }+ 7x – 6x – 42 = 0

(x + 7) (x – 6) = 0

x = 6 , – 7

y^{2 }+ 7y + 12 = 0

y^{2 }+ 4y + 3y + 12 = 0

(y + 3 ) (y + 4) = 0

y = -3 , – 4

Thus, the relationship can not be established between x and y.

**Directions (11-15): **The table given below shows the total production of mobile and chargers in six different factories and the ratio of Mobile to chargers in these six factories. Study the data carefully & answer the following question.

Factory | Total Item | Mobile : Charger |
---|---|---|

A | 1440 | 7 : 11 |

B | 1200 | 3 : 5 |

C | 1800 | 7 : 8 |

D | 1680 | 2 : 5 |

E | 1320 | 5 : 6 |

F | 1600 | 11 : 9 |

**11. Question**

A total number of mobile in F and D together is what percent of the total number of Charger in A and F together?

(a) 77.5%

(b) 80%

(c) 85%

(d) 82.5%

(e) 87.5%

**Answer: C**

**Explanation:**

Total number of Mobile in F and D factory together

=(1600 × 11/20) + (1680 × 2/7)

= (880 + 480)

= 1360

Total number of Charger in A and F factory together

=(1440 × 11/18) + (1600 × 9/20)

= 880 + 720

= 1600

Required %

=(1360/1600)× 100

= 85%

**Thus, the percentage is = 85 %**

**12. Question**

Find the average number of chargers producing in B, C, and E together?

(a) 800

(b) 810

(c) 820

(d) 830

(e) 840

**Answer: B**

**Explanation:**

Charger in B, C & E factory together

=(1200 /8)× 5 +(1800 /15)× 8 +(1320/11)× 6

= 750 + 960 + 720

= 2430

Required average

=2430/ 3

= 810

Thus, the average number of chargers produced by B, C, and E together is 810.

**13. Question**

Find the ratio of Charger in E and F together to Mobile in C and D factory together?

(a) 2 : 3

(b) 4 : 5

(c) 5 : 6

(d) 7 : 8

(e) 11 : 12

**Answer: A**

**Explanation:**

Charger producing in E and F together

= (1320/11) × 6 + (1600 / 20) × 9

=720 + 720

= 1440

Mobile producing in C and D together

= (1800 /15)× 8 + (1680 /7) ×5

= 960 + 1200

= 2160

Ratio = 1440 / 2160

= 2 : 3

**Thus, the required ratio = 2 : 3**

#### 14. Question

In another factory P, Mobile are 20% more than the mobile in F while Chargers are 10% more than Chargers in E. Total number of items in factory P is what percent more than the total number of items in factory D?

(a) 20%

(b) 15%

(c) 10%

(d) 5%

(e) 25%

**Answer: C**

**Explanation:**

Mobile producing in factory P

=1600/20 × 11 × 120/100

= 1056

Charger producing in P

=1320 /11× 6 × 11/ 10

= 792

Total items in factory P

=( 792 + 1056)

= 1848

Required %

=(1848−1680 /1680) × 100

=168 / 1680 × 100

= 10%

Thus, the required percentage is = 10 %

**15. Question**

Find the average number of mobile producing in factories A, D, E, and F together?

a) 630

b) 500

c) 550

d) 670

e) 700

**Answer: A**

**Explanation:**

Mobile in A = 1440 × 7/18

= 560

Mobile in D = 1680 × 2/7

= 480

Mobile in E = 1320 × 5/11

= 600

Mobile in F = 1600 × 11/20

= 880

Total = (560 + 480 + 600 + 880)

= 2520

Average = 2520/4

= 630

Thus, the average number of mobile producing by A, D, E and F is 630.

**16. Question**

An old lady walking at 4/5 of her usual speed, she is late by 11/2 hours. what is her usual time to taken?

a) 22 hours

b) 28 hours

C) 17 hours

d) 20 hours

e) 25 hours

**Answer: A**

**Explanation:**

Speed and time are inversely proportional.

So time is taken in the present (new) = 5/4 of the usual time

According to the question,

5/4 of the usual time – usual time = 11/2

( 5/4 – 1) of the usual time = 11/2

1/4 of the usual time = 11/2

Usual time = 11/2 × 4 = 22 hours

Thus the usual time taken by the old lady is 22 hours.

#### 17. Question

Three small cubes are melted to form a big cube. The sides of these three small cubes are 30 cm, 40 cm, and 50 cm respectively. Find the side of the big cube which is made by melting the three small cubes together.

a) 60 cm

b) 65 cm

c) 75 cm

d) 80 cm

e) 100 cm

**Answer: A**

**Explanation:**

Sum of volume of three small cubes = Volume of big cube

Let, Side of the big cube = A cm

(30) ^3 +( 40 )^3 + (50) ^3 = A^3

A^3 = (27000 + 64000 + 125000 ) cm^3

A = (216000) ^1/3 cm

A = 60 cm

Thus, the side of the big cube is = 60 cm.

#### 18. Question

The radius and height of a cylinder are decreased by 200 %. Then what percentage does the volume of a cylinder decrease?

a) 100 %

b) 200 %

c) 300 %

d) 400 %

e) none of these

**Answer: B**

**Explanation:**

Radius / height = x

Volume of the cylinder

= [{(1 + x/100) ^3 – 1 }]× 100 %

= [{(1 + -200/100)^3 – 1}]×100 %

={ (1 – 2) ^3 – 1 } × 100%

={( – 1 ) ^3 – 1 } × 100 %

= (- 1 – 1) × 100 %

= – 200 %

Negative sign indicates decrease.

Thus, the percentage of the volume of the cylinder is decreased by 200%.

**19. Question**

A shopkeeper sold book A at 25% gain and another book B at 25% loss. Find his overall gain or loss percent if the Selling price of both books were the same?

a) 4 % loss

b) 4 % gain

c) 6 % gain

d) 6.25 % gain

e) 6.25 % loss

**Answer: E**

**Explanation:**

Cost price | Selling price |
---|---|

100 (×3) = 300 | 125 (×3) = 375 |

100 (×5) = 300 | 75 (×3) = 375 |

The selling price of each book is the same.

Total C.P = 300 + 500 = 800

Total S. P = 375 + 375 = 750

Loss % = (800 – 750 ) /800 × 100 %

= 50/800 × 100 %

=6.25 %

Thus, the shopkeeper’s overall loss percentage is 6.25 %.

**20. Question**

Three friends can do a piece of work in 90 days, 40 days, and 120 days respectively. If all three works alternatively each day starting with 1st, 2nd then 3rd friend respectively. Find the total time to complete the work?

(a) 67.44 days

(b) 65.44 days

(c) 66 days

(d) 66.44 days

(e) 67 days

**Answer: A**

**Explanation:**

1st friend = 90 days

2nd friend = 40 days

3rd friend = 120 days

Total work (LCM of 90 , 40, 120) = 360

1’s one day work = 360/90 = 4

2’s one day work = 360/40 = 9

3’s one day work = 360 /120 = 3

Work done by three friends together in 3 days = 16

Work done by three friends in 66 days = 352

Remaining work = 360 – 352 = 8

Total time required to complete the work

= 66 + 1 + 4/9

= 67 + 4/9

= 67.44 days

Thus, the time taken to complete the work = 67.44 days.

**Directions (21 – 26): **Study the bar graph given below carefully and answer the questions given below.

**21. Question**.

The pensioner’s rate of which town is 25% more than that of Chennai?

a) Mumbai

b) Ranchi

c) Kolkata

d) Delhi

e) None of these

**Answer: C ****Explanation:**

Pensioner’s rate of Chennai = 16

Pensioner’s rate of Kolkata = 20

Required different % = (20 – 16)/16 × 100 %

= 25 %

Thus, Kolkata is 25 % more than that of Chennai.

**22. Question **

The pensioner’s rate of Mumbai is what percent of the pensioner’s rate of Kolkata?

a) 100 %

b) 140 %

c) 165 %

d) 155 %

e) None of these

**Answer: C****Explanation:**

Pensioner’s rate of Mumbai = 33

Pensioner’s rate of Kolkata = 20

Required percentage = 33/20 × 100 % = 165%

Thus, the pensioner’s rate of Mumbai is 165 % of Kolkata’s pensioner rate.

**23. Question**

The pensioner’s rate of Ranchi is how many times the pensioner’s rate of Chennai?

a) 4

b) 7

c) 3.8

d) 5.2

e) 2.5

**Answer: E**

**Explanation:**

Pensioner’s rate of Ranchi = 40

Pensioner’s rate of Chennai = 16

Required answer = 40/16 = 2.5

Thus, the pensioner’s rate of Ranchi is 2.5 times of Chennai.

**24. Question**

What is the ratio of pensioner’s rate of Kolkata and Ranchi together to that of Patna?

a) 9 : 2

b) 4 : 1

c) 5 : 4

d) 5 : 1

e) 7 : 2

**Answer: B**

**Explanation:**

Pensioner’s rate of Kolkata and Ranchi = ( 20 + 40) = 60

Pensioner’s rate of Patna = 15

Required ratio = 60 : 15

= 4 : 1

Thus the required ratio = 4 : 1

**25. Question **

By how much percentage is the pensioner’s rate of Kolkata less than the pensioner’s rate of Delhi?

a) 33.33%

b) 25.75%

c) 27 %

d) 37.25 %

e) None of these**Answer: A**

**Explanation: **

Pensioner’s rate of Kolkata = 20

Pensioner’s rate of Delhi = 30

Required percentage = (30 – 20)/30 × 100 %

= 100/3 % = 33.33 %

**26. Question**

What is the ratio of pensioner’s rate of Kolkata, Patna, and Delhi together to that of Ranchi and Mumbai together?

a) 55 : 43

b) 65 : 73

c) 4 : 7

d) 47 : 61

e) None of these

Answer: B

**Explanation: **

total pensioner’s rate of Kolkata,Patna and Delhi = (20+15+30)

= 65

Total pensioner’s rate of Ranchi and Mumbai = (40 + 33)

= 73

Required ratio = 65 : 73

**27. Question**

A motorboat travels 40 km upstream and 60 km downstream in 13 hours. If the motorboat travels 50 km upstream and 72 km downstream in 16 hours. Find the speed of the boat in still water.

(a) 7.5 km/hr

(b) 8.5 km/hr

(c) 9.5 km/hr

(d) 7 km/hr

(e) 6 km/hr

**Answer :- B**

**Explanation :-**

Let upstream speed = x

Downstream speed = y

Now

40/x + 60 /y= 13 …………..… (i)

And,

50/x + 72 /y= 16 ……………..(ii)

Solving equation. (i) and (ii)

x = 12

y = 5

speed of motorboat =(x + y) /2

=(12 + 5)/2

= 8.5 km/hr

**Thus, in still water, the speed of the boat = 8.5 km/hr**

**28. Question**

Two bottles A and B contain a mixture of spirit and water. The ratio of spirit and water in bottles A and B are in ratio 1 : 2 and 3 : 2 respectively. In what ratio these two mixtures should be mixed so that the new mixture contains 50% spirit and 50% water.

(a) 5 : 2

(b) 2 : 3

(c) 3 : 5

(d) 1 : 4

(e) 7 : 4

**Answer: C**

**Explanation:**

A B

1/3 3/5

\ /

\ /

1/2

/ \

/ \

(3/5 – 1/2) (1/2 – 1/3)

= 1/10 =1/6

Ratio = 1/10 : 1/6 = 3 : 5

Thus, the ratio in the new mixture is 3 : 5.

**29. Question**

The perimeter of the square is 96 cm and the length of the rectangle is four times that of breath. Rectangle’s and square’s area is the same. Then Find the perimeter of the rectangle?

a) 90 cm

b) 98 cm

c) 110 cm

d) 120 cm

e) 135 cm

**Answer: D**

**Explanation:**

Let, side of a square = a cm

According to question,

4a = 96

a = 24

Area of square = a^2 = (24) ^2 = 576 sq. cm

Let, length of a rectangle = l

Breadth of the rectangle = b

According to question,

l = 4b

l × b = 576

4b × b = 576 ( putting the value of l = 4b)

b^2 = 144

b = 12 cm

l = 12 ×4 = 48 cm

Perimeter = 2 (l + b) cm

= 2 ( 48 + 12 ) cm

= 120 cm

**Thus, the perimeter of the rectangle is = 120 cm.**

**30. Question**

Two pipes X and Y can fill a tank separately in 4 hrs and 5 hrs respectively. Another outlet pipe Z can empty the full tank in 10 hrs. Find the time taken to fill the tank when all the three pipes are opened together.

a) 20/7 hrs

b) 4 hrs

c) 19/3 hrs

d) 14/3 hrs

e) 15/4 hrs

**Answer: A**

**Explanation:**

Pipe X fill the tank in 1 hr = 1/4 part

Pipe Y fill the tank in 1 hr = 1/5 part

Pipe Z empty the tank in 1 hr = 1/10 part

Pipe (X + Y + Z) fill the tank in 1 hr

=( 1/4 + 1/5 – 1/10 )

=7/20

**Thus the total time required to fill the tank = 20/7 hrs**

**31. Question**

A boy and a girl travel from A to B, a distance of 42 km at 6 km/hr and 8 km/hr respectively. Girl reaches B first and returns immediately and meets the boy at another point C. Find the distance from point A to C.

a) 46 kms

b) 30 kms

c) 36 kms

d) 25 kms

e) 50 kms

**Answer: C**

**Explanation:**

Let, AC = x km

According to the question,

Time is taken by the boy to cover AC = time taken by the girl to cover ( AB + BC)

x/6 = [42 + (42 – x) ] / 8

4x = 3 ( 42 + 42 – x)

7x = 252

x = 36 km

Thus, the distance from A to C is 36 km.

**Direction (32 – 35):** **Read the given information of the paragraph carefully and answer the questions :**

Total 150 employees in four different departments, i.e. clerk, guard, cleaner, driver of a company. Total employees in guard are 33.33 % of total employees in the company and ratio of total employees in clerk to cleaner is 1 : 3. Total employees in the driver are 20% less than that of in guard.

**32. Question**

Find the difference between the total employees in guard and clerk?

(a) 45

(b) 25

(c) 35

(d) 5

(e) 15

**Answer: C**

**Explanation:**

Total employees in guard

= 150 × 1 /3

= 50

Let total employees in clerk and cleaner be x and 3x respectively

Total employees in driver = 50 × (1 − 20 /100) = 40

(50 + x + 3x + 40) = 150

4x = 60

x = 15

Department | Employees |
---|---|

Clerk | 15 |

Guard | 50 |

Cleaner | 45 |

Driver | 40 |

Total | 150 |

**(Note :- above explanation for question 32 – 35)**

Required difference

= (50 – 15)

= 35

Thus the difference between guard and clerk = 35

**33. Question**

The total number of employees in the clerk department is what percent less than that of in driver department?

(a) 55.5%

(b) 62.5%

(c) 60.5%

(d) 50%

(e) 55%

**Answer: B**

**Explanation:**

Required percentage

=( 40−15 )/40 × 100

= 62.5%

Thus, the required less percentage is 62.50%.

**34. Question**

If the ratio of male to female in the driver is 5 : 3, then find total male employees in the driver are what percent of total employees in the guard?

(a) 60%

(b) 80%

(c) 40%

(d) 30%

(e) 50%

**Answer: E**

**Explanation:**

Total male employees in Operation

= 40 × 5 /8

= 25

Required percentage

= 25 / 50 × 100

= 50%

Thus, total male employees in the driver are 50 % of total employees in guard.

**35. Question**

Find the ratio of total employees in clerk to that of in guard?

(a) 7:10

(b) 10: 3

(c) 3: 10

(d) 3: 8

(e) 3: 5

**Answer: C**

**Explanation:**

Required ratio

= 15 : 50

= 3 : 10

Thus, the ratio between the total employees in clerk and guard is = 3 : 10