# SBI PO Prelims Quantitative Aptitude Question Paper 2020

• Last Updated : 25 Mar, 2022

#### Direction (1-5): The following question contains two equations as I and II. Solve both equations and determine the relationship between them and give answers as:

a) x > y

b) x ≥ y

c) x = y or relationship can not determine.

d) x < y

e) x ≤ y

#### 1. Question:

I) x² – x – 210 = 0

II) y² – 13y – 198 = 0

Solution:

x²-x-210-0

x² – 15x + 14x-210 = 0

x(x – 15) + 14(x – 15) = 0

(x + 14)(x-15) = 0

x = -14, 15

y²-13y – 198=0

y²-22y +9y- 198 = 0

y(y-22) + 9(y-22) = 0

(y +9)(y-22) = 0

y = -9, 22

The relationship between x and y cannot be established.

#### 2. Question:

I) 4x² –  28x+ 40  = 0

II) 3y² – 9y – 30 = 0

a) x> y

b) x ≥y

c) x = y or relationship can not determine.

d) x<y

e) x ≤y

Solution:

4x² – 28x + 40 = 0

4x² – 20x -8x + 40 = 0

4x(x – 5)-8(x – 5) = 0

(4x-8)(x-5)=0

x = 2,5

3y²-9y-30=0

3y² – 15y + 6y-30 = 0

3y(y – 5) + 6(y-5) = 0

(3y + 6)(y-5) = 0

y = -2,5

Relationship between x and y cannot be established

#### 3. Question:

I) x²+ 12x +35=0

II) y² + 16y +63=0

a) x> y

b) x ≥y

c) x = y or relationship can not determine.

d) x< y

e) x ≤y

Solution:

x² + 12x + 35=0

x² + 7x + 5x + 35=0

x(x + 7) + 5(x + 7) = 0

(x + 5)(x + 7) = 0

x = -5, -7

y² + 16y +63 = 0

y² + 7y +9y+63 = 0

y(y + 7) + 9(y + 7) = 0

(y +9)(y + 7) = 0

y = -9, -7

x≥y

#### 4. Question:

I) x² – 22x + 121 = 0

II) y²- 23y + 132 = 0

a) x> y

b) x ≥ y

c) x = y or relationship can not determine.

d) x< y

e) x ≤y

Solution:

x² -22x + 121=0

x²-11x – 11x + 121=0

x(x – 11)-11(x-11)=0

(x-11)(x-11) = 0

x = 11, 11

y²- 23y + 132 = 0

y²- 12y – 11y + 132 = 0

y(y – 12)-11(y – 12) = 0

(y-11)(y- 12) = 0

y = 11, 12

x≤y

#### 5. Question:

I) x² – 45x + 506=0

II) y² – 50y + 621 = 0

a) x> y

b) x ≥ y

c) x = y or relationship can determine.

d) x< y

e) x ≤ y

Solution:

x² -45x + 506 = 0

x² -23x -22x + 506 = 0

x(x – 23)-22(x – 23) = 0

(x-22)(x-23) = 0

x = 22, 23

y²-50y + 621 0

y²-23y-27y + 621 = 0

y(y-23)-27(y-23) = 0

(y-27)(y-23) = 0

y = 27, 23

x≤y

#### 6. Question:

Train passes a platform in 48 seconds and the speed of the train is 54 km/hr. If the same train crosses another platform that is double the length of the first platform in 70 seconds, then find the length of the train?

a) 440

b) 450

c) 390

d) 340

e) 360

Solution:

54 km/hr = 54×5/18 m/s

= 15 m/s

Length of train = x

Length of platform=y

x+y = 48×15 = 720 m  ————(1)

x+2y=70×15 =1050 m ————- (2)

From (1) and (2),

y = 330

length of train (x) = (720 – 330) m

= 390 m

#### 7. Question:

Anuj and Bikram started the business with an investment in the ratio of 2:3. After 4 months, Bikram withdrew half of his initial investment and Anuj left the business. After 2 more months, Chetan joined with the investment of Rs.8000 which is Rs.2000 more than the initial investment of Bikram. At the end of the year, the difference between the profit share of Anuj and Chetan is Rs.3000, then find the total profit of the business?

a) Rs.10500

b) Rs.10400

c) Rs.9800

d) Rs.10700

e) None of these

Solution:

Investment of Bikram = (8000-2000) = 6000

Investment of Anuj = 6000×2/3 = 4000

Investment of Chetan = 8000

Profit ratio of Anuj, Bikram and Chetan

=(4000×4):(6000×4+6000/2×8):(8000×6)

=16:48:48

=1:3:3

Difference between profit of Anuj and Chetan

=(3-1)

= 2 unit

=(3000÷2) Rs.

=1500 Rs.

Total profit =7×1500 Rs

=10500 Rs.

#### 8. Question:

The cost price of Pen and Pencil is in the ratio 3:2. If the shopkeeper sold the pen at 10% profit and the pencil at 20% profit and offers a discount of 10% and 20% on the marked price of Pen and Pencil respectively, then what is the ratio of the marked price of Pen and Pencil?

a) 11:5

b) 21:14

c) 11:9

d) 19:14

e) None of these

Solution:

Pen’s cost price and market price ratio

CP(100 -D%) : MP(100+P%)

=90 : 110

=9 : 11

=3 : 11/3

Pencil’s CP: MP

=80 : 120

=2 : 3

The ratio of the market price of Pen and Pencil

=11/3 : 3

=11:9

#### 9. Question:

The ratio of the length of two rectangles is 4:5 and the breadth of these two rectangles is 3:2. If the perimeter of the second rectangle is 74 cm and the length of the second rectangle is 13 cm more than its breadth, then find the area of the first rectangle?

a) 270 cm²

b) 360 cm²

c) 480 cm²

d) 540 cm²

e) 620 cm²

Solution:

5x – 2y=13  ————–(1)

Perimeter of second rectangle=2(5x+2y)

2(5x+2y) =74

5x+2y=37  —————(2)

From (1) and (2),

x=5

y=6

Area of first rectangle

=(4x)×(3y)

=4×5×3×6

=360

#### 10. Question:

Three pipes A, B, and C. A and B are inlets and C is outlet pipe. Pipe A alone fills the tank in 20 hours and the efficiency of A is 75% of the efficiency of pipe B and the efficiency of pipe B is double of pipe C. If all the three pipes are opened simultaneously then in how many hours the tank will be completely filled?

a) 12 hours

b) 15 hours

c) 18 hours

d) 9 hours

e) 20 hours

Solution:

A=20 hours

B=20×75/100 =15 hours

C=15×2 =30 hours

[Efficiency and time are inversely proportional]

LCM of 20,15,30 is

=60

A fill the tank=60÷20 = 3 times

B fill the tank=60÷15 = 4 times

C empty the tank=60÷30= 2 times

The tank fully filled in

=60÷(7-2)

=12 hours

### Directions (11-15): Read the given information carefully and answer the questions given below.

The total income of five families A, B, C, D, and E together is Rs.70000. The ratio of the total savings of all the families together to the savings of C is 11:2. Expenditure of B is half of the savings of C. The expenditure of B is 25% of the expenditure of D and the expenditure of E is Rs.2000 more than the expenditure of C. Income of B is Rs.12000 which is Rs.5000 more than the savings of E. Savings of A is 20% more than the savings of D and the ratio of the expenditure of A to the savings of E is 4:7. Total Expenditure of all the families together is Rs.4000 more than the total savings of all the families together.

11. Question: What is the ratio of the income of D to E?

a) 2:1

b) 3:1

c) 1:2

d) 1:3

e) None of these

Solution:

Total savings = (70000- 4000)/2 = Rs.33000

Total expenditure = 33000 + 4000 = 37000

Savings of C = 2/11 * 33000 = 6000

Income of B = 12000

Expenditure of B = 6000/2 = 3000

Savings of B = 12000 – 3000 = 9000

Expenditure of D = 100/25 * 3000 = 12000

Savings of E = 12000 – 5000 = 7000

Savings of A and D = 33000 – 6000 – 9000 – 7000 = 11000

Ratio of the savings of A and D = 120:100 = 6:5

Savings of A = 6/11 * 11000 = 6000

Savings of D = 5/11 * 11000 = 5000

Expenditure of A = 4/7 * 7000 = 4000

Expenditure of C and E = 37000 – 4000 – 3000 – 12000

=18000

Expenditure of E is Rs.2000 more than the expenditure of C.

Expenditure of E = 10000

Expenditure of C = 8000

Required ratio = (5000 + 12000):(7000 + 10000)

= 1:1

12. Question: What is the average income of D, C and E?

a) Rs.12000

b) Rs.14000

c) Rs.16000

d) Rs.10000

e) Rs.18000

Solution:

Required average = {(8000 + 6000) + (10000 + 7000) + (5000 + 12000)}/3

= (14000 + 17000 + 17000)/3

= 48000/3

= 16000

13. Question: What is the difference between the Savings and Expenditures of B, C, and D together?

a) Rs.2000

b) Rs.2500

c) Rs.3000

d) Rs. 3500

e) Rs. 4000

Solution:

Savings of B, C and D =( 9000 + 6000 +5000 )

= 20000

Expenditure of B, C and D

= (3000 +8000+12000)

= 23000

Difference

= 23000 – 20000

= 3000

14. Question: The Savings of A is what per cent of the income of A?

a) 60%

b) 40%

c) 80%

d) 70%

e) 50%

Solution:

Required ratio

= 6000/10000 * 100

= 60%

15. Question: What is the ratio of the Savings to Expenditure of E?

a) 7:8

b) 7:10

c) 3:5

d) 14:19

e) 14:21

Solution:

Required ratio = 7000:10000

= 7:10

### Directions (16-20): In the following number series find out the wrong number.

16. Question: 1, 13, 49, 73, 133, 170

a) 73

b) 133

c) 13

d) 170

e) 49

Solution:

1 + 12 * 1 = 13

13 + 18 * 2 = 49

49 + 24 * 1 = 73

73 + 30 * 2 = 133

133 + 36 * 1 = 169

17. Question: 12, 32, 70, 132, 224, 364

a) 70

b) 224

c) 32

d) 132

e) 364

Solution:

2³+4=12

3³ +5 = 32

4³ +6=70

5³ + 7 = 132

6³ +8=224

7³ +9=352

18. Question: 20, 30, 40, 65, 100, 145

a) 30

b) 145

c) 40

d) 65

e) 100

Solution:

20+5=25

25+15=40

40+25= 65

65 +35= 100

100+ 45= 145

19. Question: 30, 34, 50, 86, 150, 220

a) 150

b) 34

c) 220

d) 86

e) 50

Solution:

30+22=34

34+4²=50

50 +6² = 86

86 +8²  =150

150 + 10²= 250

20. Question: 45, 91, 147, 183, 229, 275

a) 183

b) 229

c) 275

d) 147

e) 91

Solution :

(45*1) +0=45

(45*2) +1 = 91

(45*3) + 2 = 137

(45 * 4) + 3 = 183

(45 * 5) + 4 = 229

(45*6) + 5 = 275

21. Question: The ratio between the present age of Pritam and Shyam is 4:7 respectively. 27 years hence Shyam will be 12 years older than Pritam. What is the present age of Pritam?

a) 15 years

b) 5 years

c) 16 years

d) 3 years

e) 4 years

Solution :

Pritam : Shyam = 4 : 7

The age difference between Pritam and Shyam = 3 unit

=12 ÷ 3

=4 years

Present age of Pritam= 4 × 4 = 16 years

22. Question: A motorboat can row 15 km/hr in still water and finds that it takes him to double as much time to row up than as to row down the same distance in the river. The speed of the current is:-

a) 9 km/hr

b) 4 km/hr

c) 5 km/hr

d) 4.5 km/hr

e) None of these

Solution :

Let upstream speed = x km/hr

Speed downstream = 2x km/hr

Speed in still water

= 1/2(2x+x)

=3x/2  km/hr

3x/2 = 15

x = 10 km/hr

Downstream speed = 10 × 2 = 20 km/hr

Speed of current

=(20 – 10) /2

= 5  km/hr

23. Question: Amal, Bimal, and Chandu together can complete a work in 32 days and Chandu alone complete the work in 40 days. If Amal, Bimal, and Chandu started the work together and after 20 days Amal and Bimal left the work, in how many days did Chandu alone complete the remaining work?

a) 12 days

b) 20 days

c) 24 days

d) 18 days

e) 15 days

Solution:

Amal + Bimal + Chandu = 32 days

Chandu =40 days

Total work (LCM of 32 and 40) = 160

(Amal + Bimal + Chandu) ‘s work

= 160 ÷ 32

= 5 unit

Chandu work = 160 ÷ 40 = 4 unit

(Amal + Bimal + Chandu) ‘s 20 days work

= 5 × 20

= 100 unit

Remaining work

= (160 – 100)

= 60 unit

Remaining work done by Chandu

= 60 ÷ 4

= 15 days

24. Question: A company gives a 5% commission to its salesman up to the sale of 10,000 and a commission of 4% on the sales above 10,000. If the salesman deposited Rs. 31,100 in the company after deducting his commission then find the total sales.

a) Rs. 31,100

b) Rs. 32,000

c) Rs. 32500

d) Rs. 33,700

e) Rs. 34,100

Solution:

If he gets 4% commission overall,

then Company did not pay him  1%

= 10,000×1/100

=100

Total deposited money = (31, 100 + 100)

= 31,200

96% = 31,200

100% = 100 × 31200/96  = 32,500

25. Question: A boy bought two toys for Rs. 600. He sold one at a profit of 14% and another one at a loss of 14%. Each toy was sold at the same price. Find the cost price of two toys.

a) 258 , 342

b) 243 , 357

c) 295, 305

d) 250, 350

e) None of these

Solution:

CP1 × 114/100 = CP2 × 84/100

CP1/ = 43/57

CP1 + CP2 = (43+57) =100

CP1 + CP2 = 600 ÷ 100 = 6

Cost price of first toy = 43 × 6 = 258

Cost price of 2nd toy = 57 × 6 = 342

### Directions (26-30): Read the given information carefully and answer the questions given below. The given table shows the number of students studying in class I, class II and class III in five different schools. (some data is missing)

26. Question: If the average number of students studying in B is 105 and the ratio of the number of students studying in class I and class III in B is 3:4, then find the difference between the number of students studying in class I and class II in B?

a) 10

b) 15

c) 20

d) 25

e) 30

Solution :

Total number of students in B = 105 * 3 = 315

Number of students studying in class I in B

=  3/7 * (315 – 105)

= 90

Difference = 105 – 90 = 15

27. Question: If the total number of students studying in A and C is 200 and 300 respectively, then find the average number of students studying in class II in all the schools together?

a) 76

b) 78

c) 80

d) 82

e) 84

Solution :

Number of students studying in class II in A

= 200– (60 + 60)

= 80

Number of students studying in class II in C

= 300 –(150 + 75)

= 75

Required average

= (80 + 75 + 105 + 90 + 60)/5

= 410/5

= 82

28. Question: The ratio of the number of students studying in class I in D to the number of students studying in class III in E is 6:7. If the number of students studying in class I in D is 20% more than the number of students studying in class III in C, then what is the difference between the total number of students studying in D and E?

a) 40

b) 30

d) 60

d) 20

e) 50

Solution :

Number of students studying in class I in D

= 120/100* 75

= 90

Number of students studying in class III in E

= 7/6 * 90

= 105

Required difference

= (105 + 105 + 60) – (90 + 90 + 50)

= 270 – 230

= 40

29. Question: The average number of students studying in class I in all the schools together is 103 and the average number of students studying in class III in all the schools together is 75. What is the ratio of the number of students studying in class I in B and D together to the number of students studying in class III in B and E together?

a) 10:9

b) 1:1

c) 5:3

d) 4:3

e) None of these

Solution :

Number of students studying in class I in B and D

=103 * 5 – (60 + 150 + 105)

= 200

Number of students studying in class III in B and E

=75 * 5 – (60 + 75 + 50)

= 190

Required ratio = 200:190

= 20:19

30. Question: The number of students studying in class II in D and E together is what per cent of the number of students studying in class I in A and C together?

a) 71.42%

b) 75.62%

c) 69.02%

d) 65.42%

e) 78.92%

Solution :

Required percentage

= (90 + 60)/(150 + 60) * 100

= (150/210) * 100

= 71.42%

### Directions (31-35): Read the given information carefully and answer the questions which are given below.

The given line graph shows the percentage of employees in the two departments in four different shopping malls. The given table shows the total number of employees (garments, groceries, and accessories) and the ratio of the number of boys to girls in four different shopping malls.

31. Question: What is the ratio of the number of employees in garment and grocery departments in A to the number of employees in garment and accessory departments in C?

a) 5:6

b) 6:7

c) 3:5

d) 9:10

e) None of these

Solution (31-35):

Required ratio

= (480 + 420):(225 + 825)

= 900:1050

= 6:7

32. Question: In B, If 25% of boys are from the grocery department and 30% of boys are from the accessory department, then what is the difference between the number of girls in the garment and grocery department in B?

a) 100

b) 110

c) 120

d) 130

e) 150

Solution:

Number of boys in grocery

= 25/100 * 600

=150

Number of girls in grocery

= 400 – 150

= 250

Number of boys in garment

= 600 * (45/100)

= 270

Number of girls in garment

= 400 – 270

= 130

Difference

= 250 – 130

= 120

33. Question: If the number of boys in the garment department in A is 30% of the total number of boys in A and the number of boys in the grocery department in D is 20% more than the number of boys in the garment department in A, then find the number of girls in grocery department in D?

a) 486

b) 492

c) 504

d) 516

e) 518

Solution :

Number of boys in garment in A

= 30/100 * 900

= 270

Number of boys in grocery in D

=120/100 * 270

=  324

Number of girls in grocery in D

= 810 – 324

= 486

34. Question: The ratio of the number of girls in grocery to garment department in B is 3:2 and the ratio of the number of boys in grocery to garment department in B is 5:6, then the number of boys in grocery and garment department together in B is what per cent of the total number of boys in D?

a) 45%

b) 50%

c) 60%

d) 55%

e) Cannot be determined

Solution :

Number of girls in grocery in B = 3x

Number of girls in garment in B = 2x

Number of boys in grocery in B = 5y

Number of boys in garment in B = 6y

3x + 5y = 400—-(1)

2x + 6y = 400

x + 3y = 200 —–(2)

From (1) and (2)

4y = 200

y = 50

x = 50

Number of boys in grocery and garment in B

=11 *  50

= 550

Required percentage

= 550/1000 * 100

= 55%

35. Question: What is the average number of girls in all the shopping malls together?

a) 525

b) 545

c) 555

d) 565

e) 575

Solution :

Required average

= (300 + 400 + 800 + 800) ÷ 4

= 2300 ÷ 4

= 575

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