SBI Clerk · Topic Collection

Quantitative Aptitude PYQs

Arithmetic, number systems, percentages, ratios, profit and loss, time and work, algebra, geometry, mensuration, and trigonometry.

1495 Total Questions38 Official PapersYears: 2018–2025Search in Quantitative Aptitude

Showing questions 721–740 of 1495Page 37 of 75

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 721

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

A box contains 10 reds, 9 green and some blue balls. Find the probability of getting three different coloured balls when 3 balls are drawn from box at random if total number of balls in the box is 36.

AOption (a)
1/15
BOption (b)
2/7
COption (c)
9/14
DOption (d)
3/14
EOption (e)
None of these

View answer & explanation

Correct answer

3/14

Why this answer is correct

Shortcut Trick

Total balls = 36 | Red = 10, Green = 9, Blue = 36 − (10 + 9) = 17.

Probability of 3 different colors = (10C₁ × 9C₁ × 17C₁) ÷ 36C₃.

⇒ (10 × 9 × 17) ÷ [(36 × 35 × 34) ÷ (3 × 2 × 1)] = 1530 ÷ 7140 = 153/714.

⇒ 153/714 = 3/14 (dividing by 51).

∴ The correct answer is 3/14.

Alternate Method

Given: Red balls = 10, Green balls = 9, Total balls = 36.

Step 1: Calculate number of Blue balls.

⇒ Blue balls = 36 − (10 + 9) = 17.

Step 2: Total ways to draw 3 balls (n(S)).

⇒ 36C₃ = (36 × 35 × 34) ÷ (3 × 2 × 1) = 6 × 35 × 34 = 7140.

Step 3: Favorable ways (n(E)) to get one of each color.

⇒ 10C₁ × 9C₁ × 17C₁ = 10 × 9 × 17 = 1530.

Step 4: Probability P(E) = n(E) / n(S).

⇒ P(E) = 1530 ÷ 7140 = 153/714 = 3/14.

∴ The correct answer is 3/14.

Additional Information

Combinations Formula

Selection of r items from n is nC_r = n! / [r! × (n−r)!]. Useful for drawing items simultaneously.

Probability Basics

P(Event) = (Number of favorable outcomes) ÷ (Total number of outcomes). Range is always 0 to 1.

AND vs OR Rule

In probability, AND signifies multiplication (independent events), while OR signifies addition (mutually exclusive events).

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 722

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Directions: The bar graph shows the total number of calls (Inbound and Outbound) from six different call centers.

Note:

1).Inbound calls = International (IB) + Domestic (IB)

2). Outbound calls = International (OB) + Domestic (OB)

The given table shows the ratio between inbound calls and outbound calls

Call centres	Ratio between Inbound calls to Outbound calls
A	11: 5
B	4: a
C	19: 5
D	-
E	5: 6
F	3: 5

If the number of males to females bounded the inbound calls in C are in the ratio 10: 9. And the total number of males bounded the all the calls in C is 70.Find the difference between the number of females bounded the inbound calls and outbound calls in the centre C.

AOption (a)
50
BOption (b)
40
COption (c)
30
DOption (d)
35
EOption (e)
45

View answer & explanation

Correct answer

Why this answer is correct

General solution:

Call centre A:

Total calls in centre A = 80

Number of Inbound calls = 80/16 × 11 = 55

Number of outbound calls = (80 - 55) = 25

Call centre C:

Total calls in centre C = 120

Number of Inbound calls = 120/24 × 19 = 95

Number of outbound calls = (120 - 95) = 25

Call centre E:

Total calls in centre E = 220

Number of Inbound calls = 220/11 × 5 = 100

Number of outbound calls = (220 - 100) = 120

Call centre F:

Total calls in centre F = 160

Number of Inbound calls = 160/8 × 3 = 60

Number of outbound calls = (160 - 60) = 100

Solution:

Number of inbound calls in C = 95

Number of males bounded inbound calls = 95/19 × 10 = 50

Number of females bounded inbound calls = (95 - 50) =45

Number of males bounded outbound calls =(70 - 50) = 20

Number of outbound calls = 25

Number of females bounded outbound calls = 25 - 20 = 5

Required difference = (45 - 5) = 40

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 723

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Directions: The bar graph shows the total number of calls (Inbound and Outbound) from six different call centers.

Note:

1).Inbound calls = International (IB) + Domestic (IB)

2). Outbound calls = International (OB) + Domestic (OB)

The given table shows the ratio between inbound calls and outbound calls

Call centres	Ratio between Inbound calls to Outbound calls
A	11: 5
B	4: a
C	19: 5
D	-
E	5: 6
F	3: 5

Inbound and Outbound Calls of D are in the ratio 23:7. Then Inbound Calls of D are (Y + 3)% more than total calls made by B. If the ratio of domestic inbound calls and International inbound calls of C is (Y + 4): 3. Find International Inbound Calls of C.

AOption (a)
25
BOption (b)
15
COption (c)
18
DOption (d)
20
EOption (e)
16

View answer & explanation

Correct answer

Why this answer is correct

General solution:

Call centre A:

Total calls in centre A = 80

Number of Inbound calls = 80/16 × 11 = 55

Number of outbound calls = (80 - 55) = 25

Call centre C:

Total calls in centre C = 120

Number of Inbound calls = 120/24 × 19 = 95

Number of outbound calls = (120 - 95) = 25

Call centre E:

Total calls in centre E = 220

Number of Inbound calls = 220/11 × 5 = 100

Number of outbound calls = (220 - 100) = 120

Call centre F:

Total calls in centre F = 160

Number of Inbound calls = 160/8 × 3 = 60

Number of outbound calls = (160 - 60) = 100

Solution:

Total calls bound by D= 150

Number of Inbound calls by D = 150/30 × 23 = 115

Number of outbound calls by D = (150 - 115) = 35

Total number of calls made by B = 100

Number of outbound calls made by D is (115 - 100)/100 × 100 = 15%

(Y + 3)% = 15%

Y = 12

The ratio of domestic inbound calls to International inbound calls of C is (Y + 4): 3

(12 + 4): 3 = 16: 3

Total Inbound calls by C = 95

Number of international inbound calls = 95/19 × 3 = 15

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 724

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Directions: The bar graph shows the total number of calls (Inbound and Outbound) from six different call centers.

Note:

1).Inbound calls = International (IB) + Domestic (IB)

2). Outbound calls = International (OB) + Domestic (OB)

The given table shows the ratio between inbound calls and outbound calls

Call centres	Ratio between Inbound calls to Outbound calls
A	11: 5
B	4: a
C	19: 5
D	-
E	5: 6
F	3: 5

The total internationals by A is 41 and the ratio of Inbound domestic calls and outbound domestic calls are in the ratio 4.5: 2.Find the ratio of inbound domestic calls by A and outbound calls by E.

AOption (a)
23: 95
BOption (b)
17: 86
COption (c)
27: 120
DOption (d)
32: 99
EOption (e)
24: 89

View answer & explanation

Correct answer

27: 120

Why this answer is correct

General solution:

Call centre A:

Total calls in centre A = 80

Number of Inbound calls = 80/16 × 11 = 55

Number of outbound calls = (80 - 55) = 25

Call centre C:

Total calls in centre C = 120

Number of Inbound calls = 120/24 × 19 = 95

Number of outbound calls = (120 - 95) = 25

Call centre E:

Total calls in centre E = 220

Number of Inbound calls = 220/11 × 5 = 100

Number of outbound calls = (220 - 100) = 120

Call centre F:

Total calls in centre F = 160

Number of Inbound calls = 160/8 × 3 = 60

Number of outbound calls = (160 - 60) = 100

Solution:

Total international calls by centre A = 41

Total domestic calls b centre A = (80 - 41) = 39

Ratio between the Inbound domestic calls and outbound domestic calls = 9: 4

Inbound domestic calls = 39/13 × 9 = 27

Outbound domestic calls = 39 - 27 = 12

Number of outbound calls by E = 120

Required ratio = 27: 120

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 725

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Directions: The bar graph shows the total number of calls (Inbound and Outbound) from six different call centers.

Note:

1).Inbound calls = International (IB) + Domestic (IB)

2). Outbound calls = International (OB) + Domestic (OB)

The given table shows the ratio between inbound calls and outbound calls

Call centres	Ratio between Inbound calls to Outbound calls
A	11: 5
B	4: a
C	19: 5
D	-
E	5: 6
F	3: 5

Total International Calls (Inbound and Outbound) by B and Total International Calls (Inbound and Outbound) by D are in the ratio 26:45 and Total Domestic calls (Inbound and Outbound) by B are 20% Less than Total Domestic Calls (Inbound and Outbound) by D. Find the Difference between International Calls (Inbound and Outbound) by B and Total Inbound Calls (International and Domestic) by F.

AOption (a)
7
BOption (b)
6
COption (c)
5
DOption (d)
8
EOption (e)
4

View answer & explanation

Correct answer

Why this answer is correct

General solution:

Call centre A:

Total calls in centre A = 80

Number of Inbound calls = 80/16 × 11 = 55

Number of outbound calls = (80 - 55) = 25

Call centre C:

Total calls in centre C = 120

Number of Inbound calls = 120/24 × 19 = 95

Number of outbound calls = (120 - 95) = 25

Call centre E:

Total calls in centre E = 220

Number of Inbound calls = 220/11 × 5 = 100

Number of outbound calls = (220 - 100) = 120

Call centre F:

Total calls in centre F = 160

Number of Inbound calls = 160/8 × 3 = 60

Number of outbound calls = (160 - 60) = 100

Solution:

Total international calls by bound by B = 26a

Total international calls by bound by D = 45a

Total domestic calls by B = 4b

Total domestic calls by D = 5b

Total calls by B = 100

Total calls by D = 150

26a + 4b = 100 —----> 1

45a + 5b = 150 —-----> 2

Multiply equation 1 by 5 and equation 4 by 2

130a + 20b = 500

180a + 20b = 600

Subtracting equation 1from 2

-50a = - 100

a = 2

International calls bound by B = 26 × 2 = 52

Total Inbound calls bound by F = 60

Required difference = 60 - 52 = 8

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 726

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Directions: The bar graph shows the total number of calls (Inbound and Outbound) from six different call centers.

Note:

1).Inbound calls = International (IB) + Domestic (IB)

2). Outbound calls = International (OB) + Domestic (OB)

The given table shows the ratio between inbound calls and outbound calls

Call centres	Ratio between Inbound calls to Outbound calls
A	11: 5
B	4: a
C	19: 5
D	-
E	5: 6
F	3: 5

If total outbound calls by B are 33.33% more than (K+3). Where K is the value of Inbound Calls by F. The Find the value of b, if b is the smaller root of equation; m² - am + 108 = 0

AOption (a)
11
BOption (b)
8
COption (c)
9
DOption (d)
4
EOption (e)
5

View answer & explanation

Correct answer

Why this answer is correct

General solution:

Call centre A:

Total calls in centre A = 80

Number of Inbound calls = 80/16 × 11 = 55

Number of outbound calls = (80 - 55) = 25

Call centre C:

Total calls in centre C = 120

Number of Inbound calls = 120/24 × 19 = 95

Number of outbound calls = (120 - 95) = 25

Call centre E:

Total calls in centre E = 220

Number of Inbound calls = 220/11 × 5 = 100

Number of outbound calls = (220 - 100) = 120

Call centre F:

Total calls in centre F = 160

Number of Inbound calls = 160/8 × 3 = 60

Number of outbound calls = (160 - 60) = 100

Solution:

Inbound calls by F = 60

Total outbound calls by B = (60 + 3) × 4/3 = 84

Total calls bound by B = 100

Number of inbound to outbound calls in B are in the ratio 4: a.

Inbound calls by B = 100 - 84 = 16

16: 84

4: 21

a = 21

To find value of b smaller root of the equation

m² - 21m + 108 = 0

m² - 12m - 9m + 108 = 0

m ( m - 12) -9 (m - 12) = 0

(m - 9) (m - 12) = 0

m = 12, 9

b = 9

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 727

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

The monthly income of 'X' and 'Y' are in the ratio of 1 : 3 respectively. 'X' spends Z% of his monthly income and saves the remaining 12480 Rs. The sum of the monthly expenditure of 'X' and 'Y' is 59020 Rs. If the monthly saving and expenditure of 'Y' is in the ratio of 5 : 7 respectively then find the average monthly income of 'X' and 'Y' together?

AOption (a)
52000 Rs.
BOption (b)
54000 Rs.
COption (c)
56000 Rs.
DOption (d)
58000 Rs.
EOption (e)
60000 Rs.

View answer & explanation

Correct answer

52000 Rs.

Why this answer is correct

Shortcut Trick

Ratio of Income X : Y = 1 : 3.

Ratio of Saving : Expenditure for Y = 5 : 7 ⇒ Total Income of Y = 12 units.

Since 1 unit for X corresponds to 12 units for Y (to match 1:3 ratio), X income = 4 units.

Expenditure of Y = (7/12) × 3 × X_income. Let X_income = I. Exp_Y = 1.75I.

Total Exp = Exp_X + Exp_Y = 59020. Savings_X = 12480. Income_X = Exp_X + 12480.

Income_X + 1.75(Income_X) = 59020 + 12480 ⇒ 2.75 × Income_X = 71500 ⇒ Income_X = 26000.

∴ The correct answer is 52000 Rs.

Alternate Method

Given: Ratio of Income (X:Y) = 1:3, Savings of X = 12480 Rs, Exp_X + Exp_Y = 59020 Rs, Ratio Y_savings : Y_exp = 5:7.

⇒ Let Income of X = I and Income of Y = 3I.

⇒ For Y: Savings + Expenditure = Income ⇒ 5a + 7a = 3I ⇒ 12a = 3I ⇒ a = I/4.

⇒ Expenditure of Y = 7a = 7I/4 = 1.75I.

⇒ Expenditure of X = Income_X − Savings_X = I − 12480.

⇒ (I − 12480) + 1.75I = 59020.

⇒ 2.75I = 59020 + 12480 = 71500.

⇒ I = 71500 ÷ 2.75 = 26000 Rs.

⇒ Total Income = I + 3I = 4I = 4 × 26000 = 104000 Rs.

⇒ Average Income = 104000 ÷ 2 = 52000 Rs.

∴ The correct answer is 52000 Rs.

Additional Information

Income, Expenditure, and Savings Relationship

The fundamental equation is Income = Expenditure + Savings. When ratios are given, always convert them to a common variable to solve linear equations.

Average Calculation

Average of two quantities A and B is calculated as (A + B) ÷ 2. In ratio terms, if A:B is 1:3, average is (1+3)/2 = 2 units.

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 728

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Anuj and Bhanu started a business with the investment of Rs. 15000 and Rs 18000 respectively, after 6 months Anuj withdrawn 20% of his initial investment and Chandan started the business with the investment of Rs.

X. At the end of the year, the total profit of the business is Rs 66000 and profit share of Bhanu is Rs 24000 and find the value of X ?

AOption (a)
38000
BOption (b)
36000
COption (c)
40000
DOption (d)
28000
EOption (e)
42000

View answer & explanation

Correct answer

36000

Why this answer is correct

Shortcut Trick

Ratio of Bhanu profit to Total profit = 24000 ÷ 66000 = 4 ÷ 11.

Bhanu investment factor = 18000 × 12 = 216000.

Total investment factor = (216000 × 11) ÷ 4 = 594000.

Anuj investment factor = (15000 × 6) + (12000 × 6) = 162000.

Chandan investment factor = 594000 − 216000 − 162000 = 216000.

Since Chandan invested for 6 months, 6X = 216000 ⇒ X = 36000.

∴ The correct answer is Rs. 36000.

Alternate Method

Given → Initial Investments: Anuj = 15000, Bhanu = 18000. Total Profit = 66000, Bhanu Share = 24000.
Formulas: Profit Ratio = (Investment₁ × Time₁) : (Investment₂ × Time₂).

⇒ Anuj Profit Share = (15000 × 6) + (15000 × 80/100 × 6) = 90000 + 72000 = 162000.
⇒ Bhanu Profit Share = 18000 × 12 = 216000.
⇒ Chandan Profit Share = X × 6 = 6X.
⇒ Ratio of profits = 162000 : 216000 : 6X = 27000 : 36000 :

X. ⇒ Ratio of Bhanu profit to Total profit = 36000 ÷ (27000 + 36000 + X) = 24000 ÷ 66000.
⇒ 36000 ÷ (63000 + X) = 4 ÷ 11.
⇒ 9000 × 11 = 63000 + X ⇒ 99000 = 63000 + X.
⇒ X = 36000.

∴ The correct answer is 36000.

Additional Information

Partnership Basics

Profit is distributed based on the product of investment and time period. Profit = Investment × Time.

Variable Investment

If investment changes, total profit factor = ∑ (Investment_i × Time_i).

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 729

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

In this question, two equations are given, Answer the questions based on given equations:

I. x2 - 10x + 21 = 0

II. x2 - 24x + 144 = 0

If one of the roots of given equation is taken from both equation, a new equation is formed which is :

AOption (a)
x2 + 12x + 32 = 0
BOption (b)
x2 - 15x + 36 = 0
COption (c)
x2 -15x - 36 = 0
DOption (d)
x2 - 12x + 32 = 0
EOption (e)
x2 -16x + 63 = 0

View answer & explanation

Correct answer

x2 - 15x + 36 = 0

Why this answer is correct

Solution:

I. x2 - 10x + 21 = 0

x2 - 7x - 3x + 21 = 0

x(x - 7) - 3(x - 7) = 0

(x - 3) (x - 7) = 0

x = 3 and 7

II. x2 - 24x + 144 = 0

x2 - 12x - 12x + 144 = 0

x(x - 12) - 12(x -12) = 0

(x -12) (x -12) = 0

x = 12, 12

Solution:

Equation 1:

x2 + 12x + 32 = 0

x2 + 8x + 4x + 32 = 0

x(x + 8) + 4(x + 8) = 0

Roots of this equation = -8, -4. (which are not the roots of given equations)

Equation 2:

x2 - 15x + 36 = 0

x2 - 12x - 3x + 36 = 0

x(x- 12) - 3(x - 12) = 0

Roots of this equation = 3, 12. (which are the roots of given equations)

Similarly if we checking for all the equations we get the roots but only option 2 equation is formed by using the given equations roots.

Hence, option(2) is correct.

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 730

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

What approximate value should come in place of the question mark (?) in the following question?

? x 7.19 + 45.46% of 88.18 = 31.19 x 3.90

AOption (a)
8
BOption (b)
12
COption (c)
16
DOption (d)
20
EOption (e)
24

View answer & explanation

Correct answer

Why this answer is correct

Solution :

Concept used :

Follow the BODMAS rule to solve this question, as per the question is given below:

Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated.

⇒ ? x 7.19 + 45.46% of 88.18 = 31.19 x 3.90

⇒ ? x 7 + 45/100 x 88 = 31 x 4

⇒ ? x 7 + 40 = 124 (39.6 ≈ 40)

⇒ ? = (124 - 40)/7

⇒ ? = 84/7 = 12

Hence, the correct answer is option 2.

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 731

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

What approximate value should come in the place of the question mark (?) in the following equation?

(24.98 × 3.97) ÷ 4.99 + 12.95 + (8.02)² = ?²

AOption (a)
10
BOption (b)
8
COption (c)
9
DOption (d)
5
EOption (e)
7

View answer & explanation

Correct answer

Why this answer is correct

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

(24.98 × 3.97) ÷ 4.99 + 12.95 + (8.02)2 = ?²

⇒ (25 × 4) ÷ 5 + 13 + (8)2 = ?2

⇒ 100 ÷ 5 + 13 + (8)2 = ?2

⇒ 20 + 13 + 64 = ?2

⇒ 97 = ?²

⇒ 10

Hence, 10 will come in the place of (?).

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 732

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Read the following information carefully and answer the questions given below.

The marked price of an article B and D is 50% above its cost price and the selling price of article A is half of the marked price of article B. The ratio of the marked price of article C and the selling price of article A is 7:3. The sum of the marked price of the article C and D is Rs.810. The profit of article B is Rs.194 less than the cost price of D. The discount offered on article B is 15%.

If the Cost price of the article D is increased by 25% and the discount percentage of the article D is 20% then find which of the following is true in the given following statements?

I. New cost price of the article D is Rs. 325

II. Loss of Rs. 13 occurred

III. Selling price is Rs. 332

AOption (a)
I and II only
BOption (b)
I only
COption (c)
II only
DOption (d)
III only
EOption (e)
I and III only

View answer & explanation

Correct answer

I and II only

Why this answer is correct

General solution:

Let the cost price of the article B = 40x

Marked price of the article B = 40x × 150/100 = 60x

Selling price of the article B = 60x × ( 85/100) = 51x

Discount of the article = (60x - 51x) = 9x

Profit of the article = (51x - 40x) = 11x

Selling price of the article A = 60x/2 = 30x

Marked price of the article C = 30x/3 × 7 = 70x

Marked price of the article D = (810 - 70x)

Cost price of the article D = (11x + 194)

Marked price of article D is 50% more than the cost price of the article D

(810 - 70x)/(11x + 194) = 3/2

(1620 - 140x) = (33x + 582)

(173x = 1038)

x = 6

Let the cost price of the article B = 40x = 40 × 6 = 240

Marked price of the article B = 40x × 150/100 = 60x = 60 × 6 = 360

Selling price of the article B = 60x × ( 85/100) = 51x = 51 × 6 = 306

Discount of the article = (60x - 51x) = 9x = 9 × 6 = 54

Profit of the article = (51x - 40x) = 11x = 11 × 6 = 66

Selling price of the article A = 60x/2 = 30x = 30 × 6 = 180

Marked price of the article C = 30x/3 × 7 = 70x = 70 × 6 = 420

Marked price of the article D = (810 - 70x) = (810 - 70 × 6) = 390

Cost price of the article D = (11x + 194) = (66 + 194) = 260

Solution:

Cost price of the article D = Rs. 260

New cost price of the article D = 260 × (125/100) = 325

Marked price of the article D = Rs. 390

Selling price of the article D = 390 × (80/100) = Rs. 312

I) New Cost price is Rs. 325

II) Loss occurred = (325 - 312) = Rs. 13

III) New selling price = Rs. 312

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 733

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Read the following information carefully and answer the questions given below.

The profit obtained on article C is Rs.36 and the discount offered on article C is 20%. Find the difference between the selling price of article A and the cost price of article C.

AOption (a)
Rs. 110
BOption (b)
Rs. 120
COption (c)
Rs. 135
DOption (d)
Rs. 125
EOption (e)
Rs.140

View answer & explanation

Correct answer

Rs. 120

Why this answer is correct

General solution:

Let the cost price of the article B = 40x

Marked price of the article B = 40x × 150/100 = 60x

Selling price of the article B = 60x × ( 85/100) = 51x

Discount of the article = (60x - 51x) = 9x

Profit of the article = (51x - 40x) = 11x

Selling price of the article A = 60x/2 = 30x

Marked price of the article C = 30x/3 × 7 = 70x

Marked price of the article D = (810 - 70x)

Cost price of the article D = (11x + 194)

Marked price of article D is 50% more than the cost price of the article D

(810 - 70x)/(11x + 194) = 3/2

(1620 - 140x) = (33x + 582)

(173x = 1038)

x = 6

Let the cost price of the article B = 40x = 40 × 6 = 240

Marked price of the article B = 40x × 150/100 = 60x = 60 × 6 = 360

Selling price of the article B = 60x × ( 85/100) = 51x = 51 × 6 = 306

Discount of the article = (60x - 51x) = 9x = 9 × 6 = 54

Profit of the article = (51x - 40x) = 11x = 11 × 6 = 66

Selling price of the article A = 60x/2 = 30x = 30 × 6 = 180

Marked price of the article C = 30x/3 × 7 = 70x = 70 × 6 = 420

Marked price of the article D = (810 - 70x) = (810 - 70 × 6) = 390

Cost price of the article D = (11x + 194) = (66 + 194) = 260

Solution:

Marked price of the article C = 420

Selling price of the article C =420 × (80/100) = Rs. 336

Cost price of article C = (336 - 36) = Rs. 300

Selling price of article A = 180

Required difference = (300 - 180) = Rs. 120

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 734

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Read the following information carefully and answer the questions given below.

Selling price of article A to the selling price of article E are in the ratio 3: 5. The Marked price of article E is Rs. 50 more than its selling price. If the cost price of the article E is 10 less than the cost price of the article D. If A is sold at 20% profit. Find sum of the cost price of article E, Marked price of article E and Cost price of article A.

AOption (a)
Rs. 750
BOption (b)
Rs. 850
COption (c)
Rs. 900
DOption (d)
Rs. 650
EOption (e)
Rs. 550

View answer & explanation

Correct answer

Rs. 750

Why this answer is correct

General solution:

Let the cost price of the article B = 40x

Marked price of the article B = 40x × 150/100 = 60x

Selling price of the article B = 60x × ( 85/100) = 51x

Discount of the article = (60x - 51x) = 9x

Profit of the article = (51x - 40x) = 11x

Selling price of the article A = 60x/2 = 30x

Marked price of the article C = 30x/3 × 7 = 70x

Marked price of the article D = (810 - 70x)

Cost price of the article D = (11x + 194)

Marked price of article D is 50% more than the cost price of the article D

(810 - 70x)/(11x + 194) = 3/2

(1620 - 140x) = (33x + 582)

(173x = 1038)

x = 6

Let the cost price of the article B = 40x = 40 × 6 = 240

Marked price of the article B = 40x × 150/100 = 60x = 60 × 6 = 360

Selling price of the article B = 60x × ( 85/100) = 51x = 51 × 6 = 306

Discount of the article = (60x - 51x) = 9x = 9 × 6 = 54

Profit of the article = (51x - 40x) = 11x = 11 × 6 = 66

Selling price of the article A = 60x/2 = 30x = 30 × 6 = 180

Marked price of the article C = 30x/3 × 7 = 70x = 70 × 6 = 420

Marked price of the article D = (810 - 70x) = (810 - 70 × 6) = 390

Cost price of the article D = (11x + 194) = (66 + 194) = 260

Solution:

Selling price of the article A = 180

Selling price of the article E = 180/3 × 5 = 300

Marked price of the article E = (300 + 50) = Rs. 350

Cost price of article E = (260 - 10) = Rs. 250

Cost price of the article A = 180/120 * 100 = Rs.150

Required sum = (250 + 350 + 150) = Rs. 750

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 735

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

Read the following information carefully and answer the questions given below.

If the profit obtained on article D is Rs.25 less than the profit obtained on article B, then the discount offered on article D is

AOption (a)
14.67%
BOption (b)
26.89%
COption (c)
22.82%
DOption (d)
11.25%
EOption (e)
13.24%

View answer & explanation

Correct answer

22.82%

Why this answer is correct

General solution:

Let the cost price of the article B = 40x

Marked price of the article B = 40x × 150/100 = 60x

Selling price of the article B = 60x × ( 85/100) = 51x

Discount of the article = (60x - 51x) = 9x

Profit of the article = (51x - 40x) = 11x

Selling price of the article A = 60x/2 = 30x

Marked price of the article C = 30x/3 × 7 = 70x

Marked price of the article D = (810 - 70x)

Cost price of the article D = (11x + 194)

Marked price of article D is 50% more than the cost price of the article D

(810 - 70x)/(11x + 194) = 3/2

(1620 - 140x) = (33x + 582)

(173x = 1038)

x = 6

Let the cost price of the article B = 40x = 40 × 6 = 240

Marked price of the article B = 40x × 150/100 = 60x = 60 × 6 = 360

Selling price of the article B = 60x × ( 85/100) = 51x = 51 × 6 = 306

Discount of the article = (60x - 51x) = 9x = 9 × 6 = 54

Profit of the article = (51x - 40x) = 11x = 11 × 6 = 66

Selling price of the article A = 60x/2 = 30x = 30 × 6 = 180

Marked price of the article C = 30x/3 × 7 = 70x = 70 × 6 = 420

Marked price of the article D = (810 - 70x) = (810 - 70 × 6) = 390

Cost price of the article D = (11x + 194) = (66 + 194) = 260

Solution:

Selling price of article B = 306

Cost price of article B = 240

Profit of article B = (306 - 240) = Rs.66

Profit of article D = (66 - 25) = Rs.41

Selling price of article D = (260 + 41) = 301

Discount percentage = (89)/390 × 100 = 22.82%

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 736

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

The ratio of the cost price of two tables is 5 : 9 one table, in which cost price is less, is sold at profit of 24% and the other one is sold for Rs 2196 more than that of the first one, if the overall profit earned after selling both the tables at 6 %. What is the difference of the CP of the two table ?

AOption (a)
3150
BOption (b)
3600
COption (c)
1800
DOption (d)
2700
EOption (e)
4500

View answer & explanation

Correct answer

3600

Why this answer is correct

Shortcut Trick

Ratio of CP₁ : CP₂ = 5 : 9. Let CP₁ = 500 units and CP₂ = 900 units. Total CP = 1400 units.

SP₁ = 500 × 124% = 620 units. Total SP = 1400 × 106% = 1484 units.

SP₂ = Total SP − SP₁ = 1484 − 620 = 864 units.

Difference in SPs (SP₂ − SP₁) = 864 − 620 = 244 units.

Given 244 units = Rs 2196 ⇒ 1 unit = 2196 ÷ 244 = 9.

Difference in CP = (9 − 5) × 100 units = 400 units = 400 × 9 = Rs 3600.

∴ The correct answer is Rs 3600.

Alternate Method

Given → CP Ratio = 5 : 9, Profit₁ = 24%, Total Profit = 6%, SP₂ − SP₁ = Rs 2196.
Formulas: Profit = SP − CP | SP = CP × (100 + P%) ÷ 100.

⇒ Let Cost Price of the two tables be 5x and 9x respectively.
⇒ Total Cost Price = 5x + 9x = 14x.
⇒ Selling Price of first table (SP₁) = 5x × 1.24 = 6.2x.
⇒ Total Selling Price = 14x × 1.06 = 14.84x.
⇒ Selling Price of second table (SP₂) = 14.84x − 6.2x = 8.64x.
⇒ Difference in Selling Prices = SP₂ − SP₁ = 8.64x − 6.2x = 2.44x.
⇒ 2.44x = 2196 ⇒ x = 2196 ÷ 2.44 = 900.
⇒ Difference in CP = 9x − 5x = 4x.
⇒ Difference in CP = 4 × 900 = 3600.

∴ The correct answer is 3600.

Additional Information

Weighted Average Profit

Total Profit% = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂), where n represents the ratio of quantities/costs and p represents profit percentages.

Profit and Selling Price

When profit percentage is given, SP = CP × (1 + P/100). The difference in SP depends on both the difference in CP and the respective profit margins.

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 737

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

The ratio between the radius of the circle and the length of the rectangle is 7: 22 and the breadth of the rectangle is equal to the base of the triangle. The area and height of the triangle is 2800 cm² and 80 cm. The perimeter of the rectangle is 580 cm, then, find the perimeter of the circle?

AOption (a)
320 cm
BOption (b)
440 cm
COption (c)
550 cm
DOption (d)
650 cm
EOption (e)
None of these

View answer & explanation

Correct answer

440 cm

Why this answer is correct

Shortcut Trick

Find Triangle Base (Rectangle Breadth) using Area = 1/2 × b × h ⇒ 2800 = 0.5 × b × 80 ⇒ Breadth = 70 cm.

Use Perimeter to find Length ⇒ 2(L + 70) = 580 ⇒ L + 70 = 290 ⇒ Length = 220 cm.

Apply Ratio for Radius ⇒ 22 units = 220 cm ⇒ 7 units = 70 cm (Radius).

Perimeter of Circle = 2 × (22/7) × 70 = 440 cm.

∴ The correct answer is 440 cm.

Alternate Method

Given: Radius : Length = 7 : 22; Breadth = Triangle Base; Area (Triangle) = 2800 cm²; Height (Triangle) = 80 cm; Perimeter (Rectangle) = 580 cm.

Formula: Area (Triangle) = 1/2 × Base × Height; Perimeter (Rectangle) = 2(L + B); Perimeter (Circle) = 2πr.

⇒ Area of Triangle = 1/2 × Base × 80 = 2800

⇒ 40 × Base = 2800 ⇒ Base = 70 cm

⇒ Breadth of Rectangle (B) = 70 cm

⇒ 2 × (Length + 70) = 580

⇒ Length + 70 = 290 ⇒ Length (L) = 220 cm

⇒ Radius of circle (r) = (Length / 22) × 7 = (220 / 22) × 7 = 70 cm

⇒ Perimeter of Circle = 2 × (22/7) × 70 = 440 cm

∴ The correct answer is 440 cm.

Additional Information

Circle Properties

Area of circle = πr² and Perimeter (Circumference) = 2πr.

Rectangle Formulas

Area = Length × Breadth and Diagonal = √(L² + B²).

Triangle Area

General Area = 1/2 × Base × Height. For equilateral triangle, Area = (√3/4) × side².

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 738

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

What approximate value should come in the place of the question mark (?) in the following equation?

(17.98 × 3.99) + (25.05 ÷ 4.98) - 32.96 = ?

AOption (a)
50
BOption (b)
51
COption (c)
44
DOption (d)
40
EOption (e)
39

View answer & explanation

Correct answer

Why this answer is correct

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

(17.98 × 3.99) + (25.05 ÷ 4.98) - 32.96 = ?

⇒ (18 × 4) + (25 ÷ 5) - 33 = ?

⇒ 72 + 5 - 33 = ?

⇒ 77 - 33 = ?

⇒ ? = 44

Hence, 44 will come in the place of (?).

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 739

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

What approximate value should come in the place of the question mark (?) in the following equation?

69.99 - 19.97 × 2.02 + 35.96 × 0.99 = ?

AOption (a)
60
BOption (b)
66
COption (c)
76
DOption (d)
80
EOption (e)
56

View answer & explanation

Correct answer

Why this answer is correct

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

69.99 - 19.97 × 2.02 + 35.96 × 0.99 = ?

⇒ 70 - 20 × 2 + 36 × 1 = ?

⇒ 70 - 40 + 36 = ?

⇒ 106 - 40= ?

⇒ ? = 66

Hence, 66 will come in the place of (?).

SBI Clerk Mains Memory Based Paper (Held On: 25 Feb 2024 Shift 1) (Previous Year Paper)·Quantitative Aptitude

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Question 740

SBI Clerk 2024• Quantitative Aptitude• Quantitative Aptitude

A boat travels 1.5S km downstream in 15 hours, while a person can run S km in 24 hours. The speed of the boat in still water exceeds the running speed of the person by 4 km/h, and the speed of the current is 3 km/h. What is the speed of the boat in still water?

AOption (a)
5
BOption (b)
12
COption (c)
9
DOption (d)
6
EOption (e)
10

View answer & explanation

Correct answer

Why this answer is correct

Shortcut Trick

Let Person speed be P and Boat speed be B.

Given: B = P + 4 and Current (v) = 3 km/h.

Downstream speed = B + 3 ⇒ (P + 4) + 3 = P + 7.

From ratios: 1.5S / 15 = P + 7 and S / 24 = P.

Substitute S = 24P: 1.5(24P) / 15 = P + 7 ⇒ 2.4P = P + 7 ⇒ 1.4P = 7 ⇒ P = 5 km/h.

B = 5 + 4 = 9 km/h.

∴ The correct answer is 9 km/h.

Alternate Method

Given:

Speed of current (v) = 3 km/h

Running speed of person (P) = S/24 km/h

Speed of boat in still water (u) = P + 4 km/h

Formula Used:

Speed = Distance ÷ Time; Downstream speed = u + v

Calculations:

⇒ Downstream speed = 1.5S / 15 = 0.1S km/h

⇒ u + v = 0.1S ⇒ (P + 4) + 3 = 0.1S

⇒ P + 7 = 0.1S

⇒ From person's speed, S = 24P

⇒ P + 7 = 0.1 × (24P)

⇒ P + 7 = 2.4P

⇒ 1.4P = 7 ⇒ P = 7 / 1.4 = 5 km/h

⇒ Boat speed in still water (u) = 5 + 4 = 9 km/h

∴ The correct answer is 9 km/h.

Additional Information

Upstream Speed

Upstream speed is the net speed of a boat against the current, calculated as u − v.

Still Water Speed Calculation

Speed of boat in still water (u) = (Downstream Speed + Upstream Speed) ÷ 2.

Current Speed Calculation

Speed of current (v) = (Downstream Speed − Upstream Speed) ÷ 2.

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How should I prepare for Quantitative Aptitude in SBI Clerk?

Solving previous year questions by topic helps identify recurring concepts, question patterns, and frequently asked sub-topics. Attempt each question, review the step-by-step explanation, and note down key formulas or facts for quick revision before the exam.

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