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IBPS RRB Office Assistant (Clerk) · Topic Collection

Quantitative Aptitude PYQs

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

1935 Total Questions49 Official PapersYears: 2018–2025Search in Quantitative Aptitude
Showing questions 221–240 of 1935Page 12 of 97
RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 221

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

Directions: Study the following bar graph carefully and answer the questions that follow.

The double bar graph shows data regarding five different Managers (A, B, C, D, and E) working in an IT infrastructure company.

  • The Blue Bar represents the Total Number of Projects handled by each manager.
  • The Orange Bar represents the Average Cost per Project (in ₹ Lakhs) for that specific manager.

Note:

Total Cost = Number of Projects × Average Cost per Project.

A new Manager 'F' handles the same number of projects as the average number of projects handled by A and C. The Average Cost per project for F is equal to the average of the costs per project of B and D. Find the Total Cost of projects handled by Manager F.

  1. AOption (a)

    ₹2,697 Lakhs

  2. BOption (b)

    ₹2,745 Lakhs

  3. COption (c)

    ₹2,586 Lakhs

  4. DOption (d)

    ₹2,640 Lakhs

  5. EOption (e)

    ₹2,810 Lakhs

View answer & explanation

Correct answer

Answer: (a)A.

₹2,697 Lakhs

Why this answer is correct

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

⇒ Projects_F = (68 + 56) ÷ 2 = 62

⇒ Avg cost_F = (52 + 35) ÷ 2 = 87 ÷ 2 = 43.5 (₹ Lakhs)

⇒ Total Cost_F = 62 × 43.5 = 2697 (₹ Lakhs)

Hence, the correct answer is ₹2,697 Lakhs

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 222

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

Directions: Study the following bar graph carefully and answer the questions that follow.

The double bar graph shows data regarding five different Managers (A, B, C, D, and E) working in an IT infrastructure company.

  • The Blue Bar represents the Total Number of Projects handled by each manager.
  • The Orange Bar represents the Average Cost per Project (in ₹ Lakhs) for that specific manager.

Note:

Total Cost = Number of Projects × Average Cost per Project.

The Total Cost of projects handled by Manager B is what percentage more or less than the Total Cost of projects handled by Manager E?

  1. AOption (a)

    24.5% more 

  2. BOption (b)

    27.5% more

  3. COption (c)

    21.5% less

  4. DOption (d)

    27.5% less 

  5. EOption (e)

    32.5% more

View answer & explanation

Correct answer

Answer: (b)B.

27.5% more

Why this answer is correct

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

⇒ B = 3796 ; E = 2976

⇒ Difference = 3796 − 2976 = 820

⇒ Percent more = (820 ÷ 2976) × 100 ≈ 0.2755 × 100 = 27.55% ≈ 27.5% more

Hence, the correct answer is 27.5% more.

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 223

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

Directions: Study the following bar graph carefully and answer the questions that follow.

The double bar graph shows data regarding five different Managers (A, B, C, D, and E) working in an IT infrastructure company.

  • The Blue Bar represents the Total Number of Projects handled by each manager.
  • The Orange Bar represents the Average Cost per Project (in ₹ Lakhs) for that specific manager.

Note:

Total Cost = Number of Projects × Average Cost per Project.

Manager C achieved a total revenue of ₹4,368 Lakhs from all projects. Find the Profit earned by Manager C.

  1. AOption (a)

    ₹728 Lakhs

  2. BOption (b)

    ₹795 Lakhs

  3. COption (c)

    ₹778 Lakhs 

  4. DOption (d)

    ₹786 Lakhs

  5. EOption (e)

    ₹790 Lakhs

View answer & explanation

Correct answer

Answer: (a)A.

₹728 Lakhs

Why this answer is correct

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

⇒ C Total Cost = 3640

⇒ C Revenue = 4368 ⇒ Profit_C = 4368 − 3640 = 728 (₹ Lakhs)

Hence, the correct answer is ₹728 Lakhs

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 224

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

Directions: Study the following bar graph carefully and answer the questions that follow.

The double bar graph shows data regarding five different Managers (A, B, C, D, and E) working in an IT infrastructure company.

  • The Blue Bar represents the Total Number of Projects handled by each manager.
  • The Orange Bar represents the Average Cost per Project (in ₹ Lakhs) for that specific manager.

Note:

Total Cost = Number of Projects × Average Cost per Project.

For Manager E, 50% of the projects were "High-Priority" and the remaining were "Standard". Find the value of the number of 'Standard" projects.

  1. AOption (a)

    38 

  2. BOption (b)

    39

  3. COption (c)

    33

  4. DOption (d)

    29

  5. EOption (e)

    31

View answer & explanation

Correct answer

Answer: (e)E.

31

Why this answer is correct

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

For E — 50% High-Priority, remaining 50% Standard

Number of Standard Projects = 62 × 50 / 100 = 31

Hence, the correct answer is 31.

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 225

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

Directions: Study the following bar graph carefully and answer the questions that follow.

The double bar graph shows data regarding five different Managers (A, B, C, D, and E) working in an IT infrastructure company.

  • The Blue Bar represents the Total Number of Projects handled by each manager.
  • The Orange Bar represents the Average Cost per Project (in ₹ Lakhs) for that specific manager.

Note:

Total Cost = Number of Projects × Average Cost per Project.

What is the ratio of the Total Cost of all projects handled by Manager A to the Total Cost of all projects handled by Manager D?

  1. AOption (a)

    49 : 51 

  2. BOption (b)

    51 : 49 

  3. COption (c)

    45 : 47 

  4. DOption (d)

    15 : 14

  5. EOption (e)

    17 : 16

View answer & explanation

Correct answer

Answer: (b)B.

51 : 49 

Why this answer is correct

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

Total Cost A : Total Cost D

⇒ 3060 : 2940 = divide by 60 ⇒ 51 : 49

Hence, the correct answer is 51 : 49.

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 226

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

The total amount invested by A and B in a business is Rs. 75,000. A invested his capital for 8 months, while B invested his capital for 12 months. At the end of the term, both received an equal share of the profit. What was the amount invested by A?

  1. AOption (a)

    Rs. 30,000

  2. BOption (b)

    Rs. 40,000

  3. COption (c)

    Rs. 50,000

  4. DOption (d)

    Rs. 45,000

  5. EOption (e)

    Rs. 35,000

View answer & explanation

Correct answer

Answer: (d)D.

Rs. 45,000

Why this answer is correct

Given:

Total capital invested by A and B = Rs. 75,000

A invests for 8 months

B invests for 12 months

Both receive equal profit ⇒ Their effective capital (Capital × Time) must be equal.

Let A's capital = x

Then B's capital = 75,000 − x

Condition for equal profit:

A_capital × A_time = B_capital × B_time

x × 8 = (75,000 − x) × 12

Solve:

8x = 12(75,000 − x)

8x = 900,000 − 12x

8x + 12x = 900,000

20x = 900,000

x = 900,000 ÷ 20 = 45,000

A's investment = Rs. 45,000

Hence, the correct answer is Rs. 45,000

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 227

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

The total quantity of milk in three containers A, B, and C is 28 litres. The ratio of milk in A, B, and C is initially 4 : 1 : 2. If 4 litres of milk is added to container B, what will be the new ratio of milk in A, B, and C?

  1. AOption (a)

    3 : 2 : 1

  2. BOption (b)

    2 : 1 : 1

  3. COption (c)

    4 : 3 : 2

  4. DOption (d)

    1 : 2 : 1

  5. EOption (e)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

2 : 1 : 1

Why this answer is correct

Given:

Total milk = 28 litres

Initial ratio A : B : C = 4 : 1 : 2

Find actual quantities:

Total ratio parts = 4 + 1 + 2 = 7

⇒ 1 part = 28 ÷ 7 = 4 litres

⇒ A = 4 × 4 = 16 L

⇒ B = 1 × 4 = 4 L

⇒ C = 2 × 4 = 8 L

4 litres added to container B:

New B = 4 + 4 = 8 L

New ratio:

A : B : C = 16 : 8 : 8

Divide by common factor 8:

= 2 : 1 : 1

Hence, the correct answer is 2 : 1 : 1

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 228

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

The cost price of 12 articles is 2424. The shopkeeper sells these articles at a profit of 25%, then, find the amount gained by the shopkeeper on each article.

  1. AOption (a)

    48

  2. BOption (b)

    50

  3. COption (c)

    50.5

  4. DOption (d)

    55.5

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (c)C.

50.5

Why this answer is correct

Given:

C.P of 12 article = Rs. 2424

Profit = 25%

Concept:

Cost price × (100 + Profit %)/100 = Selling price

Where C.P = Cost price, S.P = Selling price

Calculation:

C.P of 12 article = Rs. 2424

C.P of each article:

⇒ 2424/12

⇒ 202

S.P of an article = 202 × (125/100)

⇒ 202 × 5/4

⇒ 252.5

Profit = S.P – C.P

⇒ 252.5 – 202

⇒ 50.5

∴ Rs. 50.5 gained by the shopkeeper on each article.

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 229

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in the place of the question mark ‘?’ in the following question?

37.5 × 2 + 15 ÷ 3 – 14/3 of (?) = 10

  1. AOption (a)

    6

  2. BOption (b)

    8

  3. COption (c)

    5

  4. DOption (d)

    10

  5. EOption (e)

    15

View answer & explanation

Correct answer

Answer: (e)E.

15

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.

Given expression is

37.5 × 2 + 15 ÷ 3 – 14/3 of (?) = 10 

⇒ 75 + 5 – 14/3 of (?) = 10 

⇒ 70 = 14/3 of (?) 

⇒ ? = 15

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 230

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in the place of the question mark ‘?’ in the following question?

3 × 5/9 of ( 36 ÷ 4 ) × 6 - ? + 5 = 4

  1. AOption (a)

    40

  2. BOption (b)

    76

  3. COption (c)

    91

  4. DOption (d)

    26

  5. EOption (e)

    35

View answer & explanation

Correct answer

Answer: (c)C.

91

Why this answer is correct

Concept:

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

CALCULATION :

Given expression :

3 × 5/9 of ( 36 ÷ 4 ) × 6 - ? + 5 = 4

⇒ 3 × 5/9 of (9) × 6 - ? + 5 = 4

⇒ 3 × 5 × 6 - ? + 5 = 4

⇒ 90 - ? + 5 = 4 

⇒ 90 - 4 + 5 = ?

⇒ ? = 91 

∴ ? = 91

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 231

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in the place of the question mark ‘?’ in the following question?

4 × 22.5 ÷ 1.5 + ? –  6 = 30

  1. AOption (a)

    20

  2. BOption (b)

    -24

  3. COption (c)

    -15

  4. DOption (d)

    25

  5. EOption (e)

    18

View answer & explanation

Correct answer

Answer: (b)B.

-24

Why this answer is correct

Given expression : 

4 × 22.5 ÷ 1.5 + ? –  6 = 30

⇒ 4 × 15 + ? – 6 = 30

⇒ 60 + ? - 6 = 30

 ∴ ? = 30 - 60 + 6 = -24

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 232

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What should come in place of the question mark '?' in the following number series?

[(275 - 124) + 42 - 92 - 36)] + 14 = ?

  1. AOption (a)

    64

  2. BOption (b)

    36

  3. COption (c)

    52

  4. DOption (d)

    32

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

64

Why this answer is correct

Concept used:

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

Calculation:

Now,

[(275 - 124) + 42 - 92 - 36)] + 14 = ?

⇒ [151 + 16 - 81 - 36] + 14 = ? 

⇒ 50 + 14

⇒ 64 = ?

⇒ ? = 64

∴ The required value  is 64

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 233

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in the place of the question mark ‘?’ in the following question?

75 - 3 × {(62 ÷ 6) ÷ 3} ÷ \(\frac{1}{3}\) + 13 =?

  1. AOption (a)

    50

  2. BOption (b)

    40

  3. COption (c)

    60

  4. DOption (d)

    70

  5. EOption (e)

    80

View answer & explanation

Correct answer

Answer: (d)D.

70

Why this answer is correct

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

75

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 parts of the equation that contain 'Division' and 'Multiplication' are calculated,

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

Now,

75 - 3 × {(62 ÷ 6) ÷ 3} ÷ \(\frac{1}{3}\) + 13 =?

⇒ 75 - 3 × {(62 × 1/6) ÷ 3} ÷ \(\frac{1}{3}\) + 13 = ?

⇒ 75 - 3 × {6 ÷ 3} ÷ \(\frac{1}{3}\) + 13 = ?

⇒ 75 - 3 × 2 × 3/1 + 13 = ?

⇒ 75 - 18 + 13 = ?

⇒ 88 - 18 = ?

⇒ ? = 70

∴ The required value in the place of the question mark (?) is 70

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 234

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in place of the question mark?

220 ÷ 5 ÷ 0.4 + 20 × 3 = ?

  1. AOption (a)

    170

  2. BOption (b)

    175

  3. COption (c)

    189

  4. DOption (d)

    128

  5. EOption (e)

    146

View answer & explanation

Correct answer

Answer: (a)A.

170

Why this answer is correct

Solution:

We have to follow the BODMAS rule

Calculation:

220 ÷ 5 ÷ 0.4 + 20 × 3 = ?

⇒ 44 ÷ 0.4 + 20 × 3 = ?

⇒110 + 20 × 3 = ?

⇒ 110 + 60 = ?

⇒ ? = 170

∴ 170 will come in place of question mark

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 235

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in the place of the question mark ‘?’ in the following question?

(42 × 6 × 6) ÷ 36 × 9 = 20% of ?

  1. AOption (a)

    340

  2. BOption (b)

    550

  3. COption (c)

    270

  4. DOption (d)

    720

  5. EOption (e)

    27

View answer & explanation

Correct answer

Answer: (d)D.

720

Why this answer is correct

Given:

(42 × 6 × 6) ÷ 36 × 9 = 20% of ?

Solution:

Follow the BODMAS rule according to the table given below:

Calculation:

(42 × 6 × 6) ÷ 36 × 9 = 20% of ?

⇒ (96 × 6) ÷ 36 × 9 = 20% of ?

⇒ 576 ÷ 36 × 9 = 20% of ?

⇒ 16 × 9 = 20% of ?

⇒ (144 × 100)/20 = ?

⇒ ? = 720

∴ 720 will come in place of the question mark

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 236

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in the place of the question mark ‘?’ in the following question?

\(13\frac{6}{7} + 8\frac{2}{5} - 1\frac{5}{7} \times 2\frac{1}{3} = ;?\)

  1. AOption (a)

    \(2\frac{7}{{35}}\)

  2. BOption (b)

    \(14\frac{7}{{35}}\)

  3. COption (c)

    \(12\frac{7}{{25}}\)

  4. DOption (d)

    \(8\frac{9}{{35}}\)

  5. EOption (e)

    \(18\frac{9}{{35}}\)

View answer & explanation

Correct answer

Answer: (e)E.

\(18\frac{9}{{35}}\)

Why this answer is correct

Concept used:

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

Calculation:

Considering the given equation

\(13\frac{6}{7} + 8\frac{2}{5} - 1\frac{5}{7} × 2\frac{1}{3} = ;?\)

We can transform each term within the given equation in the p/q form as follows: -

⇒ 97/7 + 42/5 - 12/7 × 7/3 = ?

⇒ 97/7 + 42/5 - 12/3 = ?

⇒ 97/7 + 42/5 - 4 = ?

⇒ 779/35 - 4 = ?

⇒ (779 - 140)/35 = ?

⇒ 639/35 = ? 

⇒ ? = 639/35

⇒ ? = \(18\frac{9}{{35}}\)

∴ The required value of "?" in the given equation is ​\(18\frac{9}{{35}}\)​

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 237

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

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

[(121)2 ÷ 11 × 2] ÷ 44 × 2 – 57 = ?2

  1. AOption (a)

    4

  2. BOption (b)

    16

  3. COption (c)

    9

  4. DOption (d)

    8

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (d)D.

8

Why this answer is correct

Concept used:

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

Calculation:

Considering the given equation

[(121)2 ÷ 11 × 2] ÷ 44 × 2 – 57 = ?2

⇒ [14641 ÷ 11 × 2] ÷ 44 × 2 – 57 = ?2

⇒ [1331 × 2] ÷ 44 × 2 – 57 = ?2

⇒ [2662] ÷ 44 × 2 – 57 = ?2

⇒ 2662 ÷ 44 × 2 – 57 = ?2

⇒ 60.5 × 2 – 57 = ?

⇒ 121 – 57 = ?

⇒ 64 = ?2

⇒ 8 = ?

∴ ? = 8

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 238

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

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

12167 ÷ 23 × 2 + 386 + ?2 = 2888

  1. AOption (a)

    48

  2. BOption (b)

    58

  3. COption (c)

    43

  4. DOption (d)

    38

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (d)D.

38

Why this answer is correct

Concept used:

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

Calculation:

Considering the given equation

12167 ÷ 23 × 2 + 386 + ?2 = 2888

⇒ 529 × 2 + 386 + ?2 = 2888

⇒ 1058 + 386 + ?2 = 2888

⇒ 1444 + ?2 = 2888

⇒ ?2 = 2888 – 1444

⇒ ?2 = 1444

⇒ ? = 38

∴ ? = 38

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 239

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

What will come in the place of the question mark ‘?’ in the following question?

∛6859 + √441 – ∛4096 – √576 = ?

  1. AOption (a)

    1

  2. BOption (b)

    48

  3. COption (c)

    0

  4. DOption (d)

    -42

  5. EOption (e)

    None of these.

View answer & explanation

Correct answer

Answer: (c)C.

0

Why this answer is correct

Concept used:

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

Calculations:

∛6859 + √441 – ∛4096 – √576 = ?

⇒ 19 + 21 – 16 – 24 = ?

⇒ 19 + 21 – 16 – 24 = ?

⇒ 40 – 40 = ?

⇒ ? = 0

∴ The value of ? is 0

RRB Clerk Prelims Memory Based Paper (Held On: 7 Dec 2025 Shift 2) (Previous Year Paper)·Numerical Ability
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Question 240

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability• Numerical Ability

Find the value of ‘?’ in the following question.

[122 + (142 – 40)] × 2.5 = ? × 25

  1. AOption (a)

    25

  2. BOption (b)

    35

  3. COption (c)

    15

  4. DOption (d)

    30

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (d)D.

30

Why this answer is correct

Concept used:

Follow BODMAS rule to solve the question as per the order given below:

Step 1: Part 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 parts of the equation that contain ‘Division’ and ‘Multiplication’ are calculated.

Step 4: Last but not least, the part of the equation that contain ‘Addition’ and ‘Subtraction’ should be calculated.

Now,

Calculations:

Considering the given equation,

[122+ (142 – 40)] × 2.5 = ? × 25

⇒ [122 + (196 – 40)] × 2.5 = ? × 25

⇒ [122 + 156] × 2.5 = ? × 25

⇒ [144 + 156] × 2.5 = ? × 25

⇒ 300 × 2.5 = ? × 25

⇒ 750 = ? × 25

∴ ? = 30

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How should I prepare for Quantitative Aptitude in IBPS RRB Office Assistant (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.