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 861–880 of 1935Page 44 of 97
RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 861

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

What approximate value should come in place of x in the following question?

\(23\frac{2}{3}{\rm{% ;of;}}\sqrt {323.91} + \left( {{{11.97}^2} \div \sqrt[3]{{26.949}}} \right){\rm{% ;of;}}\sqrt {1679} = 30{\rm{% ;of;x}} + 2.87\)

  1. AOption (a)

    57

  2. BOption (b)

    70

  3. COption (c)

    65

  4. DOption (d)

    76

  5. EOption (e)

    81

View answer & explanation

Correct answer

Answer: (b)B.

70

Why this answer is correct

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

\(23\frac{2}{3}\)% of \(\sqrt {323.91}\) + (11.972 ÷ \(\sqrt[3]{{26.949}}\))% of \(\sqrt {1679} \) = 30% of x + 2.87

\(\Rightarrow \frac{{71}}{3}\) % of \(\sqrt {324}\) + (122 ÷ \(\sqrt[3]{{27}}\))% of \(\sqrt {1681}\) = 30% of x + 2.87

\(\Rightarrow\) \(\frac{{71}}{{3 \times 100}}\) × 18 + (144 ÷ 3)% of 41 = 30% of x + 2.87

\(\Rightarrow\) \(\frac{{71}}{{100}}\) × 6 + 48% of 41 = 30% of x + 2.87

\(\Rightarrow\) 4.26 + \(\frac{{48}}{{100}}\) × 41 = \(\frac{{30}}{{100}}\) x + 2.87

\(\Rightarrow\) 4.26 + 19.68 = \(\frac{{30}}{{100}}\) x + 2.87

\(\Rightarrow\) 4.26 + 19.68 – 2.87 = \(\frac{{30}}{{100}}\) x

\(\Rightarrow\) 21.07 = \(\frac{{30}}{{100}}\) x

\(\Rightarrow\) 21 = \(\frac{{30}}{{100}}\) x

\(\Rightarrow\) x = \(\frac{{2100}}{{30}}\)

\(\Rightarrow\) x = 70

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 862

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

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

\(\left( \frac{\sqrt{120.77}\div 11+\sqrt{64.03}}{2.98} \right)\frac{14.98}{2.94}=~?\)

  1. AOption (a)

    15

  2. BOption (b)

    20

  3. COption (c)

    16

  4. DOption (d)

    12

  5. EOption (e)

    9

View answer & explanation

Correct answer

Answer: (a)A.

15

Why this answer is correct

Given

[(√120.77 ÷ 11 + √64.03)/2.98] × (14.98/2.98) = ?

⇒ [(√121 ÷ 11 + √64)/3] × (15/3) = ? 

⇒ (11/11 + 8)/3 × 5 = ?

⇒ (1 + 8)/3 × 5 = ?

∴ ? = 9/3 × 5 = 3 × 5 = 15.

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 863

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

What approximate value should come in the place of the question mark ‘?’ in the following question?

124.83 × 63.82 ÷ 24.94 + 168.79 × 13.96 ÷ 12.98 + 122.96 = ?2

  1. AOption (a)

    20

  2. BOption (b)

    22

  3. COption (c)

    25

  4. DOption (d)

    26

  5. EOption (e)

    23

View answer & explanation

Correct answer

Answer: (c)C.

25

Why this answer is correct

Given: 

124.83 × 63.82 ÷ 24.94 + 168.79 × 13.96 ÷ 12.98 + 122.96 = ?2

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

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

124.83 × 63.82 ÷ 24.94 + 168.79 × 13.96 ÷ 12.98 + 122.96 = ?2

⇒ 125 × 64 ÷ 25 + 169 × 14 ÷ 13 + 123 = ?2

⇒ 5 × 64 + 13 × 14 + 123 = ?2

⇒ 320 + 182 + 123 = ?2

⇒ 625 = ?2

⇒ ? = √625

⇒ ? = 25

∴ The value of ‘?’ Is 25.

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 864

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Find the value of x in the following question given below

(√440.98 /7 × 11.98) - 24.98% of 399.98 + (15.89 × 17.89) - (∛1727.79 × 5.09) - (16.09 × 34.98)/7 = x

  1. AOption (a)

    108

  2. BOption (b)

    96

  3. COption (c)

    84

  4. DOption (d)

    72

  5. EOption (e)

    60

View answer & explanation

Correct answer

Answer: (c)C.

84

Why this answer is correct

Solution:

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

(√440.98 /7 × 11.98) - 24.98% of 399.98 + (15.89 × 17.89) - (∛1727.79 × 5.09) - (16.09 × 34.98)/7 = x

(√441 /7 × 12) - 25% of 400 + (16 ×  18) - (∛1728 × 5) - (16 ×  35)/7 = x

(21/7) × 12 - (0.25 ×  400) + 288 - (12 ×  5) - (16 ×  35)/7 = x

(3 × 12) - 100 + 288 - 60 - 80 = x

36 - 100 + 288 - 60 - 80 = x

x = 84

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 865

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Which of the following approximate value will replace the question mark?

(516.92 + 37.04 + 471.93)/113.98*?= 576.05

  1. AOption (a)

    64

  2. BOption (b)

    88

  3. COption (c)

    125

  4. DOption (d)

    50

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

64

Why this answer is correct

Calculation:

(516.92 + 37.04 + 471.93)/113.98×?= 576.05

(517 + 37 + 472)/114 ×? = 576

1026/114 ×? = 576

? = (576 x 114)/1026

? = 65664/1026

? = 64

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 866

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Find the missing term in the given series.

2000, 1800, 1700, 1600, ?, 600

  1. AOption (a)

    1450

  2. BOption (b)

    1800

  3. COption (c)

    1400

  4. DOption (d)

    1200

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (c)C.

1400

Why this answer is correct

2000, 1800, 1700, 1600, ?, 600

The value of ? is 1400

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 867

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

A, B, and C enter into a partnership with their total initial investment of Rs.16200. A invested for 1 year while B and C invested for same time period, Find the time period for which B and C invested their capital.

Statement I: Profit share of B is Rs.2400 out of total annual profit.

Statement II: Ratio of investment of A, B and C is 5: 3: 10 respectively and profit share of person A is Rs.2400.

  1. AOption (a)

    Statement I is alone sufficient 

  2. BOption (b)

    Statement II is alone sufficient

  3. COption (c)

    Either I or II

  4. DOption (d)

    Both the Statement I and II together sufficient

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (d)D.

Both the Statement I and II together sufficient

Why this answer is correct

Solution:

Statement I:

Profit share of B is Rs. 2400 and the total investment is Rs. 16200

Statement I is not sufficient to answer

Statement II:

Investment of A, B and C are 5a: 3a: 10a respectively.

Investment amount of A = 16200/18 × 5 = Rs. 4500

Investment amount of B = 16200/18 × 3 = Rs. 2700

Investment amount of C = 16200/18 × 10 =Rs. 9000

Investment period of B and C be x respectively.

Statement II is not sufficient to answer

From statement I and II

Profit share of A = Rs. 2400

Profit share of B = Rs. 2400

Profit share of C = y

To find the profit share C

(5a × 1)/(3a × x) = 2400/2400

x = 5/3 × 12

Investment period of B and C will be 1 year and 8 months respectively.

So statements I and II are sufficient to answer the question.

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 868

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

A and B are positive integers. Is (A × B)/10 is an integer?

Statement I: Both A and B are multiple of 5.

Statement II: A × B is a multiple of 20.

  1. AOption (a)

    Statement I is alone sufficient

  2. BOption (b)

    Statement I and II both are sufficient 

  3. COption (c)

    Either Statement I or II

  4. DOption (d)

    Statement II is alone sufficient

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (d)D.

Statement II is alone sufficient

Why this answer is correct

Solution:

Statement I:

Multiples of A and  B are 5

The values of A and B can any of the values of  5, 10, 15, 20 and 25 respectively.

So statement I is not sufficient to answer

Statement II:

A × B are the multiples 20

The values of (A × B) can be any of the values of 20, 40, and 60 so on.

Let (A × B) ​= 20x

So, (A × B)/10 = 20x/10 = 2x where x is an integer so (A × B)/10 is an integer.

Statement II is alone sufficient to give answer.

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 869

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

A train of length 'a + 100' m crosses a person running with speed 5 km/h in the same direction in 20 seconds. If speed of train is 50 km/h, then find the time taken by train to cross a platform of length 150 meter.(approx)

  1. AOption (a)

    29 seconds

  2. BOption (b)

    32 seconds

  3. COption (c)

    44 seconds

  4. DOption (d)

    38 seconds

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

29 seconds

Why this answer is correct

Solution:

Length of the train = (a + 100)

Speed of the train = 50 kmph

Speed of the running man = 5 kmph

Time taken by the train to cross the running man = 20 seconds

(a + 100)/(50 - 5) ×  5/18 = 20

(a + 100) = 12.5 × 20 

(a + 100) = 250

Length of the train is 250m

Time taken by the train to cross the 150m platform

(150 + 250)/50 × 5/18 = 28.8 seconds = 29 seconds

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 870

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Downstream speed of a boat is 24 km/h while it takes a total of 3.75 hours to cover a certain upstream and comes back to the same point. If the speed of the stream is 4 km/h, then find the time taken by the boat with the speed of the boat in still water to the same distance covered upstream and downstream?

  1. AOption (a)

    3.6 hours

  2. BOption (b)

    4 hours

  3. COption (c)

    3 hours

  4. DOption (d)

    5 hours

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

3.6 hours

Why this answer is correct

Solution:

Downstream speed of the boat = 24 kmph

Speed of the stream = 4 kmph

Speed of boat in still water = 24 - 4 = 20 kmph

Distance covered = D km

D/(24) + D/(16) = 3.75

D/24 + 1.5D/24 = 3.75

2.5D = 3.75 ×  24

D = 90/2.5 = 36 km

Distance covered in upstream and downstream = 36 * 2 = 72km

72/(20) = 3.6 hours

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 871

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

A sum '2P' is invested in scheme A for 4 years at 10% simple rate of interest and amount received from scheme A is invested in scheme B for 5 years at 8% simple rate of interest. Difference between total amount received from scheme A and scheme B is 201.60, then find the total amount invested in scheme B.

  1. AOption (a)

    504

  2. BOption (b)

    484

  3. COption (c)

    552

  4. DOption (d)

    450

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

504

Why this answer is correct

Solution:

Investment in scheme A = 2P

Number of years = 4

Rate of interest = 10%

SI = (PNR/100)

SI = (2P ×  4 ×  10)/100

SI in scheme A = (2P ×  4)/10 = 8P/10 = 0.8P

If the amount of (2P + 0.8P) is invested in scheme A for 5 years at 8% rate of interest

SI = (2.8P ×  5 ×  8)/100 = 1.12P

SI in scheme B = 1.12P

Difference between the amount received = 201.60

(2.8P + 1.12P) - (2.8P) = 201.60

3.92P - 2.8P = 201.60

1.12P = 201.60

P = 180

Amount invested in scheme B = 180 ×  2.8 = Rs. 504

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 872

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

In 20 L milk, a mixture that contain equal quantity of milk and water is mixed such that quantity of milk becomes 75% of total quantity. Now, 25% of total quantity of this mixture is taken out and replaced by same quantity of water. Find the quantity of water in the final mixture.

  1. AOption (a)

    15 litres

  2. BOption (b)

    10 litres

  3. COption (c)

    17.5 litres

  4. DOption (d)

    12.5 litres

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (c)C.

17.5 litres

Why this answer is correct

Shortcut Trick

Initial Milk = 20 L. Let added mixture be 2x (x Milk + x Water).

New Milk/Total = (20 + x)/(20 + 2x) = 3/4 ⇒ 80 + 4x = 60 + 6x ⇒ 2x = 20 L.

Total mixture = 40 L (30 L Milk, 10 L Water). 25% removal leaves 75% of Milk.

Final Milk = 30 × 0.75 = 22.5 L. Final Water = 40 − 22.5 = 17.5 L.

∴ The correct answer is 17.5 litres.

Alternate Method

Given: Initial Pure Milk = 20 L. Added mixture has Milk : Water = 1 :

  1. ⇒ Let Water added = x L, then Milk added = x L. Total added = 2x L.

⇒ Total Milk = 20 + x; Total Mixture = 20 + 2x.

⇒ (20 + x) / (20 + 2x) = 75/100 = 3/4.

⇒ 4(20 + x) = 3(20 + 2x) ⇒ 80 + 4x = 60 + 6x ⇒ 2x = 20 ⇒ x = 10 L.

⇒ Total mixture = 40 L. Initial Milk in this mixture = 30 L, Water = 10 L.

⇒ 25% of mixture (10 L) is removed. Milk removed = 25% of 30 = 7.5 L.

⇒ Remaining Milk = 30 − 7.5 = 22.5 L.

⇒ Final Water = Total Volume − Final Milk = 40 − 22.5 = 17.5 L.

∴ The correct answer is 17.5 litres.

Additional Information

Replacement Formula

When a portion of mixture is replaced by water, the final quantity of original liquid = Initial × (1 − replaced/total)n.

Alligation Method

Can be used to find the ratio of two liquids mixed to get a final concentration: (Higher value − Mean) / (Mean − Lower value).

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 873

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Ratio of ages of Q and R before 8 years was 5: 6 and average of present ages of P and R is 36 years. If age of P after 5 years will be same as the age of Q before 5 years, find the present age of Q.

  1. AOption (a)

    42

  2. BOption (b)

    38

  3. COption (c)

    44

  4. DOption (d)

    40

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (b)B.

38

Why this answer is correct

Solution:

Present age of Q = (5x + 8) 

Present age of R = (6x + 8)

Average present ages of P and R = 36

Total present ages of P and R = 72

Age of P after 5 years is same as the age of Q before 5 years 

Age of P after 5 years = (5x + 8) - 5 = (5x + 3)

Present age of P = (5x + 3) - 5 = 5x - 2

(5x - 2) + (6x + 8) = 72

11x + 6 = 72

11x = 66

x = 6

Present age of Q = (5 * 6 + 8) = 38 years

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Question 874

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Two persons A and B enter into partnership with total investment of 'X'. If A left the business after 3 months and at the same time C joins with 1000 investment. If ratio of profit share of A, B, and C at end of a year is 3: 8: 3, then find the difference between initial investment of A and B.

  1. AOption (a)

    1000

  2. BOption (b)

    1200

  3. COption (c)

    2200

  4. DOption (d)

    1800

  5. EOption (e)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

1000

Why this answer is correct

Solution:

Profit = Investment ×  Time

Investment Period of A = 3

Investment Period of B = 12

Investment Period of C = 9

3: 8: 3 = Investment ratio ×  (1 :4: 3)

(3/1: 8/4: 3/3) = Investment ratio

3: 2: 1

3a: 2a: a

a = 1000

Difference between initial investment of A and B = (3a - 2a) = 1000

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Question 875

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

The average of the height of the 20 employee of the company is 1.5 m.In the month of June, two employees named Rita and Sumit left the company whose height are 1.4 m and 1.6 m respectively. In August, two more people left the company whose average of height is 1.7 m and one person join the company whose height is 87.5% of height of Sumit. Before the December, more four people left the company whose height are 1.3 m, 1.1 m, 1.32 m and 1.25 m respectively and more 5 people join the company whose height are 1.26 m, 1.18 m, 1.39 m and 1.4 m respectively. The average of height in the month of September is ….

  1. AOption (a)

    1.2

  2. BOption (b)

    1.3

  3. COption (c)

    1.47

  4. DOption (d)

    1.62

  5. EOption (e)

    1.8

View answer & explanation

Correct answer

Answer: (c)C.

1.47

Why this answer is correct

Sum of the height of 20 employee = 1.5 × 20 = 30 m

For the average in month of June,

Number of people available = 20 – 2 = 18

Sum of the height of 18 people = 30 – 1.4 – 1.6 = 27 m

Average height = 27/18 = 1.5 m

For the average in month of September,

Number of people available = 18 – 2 + 1 = 17

Sum of height of two person who left the company = 1.7 × 2 = 3.4 m

Height of person who join the company = 87.5% of height of Sumit = (7/8) × 1.6 = 1.4 m

Sum of the height of 17 people = 27 – 3.4 + 1.4 = 25 m

∴ Average of height in the month of September = 25/17 = 1.47 m

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Question 876

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Directions: The below given line graph shows the total number of professors who teach (Literature + Commerce) and the number of professors who teach Commerce in five different colleges A, B, C, D and E.

All the professors in college D who teaches commerce takes  2 hours of class each and all the professors in college D who teaches Literature takes m hours of class. If the total class taken by all the professors in college D is 590 hours. Find the value of m.

  1. AOption (a)

    4

  2. BOption (b)

    6

  3. COption (c)

    8

  4. DOption (d)

    3

  5. EOption (e)

    5

View answer & explanation

Correct answer

Answer: (d)D.

3

Why this answer is correct

General Solution:

 

College A:

Total number of professor (Literature + Commerce) = 200

Number of professor in commerce = 80

Number of professor in Literature = (200 - 80) = 120

 

College B:

Total number of professor (Literature + Commerce) = 180

Number of professor in commerce = 50

Number of professor in Literature = (180 - 50) = 130

 

College C:

Total number of professor (Literature + Commerce) = 220

Number of professor in commerce = 80

Number of professor in Literature = (220 - 80) = 140

 

College D:

Total number of professor (Literature + Commerce) = 240

Number of professor in commerce = 130

Number of professor in Literature = (240 - 130) = 110

 

College E:

Total number of professor (Literature + Commerce) = 160

Number of professor in commerce = 90

Number of professor in Literature = (160 - 90) = 70

 

Name of the collegeNumber of professors in commerceNumber of professors in Literature
A80120
B50130
C80140
D130110
E9070

 Solution:

Number of hours teached by the commerce professors in college D = 2 hours 

Number of hours teached by the literature professors in college D = m hours

If the total class hours  teached by the both the professors = 590 hours

590 = 2 ×  130 + m × 110 

590 - 260 = 110m

330 = 110m

m = 3

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Question 877

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Directions: The below given line graph shows the total number of professors who teach (Literature + Commerce) and the number of professors who teach Commerce in five different colleges A, B, C, D and E.

What is the sum of the average number of commerce professors teaches in all the colleges and the average number of literature professors teaches in all the colleges? 

  1. AOption (a)

    250

  2. BOption (b)

    240

  3. COption (c)

    210

  4. DOption (d)

    200

  5. EOption (e)

    220

View answer & explanation

Correct answer

Answer: (d)D.

200

Why this answer is correct

General Solution:

 

College A:

Total number of professor (Literature + Commerce) = 200

Number of professor in commerce = 80

Number of professor in Literature = (200 - 80) = 120

 

College B:

Total number of professor (Literature + Commerce) = 180

Number of professor in commerce = 50

Number of professor in Literature = (180 - 50) = 130

 

College C:

Total number of professor (Literature + Commerce) = 220

Number of professor in commerce = 80

Number of professor in Literature = (220 - 80) = 140

 

College D:

Total number of professor (Literature + Commerce) = 240

Number of professor in commerce = 130

Number of professor in Literature = (240 - 130) = 110

 

College E:

Total number of professor (Literature + Commerce) = 160

Number of professor in commerce = 90

Number of professor in Literature = (160 - 90) = 70

 

Name of the collegeNumber of professors in commerceNumber of professors in Literature
A80120
B50130
C80140
D130110
E9070

 

Solution:

Average number of commerce professors teaches in all the colleges A, B, C, D and E = (80 + 50 + 80 + 130 + 90)/5 = 86

Average number of literature professors teaches in all the colleges A, B, C, D and E = (120 + 130 + 140 + 110 + 70)/5 = 114

Required sum = (114 + 86) = 200

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 878

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Directions: The below given line graph shows the total number of professors who teach (Literature + Commerce) and the number of professors who teach Commerce in five different colleges A, B, C, D and E.

 Number of professors who teaches literature in college A has increased by 33.3% and the number of professors who teach commerce in college B has decreased by 20%. Find the ratio between increased professors in literature in college A and decreased commerce professors in college B.

  1. AOption (a)

    5: 3

  2. BOption (b)

    2: 1

  3. COption (c)

    4: 1

  4. DOption (d)

    3: 2

  5. EOption (e)

    4: 5

View answer & explanation

Correct answer

Answer: (c)C.

4: 1

Why this answer is correct

General Solution:

 

College A:

Total number of professor (Literature + Commerce) = 200

Number of professor in commerce = 80

Number of professor in Literature = (200 - 80) = 120

 

College B:

Total number of professor (Literature + Commerce) = 180

Number of professor in commerce = 50

Number of professor in Literature = (180 - 50) = 130

 

College C:

Total number of professor (Literature + Commerce) = 220

Number of professor in commerce = 80

Number of professor in Literature = (220 - 80) = 140

 

College D:

Total number of professor (Literature + Commerce) = 240

Number of professor in commerce = 130

Number of professor in Literature = (240 - 130) = 110

 

College E:

Total number of professor (Literature + Commerce) = 160

Number of professor in commerce = 90

Number of professor in Literature = (160 - 90) = 70

 

Name of the collegeNumber of professors in commerceNumber of professors in Literature
A80120
B50130
C80140
D130110
E9070

 

Solution:

Number of professors who teaches literature initially in college A = 120

Number of professors who teaches literature at present in college A = 120 × (133.3/100) = 160

Number of professors who teaches commerce initially in college B = 50

Number of professors who teaches commerce at present in college B = 50 ×  (80/100) = 40

Required ratio = 160: 40 = 4: 1

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
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Question 879

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Directions: The below given line graph shows the total number of professors who teach (Literature + Commerce) and the number of professors who teach Commerce in five different colleges A, B, C, D and E.

If 3/7 th of the professors who teach literature in college C are females and the total number of female professors in college C is 80, find the difference between the number of male professors who teach literature in college C and the number of male professors who teach commerce in college C.

  1. AOption (a)

    30

  2. BOption (b)

    20

  3. COption (c)

    25

  4. DOption (d)

    35

  5. EOption (e)

    15

View answer & explanation

Correct answer

Answer: (b)B.

20

Why this answer is correct

General Solution:

 

College A:

Total number of professor (Literature + Commerce) = 200

Number of professor in commerce = 80

Number of professor in Literature = (200 - 80) = 120

 

College B:

Total number of professor (Literature + Commerce) = 180

Number of professor in commerce = 50

Number of professor in Literature = (180 - 50) = 130

 

College C:

Total number of professor (Literature + Commerce) = 220

Number of professor in commerce = 80

Number of professor in Literature = (220 - 80) = 140

 

College D:

Total number of professor (Literature + Commerce) = 240

Number of professor in commerce = 130

Number of professor in Literature = (240 - 130) = 110

 

College E:

Total number of professor (Literature + Commerce) = 160

Number of professor in commerce = 90

Number of professor in Literature = (160 - 90) = 70

 

Name of the collegeNumber of professors in commerceNumber of professors in Literature
A80120
B50130
C80140
D130110
E9070

Solution:

Number of professors who teaches literature in college C = 140

Number of female professors who teaches literature in college C = 140 ×  3/7 = 60

Number of male professors who teaches literature in college C = (140 - 60) = 80

Number of female professors who teaches commerce in college C = 80 - 60 = 20

Number of male professors who teaches commerce in college C = 80 - 20 = 60

Required difference = (80 - 60) = 20

RRB Clerk Mains Memory Based Paper (Held On: 16 September 2023) (Previous Year Paper)·Numerical Ability
View full paper

Question 880

IBPS RRB Office Assistant (Clerk) 2023• Numerical Ability• Numerical Ability

Directions: The below given line graph shows the total number of professors who teach (Literature + Commerce) and the number of professors who teach Commerce in five different colleges A, B, C, D and E.

Number of professors who teach literature in college D has increased by x and the number of professors who teach commerce in college D has decreased by x. Then the ratio between the number of professors who teach literature to the number of professors who teach commerce in college D is 7: 5. Find the value of x.

  1. AOption (a)

    20

  2. BOption (b)

    30

  3. COption (c)

    25

  4. DOption (d)

    15

  5. EOption (e)

    35

View answer & explanation

Correct answer

Answer: (b)B.

30

Why this answer is correct

General Solution:

 

College A:

Total number of professor (Literature + Commerce) = 200

Number of professor in commerce = 80

Number of professor in Literature = (200 - 80) = 120

 

College B:

Total number of professor (Literature + Commerce) = 180

Number of professor in commerce = 50

Number of professor in Literature = (180 - 50) = 130

 

College C:

Total number of professor (Literature + Commerce) = 220

Number of professor in commerce = 80

Number of professor in Literature = (220 - 80) = 140

 

College D:

Total number of professor (Literature + Commerce) = 240

Number of professor in commerce = 130

Number of professor in Literature = (240 - 130) = 110

 

College E:

Total number of professor (Literature + Commerce) = 160

Number of professor in commerce = 90

Number of professor in Literature = (160 - 90) = 70

 

Name of the collegeNumber of professors in commerceNumber of professors in Literature
A80120
B50130
C80140
D130110
E9070

Solution:

Number of professors who teaches literature in college D initially = 110

Number of professors who teaches literature at present in college D = (110 + x)

Number of professors who teaches commerce in college D initially = 130

Number of professors who teaches commerce at present in college D = (130 - x)

Then the ratio between the number of professors who teach literature to the number of professors who teach commerce in college D is 7: 5.

(110 + x)/(130 - x) = 7/5

(550 + 5x) = (910 - 7x)

360/12 = x

x = 30

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.