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Question 401

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

The bar graph shows the profit amount of two companies [ FIAT and AMCAT] in four different months [ March, April, May and June].

Find the ratio of total profit of FIAT to total profit of AMCAT for the given months.

  1. AOption (a)

    5:11

  2. BOption (b)

    55:57

  3. COption (c)

    57:59

  4. DOption (d)

    7:9

  5. EOption (e)

    4:5

View answer & explanation

Correct answer

Answer: (c)C.

57:59

Why this answer is correct

CALCULATION

Total profit of FIAT

= 1500 + 1600 + 1200 + 1400

= 5700

Total profit of AMCAT

= 1400 + 1800 + 1600 + 1100

= 5900

Required ratio

= 5700 : 5900

= 57 : 59

Question 402

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

The bar graph shows the profit amount of two companies [ FIAT and AMCAT] in four different months [ March, April, May and June].

Find the average monthly profit of AMCAT.

  1. AOption (a)

    1845

  2. BOption (b)

    1665

  3. COption (c)

    1454

  4. DOption (d)

    1475

  5. EOption (e)

    1195

View answer & explanation

Correct answer

Answer: (d)D.

1475

Why this answer is correct

CALCULATION

Total profit of AMCAT = 5900

Average profit

= 5900 / 4

= 1475

Answer: 1475

Question 403

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

The bar graph shows the profit amount of two companies [ FIAT and AMCAT] in four different months [ March, April, May and June].

What is the percentage decrease in FIAT’s profit from April to May?

  1. AOption (a)

    18%

  2. BOption (b)

    32%

  3. COption (c)

    25%

  4. DOption (d)

    20%

  5. EOption (e)

    12%

View answer & explanation

Correct answer

Answer: (c)C.

25%

Why this answer is correct

CALCULATION

Profit in April = 1600

Profit in May = 1200

Decrease = 1600 − 1200 = 400

Percentage decrease

= (400 / 1600) × 100

= 25%

Answer: 25%

Question 404

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

Simple interest for Rs.  P at 20% rate of interest in three years is Rs. 3420. Find the simple interest for Rs. [P – 200] at same interest rate for two years?

  1. AOption (a)

    2000

  2. BOption (b)

    2800

  3. COption (c)

    2200

  4. DOption (d)

    2400

  5. EOption (e)

    2600

View answer & explanation

Correct answer

Answer: (c)C.

2200

Why this answer is correct

Given:

Simple Interest (SI) = Rs. 3420

Rate (R) = 20%

Time (T) = 3 years

Formula used:

Simple Interest = (P × R × T) ÷ 100

Calculations:

SI = (P × 20 × 3) ÷ 100

⇒ 3420 = (60P) ÷ 100

⇒ 3420 = 0.6P

⇒ P = 3420 ÷ 0.6 = 5700

New principal = P − 200 = 5700 − 200 = 5500

Time = 2 years, Rate = 20%

SI = (5500 × 20 × 2) ÷ 100

⇒ SI = 2200

∴ The simple interest is Rs. 2200.

Question 405

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

Ratio of length and breadth of a rectangle is 8:3. If area is 384 m2. Find the breadth of the rectangle?

  1. AOption (a)

    15

  2. BOption (b)

    12

  3. COption (c)

    10

  4. DOption (d)

    8

  5. EOption (e)

    9

View answer & explanation

Correct answer

Answer: (b)B.

12

Why this answer is correct

Given:

Ratio of length and breadth = 8 : 3

Area of rectangle = 384 m2

Formula used:

Area of rectangle = Length × Breadth

Calculations:

Let length = 8x

Let breadth = 3x

⇒ Area = 8x × 3x = 24x2

⇒ 24x2 = 384

⇒ x2 = 384 ÷ 24 = 16

⇒ x = 4

Length = 8 × 4 = 32 m

Breadth = 3 × 4 = 12 m

∴ Length = 32 m and Breadth = 12 m.

Question 406

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

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

{(8 + 8 ÷ 2 × 3) of 8}/2 – 3 = 2.5 × ?

  1. AOption (a)

    20.8

  2. BOption (b)

    25.8

  3. COption (c)

    30.8

  4. DOption (d)

    35.8

  5. EOption (e)

    40.8

View answer & explanation

Correct answer

Answer: (c)C.

30.8

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 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.

{(8 + 8 ÷ 2 × 3) of 8}/2 – 3 = 2.5 × ?

⇒ {(8 + 4 × 3) × 8)}/2 – 3 = 2.5 × ?

⇒ 160/2 - 3 = 2.5 × ?

⇒ ? = 77/2.5

⇒ ? = 30.8

Question 407

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

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

\(42{\rm{}} \times {\rm{}}\left[ {\frac{1}{7} \times \left( {91 + 49} \right) + 17\frac{1}{3}} \right]{\rm{}} = {\rm{}}?\)

  1. AOption (a)

    1568

  2. BOption (b)

    1484

  3. COption (c)

    1674

  4. DOption (d)

    1878

  5. EOption (e)

    2106

View answer & explanation

Correct answer

Answer: (a)A.

1568

Why this answer is correct

Concept used:

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

Calculation:

\(42{\rm{}} \times {\rm{}}\left[ {\frac{1}{7}; \times ;\left( {91 + 49} \right) + 17\frac{1}{3}} \right]{\rm{}} = {\rm{}}?\)

⇒ 42 × [(1/7) × 140 + (52/3)] = ?

⇒ 42 × [20 + (52/3)] = ?

⇒ ? = 42 × (112/3)

⇒ ? = 1568

∴ The value of ? is 1568

Question 408

IBPS RRB Office Assistant (Clerk) 2025• Numerical Ability

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

55% of 6620 + ∛681472 × (1/11) – 322 = ?

  1. AOption (a)

    2625

  2. BOption (b)

    2764

  3. COption (c)

    2714

  4. DOption (d)

    2845

  5. EOption (e)

    2645

View answer & explanation

Correct answer

Answer: (a)A.

2625

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.

55 % of 6620 +∛681472 × (1/11) – 322 = ?

⇒ (55/100) × 6620 + ∛681472 × (1/11) - 322  = ?

⇒ (11/20) × 6620 + ∛681472 × (1/11) – 1024 = ?

⇒ (11 × 331) + 88 × (1/11) – 1024 = ?

⇒ 3641 + 8 – 1024 = ?

⇒ 3649 – 1024 = ?

∴ ? = 2625

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