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IBPS PO Mains 30 Nov 2024 Shift 2 Paper

Question Numbers: 134-137

Directions: Read the following information carefully and answer the questions based on it.

There are three sellers – Arun, Binny, and Chanda – dealing in hats and caps.

The number of hats sold by Binny is x% more than the number of caps sold by Arun. The total items (hats + caps) sold by Arun are 50% higher than the number of caps sold by him. The ratio of Binny’s caps to Chanda’s hats is 7 : 5. The number of caps sold by Chanda is 20% less than the caps sold by Arun. The total sales of Chanda (hats + caps) equals (Y × Binny’s hats) + 1. Chanda’s caps are also 36 more than Arun’s hats. The average number of hats sold by Arun and Chanda is 80. Altogether, Arun, Binny, and Chanda sold 672 hats and caps.

Question : 135 of 155

Marks: +1, -0

If the number of hats sold by Darsha

n = 48

Y, then find the sum of Arun’s caps and Darshan’s hats.

180
215
145
230
None of these

Validate

Solution:

Concept:

Use the given relations between hats and caps to form equations, solve for

Y

, then compute Darshan's hats and Arun's caps.

Explanation:

Let

H_A, H_B, H_C

be hats sold by Arun, Binny, and Chanda.

Let

C_A, C_B, C_C

be caps sold by Arun, Binny, and Chanda.

The given conditions become:

H_B = C_A\left(1 + \frac{x}{100}\right)

H_A + C_A = 1.5 C_A

C_B : H_C = 7 : 5

C_C = 0.8 C_A

H_C + C_C = Y \cdot H_B + 1

C_C = H_A + 36

\frac{H_A + H_C}{2} = 80

8. Total items

= 672

From (2):

H_A = 1.5C_A - C_A = 0.5C_A

Using (6):

C_C = 0.5C_A + 36

Equating with (4):

0.8C_A = 0.5C_A + 36 \Rightarrow 0.3C_A = 36 \Rightarrow C_A = 120

H_A = 0.5 \times 120 = 60

and

C_C = 0.8 \times 120 = 96

From (7):

H_A + H_C = 160 \Rightarrow 60 + H_C = 160 \Rightarrow H_C = 100

From (3):

\frac{C_B}{H_C} = \frac{7}{5} \Rightarrow \frac{C_B}{100} = \frac{7}{5} \Rightarrow C_B = 140

From (8):

(H_A + C_A) + (H_B + C_B) + (H_C + C_C) = 672

Substitute values:

(60+120) + (H_B+140) + (100+96) = 672

180 + H_B + 140 + 196 = 672 \Rightarrow H_B + 516 = 672 \Rightarrow H_B = 156

From (1):

156 = 120\left(1 + \frac{x}{100}\right) \Rightarrow 1.3 = 1 + \frac{x}{100} \Rightarrow x = 30

From (5):

100 + 96 = Y \times 156 + 1 \Rightarrow 196 = 156Y + 1 \Rightarrow 156Y = 195 \Rightarrow Y = \frac{195}{156} = 1.25

Darshan's hats

= 48Y = 48 \times 1.25 = 60

Arun's caps

= C_A = 120

Required sum

= 120 + 60 = 180

Answer:

180

(Option A).

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