Question

In a bag there are a total of 150 coins in three denominations – Re.1, ₹2 and ₹5 – with at least one coin of each denomination. The total value of Re.1 coins is at least 50% of the total value of the coins. There are 23 ₹5 coins and the total value of ₹2 coins is at least 3% of the total value of coins. Find the number of ₹2 coins in the bag.

• A. 2
• B. 3
• C. 4
• D. 1

Hint:

Translate value constraints into equations or inequalities and substitute values logically. Use the total coin constraint to simplify.

Answer 1

The Correct Option is B

Let the number of ₹1 coins = \( x \)
Let the number of ₹2 coins = \( y \)
Number of ₹5 coins is given = 23
Total number of coins:
\[ x + y + 23 = 150 \Rightarrow x + y = 127 \text{(1)} \] Total value of the coins:
\[ \text{Value} = 1 x + 2 y + 5 23 = x + 2y + 115 \] Value of ₹1 coins = \( x \)
It is at least 50% of total value:
\[ x \geq 0.5(x + 2y + 115) \] Multiply both sides by 2:
\[ 2x \geq x + 2y + 115 \Rightarrow x \geq 2y + 115 \text{(2)} \] From (1), \( x = 127 - y \). Substituting in (2):
\[ 127 - y \geq 2y + 115 \Rightarrow 127 - 115 \geq 3y \Rightarrow 12 \geq 3y \Rightarrow y \leq 4 \] Also, total value of ₹2 coins is at least 3% of total value:
\[ 2y \geq 0.03(x + 2y + 115) \] Substitute \( x = 127 - y \) into the RHS:
\[ 2y \geq 0.03(127 - y + 2y + 115) = 0.03(127 + y + 115) = 0.03(242 + y) \] Multiply both sides by 100 to simplify:
\[ 200y \geq 3(242 + y) \Rightarrow 200y \geq 726 + 3y \Rightarrow 197y \geq 726 \Rightarrow y \geq \frac{726}{197} \approx 3.68 \Rightarrow y \geq 4 \] So from earlier: \( y \leq 4 \), and from this: \( y \geq 4 \)
\(\Rightarrow y = 4 \) But verify again – we must have both conditions satisfied. Let’s try \( y = 3 \):
Then \( x = 127 - 3 = 124 \)
Total value = \( 124 + 2 3 + 115 = 244 \)
Value of ₹1 coins = 124 \(\Rightarrow \frac{124}{244} \approx 50.8% \)
Value of ₹2 coins = 6 \(\Rightarrow \frac{6}{244} \approx 2.45% \)
Try \( y = 4 \Rightarrow x = 123 \Rightarrow \) Total value = \(123 + 8 + 115 = 246\)
Re.1 coins = 123 → \( \frac{123}{246} = 50% \)
₹2 coins = 8 → \( \frac{8}{246} \approx 3.25% \)
Final Answer: \( \boxed{4} \) Wait — final working shows \( y = 4 \) is valid and satisfies both constraints. But options only go up to 4, and from above, only \( y = 4 \) works. Hence, correct answer is: \(\boxed{4}\) — Option
(c)