Raju went to the market to purchase three pens. The cost of the first pen was more than that of the second by 25% , which was more than that of the third by 20% . If the average cost of the first two pens was Rs. 15 more than of the last two pens, find the cost of the costliest pen.

Answer

To solve the problem step by step, we will denote the costs of the three pens as follows:

- Let the cost of the third pen (C) be \( x \).
- The cost of the second pen (B) is 20% more than the cost of the third pen:
\[ B = x + 0.2x = 1.2x \]
- The cost of the first pen (A) is 25% more than the cost of the second pen:
\[ A = B + 0.25B = 1.25B = 1.25(1.2x) = 1.5x \]

Now we have the costs of the three pens in terms of \( x \):
- Cost of the first pen (A) = \( 1.5x \)
- Cost of the second pen (B) = \( 1.2x \)
- Cost of the third pen (C) = \( x \)

Next, we need to find the average cost of the first two pens and the last two pens:

1. **Average cost of the first two pens (A and B)**:
\[ \text{Average of A and B} = \frac{A + B}{2} = \frac{1.5x + 1.2x}{2} = \frac{2.7x}{2} = 1.35x \]

2. **Average cost of the last two pens (B and C)**:
\[ \text{Average of B and C} = \frac{B + C}{2} = \frac{1.2x + x}{2} = \frac{2.2x}{2} = 1.1x \]

According to the problem, the average cost of the first two pens is Rs. 15 more than the average cost of the last two pens:
\[ 1.35x = 1.1x + 15 \]

Now, we can solve for \( x \):
\[ 1.35x - 1.1x = 15 \]
\[ 0.25x = 15 \]
\[ x = \frac{15}{0.25} = 60 \]

Now that we have the cost of the third pen (C), we can find the costs of all three pens:
- Cost of the third pen (C) = \( x = 60 \)
- Cost of the second pen (B) = \( 1.2x = 1.2 \times 60 = 72 \)
- Cost of the first pen (A) = \( 1.5x = 1.5 \times 60 = 90 \)

Finally, the cost of the costliest pen (A) is:
\[ \text{Cost of the costliest pen} = 90 \]

### Summary of the Solution:
The cost of the costliest pen is Rs. 90.