A shopkeeper marks up the price of an LED 60\% above its cost price and gives three successive discounts of 12\frac{1}{2}\%, d\%, and 25\%. If the shopkeeper made a loss of 16\% on the LED, find the second discount allowed by the shop.

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**Solution:**

1. Let the cost price (C.P.) of the LED be $100x$ Rs.

2. The marked price (M.P.) is 60\% above cost price:
   $$
   \text{M.P.} = 100x + 60\% \times 100x = 160x
   $$

3. The first discount is 12\frac{1}{2}\% = $12.5\% = \frac{1}{8}$:
   $$
   \text{First discount amount} = 160x \times \frac{1}{8} = 20x
   $$

4. Price after first discount:
   $$
   160x - 20x = 140x
   $$

5. Let the second discount be $d\%$, and the third discount is 25\%.

6. After applying the second and third discounts successively, the final selling price (S.P.) is given by:
   $$
   140x \times \frac{100 - d}{100} \times \frac{100 - 25}{100} = \text{S.P.}
   $$

7. The shopkeeper made a loss of 16\%, so:
   $$
   \text{S.P.} = 100x - 16\% \times 100x = 84x
   $$

8. Substitute S.P. in the equation:
   $$
   140x \times \frac{100 - d}{100} \times \frac{75}{100} = 84x
   $$

9. Simplify:
   $$
   140x \times \frac{100 - d}{100} \times \frac{3}{4} = 84x
   $$

10. Divide both sides by $x$:
    $$
    140 \times \frac{100 - d}{100} \times \frac{3}{4} = 84
    $$

11. Calculate constants:
    $$
    140 \times \frac{3}{4} = 140 \times 0.75 = 105
    $$

12. So:
    $$
    105 \times \frac{100 - d}{100} = 84
    $$

13. Multiply both sides by 100:
    $$
    105 (100 - d) = 8400
    $$

14. Divide both sides by 105:
    $$
    100 - d = \frac{8400}{105} = 80
    $$

15. Solve for $d$:
    $$
    d = 100 - 80 = 20\%
    $$

**Answer:** $20\%$

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