Question

Jaideep was given some money by his mother on his birthday. Jaideep spent all of it in five stores. In each store he spent one rupee more than half of what he had when he came in. How much did he get from his mother?

• A. Rs. 25
• B. Rs. 30
• C. Rs. 50
• D. Rs. 62

Hint:

Work backwards from the fact that nothing is left after the fifth store. If he leaves each store with \(\frac{y}{2}-1\), then what he had entering that store is \(2(\text{amount left}+1)\).

Answer 1

The Correct Option is D

Instead of working backwards from the last store, we can write the amount left after each store as a single algebraic expression in terms of the starting amount $x$, and solve directly.

1. After store 1: remaining $= \dfrac{x}{2} - 1$.
2. After store 2: remaining $= \dfrac{1}{2}\left(\dfrac{x}{2}-1\right) - 1 = \dfrac{x}{4} - \dfrac{3}{2}$.
3. After store 3: remaining $= \dfrac{1}{2}\left(\dfrac{x}{4}-\dfrac{3}{2}\right) - 1 = \dfrac{x}{8} - \dfrac{7}{4}$.
4. After store 4: remaining $= \dfrac{1}{2}\left(\dfrac{x}{8}-\dfrac{7}{4}\right) - 1 = \dfrac{x}{16} - \dfrac{15}{8}$.
5. After store 5: remaining $= \dfrac{1}{2}\left(\dfrac{x}{16}-\dfrac{15}{8}\right) - 1 = \dfrac{x}{32} - \dfrac{31}{16}$.

Since he has spent everything after the fifth store, this final remainder equals 0: $\dfrac{x}{32} - \dfrac{31}{16} = 0$, so $\dfrac{x}{32} = \dfrac{31}{16}$, giving $x = \dfrac{31}{16} \times 32 = 62$.

Notice the pattern in the fractions subtracted at each stage: $1, \frac{3}{2}, \frac{7}{4}, \frac{15}{8}, \frac{31}{16}$. Each numerator is one less than a power of 2 (1, 3, 7, 15, 31), a useful shortcut for repeat-halving problems like this one.

Let's summarize:

• Building one combined formula for the amount left after $n$ stores avoids repeated backward substitution.
• Setting that formula to 0 for $n=5$ gives $x = 62$ directly.
• Rs 30 fails the problem's own condition, since starting from Rs 30 he runs out of money after only four stores, not five.

So Jaideep received Rs 62 from his mother.