A person loses 75% of his money in the first bet, 75% of the remaining money in the second bet, and 75% of the remaining money again in the third bet, and finally returns home with Rs 2. What was his initial amount of money?

Introduction / Context:
This question deals with repeated percentage losses applied successively to a quantity, in this case money used for betting. Instead of a single reduction, the person repeatedly loses 75% of the current amount. At the end, we are given the final amount left and asked to work backwards to find the original amount. These types of questions highlight how repeated losses compound and how to reverse such processes using inverse operations.

Given Data / Assumptions:

• A person starts with some initial amount of money, say X rupees.
• In the first bet, he loses 75% of X, so 25% of X remains.
• In the second bet, he loses 75% of the remaining amount, so only 25% of that remains.
• In the third bet, he again loses 75% of the current amount, so 25% of that remains.
• After the third bet, he is left with Rs 2.

Concept / Approach:
Each loss of 75% means the remaining fraction is 25% or 1/4 of the previous amount. So after the first bet he has X * 1/4, after the second bet he has X * (1/4)^2 and after the third bet he has X * (1/4)^3. This final amount is given as Rs 2. We set up an equation X * (1/4)^3 = 2 and solve for X. Alternatively, we can work backwards by reversing the losses: if 2 is one quarter of the amount before the third bet, then the previous amount was 4 times 2, and so on.

Step-by-Step Solution:
Step 1: Let initial money be X rupees. Step 2: After first bet, he loses 75%, so he keeps 25%: amount = X * 25/100 = X/4. Step 3: After second bet, again he keeps only 25% of current amount: amount = (X/4) * 1/4 = X/16. Step 4: After third bet, again he keeps only 25%: amount = (X/16) * 1/4 = X/64. Step 5: Given final amount after third bet is X/64 = 2. Step 6: Solve for X: X = 2 * 64 = 128.

Verification / Alternative check:
Check by forward simulation. Start with Rs 128. After first loss of 75%, remaining money = 25% of 128 = 0.25 * 128 = 32. After second loss of 75%, remaining money = 25% of 32 = 8. After third loss of 75%, remaining money = 25% of 8 = 2. This matches the final amount given in the question, so the initial amount must indeed be Rs 128.

Why Other Options Are Wrong:

• Rs 64 would produce 64 → 16 → 4 → 1, which does not match the final amount of 2.
• Rs 256 would produce 256 → 64 → 16 → 4, again inconsistent with the required final amount.
• Rs 512 would give even larger values and still not end at 2 after three 75% losses.

Common Pitfalls:
Students sometimes try to subtract 75% three times from the original amount as if the loss were always calculated on X instead of the remaining amount. Others misinterpret 75% as leaving 75% instead of 25%. Another mistake is incorrect handling of powers of 1/4, for example using (1/4)^2 when three losses are given. Working with fractions carefully and checking with a forward simulation helps avoid these errors.

Final Answer:
The person initially had Rs 128.