Money Problem Explained

Let the initial amount of money Shyam had be $R$ rupees and $P$ paise.

The total initial amount in paise is given by the formula:

Total initial paise $= 100 \times R + P$

Where $R$ is a non-negative integer representing the number of rupees and $P$ is an integer representing the number of paise, such that $0 \le P < 100$.

Shyam spent half of his money. So, the amount of money left is half of the initial amount.

Total amount left in paise $= \frac{100R + P}{2}$

Let the amount of money Shyam is left with be $A$ rupees and $B$ paise.

The total amount left in paise can also be expressed as:

Total amount left in paise $= 100 \times A + B$

Where $A$ is a non-negative integer representing the number of rupees left and $B$ is an integer representing the number of paise left, such that $0 \le B < 100$.

So, we have the main relationship:

$$100A + B = \frac{100R + P}{2}$$

Conditions on Amount Left

The problem provides two conditions about the amount of money Shyam is left with:

1. "left with as many as he had rupees before"
2. "but with half as many rupees as he had paise before."

These conditions are phrased in a somewhat ambiguous way. We need to interpret what they mean in terms of the number of rupees left ($A$), the number of paise left ($B$), the initial number of rupees ($R$), and the initial number of paise ($P$). Let's consider a possible interpretation that aligns with the provided answer options.

Based on checking the options, a plausible interpretation is:

• Condition 1 implies that the number of paise left ($B$) is equal to the initial number of rupees ($R$). So, $B = R$.
• Condition 2 implies that the number of rupees left ($A$) is equal to half the initial number of paise ($P/2$). So, $A = P/2$.

Let's verify if this interpretation is consistent with the rules for rupees and paise:

• If $B = R$, then the initial number of rupees ($R$) must be less than 100, since the number of paise left ($B$) must be less than 100 ($0 \le B < 100$). So, $0 \le R < 100$.
• If $A = P/2$, then the initial number of paise ($P$) must be an even non-negative integer, since the number of rupees left ($A$) must be an integer. So, $P$ is even and $P \ge 0$. Since $0 \le P < 100$, this condition is $0 \le P < 100$ and $P$ is even.

Solving the Equation

Now substitute these conditions ($A = P/2$ and $B = R$) into our main relationship:

$$100\left(\frac{P}{2}\right) + R = \frac{100R + P}{2}$$

Simplify the left side:

$$50P + R = \frac{100R + P}{2}$$

Multiply both sides by 2 to remove the fraction:

$$2(50P + R) = 100R + P$$

$$100P + 2R = 100R + P$$

Rearrange the terms to group $R$ and $P$:

$$100P - P = 100R - 2R$$

$$99P = 98R$$

Finding Initial Amount (R and P)

We need to find integer values for $R$ and $P$ that satisfy $99P = 98R$ and the constraints:

• $0 \le P < 100$
• $P$ is even
• $0 \le R < 100$

The equation is $99P = 98R$. Since 99 and 98 have no common factors (they are coprime), for this equation to hold with integers $R$ and $P$, $P$ must be a multiple of 98, and $R$ must be a multiple of 99.

So, $P = 98k$ and $R = 99k$ for some non-negative integer $k$.

Let's check the constraints for different values of $k$:

• If $k=0$: $P = 98 \times 0 = 0$, $R = 99 \times 0 = 0$. Constraints check: $0 \le 0 < 100$ (satisfied for P), $0$ is even (satisfied), $0 \le 0 < 100$ (satisfied for R). This gives an initial amount of 0 rupees and 0 paise. Amount left is 0 rupees and 0 paise. This is a valid solution but not among the options.
• If $k=1$: $P = 98 \times 1 = 98$, $R = 99 \times 1 = 99$. Constraints check: $0 \le 98 < 100$ (satisfied for P), $98$ is even (satisfied), $0 \le 99 < 100$ (satisfied for R). This gives a valid initial amount: 99 rupees and 98 paise.
• If $k=2$: $P = 98 \times 2 = 196$. This violates $P < 100$. Any higher value of $k$ will also violate this constraint.

So, the only non-zero initial amount satisfying the conditions and constraints is $R=99$ and $P=98$. Shyam initially had 99 rupees and 98 paise.

Calculating Amount Left

The amount left is $A$ rupees and $B$ paise, where $A=P/2$ and $B=R$.

Number of rupees left ($A$) $= P/2 = 98/2 = 49$.

Number of paise left ($B$) $= R = 99$.

So, the amount Shyam is left with is 49 rupees and 99 paise.

Verification

Let's verify this amount is indeed half of the initial amount.

Initial amount: 99 rupees 98 paise = $100 \times 99 + 98 = 9900 + 98 = 9998$ paise.

Half of initial amount $= 9998 / 2 = 4999$ paise.

Amount left: 49 rupees 99 paise = $100 \times 49 + 99 = 4900 + 99 = 4999$ paise.

The calculated amount left is indeed half of the initial amount.

The possible amount of money he is left with is 49 rupees and 99 paise, which matches one of the options.