A man paid one half money in a shop. Coming out of the shop he discovered that he had just as many paise as he had rupees before he went in and half as many rupees as he had paise when he went in.

How much money did he have with him before he went into the shop?

You must of course is debarred from using the math tricks of standard algebra. Solution must be by math reasoning and other means.

**Recommended time to solve:** 20 minutes.

### Solution to half and half money math puzzle

Let us name the Rupees and Paise before and after as $R_{before}$, $P_{before}$, $R_{after}$ and $P_{after}$. This is not algebra, it is just a way of calling the four numbers we are to discover.

Two conditions are given,

- $R_{before}$ equals $P_{after}$ and,
- $R_{after}$ is half of $P_{before}$.

You can easily draw your first conclusion,

**Conclusion 1:**$P_{before}$ must be even and also all four of them must be less than 100.

What is the basis of the second part of the conclusion?

**Reason 1:** $P_{before}$ and $P_{after}$ both are less than 100 by the rule that 100 paise make 1 rupee. With this in mind when we consider that $R_{before}$ equals $P_{after}$ and $R_{after}$ is half of $P_{before}$ it dawns upon us that all four of them must be less than 100.

#### A trial with assumed value

To see how the four values behave with respect to one another, let us make a simple trial with $P_{before}$ as 40 paise.

By the second condition then, $R_{after}$ is 20 paise and also let us not forget the **zeroeth condition that money after was half of money before.**

For this overriding condition to be true, double of $P_{after}$ must create 40 paise. On two instances it can happen. $P_{after}$ may either be 20 paise or 70 paise. This implies, $R_{before}$ must be 20 rupees or 70 rupees **violating the zeroeth condition of money before double of money after.**

What can we conclude from this little experiment?

**Conclusion 2:** For all values of $P_{after}$ from 1 to 49 (and values of $P_{before}$ from 2 to 98) this violation of zeroeth condition will remain.

**Conclusion 3:** The only way this barrier can be overcome is to **create a carry of 1 rupee by doubling $P_{after}$** leading to the next realization that, $P_{after}$ must be more than 50.

What **should then be the mechanism** that will satisfy all three given conditions?

As money after when doubled must create a carry of 1 rupee by paise after doubling,

**Conclusion 4:** $R_{before}$ must be twice of $R_{after}$ plus 1.

But twice of $R_{after}$ equals $P_{before}$. So,

**Conclusion 5:** $R_{before}$ must be $P_{before}$ plus 1.

Before the man entered the shop, his number of rupees must have been 1 more than the number of paise. You sense that solution is very near and take up the **analysis of the mechanism of doubling the money and creating the carry of 1 rupee.**

By doubling $P_{after}$ a carry of 1 rupee is created. What does it mean? It means simply, twice of $P_{after}$ is 100 plus $P_{before}$ (as 1 rupee is 100 paise).

As $P_{after}$ equals $R_{before}$, we can conclude,

**Conclusion 6:** Twice of $R_{before}$ is 100 plus $P_{before}$.

Subtract the relation in Conclusion 5 from the relation in Conclusion 6. Result is,

**Conclusion 7:** $R_{before}$ equals 100 minus 1, that is 99. And so, $P_{before}$ is 98.

**Answer:** The amount of money the man had before he entered the shop was 99 rupees 98 paise.

You may now verify the result by algebraic means easily.

### Algebraic solution of half and half money math riddle

Let $R_1$, $P_1$, $R_2$, $P_2$ be the before and after rupee and paisa figures.

By the two give conditions 1 and 2,

$R_2 = \displaystyle\frac{P_1}{2}$ and $R_1=P_2$.

But money before must be double of money after. So,

$2(100R_2+P_2) = 100R_1 + P_1$

Or, $2(50P_1+R_1) = 100R_1 + P_1$

Or, $99P_1=98P_2$.

As $P_1$ and $P_2$ must both be less than 100 and 99 and 98 do not have any common factors,

$P_1 = 98$ and $P_2 = 99 = R_1$.

If you use algebra, there won't be any puzzle.

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