Question

A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupees tip. She spent half of the remaining amount on books and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?

Answer 1

In such types of questions, the solution is obtained by simple unitary operations. Thus the students are advised to carefully note down the given data and establish a relationship between them. For the ease of understanding it is always better to assume the unknown quantity with a variable.

Formula used: To answer this question, we must have an idea about simple equations. This question tells that the expression on the left-hand side is equal to that on the right-hand side.
Hence, we can say $LHS{\text{ }} = {\text{ }}RHS$.
Further using the unitary operations to establish the relationship.

Complete step by step solution:
let us understand what the question provides and what it demands,
The question provides, A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupees tip. She spent half of the remaining amount on books and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left.
The question demands us to find,How much money did she start with?
Let us consider that the lady started with Rs x.
She spent half of what she had on buying hankies, that is Rs.$\dfrac{x}{2}$.
Hence, she is left with Rs.$x - \dfrac{x}{2} = \dfrac{x}{2}$.
Now, She gave $1$rupee to beggars outside the shop.
So now the remaining amount with her is Rs.$\left( {\dfrac{x}{2} - 1} \right)$.
She spent half of remaining at lunch followed by 2 rupees tip.
So, in total she she spent Rs.$\left( {\dfrac{{\dfrac{x}{2} - 1}}{2} + 2} \right)$
Money left with her after lunch is Rs.$\left( {\dfrac{{\dfrac{x}{2} - 1}}{2} - 2} \right)$
She spent half of the remaining on book and $3$ rupees on bus fare.
Thus, money left with her is Rs.$\left( {\dfrac{{\dfrac{{\dfrac{x}{2} - 1}}{2} - 2}}{2} - 3} \right)$
Finally, she has Rs. $1$left with her.
So, $\dfrac{{\dfrac{{\dfrac{x}{2} - 1}}{2} - 2}}{2} - 3 = 1$
$\Rightarrow \dfrac{{\dfrac{x}{2} - 1}}{2} - 2 = 8 \\\ \Rightarrow \dfrac{x}{2} - 1 = 20 \\\ \Rightarrow x = 42$

So, she started with Rs. 42.

Note: In simple equations, we involve letters as well numbers. Letters are used to replace some of the numbers where a numerical expression would be too complicated or where we want to generalize rather than use of specific numbers.
Students should approach such examples with proper steps to avoid mistakes in calculations.
Students are advised to mention units without fail.