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# 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 lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book 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? Verified
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Hint: In this problem the lady spends money sequentially at different spaces. She spends a fraction of what is left on next purchase. This is the problem that will take us into a fractional equation giving the last solution that is money left one rupee.
Try to solve the problem along with the journey of the lady.

Consider the amount lady had initially $= x\,\,Rupees$
First, she spent half she had for hankie and gave one rupee to the beggar
Money spent:
= $\dfrac{x}{2} + 1$
Money left:
$= x - \left( {\dfrac{x}{2} + 1} \right) \\ \Rightarrow \dfrac{x}{2} - 1 \;$
Second, she spent half left on lunch with two rupee tip
Money spent:
$\Rightarrow \dfrac{{\dfrac{x}{2} - 1}}{2} + 2 \\ \Rightarrow \dfrac{x}{4} + \dfrac{3}{2} \;$
Money left:
 $\Rightarrow \dfrac{x}{2} - 1 - \left( {\dfrac{x}{4} + \dfrac{3}{2}} \right) \\ \Rightarrow \dfrac{x}{4} - \dfrac{5}{2} \;$
Third, purchase she spent half of what was left on a book and 3 rupees on bus fare
Money Spent:
$\Rightarrow \left( {\dfrac{{\dfrac{x}{4} - \dfrac{5}{2}}}{2} + 3} \right) \\ \Rightarrow \dfrac{x}{8} + \dfrac{7}{4} \;$
Money left:
$\Rightarrow \dfrac{x}{4} - \dfrac{5}{2} - \left( {\dfrac{x}{8} + \dfrac{7}{4}} \right) \\ \Rightarrow \dfrac{x}{8} - \dfrac{{17}}{4} \;$
Money she left with is one rupee
$\Rightarrow \dfrac{x}{8} - \dfrac{{17}}{4} = 1 \\ \Rightarrow \dfrac{x}{8} = \dfrac{{21}}{4} \\ \Rightarrow x = 42\,Rupees \;$
Hence, she had 42 rupees at the start of the journey.
So, the correct answer is “42 rupees”.

Note: The concept of fraction is widely used in many areas like calculating age, estimating profit loss, getting the share of profit of one among the three, etc. The simplification involves all mathematical operations on one variable. The use of multiple operations makes the problem tricky.