Question

From a number of mangoes, man sells half the number of existing mangoes plus 1 to the first customer, then he sells ${{\dfrac{1}{3}}^{rd}}$ of the remaining mangoes plus 1 to the second customer.
Then ${{\dfrac{1}{4}}^{th}}$ of the remaining number of mangoes plus 1 to the third customer and ${{\dfrac{1}{5}}^{th}}$ of the remaining mangoes plus 1 to the fourth customer. He then finds that he has no mangoes left.
How many mangoes were originally present?

Answer 1

Hint: Now let us say that the shopkeeper has x mangoes initially. Then the man sells half the number of existing mangoes plus 1 to the first customer. Hence we will form an expression and now the remaining mangoes will be the original number of mangoes – mangoes he sold. Then he sells ${{\dfrac{1}{3}}^{rd}}$ of the remaining mangoes plus 1 to the second customer. . Hence we will form an expression and now the remaining mangoes will be mangoes left with him before – mangoes he sold. Similarly we will calculate the remaining mangoes after selling it to third and fourth. Now he has no mangoes left after he sells to his fourth customer hence the number of mangoes remaining after the fourth customer should be equated to 0. Hence we will get an equation in x,
Solving this we will find the value of x.

Complete step-by-step answer:
Now let us say the man had x mangoes originally.
Now consider the mangoes sold to first customer
Man sells half the number of existing mangoes plus 1 to the first customer
Now man had originally x mangoes. Hence he now sells $\dfrac{x}{2}+1$
Now he had x mangoes and he sold $\dfrac{x}{2}+1$ mangoes. Hence he is left with $x-\left( \dfrac{x}{2}+1 \right)$ mangoes
Hence he has $x-\dfrac{x}{2}-1=\dfrac{2x-x-2}{2}=\dfrac{x-2}{2}$ mangoes remaining.
Now consider the mangoes sold to the second customer.
Man sells ${{\dfrac{1}{3}}^{rd}}$ of the remaining mangoes plus 1 to the second customer
Now the remaining mangoes were $\dfrac{x-2}{2}$ hence he sells $\dfrac{1}{3}\left( \dfrac{x-2}{2} \right)+1$ mangoes
Hence now the remaining mangoes are $\dfrac{x-2}{2}-\left( \dfrac{1}{3}\left( \dfrac{x-2}{2} \right)+1 \right)$
Let us simplify the expression $\dfrac{x-2}{2}-\dfrac{1}{3}\left( \dfrac{x-2}{2} \right)-1$
Now taking LCM we get
\[\begin{align}
  & \dfrac{3\left( x-2 \right)}{6}-\left( \dfrac{x-2}{6} \right)-\dfrac{6}{6} \\
 & \Rightarrow \dfrac{3x-6-x+2-6}{6} \\
 & \Rightarrow \dfrac{2x-10}{6} \\
\end{align}\]
Hence the shopkeeper is left with $\dfrac{2x-10}{6}$ mangoes.
Now consider the mangoes sold to third customer
Man sells ${{\dfrac{1}{4}}^{th}}$ of the remaining number of mangoes plus 1 to the third customer
The Man had $\dfrac{2x-10}{6}$ remaining mangoes. Hence he sold $\dfrac{1}{4}\left( \dfrac{2x-10}{6} \right)+1$ mangoes.
Now the man has $\dfrac{2x-10}{6}-\left( \dfrac{1}{4}\left( \dfrac{2x-10}{6} \right)+1 \right)$ mangoes
Again let us simplify the expression $\dfrac{2x-10}{6}-\dfrac{1}{4}\left( \dfrac{2x-10}{6} \right)-1$
Taking LCM we get
\[\begin{align}
  & \dfrac{8x-40}{24}-\left( \dfrac{2x-10}{24} \right)-\dfrac{24}{24} \\
 & \Rightarrow \dfrac{8x-2x-40+10-24}{24} \\
 & \Rightarrow \dfrac{6x-54}{24} \\
\end{align}\]
Hence now the man is left with \[\dfrac{6x-54}{24}\] mangoes.
Now let us consider the mangoes sold to the fourth customer.
Now he sells ${{\dfrac{1}{5}}^{th}}$ of the remaining mangoes plus 1 to the fourth customer.
The remaining mangoes were \[\dfrac{6x-54}{24}\] hence the mangoes he sold was \[\dfrac{6x-54}{24}-\left( \dfrac{1}{5}\left( \dfrac{6x-54}{24} \right)+1 \right)\]
Now let us simplify the expression \[\dfrac{6x-54}{24}-\left( \dfrac{1}{5}\left( \dfrac{6x-54}{24} \right)+1 \right)\]
Now taking LCM we get
\[\begin{align}
  & \dfrac{30x-270}{120}-\left( \dfrac{6x-54}{120} \right)-\dfrac{120}{120} \\
 & \Rightarrow \dfrac{30x-6x-270+54-120}{120} \\
 & \Rightarrow \dfrac{24x-336}{120} \\
\end{align}\]
Hence the man has \[\dfrac{24x-336}{120}\] remaining mangoes.
Now after this he was left with no mangoes.
Hence this must be equal to 0
Hence we get
\[\dfrac{24x-336}{120}=0\]
Hence we have $24x=336$
Dividing by 24 on both sides we get x = 14.
Hence the man had 14 mangoes originally.

Note: First remember that the mangoes remaining after each transaction is the mangoes that were there before – mangoes sold. And in each step check that the remaining mangoes will change and the ratio will be taken with respect to remaining mangoes.