Question

From a number of apples a man sells half number existing apples plus 1 to first customer,sells 1/3rd of the remaining apples plus 1 to second customer and 1/5th of the remaining apples plus 1 to third customer. He finds that he has 3 apples remaining. How many apples he originally had

Answer 1

Answer:

20 apples

Step-by-step explanation:

Let the number of apples = x

First customer:

Sold apples =  (1/2) of x + 1

$=\frac{x}{2}+1$

Remaining apples

$=x -(\frac{x}{2}+1)\\\\=x - \frac{x}{2}-1\\\\=\frac{x}{2}-1$

Second customer:

Sold apples = (1/3) of remaining apples + 1

$=\frac{1}{3}*(\frac{x}{2}-1)+1\\\\=[\frac{1}{3}*\frac{x}{2}]- \frac{1}{3}*1+1\\\\=\frac{x}{6}-\frac{1}{3}+1\\\\=\frac{x}{6}-\frac{1}{3}+\frac{3}{3}\\\\=\frac{x}{6}+\frac{2}{3}$

Remaining apples =

$=(\frac{x}{2}-1)-(\frac{x}{6}+\frac{2}{3})\\\\=\frac{x}{2}-1-\frac{x}{6}-\frac{2}{3}\\\\=\frac{x}{2}-\frac{x}{6}-1-\frac{2}{3}\\\\=\frac{3x}{6}-\frac{x}{6}-\frac{3}{3}-\frac{2}{3}\\\\=\frac{2x}{6}-\frac{5}{3}\\\\=\frac{x}{3}-\frac{5}{3}$

Third customer:

Sold apples = (1/5) of reaming apples +1

$=\frac{1}{5}*[\frac{x}{3}-\frac{5}{3}]+1\\\\=\frac{1}{5}*\frac{x}{3}-\frac{1}{5}*\frac{5}{3}+1\\\\=\frac{x}{15}-\frac{1}{3}+1\\\\=\frac{x}{15}-\frac{1}{3}+\frac{3}{3}\\\\=\frac{x}{15}+\frac{2}{3}$

Remaining apples =

$=\frac{x}{3}-\frac{5}{3}-[\frac{x}{15}+\frac{2}{3}]\\\\=\frac{x}{3}-\frac{5}{3}-\frac{x}{15}-\frac{2}{3}\\\\=\frac{x}{3}-\frac{x}{15}-\frac{5}{3}-\frac{2}{3}\\\\=\frac{5x}{15}-\frac{x}{15}-\frac{7}{3}\\\\=\frac{4x}{15}-\frac{7}{3}$

Remaining apples with the man = 3

$\frac{4x}{15}-\frac{7}{3}=3\\\\\frac{4x}{15}=3+\frac{7}{3}\\\\\frac{4x}{15}=\frac{9}{3}+\frac{7}{3}\\\\\frac{4x}{15}=\frac{16}{3}\\\\x=\frac{16}{3}*\frac{15}{4}\\\\x=4*5$

x = 20

He had 20 apples