Question

A man whose age is 45 yrs has 3 sons named john, jill, jack. He went to a park weekly twice. He loves his sons ver y much. On a certain day he found the shop keepers selling different things. An apple cost 1penny, 2chocalate costs 1penny & 3 bananas cost 1 penny. He has bought equal number of apple, chocolate & banana for each son. If the total amount he invest is 7 penny then how many he has bought from each piece for his son?

Answer 1

Answer:

The man has bought 3 apple, 6 chocolates and 3 bananas (that is, one apple, two chocolates and one banana for each son).

Step-by-step explanation:

First, we are assuming that he buys at least one of each items (if not, we have had many possible ways; for example, buying 21 bananas or 3 apples and 12 bananas).

We need the number of each item to be a multiple of 3 so he could share equally between his sons. Then, he needed to buy at least 3 apples (and spended at least 3 pennies) and at least 3 pairs of chocolates (and spended at least 3 more pennies). A priori, he could have spent as much money as he wanted in bananas, once it is sold in packs of 3.

If $a$ represents how many pennies he spent with apples, $b$ with bananas and $c$ with chocolates, then we have, in particular,

$a+b+c = 7$ and $a \geq 3, b\geq 3$ and $c\geq 1$.

So the only possible way to solve this problem is with $a = b = 3$ and $c=1$.