Question

A greedy worker at a honey manufactory steals 10 litres from a full vat of honey, replacing it by an equal quantity of water to hide the crime. After some time he steals another 10 litres from the same vat, replacing it again by water. The ratio of honey to water is now 49:32. What is the capacity of the vat?​

Answer 1

Given:

A greedy worker steals 10 litres from a vat full of honey and adds water to it to hide his crime

He then repeats the same process once again

The ratio of honey to water now becomes 49:32

To find:

The capacity of the vat

Solution:

Let "x" litres be the capacity of the vat.

So,

The quantity of honey left in the vat after 2 operations = $x[1\:-\:\frac{10}{x} ]^2$

The new ratio of honey to water is given as = 49 : 32

Now, in order to find the capacity of the vat we will form an eq. as,

$\frac{Quantity\:of\: honey\: left\: in\: the\: vessel}{Capacity\: of\: the\: vat} = \frac{New\: quantity\: of \:honey}{Total\: quantity\: of\: mixture}$

∴ $\frac{x[1\:-\:\frac{10}{x} ]^2}{x} = \frac{49}{49+32}$

⇒ $[1\:-\:\frac{10}{x} ]^2 = \frac{49}{81}$

taking the square roots on both sides

⇒ $[1\:-\:\frac{10}{x} ] = \frac{7}{9}$

⇒  $[\frac{x\:-\:10}{x} ] = \frac{7}{9}$

⇒ 9[x - 10] = 7x

⇒ 9x - 90 = 7x

⇒ 9x - 7x = 90

⇒ 2x = 90

⇒ x = 45 litres

Thus, the capacity of the vat is 45 litres.

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