1.

A bottle full of Brandy contains 40% alcohol. A part of this Brandy was replaced by another one having 19% alcohol and the percentage now became 26%. What was the quantity of Brandy replaced?1). 2/32). 1/33). 1/24). 5/6

Answer»

Concentration of Alcohol in the FIRST BOTTLE = 40%

Concentration of Alcohol in the SECOND bottle = 19%

We know that,

$(\frac{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;cheaper\;substance}}\left( {\rm{c}} \right)}}{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;dearer\;substance}}\left( {\rm{d}} \right)}} = \frac{{{\rm{Mean\;price\;}} - {\rm{\;c\;}}}}{{{\rm{d\;}} - {\rm{Mean\;price}}}})$

For first bottle,

26 - 19 = 7

For second bottle,

40 - 26 = 14

Hence the RATIO is 1 ? 2.

The part of BRANDY replaced = $(\frac{2}{3})$

∴ The part of Brandy replaced is 2/3


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