Question

A jar full of whisky contains 40% alcohol. A part of Whisky is replaced by another containing 19% of alcohol and now the percentage of alcohol is found to be 26%. If the quantity of whisky replaced is k. Find 3k.

Answer 1

Hint: Initially take the volume of the jar as 100 ml and let’s denote whisky containing 40% alcohol as \[whisk{y_1}\] and whisky containing 19% of alcohol as \[whisk{y_2}\] .

Complete step-by-step answer:
 Initially the jar contains 100ml of \[whisk{y_1}\]. Given k ml of \[whisk{y_1}\] is replaced by \[whisk{y_2}\].
Let's take the jar of 100ml.
Initial condition: Jar contains 100 ml of whisky (containing 40% of alcohol) \[ \Rightarrow \] 40ml of alcohol is present in 100 ml of mixture .
Now k ml of\[whisk{y_1}\]k ml of\[whisk{y_2}\].
Final condition: Jar contains 100 ml of new whisky which is mixture of (100-k)ml \[whisk{y_1}\] and k ml of \[whisk{y_2}\] and it is given 26% of alcohol is present in new whisky in the jar.
Therefore , the total amount of alcohol present in new whisky is the amount of alcohol in \[whisk{y_1}\] and amount of alcohol in \[whisk{y_2}\] .
 I.e, Total amount of alcohol present in new whisky = amount of alcohol in \[whisk{y_1}\] + amount of alcohol in \[whisk{y_2}\] \[....\left( 1 \right)\]
Amount of alcohol in \[whisk{y_1}\]:
Amount of \[whisk{y_1}\]present in new whisky is (100-k) ml, given that \[whisk{y_1}\] contains 40% of alcohol.
Therefore, amount of alcohol in \[whisk{y_1}\]=
 (amount of \[whisk{y_1}\])(percentage of alcohol in \[whisk{y_1}\] ) \[\dfrac{1}{{100}}\]\[ = \left( {100 - k} \right)\left( {\dfrac{{40}}{{100}}} \right)......\left( 2 \right)\]
Amount of alcohol in \[whisk{y_2}\]:
Amount of \[whisk{y_2}\]present in new whisky is k ml, given \[whisk{y_2}\]contains 19% alcohol.
Therefore, amount of alcohol in \[whisk{y_2}\]=
 (amount of \[whisk{y_2}\])(percentage of alcohol in \[whisk{y_2}\] ) \[\dfrac{1}{{100}}\]\[ = \left( k \right)\left( {\dfrac{{19}}{{100}}} \right)......\left( 3 \right)\]
Given that 26% of alcohol is contained in the final mixture.
\[ \Rightarrow \]26 ml of alcohol in 100 ml of the new whisky. \[....\left( 4 \right)\]
substitute equation 2 , 3, 4 in the equation 1.
\[26 = \left( {100 - k} \right)\left( {\dfrac{{40}}{{100}}} \right) + \left( k \right)\left( {\dfrac{{19}}{{100}}} \right)\]
 \[26 = \dfrac{{\left( {4000 - 40x + 19k} \right)}}{{100}}\]
\[2600 = 4000 - 21k\]

\[ \Rightarrow 21x = 1400\]
\[ \Rightarrow 3k = 200\]
Hence, the value of 3k= 200.
\[ \Rightarrow 3k = 200\]

Note: We can take the jar of any volume according to our interest . Here 100 is taken for the convenience of reducing the calculation part and giving less scope to the calculation errors.
Students can take the volume of the jar as a variable and solve the question it does not disturb or interrupt the final answer