Blog

Parin Sir > Quantitative Aptitude (Maths) > ALLIGATION OR MIXTURE > Examples(2) > Example(2)

Example(2)

No Comments

click to previous

(11) Two vessels  A and B contain milk and water mixed in the ratio  7 : 2 and  7 : 5 respectively. Find the ratio in which these mixtures are to be mixed to get a new mixture containing milk and water in the ratio  2 : 1.

Solution.

Thus, mixture of  vessel  A  to  vessel  B  =  \frac{15}{180}:\frac{20}{180}   :    =  3  :  4

 (12) A container contains 100 kg. of  milk. From this container, 10 kg. of  milk was taken out and replaced by water. This process was further repeated two times. How much milk is now contained by the container ?

Soln. Here,  x = 100, y = 10 and n = 3

Amount of  milk left (a)

=    x\left (1-\frac{10}{100} \right )^{n}

=    100\left (1-\frac{10}{100} \right )^{3} kg.

=  100 \left (\frac{9}{10} \right )^{3}  kg.     =     72.9  kg.

(13) 7 litres are drawn from a cask full of  wine, and it is then filled with water. This process was further repeated once.  Now the quantity of  wine left and the water in it are in the ratio  16 : 9. Find the capacity of  the cask.

Soln. Here,  wine left  :  the water   =   16 : 9

\therefore Wine left (a)  :  Total mixture ( at the end ) (x)   =   16  :  25   ( \because 16 + 9  =  25 )

\therefore    a  :  x   =   16  :  25

\therefore  Total mixture ( at the end )  =  Total wine  ( initially )  =   x   =   25k,

And  we are given,  y  =  7  and  n  =  2

Amount of  milk left (a) =   x1-\left (\frac{y}{x} \right )^{n}

\therefore \frac{1}{x}= \left (\frac{x-y}{x} \right )^{2}

\therefore \frac{16}{25}= \left (\frac{25k-7}{25k} \right )^{2}

\therefore \frac{4}{5}= \frac{25k-7}{25k}

\therefore 4 ( 25 k )   =   5 ( 25 k – 7 )

\therefore 100 k   =   125 k – 35

\therefore 35   =   125 k  –  100 k

\therefore 25 k   =   35

\therefore x   =   25k   =   35

 

Share

Leave a Reply