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Examples 6 to 11

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(6) A bag contains rupee,  50 paise coins and 25 paise coins in the ratio 10 : 17 : 16.  If the total amount is Rs. 90, find the number of  coins of  each kind.

Soln. Ratio of  values of  the coins  =  \frac{10}{1}:\frac{17}{2}:\frac{16}{4}=\frac{10}{1}:\frac{17}{2}:4   =      20 : 17 : 8.

Value of  1 rupee coins  =  Rs. \left (90 \times \frac{20}{45} \right )   =   Rs. 40.

Value of  50-paise coins  =  Rs. \left (90 \times \frac{17}{45} \right )   =   Rs. 34.

Value of  25-paise coins  =  Rs. \left (90 \times \frac{8}{45} \right ) =  Rs.  16.

\therefore Number of  1-rupee coins  =  40 ;

Number of  50-paise coins  =  ( 34 \times 2 )  =  68 ;

Number of  25-paise coins  =  ( 16 \times 4 )  =  64.

(7) In a mixture of  35 litres, the ratio of  milk and water is 5 : 2.  If  5  litres of  water is added to the mixture, find the ratio of  milk and water in the new mixture.

Soln. Milk  in given mix. = \left (\frac{5}{7}\times 35 \right ) = 25 litres, water in this mix. = 35 – 25  = 10 litres.

After adding 5 litres of  water to this mixture, it contains 25 litres of  milk and 15 litres of  water.

So, ratio of  milk and water in new mix.  =  25 : 15  =  5 : 3.

(8) A and B are two alloys of  gold and copper prepared by mixing metals in the ratio 5 : 2 and 4 : 3 respectively.  If  equal quantities of  alloys are melted to form a third alloy  C, find the ratio of  gold and copper in  C.

Soln. Let  1 kg of  each one of  A  and  B be taken to form  2 kg of  C.

Ratio of  gold and copper in C  = \frac{9}{7}: \frac{5}{7}  :   =  9 : 5.

(9) Find three numbers in the ratio  3 : 2 : 5, the sum of  whose squares is 342.

Soln. Let the numbers be  3k,  2k  and  5k.

Here,  (3k)^{2} +  (2k)^{2}  + (5k)^{2}   =  342

38K^{2}  =   342

k^{2}   =   9

k  =  \pm  3

The numbers are  9, 6  and  15   or  – 9, – 6 and  – 15.

(10) A mixture contains alcohol and water in the ratio  5 : 2.  If  42  litres of  water is added to it, the ratio of  alcohol and water becomes  2 : 5. Find the quantity of  alcohol  in the mixture.

Soln. Let alcohol  =  5k   and   water  =  2k.

Then    =  \frac{5k}{2k+42}=\frac{2}{5}\Leftrightarrow    25k  =  2 ( 2k + 42 ) \large \Leftrightarrow  k  =  4.

Alcohol  =  5k  =  20 litres.

(11) A quantity x varies as a cube of  y and  inversely as square of  z. Given that x = 256 when y = 2 and z = 3. Find the value of  z when x = 32 and  y = 4.

Soln. Here,  x \propto \frac{y^{3}}{z^{2}}

x   =  \frac{ky^{3}}{z^{2}}   –––  ( 1 )   ;    k  \ne  0.

 

Best Of Luck

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