Question

(i) In a mixture of 45 litres, the ratio of milk to water is 13: 2. How much water must be added to this mixture to make the ratio of milk to water as 3: 1?

(ii) The ratio of the number of boys to the numbers of girls in a school of 560 pupils is 5: 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3: 2.

Answer 1

(i) It is given that

Mixture of milk to water $=45litres$

Ratio of milk to water $=13:2$

Sum of ratio $=13+2=15$

Here the quantity of milk $=(45\times 13)/15=39litres$

Quantity of water $=45\times 2/15=6litres$

Consider x litre of water to be added, then water $=(6+x)litres$

Here the new ratio $=3:1$

$39:(6+x)=3:1$

We can write it as

$\frac{39}{(6+x)}=\frac{3}{1}$

By cross multiplication

$39=18+3x$

$3x=39–18=21$

$x=21/3=7litres$

Hence, $7litres$ of water is to be added to the mixture.

(ii) It is given that

Ratio between boys and girls $=5:3$

Number of pupils $=560$

So the sum of ratios $=5+3=8$

We know that

Number of boys $=\frac{5}{8}\times 560=350$

Number of girls $=\frac{3}{8}\times 560=210$

Number of new boys admitted $=10$

So the total number of boys $=350+10=360$

Consider x as the number of girls admitted

Total number of girls $=210+x$

Based on the condition

$360:210+x=3:2$

We can write it as

$\frac{360}{210+x}=\frac{3}{2}$

By cross multiplication

$630+3x=720$

$3x=720–630=90$

So we get

$x=90/3=30$

Hence, $30$ new girls are to be admitted.