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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.


Solution

(i) It is given that

Mixture of milk to water $$= 45\ litres$$

Ratio of milk to water $$= 13: 2$$

Sum of ratio $$= 13 + 2 = 15$$

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

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

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

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

By cross multiplication

$$39 = 18 + 3x$$

$$3x = 39 – 18 = 21$$

$$x = 21/3 = 7\ litres$$

Hence, $$7\ litres$$ 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 $$= \dfrac{5}{8} \times 560 = 350$$

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

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

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

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.


Maths

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