Arithmetic

Mixture & Alligation

Medium

applied

The water and milk in two vessels are in the ratio:1:1 and 3:8 respectively. In what ratio, the mixtures in the vessels be mixed to obtain a new mixture containing water and milk in the ratio 4:7?

1:4

2:3

3:4

4:1

✅ Correct Option: 2

Vessel 1 has water and milk in the ratio 1:1 (equal parts water and milk)

Vessel 2 has water and milk in the ratio 3:8 (less water, more milk)

The target mixture needs water and milk in the ratio 4:7

For Vessel 1 (1:1):

Total parts = 1 + 1 = 2

Water fraction = $\frac{1}{2}$

Milk fraction = $\frac{1}{2}$

For Vessel 2 (3:8):

Total parts = 3 + 8 = 11

Water fraction = $\frac{3}{11}$

Milk fraction = $\frac{8}{11}$

For Target Mixture (4:7):

Total parts = 4 + 7 = 11

Water fraction = $\frac{4}{11}$

Milk fraction = $\frac{7}{11}$

Let Vessel 1 and Vessel 2 be mixed in the ratio $x:y$

The water fraction in the final mixture:

$x \times \frac{1}{2} + y \times \frac{3}{11} = \frac{4}{11} \times (x + y)$

$\frac{x}{2} + \frac{3y}{11} = \frac{4(x + y)}{11}$

$11 \times \left(\frac{x}{2} + \frac{3y}{11}\right) = 4(x + y)$

$\frac{11x}{2} + 3y = 4x + 4y$

$\frac{11x}{2} - 4x = 4y - 3y$

$\frac{11x - 8x}{2} = y$

$\frac{3x}{2} = y$

$\frac{x}{y} = \frac{2}{3}$

The mixtures should be mixed in the ratio 2:3

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Solution

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