A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio Option 1: 10 : 3 Option 2: 1 : 1 Option 3: 3 : 10 Option 4: 10 : 13

Answer: 1 : 1. Solution: The ratio is 1:1 Initial setup: Glass: 500cc pure milk Cup: 500cc pure water Transfer 150cc milk from glass to cup After transfer: Glass: 500 - 150 = 350cc milk Cup: 500cc water + 150cc milk = 650cc mixture The cup now contains a mixture, not pure water. Find the mixture composition in the cup: In the 650cc mixture: Milk = 150cc Water = 500cc Milk fraction = 150/650 = 3/13 Water fraction = 500/650 = 10/13 Transfer 150cc mixture from cup back to glass Since we're moving 150cc of mixture: Milk transferred back = 150 x 3/13 = 450/13cc Water transferred = 150 x 10/13 = 1500/13cc Final amounts: Water in glass = 1500/13cc Milk in cup: Started with 150cc milk Lost 450/13cc milk (went back to glass) Remaining = 150 - 450/13 = (1950 - 450)/13 = 1500/13cc Water in glass : Milk in cup = 1500/13 : 1500/13 = 1:1 In mixture problems where equal volumes are transferred both ways, the amount of liquid A ending up in container B always equals the amount of liquid B ending up in container A.

Arithmetic

Mixture & Alligation

Easy

A glass contains $500 \mathrm{cc}$ of milk and a cup contains $500 \mathrm{cc}$ of water. From the glass, $150 \mathrm{cc}$ of milk is transferred to the cup and mixed thoroughly. Next, $150 \mathrm{cc}$ of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio

$10 : 3$

$1 : 1$

$3 : 10$

$10 : 13$

✅ Correct Option: 2

The ratio is $1:1$

Initial setup:

Glass: $500\text{cc}$ pure milk

Cup: $500\text{cc}$ pure water

Transfer $150\text{cc}$ milk from glass to cup

After transfer:

Glass: $500 - 150 = 350\text{cc}$ milk

Cup: $500\text{cc}$ water $+ 150\text{cc}$ milk $= 650\text{cc}$ mixture

The cup now contains a mixture, not pure water.

Find the mixture composition in the cup:

In the $650\text{cc}$ mixture:

Milk $= 150\text{cc}$

Water $= 500\text{cc}$

Milk fraction $= \dfrac{150}{650} = \dfrac{3}{13}$

Water fraction $= \dfrac{500}{650} = \dfrac{10}{13}$

Transfer $150\text{cc}$ mixture from cup back to glass

Since we're moving $150\text{cc}$ of mixture:

Milk transferred back $= 150 \times \dfrac{3}{13} = \dfrac{450}{13}\text{cc}$

Water transferred $= 150 \times \dfrac{10}{13} = \dfrac{1500}{13}\text{cc}$

Final amounts:

Water in glass $= \dfrac{1500}{13}\text{cc}$

Milk in cup:

Started with $150\text{cc}$ milk

Lost $\dfrac{450}{13}\text{cc}$ milk (went back to glass)

Remaining $= 150 - \dfrac{450}{13} = \dfrac{1950 - 450}{13} = \dfrac{1500}{13}\text{cc}$

Water in glass : Milk in cup $= \dfrac{1500}{13} : \dfrac{1500}{13} = 1:1$

In mixture problems where equal volumes are transferred both ways, the amount of liquid A ending up in container B always equals the amount of liquid B ending up in container A.

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