Arithmetic

Mixture & Alligation

Easy

A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture $P$ and replaced with same amount of cocoa powder to form a new mixture $Q$ . If the ratio of coffee and cocoa in the mixture $Q$ is $16: 9$ , then the ratio of cocoa in mixture $P$ to that in mixture $Q$ is

$5:9$

$4:9$

$1:3$

$1:2$

✅ Correct Option: 1

We start with pure coffee, then create two mixtures by repeatedly removing some amount and replacing it with cocoa powder. Let's work backwards from the final result to find our answer.

In mixture Q, coffee : cocoa = 16 : 9

This means:

Out of every 25 parts of mixture Q, 16 parts are coffee and 9 parts are cocoa

Coffee fraction in mixture Q = $\tfrac{16}{25}$

Cocoa fraction in mixture Q = $\tfrac{9}{25}$

When we remove a fraction of mixture and replace it with a pure substance, the concentration of the original substance gets multiplied by (1 - fraction removed).

Since this process happens twice with the same fraction removed each time:

Let's call the fraction removed each time = k

After first removal: Coffee fraction = (1 - k) × 1 = (1 - k)

After second removal: Coffee fraction = (1 - k) × (1 - k) = (1 - k)²

We know the final coffee fraction is $\tfrac{16}{25}$ , so:

$(1 - k)^2 = \tfrac{16}{25}$

$1 - k = \tfrac{4}{5}$

Therefore:

$k = 1 - \tfrac{4}{5} = \tfrac{1}{5}$

This means $\tfrac{1}{5}$ of the mixture is removed each time.

After the first removal and replacement:

Coffee remaining = $1 - \tfrac{1}{5} = \tfrac{4}{5}$

Cocoa added = $\tfrac{1}{5}$

So in mixture P: Cocoa fraction = $\tfrac{1}{5}$

Cocoa in mixture P = $\tfrac{1}{5}$

Cocoa in mixture Q = $\tfrac{9}{25}$

Ratio = $\tfrac{\text{Cocoa in P}}{\text{Cocoa in Q}} = \tfrac{\tfrac{1}{5}}{\tfrac{9}{25}}$

$\tfrac{1}{5} \div \tfrac{9}{25} = \tfrac{1}{5} \times \tfrac{25}{9} = \tfrac{25}{45} = \tfrac{5}{9}$

The ratio of cocoa in mixture P to that in mixture Q is 5:9.

For repeated mixture problems: If you remove fraction 'k' of mixture 'n' times and replace with pure substance, the original substance's final fraction = $(1-k)^n$ × initial fraction.

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