Q.

In Mrs. White’s kitchen, Mr. White accidentally spilled some milk into a jar of honey. In order to compensate for the spilling, he decided to replace 100 ml of the mixture with 100 ml of pure honey such that the ratio of milk to honey in the mixture becomes 5 : 19. If the jar initially had 450 ml of pure honey, then how much milk was spilled in the jar?

A. 25 ml

B. 150 ml

C. 50 ml

D. None of the above

B. 150 ml

This is a tough question. We will either have to proceed with using a variable. Or if we want to avoid variable & equations, then we will have to use options.

Approach 1: Using options …
[A]

If 25 ml was spilled, then the volume at end will be 450+25–100+100 = 475.

Dividing this in ratio 5 : 19, we will get fractions. So we will not continue with this option for now and check if any other option is easier.

[B]

If 150 ml was spilled, then the volume at end will be 450+150–100+100 = 600.

Dividing this in ratio 5 : 19, we get, milk = 125 and honey = 475.

These are the volumes after 100 ml of pure honey was added.

So the volumes just before adding honey would have been, milk = 125 and honey = 475 – 100 = 375 and total = 500

i.e. ratio of milk to honey is 1 : 3, before adding honey but after removing 100 ml of solution.

Removing a part of a solution does not change the ratio of milk to honey. Hence, ratio of milk and honey before removing 100 ml of solution would also have been 1 : 3 and the solution would have been 600 ml then. So, honey in the solution would have been 3/4 * 600 = 450 ml and milk in the solution would have been 1/4 * 600 = 150 (the spilled milk). All of this fits in with the data given.

Thus, this is the correct option.

Approach 2: Using variable.

To reduce our work, we will work on milk and total. As the working below shows, the work on total volume is very easy. And between milk and honey, milk has fewer changes … it is removed but is not added … whereas honey has more changes … honey is removed also and added also. As practice, we suggest you work with honey also and experience for yourself how the volume of honey adds to the work-load.

Milk Total
At start 0 450
After x ml of milk is spilled x 450+x
Now 100 ml of solution is removed from (450+x) ml. The same fraction, i.e. 100/(450+x), of milk will also be removed.
But we are interested in volume left in jar. So the fraction of volume left in jar will be (1 – fraction removed) i.e. (350+x)/(450+x), for both milk and total.
After 100 ml of solution is removed \(x \times \frac{350+x}{450+x}\) 350+x
After 100 ml of honey is added \(x \times \frac{350+x}{450+x}\)  450+x

And this ratio of milk to total is given as 5 : (5 + 19) i.e. 5 : 24.

So, we need to solve, \(\frac{x\left(350+x\right)}{{\left(450+x\right)}^2} = \frac{5}{24}\).

It is best that we now plug options and confirm which satisfies rather than solving this.

[A] (50 × 400) : (500 × 500) is 2 : 25

[B] (150 × 500) : (600 × 600) is 5 : 24. So this is the answer.