Let’s start with an example:

**Two boxes A and B each hold a few balls. The weight of the heaviest ball in A is less than the weight of the lightest ball in B. If the balls from both the boxes are put together, is the median weighted ball’s weight less than 327 grams?**

**1. A contains 47 balls and its heaviest ball weighs 330 grams**

**2. A and B put together contain 95 balls**

Readers, let’s try to take 30 seconds – 1 min to crack this 🙂

**Trick: For questions relating to combined median in two or more sets, sort the items spatially in an ascending order left to right.**

Algorithm : Let’s assume A_{1}, A_{2}, A_{3}, A_{4}, A_{5} are balls in A arranged in an ascending order and B_{1}, B_{2}, B_{3}, B_{4}, B_{5, }B_{6} are balls in B arranged in an ascending order. If we assume A_{5} < B_{1} i.e., heaviest ball in A is lighter than lightest ball in B and sort balls from both these boxes in ascending order, we will get something:

A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, B_{1}, B_{2}, B_{3}, B_{4}, B_{5, }B_{6} – we see that the median of this set is B_{1} i.e the item in the middle. In this example, we assumed A has 5 balls and B has 6 balls.

Let’s use the above simple representation to solve the original question.

Essentially, the question is asking us to establish:

M (median balls weight) < 327 grams.

In order to establish or refute the above, we need some way to approximate the value of the median i.e., the LHS.

In our algorithm above, if we know the # of balls in A and # of balls in B – we can spatially distribute them and figure out the middle number. Still does not give us the value of the median but may help us orient ourselves.

Statement (1) by itself does not help – to find out where the median of the combined balls falls – we will need to know the count of the balls in A AND in B. A containing 47 balls may contain the median ball or it may not.

If we do not know where the median balls falls, we cannot establish if it is less than 327 grams

Statement (2) by itself is not helpful. It tells us nothing about the position of median or value of the magnitude of weights.

Statement (1) and (2) combined tells us that B has (95 – 47) = 48 balls.

If we spatially distribute A and B balls:

A_{1}, A_{2}, …., A_{47}, B_{1}, B_{2}, …., B_{48} we know B_{1} is the median which also happens to be the lightest ball in box B. We also know A_{47} = 330 grams from statement (1)

Now, B_{1} (i.e median) > 330 grams means the median cannot be less than 327 grams. Hence, both statement (1) and (2) put together allow us to refute the question statement. We go with **option (C)**