## Algebra - Hidden Constraints in Inequalities

Many of us are hooked on to using algebra in Quant questions. The thought probably is that how can it be a Quant question if one did not need to take a couple of variables and make a couple of equations/inequalities. I love to harp on about how algebra is time consuming and unnecessary in most cases. But today we will go one step further and discuss how indiscriminate use of algebra can actually result in incorrect answers. Surprised, eh?

Of course if you make correct equations and solve them correctly, there is no reason you shouldn’t get the correct answer. The problem that arises is in making correct equations/inequalities. Now I am sure you are thinking that you know how to make equations and so probably this post is a waste of your time. Hold on to that thought – Let me give you a statement:

The total number of apples is more than 20.

How will you convert it in terms of algebra? Will you write it as ‘Na > 20’? If this is what you did, please do go through the post. I am sure there will be a couple of things you will find interesting.

Let me take an example to show you why something like this may not be enough.

Question: Harry bought some red books that cost \$8 each and some blue books that cost \$25 each. If Harry bought more than 10 red books, how many blue books did he buy?

Statement 1: The total cost of blue books that Harry bought was at least \$150.
Statement 2: The total cost of all books that Harry bought was less than \$260.

Solution: The first thing we will do is look at the most common algebra solution.

Let the number of blue books be B and red books be R.

He bought more than 10 red books so R > 10.

Statement 1: The total cost of blue books that Harry bought was at least \$150

25B >= 150

So B >= 6

This statement alone is not sufficient to give the actual value of B.

Statement 2The total cost of all books that Harry bought was less than \$260

8R + 25B < 260 …………. (I)
Also, 10 < R (from above) which gives 80 < 8R …………..(II)
Adding (I) and (II), we get (note that the two inequalities have the same sign ‘<‘ so they can be added)
8R + 25B + 80 < 260 + 8R

B < 7.2

This statement alone is not sufficient to give the actual value of B.

Using both statements together, we get that B >= 6 and B < 7.2

So B could be 6 or 7 (since number of blue books must be an integer value)

This is incorrect and actually, answer is (C). The question is how? There is no calculation mistake in the above given solution. Then why do we get the incorrect answer? Let me give you the logical solution and prove that answer is actually (C). Then we will discuss why algebra fails us here.

Logical Solution:

Red Books – \$8 each

Blue Books – \$25 each

No of Red books is more than 10.

Statement 1: The total cost of blue books that Harry bought was at least \$150

Blue books cost \$25 each so he bought at least 6 books. He could have bought more too. This statement alone is not sufficient.

Statement 2The total cost of all books that Harry bought was less than \$260

He bought more than 10 red books so he bought at least 11 red books. He spent at least \$8*11 = \$88 on the red books. Out of the total 260, he is left with 260 – 88 = \$172 for  the blue books. Since each blue book costs \$25, he could have bought at most 6 blue books. He could have bought fewer too. This statement alone is not sufficient.

Using both statements together: He bought at least 6 and at most 6 blue books. So he must have bought 6 blue books.

I think you would have figured out the problem with the algebra solution by now. In the algebra solution, the inequality 10 < R does not include all the information you have available. You know that R cannot be 10.5 or 10.8 i.e. a decimal. It must be an integer since it represents the number of red books. So you might want to use 11 <= R to get a tighter value for B. Mind you, it is true that R is greater than 10. Important is that it is equal to or greater than 11 too.

Hence the analysis of statement 2 changes a little:

8R + 25B < 260 ………(I)
11 <= R which gives us 88 <= 8R …………. (II)

When you add (I) and (II) now, you get 25B < 172 i.e. B < 6.9. So B must be 6 or less.

This gives us enough information such that when considering both statements together, we get B = 6.

So when you use algebra, be mindful of the hidden constraints.