## Quant - The Cunning of Question Makers (Variables Needn't be Integers)

We know that most general standardized tests focus on the simple essentials of Quant. Advanced topics such as derivatives, complex numbers, matrices etc are usually not included. All fundamentals are from the high school Math books which each one of us has gone through. So it would be natural to think that  Quant should not pose much problem for most test-takers and for engineering students, it should be particularly easy - they have actually covered far more advanced Math during their studies. Hence it comes as a shock when many test-takers, including engineering students, get a dismal score in the first practice test they take! Of course, with practice, they wise up to the cunning of question makers but till then, the Quant section is responsible for many a nightmare!

Today, let’s see what kind of treachery we are talking about - it makes some people laugh out loud and others pull at their hair! We start with a Data Sufficiency question.

Question: Is the product pqr divisible by 12?

Statement 1: p is a multiple of 3

Statement 2: q is a multiple of 4

It seems like an easy (C), doesn’t it? p is a multiple of 3 and q is a multiple of 4. So together, p*q is a multiple of 3*4 = 12.

If p * q is already a multiple of 12, then obviously, it would seem, that p*q*r would be a multiple of 12 too.

But here is the catch - where is it mentioned that r must be an integer? Just because p and q are integers (multiples of 3 and 4 respectively), it does not imply that r must be an integer too.

If r is an integer, then sure, p*q*r will be divisible by 12.

Now imagine that p = 3, q = 4 and r = 1/12

Then the product p*q*r = 3*4*(1/12) = 1

1 is not divisible by 12 i.e. in this case, pqr is not divisible by 12.

Answer here is both statements together are not sufficient to answer the question!

The question is very basic but tricks us because we assume that p, q and r are clean integer values.

On the same lines, try the next one.

Question: If 10^a * 3^b * 5^c = 450^n, what is the value of c?

Statement 1: a is 1.

Statement 2:  b is 2.

The first thing most of us will do is split 450 into its prime factors:

450 = 2 * 3^2 * 5^2

450^n = 2^n * 3^2n * 5^2n

and do the same thing with the left hand side

10^a * 3^b * 5^c = 2^a * 3^b * 5^(a+c)

Bringing the equation back, we get: 2^a * 3^b * 5^(a+c) = 2^n * 3^2n * 5^2n

Statement 1: a is 1.

Equating the power of 2 on both sides, we see that a = n = 1

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

Statement 2:  b is 2.

Equating the power of 3 on both sides, we see that b = 2n = 2 so n = 1.

If n = 1, a = 1 by equating the powers of 2 on both sides.

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

So it seems that both statements are separately sufficient. But hold on - again, the variables don’t need to be cleanly fitting integers.

The variables could pan out the way discussed or very differently.

Say, n = 1.

When statement 1 gives you that a = 1, you get 10^1 * 3^b * 5^c = 450^1

3^b * 5^c = 45

Now note that value of c depends on the value of b, which needn’t be 2.

If b  = 3,

3^3 * 5^c = 45

5^c = 45/27

c will take a non integer value here.

c = .3174 (using a calculator)

The question does not mention that all variables are integers and hence there are infinite values that c can take depending on the values of b.

Similarly, statement 2 alone is also not sufficient.

Using both statements together, you get

2^a * 3^b * 5^(a+c) = 450^n

2^1 * 3^2 * 5^(1 + c) = 450^n

5^(1 + c) = 450^n/18

We are sure you must have realised by now that depending on the value of n, c can take infinite different values.

If n = 1, c = 1

If n = 2, c = 4.8 (Using a calculator)

and so on…

We don’t need to actually find these values. It is enough to know that different values of n will give different values of c.

Hence again, both statements together are not sufficient.

Hopefully, in future, you will not make the mistake of assuming that variables must be integers.