When Not to Use Number Plugging | Number Properties

Today we discuss the kind of questions which beg you to stay away from number plugging (but somehow, people still insist on using it because they see variables).

Not every question with variables is suitable for number plugging. If there are too many variables, it can be confusing and error prone. Then there are some other cases where number plugging is not suitable. Today we discuss an official question where you face two of these problems.

Question: If ${m}$ , ${p}$ , ${s}$ and ${v}$ are positive, and ${\frac{m}{p} < \frac{s}{v}}$ , which of the following must be between ${\frac{m}{p}}$ and ${\frac{s}{v}}$ ?

I. ${\frac{m+s}{p+v}}$

II. ${\frac{ms}{pv}}$

III. ${\frac{s}{v} - \frac{m}{p}}$

(A) None

(B) I only

(D) III only

(E) I and II both

Solution: The moment people see $m$ , $p$ , $s$ and $v$ variables, they jump to $m = 1$ , $p = 2$ etc.

But two things should put you off number plugging here:

– There are four variables – just too many to plug in and manage.

– The question is a “must be true” question. Plugging in numbers is not the best strategy for ‘must be true’ questions. If you know that say, statement 1 holds for some particular values of $m$ , $p$ , $s$ and $v$ (say, 1, 2, 3 and 4), that’s fine but how do you know that it will be true for every set of valid values of $m$ , $p$ , $s$ and $v$ ? You cannot try every set because the variables can take an infinite variety of values. If you find a set of values for the variables such that statement 1 does not hold, then you know for sure that it may not be true. In this case, number plugging does have some use but it may be a while before you can arrive at values which do not satisfy the conditions. In such questions, it is far better to take the conceptual approach.

We can solve this question using some number line and averaging concepts.

We are given that $\frac{m}{p} < \frac{s}{v}$

This means, this is how they look on the number line:

…………. 0 ……………….. $\frac{m}{p}$ …………………… $\frac{s}{v}$ ……………..

(since $m$ , $p$ , $s$ and $v$ are all positive (not necessarily integers though) so $\frac{m}{p}$ and $\frac{s}{v}$ are to the right of 0)

Let’s look at statement II and III first since they look relatively easy.

II. $\frac{ms}{pv}$

Think of the case when $\frac{m}{p}$ and $\frac{s}{v}$ are both less than 1. When you multiply them, they will become even smaller.

Say $.2\times.3 = .06$ . So the product $\frac{ms}{pv}$ may not lie between $\frac{m}{p}$ and $\frac{s}{v}$ .

Tip: When working with number properties, you should imagine the number line split into four parts:

– less than -1

– between -1 and 0

– between 0 and 1

– greater than 1

Numbers lying in these different parts behave differently. You should have a good idea about how they behave.

III. $\frac{s}{v} - \frac{m}{p}$

Think of a case such as this:

…………. 0 ………………………… $\frac{m}{p}$ … $\frac{s}{v}$ ……….

$\frac{s}{v} - \frac{m}{p}$ will be much smaller than both $\frac{m}{p}$ and $\frac{s}{v}$ and will lie somewhere “here”:

…………. 0 ……… here ………………… $\frac{m}{p}$ … $\frac{s}{v}$ ……….

So the difference between them needn’t actually lie between them on the number line.

Hence $\frac{s}{v} - \frac{m}{p}$ may not be between $\frac{m}{p}$ and $\frac{s}{v}$ .

I. $\frac{m+s}{p+v}$

This is a little tricky. Think of the four numbers as $N_1$ , $N_2$ , $D_1$ , $D_2$ for ease and given fractions as $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ .

$\frac{m+s}{p+v}$

$= \frac{\frac{m+s}{2}}{\frac{p+v}{2}}$

$= \frac{Average\ of\ N_1\ and\ N_2}{Average\ of\ D_1\ and\ D_2}$

Now average of the numerators will lie between $N_1$ and $N_2$ and average of the denominators will lie between $D_1$ and $D_2$ . So $\frac{Average\ of\ N_1\ and\ N_2}{Average\ of\ D_1\ and\ D_2}$ will lie between $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ . Try to think this through.

We will try to explain this but you must take some examples to ensure that you understand it fully. When is one fraction smaller than another fraction?

When $\frac{N_1}{D_1} < \frac{N_2}{D_2}$ , one of these five cases will hold:

– $N_1 < N_2$ and $D_1 = D_2$ . For example: $\frac{2}{9}$ and $\frac{4}{9}$

$\frac{Average\ of\ numerators}{Average\ of\ denominators} = \frac{3}{9}$ (between $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ )

– $N_1 < N_2$ and $D_1 > D_2$ . For example: $\frac{2}{11}$ and $\frac{4}{9}$

$\frac{Average\ of\ numerators}{Average\ of\ denominators} = \frac{3}{10}$ (between $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ )

– $N_1 << N_2$ and $D_1 < D_2$ . For example: $\frac{2}{9}$ and $\frac{20}{19}$ i.e. $N_1$ is much smaller than $N_2$ as compared with $D_1$ to $D_2$ .

$\frac{Average\ of\ numerators}{Average\ of\ denominators} = \frac{11}{14}$ (between $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ )

– $N_1 = N_2$ but $D_1 > D_2$ . For example: $\frac{2}{9}$ and $\frac{2}{7}$

$\frac{Average\ of\ numerators}{Average\ of\ denominators} = \frac{2}{8}$ (between $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ )

– $N_1 > N_2$ but $D_1 >> D_2$ . For example: $\frac{4}{9}$ and $\frac{2}{1}$

$\frac{Average\ of\ numerators}{Average\ of\ denominators} = \frac{3}{5}$ (between $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ )

In each of these cases, $\frac{average\ of\ N_1\ and\ N_2}{average\ of\ D_1\ and\ D_2}$ will be greater than $\frac{N_1}{D_1}$ but smaller than $\frac{N_2}{D_2}$ . Take some more numbers to understand why this makes sense. Note that you are not expected to conduct this analysis during the test. The following should be your takeaway from this question:

Takeaway: $\frac{Average\ of\ N_1\ and N_2}{Average\ of\ D_1\ and\ D_2}$ will lie somewhere in between $\frac{N_1}{D_1}$ and $\frac{N_2}{D_2}$ (provided $N_1$ , $N_2$ , $D_1$ and $D_2$ are positive)

$\frac{m+s}{p+v}$ must lie between $\frac{m}{p}$ and $\frac{s}{v}$ .

Answer (B).

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