Why Multiple Methods are Important (Going from 48 to 51) | Algebra

A point to keep in mind while targeting Q50+ in GMAT: don’t buy complex official solutions. Most GMAT questions can be solved in a few steps. The point is that sometimes it is hard to identify those “few steps” and we keep going round and round in circles for a while till we arrive at the answer. The way to hit 51 is to look for simple solutions for difficult questions. The best example of this would be question number 148 of Official Guide 12. The question is tough, no doubt about it but just because it is tough, don’t think that the solution needs to be tough too – you don’t have to live with the solution provided.

If, even after reading the solution a couple of times, you know that if you try the question again in a week, you won’t be able to solve it on your own, this means you need to review either the concept or the solution. If the given solution is too complex and you almost have to learn it up step by step, it means you need a better solution. The next step of the solution should be apparent to you – you should be able to solve it on your own within two minutes.

Also, even if one method looks good, try to find other ways of solving the question. Often, there are multiple good ways of solving a particular question.

Here is a question similar to question number 148 of OG12. Let me give a few good methods of solving it:

Question: If ${x}$ , ${y}$ , and ${k}$ are positive numbers such that ${\frac{x}{(x+y)}\times20 + \frac{y}{(x+y)}\times40 = k}$ , and if ${x < y}$ , which of the following could be the value of ${k}$ ?

(A) 15

(B) 20

(D) 35

(E) 40

Solution: One solution you have in the OG. Three more are provided here:

Method 2: Algebra

Note the “could be” in the question. This means that $k$ can take multiple values and one of them is provided here.

$20\times \frac{x}{(x+y)} + 40\times \frac{y}{(x+y)} = k$

$20\times\frac{(x+2y)}{(x+y)} = k$

$20\times\frac{(x + y)}{(x + y)} + \frac{y}{(x + y)} = k$

$20\times[1 + \frac{y}{(x + y)}] = k$

Now since $y$ is greater than $x$ , $\frac{y}{x+y}$ will be more than $\frac{1}{2}$ but definitely less than 1 ( $x$ and $y$ are positive numbers).

So the value of $k$ will lie in the range $20\times[1 + \frac{1}{2}] < k < 20\times[1 + 1]$

i.e. $30 < k < 40$

Only option (D) falls in this range.

Answer (D)

Method 3: Weighted Average

Does this equation remind you of something: $20\times \frac{x}{(x+y)} + 40\times \frac{y}{(x+y)} = k$ ?

If you are a weighted average fan like me, you will notice that this is just the weighted average formula applied:

$C_{avg} = \frac{(C_1 \times w_1 + C_2 \times w_2)}{(w_1 + w_2)}$

Where $C_{avg} = k$

$C_1 = 20$

$C_2 = 40$

$w_1 = x$

$w_2 = y$

$k = \frac{(20\timesx + 40\timesy)}{(x + y)}$

It might be hard to see this on your own but the point is that if you do see it, the return is very high.

We know that the average of two quantities will lie in between them. So $k$ must lie between 20 and 40. Also, we are given that $x$ is less than $y$ i.e. weight given to 20 is less than the weight given to 40. So the weighted average will lie toward 40. Between 30 and 40, there is only option (D)

Hence, answer (D).

Method 4: Plugging Numbers

Now, what if neither of the above given methods work for you during the test and your mind goes blank? Then you can pick some numbers to get an idea of the kind of values you will get. This is absolute brute force and may not always work out but it will give you a fighting chance of getting the correct answer.

$\frac{20\timesx}{(x+y)} + \frac{40\timesy}{(x+y)} = k$

– Say, $x = 1$ , $y = 3$ ( $x$ and $y$ are positive numbers and $x < y$ )

Then $\frac{20\times1}{(1+3)} + \frac{40\times3}{(1+3)} = k = 35$

– Say, $x = 2$ , $y = 3$ (when you assume numbers, assume those which make the denominator a factor of 20 and 40 for ease of calculations. So assume numbers such that $x+y$ is 4 or 5 or 10 etc)