To take a look at the previous posts of this thread, check: Part I, Part II and Part III.

Another 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 {x/(x+y)}*20 + {y/(x+y)}*40 = k, and if x < y, which of the following could be the value of k?

(A) 15

(B) 20

(C) 25

(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*x/(x+y) + 40*y/(x+y) = k

20(x+2y)/(x+y) = k

20*{(x + y)/(x + y) + y/(x + y)} = k

20*{ 1 + y/(x + y)} = k

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

So the value of k will lie in the range 20*{1 + 1/2} < k < 20*{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*x/(x+y) + 40*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:

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

Where Cavg = k

C1 = 20

C2 = 40

w1 = x

w2 = y

k = (20*x + 40*y)/(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.

20*x/(x+y) + 40*y/(x+y) = k

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

Then 20*1/(1+3) + 40*3/(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)

Then 20*2/(2+3) + 40*3/(2+3) = k = 32

– Say, x = 1, y = 4

Then 20*1/(1+4) + 40*4/(1+4) = k = 36

Even if you do not get 35, note that the other values of k lie in the 30s. So your best bet would be to mark answer as (D).

Hope you see that there are many different ways of solving a given question, so you don’t usually require complex solutions. Practice on!

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Nice. In a blinding hurry to solve the problem, I missed the weighted avg approach which should be the best choice. I was thinking number line instead under two extreme conditions – If x is almost the same as y, then we get 30 as the lower end and if y is significantly larger than x, we can ignore x/(x+y) and just consider y/(x+y) which will be almost 1, thereby resulting in the upper end just less than 40. Hence 35 fits the bill.

Perfect! Note that weighted average is just another name for what you did. You allotted weight to y relative to x – ‘almost equal to x’ and ‘much larger than x’ and then found the range for k.

I would say that it’s even better than actually arriving at the weighted average method since this means that even if the question were worded differently and the weighted average method weren’t applicable as cleanly as here, still you would have found a way!

Yes, you’re right. Now that I mulled over it a tad more, the ‘x almost equal to y’, is akin to a 1:1 weighted split, which would result in a mean value of 30. The rest, as you say, follows from the fact that x30 and we have 35. No calculations needed!

Some formatting issue with the less than symbol. the x30 should read x ‘less than’ 30.

Sorry, meant, x ‘less than’ y.

Questions that appear difficult in first glance are really simple upon analysis. Visual thinking is a great tool in interpreting problems on weighted average.

Hi,

If we’re required to find the weighted average of a list {a,b,c} with corresponding weights {x,y,z}, can we find the weighted average of a,b using x,y and then use this intermediate result say ‘r’ with weight (x+y)/2 and c with weight z to finish the computation?

Thanks.

By using the weight as (x+y)/2, you are reducing the weight of the result ‘r’ as compared with the weight of z.

r is the number that could replace all a and all b and the total weight of r is (x+y).

Try to figure out what weighted average you get using your method in terms of variables. You know that the weighted average in case of 3 weights should come out to be (xa + yb + zc)/(x + y + z)

Thanks. And yes, you’re right – indeed, the intermediate weight is x+y. Confirmed this via algebra.