This is hard to confess publicly but I must because it is a prime example of how GMAT takes advantage of our weaknesses – A couple of days back, I answered a 650 level question of weighted averages incorrectly. Those of you who have been following my blog would understand that it was an unpleasant surprise – to say the least. I know my weighted averages quite well, thank you! For this comedown, I blame the treachery of GMAT because it knows how to get you when you become too complacent. The takeaway here is – no matter how easy and conventional the question seems, you MUST read it carefully.

Let me share that particular question with you. I will also share two solutions which give you two different answers. It is an exercise for you to figure out which one is the correct solution (that is, if one of them is the correct solution). Needless to say, the error in the solution(s) is conceptual and very easy to see (not some sly calculation mistake). It’s just that in your haste, it’s very easy to miss this important point. I hope to see some comments with some good explanations.

**Question**: The price of each hair clip is ¢ 40 and the price of each hair band is ¢ 60. Rashi selects a total of 10 clips and bands from the store, and the average (arithmetic mean) price of the 10 items is ¢ 56. How many bands must Rashi put back so that the average price of the items that she keeps is ¢ 52?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Solution 1:

Price of each clip (Pc) = 40

Price of each band (Pb) = 60

Average price of each item (Pavg) = 56

Wc/Wb = (Pb – Pavg)/(Pavg – Pc) = (60 – 56)/(56 – 40) = 1/4 (our weighted average formula)

Since the total number of items is 10, number of clips = 1*2 = 2 and number of bands = 4*2 = 8

If the average price is changed to 52,

Wc/Wb = (Pb – Pavg)/(Pavg – Pc) = (60 – 52)/(52 – 40) = 2/3

Now the ratio has changed to 2:3. This gives us number of clips as 4 and number of bands as 6.

Since previously she had 8 bands and now she has 6 bands, she must have put back 2 bands.

Answer (B)

Solution 2:

Say the number of hair clips is C and the number of hair bands is 10 – C.

(40C + 60(10 – C))/10 = 56 (Using the formula: Average = Sum/Number of items)

On solving, you get C = 2

Number of clips is 2 and number of bands is (C – 2) = 8.

Now, let’s consider the scenario when she puts back some bands, say x.

(2*40 + (8 – x)*60)/(10 – x) = 52

On solving, you get x = 5

So she puts back 5 bands so that the average price is 52.

Answer (E)

Obviously, there is only one correct answer. It’s your job to figure out whether it is (B) or (E) or some third option. Also what’s wrong with one or both of these solutions?

*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!*

I would say solution #2 is correct, because the problem does not say that she actually buys 10 items. It says she selects 10 but puts some back, so it seems to me that she’d actually end up buying fewer than 10.

avg price of 56cents

=> 8 bands, 2 clips

=>(560-x*60)/(10-x) = 52

=> x = 5

=> (E)

correct?

I believe the correct answer is option(E),which is solution 2.

In Solution 1,when the scenario where she puts back some bands is considered, it is assumed that the total number of clips and bands remain as 10.Hence the assumption leads to the wrong answer.

The question stem says she puts some bands back but doesn’t state that the total number of clips and bands would continue to remain as 10.

Yes, you all are absolutely correct! She selects 10 items but doesn’t keep 10 i.e. when the average is 52, the total number of items is less than 10. Hence solution 1 is incorrect.

Now, could someone provide the correct solution using weighted averages?

Here is how I looked at it. What she ends up buying is less than 10 because there is no mention of her replacing bands with clips to keep total qty constant. The total initial cost is 560. Irrespective of how many she puts back the total cost always has a units digit of 0. Given that the average is 52 the new total qty has to be 5 for the total cost to have a units digit of 0. So she puts bak 5.

Great solution!

I have been following your articles for quite some time now.. Nice articles that explain the concepts in a simple way…

Your solution using weighted averages is right until after you derive the new ratio, which is 2:3.

Rashi had 2 hair clips and 8 hair bands. The ratio of the two is 1:5. Now, the new ratio should be 2:3. We know that Rashi has 2 hair clips. So, she should retain only 3 hair bands to get the average to 52. As, she has 8 hair bands with her now, she has to return 5.

I didn’t understand D’s solution above. Why should the total cost have a units digit of 0? If the number of items are less than 10, then the units digit doesn’t have to be 0, right?

“Irrespective of how many she puts back the total cost always has a units digit of 0.”

“The price of each hair clip is ¢ 40 and the price of each hair band is ¢ 60″

When you pick some clips and bands, the total cost will always be multiple of 10 since the costs of both articles are multiples of 10.

If the average cost of the articles is 52, the number of articles cannot be 1/2/3/4 since they do not give you a total cost which is a multiple of 10. The number of articles must be 5/10/15… etc. Only 5 is possible.

In solution 1, the 2:3 should not be interpreted as 4 clips and 6 bands as there is no increase in the number of clips. There is no condition requiring the total number of items to stay constant.

We know the number of clips is initially 2 and that it stays the same. The number of bands must be reduced from 8 to 3 to obtain the ratio 2:3. Thus, she needs to put back 5 bands.

hi..i didn’t use the weighted avg. For 10 items the cost is 560 cents and she needs to put back the bands. so we keep on reducing the total amount by 60 cents till the time the avg is 52.

Total items Sum Avg

10 560 56

9 500 55

8 440 55

5 260 52

Quite a crude method :(

Nonetheless, it’s correct. If all logic fails you in the test, use brute force. The test doesn’t care about the method you use to get your answer as long as you do it within the time frame.