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

Total price for 10 pieces = 560

40(2)+60(8)=560

Let x be the number of bands that needs to be put back.

\(\frac{(560-60x)}{(10-x)}\) = 52
x=5

Answer: E

Acc to the given info : C + B = 10,
and avg of 10 = 56 => Total = 560
Thus 40C + 60B = 560.
Thus solving for C,B we get B = 8 and C=2.

Now, (8-x)60 + 80 = 52*x
=> x=5.
Thus 5 bands should be put down.
Ans E.

Weighted avg. question :
as the weighted average (56 ) is 16 units far from 40 and 4units far from 60 so right now the ratio of Clip:Band 4:16 . i.e. 2 clips 8Bands .
as new weighted average is 52 she has to drop band or add clips , we are told that she has to drop band. so new C:B ratio will be 8:12 or 2:3 . as we know there are 2 clips, there will be 3 bands.


8-3 = 5 bands is the answer

Representing the cost and average in a line

---- \(¢40_{clip}\)---|-----------------------\((¢16)\)-----------------------------------------|\(¢56_{avg}\)-------\((¢4)\)---------|\(¢60_{band}\)--------

Since the distance is INVERSE to the amount present in the mixture,
so the ratio of Clip:Band = 4:16
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\)= 2:8 (since the total # of clips and bands present is 10)

After Rashi drops off some hair bands to make average ¢ 52

----\(¢40_{clip}\)---|-------------------------\((¢12)\)-----------------------| \(¢52_{avg}\)------------\((¢8)\)--------------------|\(¢60_{band}\)--------

New ratio of Clip:Band = 8:12
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\)= 2:3 (since the number of Clips is the same as before)

So the number of hair bands that Rashi put back is [8(Original ratio) - 5 = 3 (new ratio)] are 5

Answer E

Shouldn't your final formula should be (8-x)60 + 80 = 52*(10-x) [not 52*x]?

VERITAS PREP OFFICIAL SOLUTION

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?
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Hi Bunuel,

IMHO, the problem with solution 1 is the it takes the number of items as 10 even after Rashi puts back some of the bands. The question doesn't say that the bands are replaced by clips. So the new quantity should be less than 10.

Regards,
Gaurav

Hi All,

This question is a directly 'lift' of a question from the GMAT2016 Book (page 172, number 137). A number of the *new* questions in the GMAT2016 are actually drawn from older materials (the pencil-and-paper Tests and other resources), so it's almost certain that the original 'source' of this question is the same.

GMAT assassins aren't born, they're made,
Rich

56*10=560$

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

560-60x=520-52x => x=5

E

I used a different approach.

We know that the current sum of the items is 560 and we're told that the sum of the items should be some multiple of 52 (52 x number of items). We will always get a digit of 0 when substracting a multiple of 60 from 560. The only way we can get a digit of 0 is when she's keeping 5 bands. Thus, answer is E.

Lets Clips = C and Bands = B

So, 40C+60B=560 -----Equation 1
And C+B = 10 ----Equation 2

Solving this B=8 and C=2

Now, using options if she drops 1 B, then Total will 560-60 = 500
Average = 500/(10-1) = 500/9 which is NOT 52

Hence, we go for higher value of B to be dropped, lets see option E, 560-5*60 = 260
Average = 260/5 = 52

Hence Option E
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[40a+60(10-a)]/10=56
a=2
band=10-2=8
[40a+60(10-a)]/10=56
(40*2+60b)/(2+b)=52
b=3
8-3=5(ans)

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