The price of each hair clip is ¢ 40 and the price of each hairband 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
[spoiler]OA=E[/spoiler]
Source: Manhattan GMAT
The price of each hair clip is ¢ 40 and the price of each hairband is ¢ 60. Rashi selects a total of 10 clips and bands
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Given that:
 price of each clip = ¢ 40
 price of each hairband = ¢ 60
 the average price of (10 clip and band) = ¢ 56
Let the number of clips purchased = c
Number of hairband = 10  c
Price of total clips = 40 * c = 40c
Price of total hairbands = 60 * (10  c)= 600  60c
$$Average\ price\ =\frac{total\ price\ of\ clip+total\ price\ of\ hairband}{total\ number\ of\ hairband\ and\ clip}$$
$$56=\frac{40c+\left(60060c\right)}{10}$$
$$560=40c60c+600$$
$$\frac{560600}{20}=\frac{20c}{20}$$
$$c=2\ and\ total\ hairbands\ =\ 10c$$
$$Total\ hairbands\ =\ 102=8$$
How many hairbands must be returned for the average price to be ¢ 52?
$$Let\ the\ number\ of\ bands\ to\ be\ returned=b$$
$$\frac{\left(40\cdot2\right)+60\left(8b\right)}{\left(10b\right)}=52$$
$$=>\frac{80+48060b}{\left(10b\right)}=52$$
$$=>56060b=52\left(10b\right)$$
$$=>56060b=52052b$$
$$=>560520=52b+60b$$
$$=>\frac{40}{8}=\frac{8b}{8}\ \ \ \ \ \ b=5$$
$$5\ bands\ must\ be\ returned\ for\ average\ price\ to\ be\ =\ 52$$
$$Answer\ =\ E$$
 price of each clip = ¢ 40
 price of each hairband = ¢ 60
 the average price of (10 clip and band) = ¢ 56
Let the number of clips purchased = c
Number of hairband = 10  c
Price of total clips = 40 * c = 40c
Price of total hairbands = 60 * (10  c)= 600  60c
$$Average\ price\ =\frac{total\ price\ of\ clip+total\ price\ of\ hairband}{total\ number\ of\ hairband\ and\ clip}$$
$$56=\frac{40c+\left(60060c\right)}{10}$$
$$560=40c60c+600$$
$$\frac{560600}{20}=\frac{20c}{20}$$
$$c=2\ and\ total\ hairbands\ =\ 10c$$
$$Total\ hairbands\ =\ 102=8$$
How many hairbands must be returned for the average price to be ¢ 52?
$$Let\ the\ number\ of\ bands\ to\ be\ returned=b$$
$$\frac{\left(40\cdot2\right)+60\left(8b\right)}{\left(10b\right)}=52$$
$$=>\frac{80+48060b}{\left(10b\right)}=52$$
$$=>56060b=52\left(10b\right)$$
$$=>56060b=52052b$$
$$=>560520=52b+60b$$
$$=>\frac{40}{8}=\frac{8b}{8}\ \ \ \ \ \ b=5$$
$$5\ bands\ must\ be\ returned\ for\ average\ price\ to\ be\ =\ 52$$
$$Answer\ =\ E$$
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Solution:Gmat_mission wrote: ↑Wed Jun 24, 2020 7:55 amThe price of each hair clip is ¢ 40 and the price of each hairband 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
[spoiler]OA=E[/spoiler]
The total price for the items Rashi bought is 56 x 10 = ¢ 560. We can let n = the number of hair bands that must be put back. Then, the total price is reduced by 60n in total, and the total number of items is 10  n. We have the following equation:
52 = (560  60n) / (10  n)
52(10  n) = 560  60n
520  52n = 560  60n
8n = 40
n = 5
Answer: E
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