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
$$Average\ price\ =\frac{total\ price\ of\ clip+total\ price\ of\ hairband}{total\ number\ of\ hairband\ and\ clip}$$
$$56=\frac{40c+\left(600-60c\right)}{10}$$
$$560=40c-60c+600$$
$$\frac{560-600}{-20}=\frac{-20c}{-20}$$
$$c=2\ and\ total\ hairbands\ =\ 10-c$$
$$Total\ hairbands\ =\ 10-2=8$$
How many hairbands must be returned for the average price to be ¢ 52?
$$Let\ the\ number\ of\ bands\ to\ be\ returned=b$$
$$\frac{\left(40\cdot2\right)+60\left(8-b\right)}{\left(10-b\right)}=52$$
$$=>\frac{80+480-60b}{\left(10-b\right)}=52$$
$$=>560-60b=52\left(10-b\right)$$
$$=>560-60b=520-52b$$
$$=>560-520=-52b+60b$$
$$=>\frac{40}{8}=\frac{8b}{8}\ \ \ \ \ \ b=5$$
$$5\ bands\ must\ be\ returned\ for\ average\ price\ to\ be\ =\ 52$$
$$Answer\ =\ E$$
- 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
$$Average\ price\ =\frac{total\ price\ of\ clip+total\ price\ of\ hairband}{total\ number\ of\ hairband\ and\ clip}$$
$$56=\frac{40c+\left(600-60c\right)}{10}$$
$$560=40c-60c+600$$
$$\frac{560-600}{-20}=\frac{-20c}{-20}$$
$$c=2\ and\ total\ hairbands\ =\ 10-c$$
$$Total\ hairbands\ =\ 10-2=8$$
How many hairbands must be returned for the average price to be ¢ 52?
$$Let\ the\ number\ of\ bands\ to\ be\ returned=b$$
$$\frac{\left(40\cdot2\right)+60\left(8-b\right)}{\left(10-b\right)}=52$$
$$=>\frac{80+480-60b}{\left(10-b\right)}=52$$
$$=>560-60b=52\left(10-b\right)$$
$$=>560-60b=520-52b$$
$$=>560-520=-52b+60b$$
$$=>\frac{40}{8}=\frac{8b}{8}\ \ \ \ \ \ b=5$$
$$5\ bands\ must\ be\ returned\ for\ average\ price\ to\ be\ =\ 52$$
$$Answer\ =\ E$$
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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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