BTGmoderatorDC wrote: ↑

Wed Jun 17, 2020 9:21 pm

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


OA E

Solution:

We are given that the price of each hair clip is 40¢ and the price of each hair band is 60¢. We are also given that Rashi selects 10 clips and bands from the store, with an average price of 56¢.

We can let the number of hair clips = h and the number of hair bands = b, and we can create the following equations:

56 = (40h + 60b) / (h + b)

56h + 56b = 40h + 60b

16h = 4b

4h = b

and

h + b = 10

Substituting 4h for b, we have:

h + 4h = 10

5h = 10

h = 2

Since h = 2, b = 8.

Next we must determine how many bands Rashi must put back so that the average price of the items that she keeps is 52¢. We can let n = the number of hair bands that must be put back:

52 = [(40(2) + 60(8 - n)] / (10 - n)

52(10 - n) = 80 + 480 - 60n

520 - 52n = 560 - 60n

8n = 40

n = 5

Alternate Solution:

Since the average price of the 10 items Rashi selects is ¢ 56, the total price for the 10 items is 56 x 10 = ¢ 560. Let Rashi put n bands back. Then, the total price will reduce to 560 - 60n, and the average price will become (560 - 60n) / (10 - n). We want this quantity to equal 52, so let’s create the following equation:

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

560 - 60n = 520 - 52n

40 = 8n

n = 5

Answer: E