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# Lady Edith bought several necklaces at the jewelry store, and each nec

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Status: Preparing for the GMAT
Joined: 02 Nov 2016
Posts: 1236

Kudos [?]: 1042 [1], given: 538

Location: Pakistan
GPA: 3.4
Lady Edith bought several necklaces at the jewelry store, and each nec [#permalink]

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17 Sep 2017, 11:40
1
KUDOS
00:00

Difficulty:

25% (medium)

Question Stats:

78% (01:16) correct 22% (00:29) wrong based on 23 sessions

Lady Edith bought several necklaces at the jewelry store, and each necklace cost 16 dollars. Lady Mary also purchased several necklaces, at a cost of $20 each. If the ratio of the number of necklaces Lady Edith purchased to the number of necklaces Lady Mary purchased is 3 to 2, what is the average cost of the necklaces purchased by Lady Edith and Lady Mary? A 16.7 B 17.1 C 17.6 D 17.9 E 18.2 [Reveal] Spoiler: OA _________________ Official PS Practice Questions Press +1 Kudos if this post is helpful Kudos [?]: 1042 [1], given: 538 Director Joined: 22 May 2016 Posts: 814 Kudos [?]: 264 [2], given: 552 Lady Edith bought several necklaces at the jewelry store, and each nec [#permalink] ### Show Tags 18 Sep 2017, 11:10 2 This post received KUDOS SajjadAhmad wrote: Lady Edith bought several necklaces at the jewelry store, and each necklace cost 16 dollars. Lady Mary also purchased several necklaces, at a cost of$20 each. If the ratio of the number of necklaces Lady Edith purchased to the number of necklaces Lady Mary purchased is 3 to 2, what is the average cost of the necklaces purchased by Lady Edith and Lady Mary?

A 16.7
B 17.1
C 17.6
D 17.9
E 18.2

Great question. A bit tricky. I used weighted average.

Average cost overall =
(Cost$$_{E}$$)(Weight$$_{E}$$) + (Cost$$_{M}$$)(Weight$$_{M}$$) / (Total Weight)

Cost = cost per necklace, for E and M

Weight = number of necklaces as portion of all necklaces purchased

To derive weight for each person, use ratio of number of necklaces purchased:

$$\frac{E}{M} = \frac{3x}{2x}$$

You could use this ratio a few ways. Easier for me: let x = 1. E buys 3 necklaces. M buys 2. Total # bought = total weight = 5:

$$\frac{16(3) + 20(2)}{5}$$ =

$$\frac{88}{5}$$ = 17.6

If substitution is preferred, use E = $$\frac{3}{2}$$M (Ms will cancel)

Total weight is M + $$\frac{3}{2}$$M = $$\frac{5}{2}$$M

$$\frac{(16)(3/2) + 20(1)}{2.5}$$ = $$\frac{44}{2.5}$$ = 17.6

Kudos [?]: 264 [2], given: 552

Lady Edith bought several necklaces at the jewelry store, and each nec   [#permalink] 18 Sep 2017, 11:10
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