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Ms. Jones has twice as much invested in stocks as in bonds. Last year,

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Ms. Jones has twice as much invested in stocks as in bonds. Last year, [#permalink]

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New post 31 Oct 2017, 09:16
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Ms. Jones has twice as much invested in stocks as in bonds. Last year, the stock investments paid 7.5 percent of their value and the bonds paid 10 percent of their value. If the total that both investments paid last year was $1,000, how much did Ms. Jones have invested in stocks?

(A) $3,636
(B) $4,000
(C) $7,500
(D) $8,000
(E) $10,000
[Reveal] Spoiler: OA

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Re: Ms. Jones has twice as much invested in stocks as in bonds. Last year, [#permalink]

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New post 31 Oct 2017, 09:23
Option A

Let investment in stock be S and bonds be B
So B=2S.
0.075S+0.10B=1000
0.075S+0.20S=1000
Solving S=3636

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Re: Ms. Jones has twice as much invested in stocks as in bonds. Last year, [#permalink]

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New post 01 Nov 2017, 09:10
x=bonds
2x=stocks
2x(.075)+.1x=1000
=>x=4000 =>2x=8000
hence D

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Re: Ms. Jones has twice as much invested in stocks as in bonds. Last year, [#permalink]

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New post 02 Nov 2017, 16:05
Bunuel wrote:
Ms. Jones has twice as much invested in stocks as in bonds. Last year, the stock investments paid 7.5 percent of their value and the bonds paid 10 percent of their value. If the total that both investments paid last year was $1,000, how much did Ms. Jones have invested in stocks?

(A) $3,636
(B) $4,000
(C) $7,500
(D) $8,000
(E) $10,000


We can let the amount invested in bonds = n and the amount invested in stocks = 2n. Thus, the interest earned from stocks was 0.075(2n) = 0.15n and the interest earned from bonds was 0.1n.

Since the total interest made from both investments was 1,000, we have:

0.15n + 0.1n = 1,000

0.25n = 1,000

n =1,000/0.25 = 100,000/25 = 4,000

The amount invested in bonds is n = 4,000. The amount invested in stocks is 2n = 8,000.

Answer: D
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Re: Ms. Jones has twice as much invested in stocks as in bonds. Last year,   [#permalink] 02 Nov 2017, 16:05
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