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# A portion of \$6600 is invested at a 5% annual return, while

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Joined: 30 Aug 2003
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A portion of \$6600 is invested at a 5% annual return, while [#permalink]

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15 Dec 2003, 12:27
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A portion of \$6600 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from the portion earning a 5% return is twice that of the other portion, what is the total income from the two investments after one year?
(A) \$180
(B) \$270
(C) \$300
(D) \$320
(E) \$360

any shortcuts??
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shubhangi

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15 Dec 2003, 12:55
x = portion invested at 5%

0.05x = 2(0.03(6600 - x))
0.05x = 396 - 0.06x
x = 3600

Return on portion invested at 5% = 0.05(3600) = 180, so the rest of the portion should return 90.

Total return income = 180 + 90 = 270

You can do trial and error methold to get the values too but that may take longer... well depends. To me this theoretical approach was faster. Less than a minute.

Kudos [?]: 13 [0], given: 0

15 Dec 2003, 12:55
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