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investment problem

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by pradeepkaushal9518 » Mon Apr 19, 2010 2:36 am
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. 300


B. 180


C. 360


D. 270


E. 320

please sumbody
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Source: — Problem Solving |

by liferocks » Mon Apr 19, 2010 2:48 am
Let amount invested at a 5% annual return is 100x
and
amount invested at a 3% annual return is 100y

according to the question 5x=2*3y or x/y=6/5

so amount invested at a 5% annual return is 6600*6/11=$3600
amount invested at a 3% annual return is 6600*5/11=$3000

so,total income from the two investments after one year =( 3600*5/100)+(3000*3/100)=180+90=270

Ans option D 270
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by outreach » Mon Apr 19, 2010 10:08 am
income at 5% be x
income at 3% be (6600-x)

as per statement given in the equation
.05*x=2*.03*(6600-x)

solving we get x=3600
now .05*x=180 - 1
.03*(6600-x)=.03*(6600-3600)=.03*3000=90 -2

add 1 and 2
ans is 270
D
pradeepkaushal9518 wrote: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. 300


B. 180


C. 360


D. 270


E. 320

please sumbody
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