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26 Jan 2017, 09:24
00:00

Difficulty:

45% (medium)

Question Stats:

71% (02:08) correct 29% (02:11) wrong based on 99 sessions

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Lorna invests $6,000, some at a 6% annual interest rate and some at 11% annual rate. If she receives a total of$580 at the end of a year from these investments, how much was invested at the 6% interest rate?

A. $160 B.$1,600
C. $2,200 D.$4,400
E. $5,840 _________________ VP Joined: 05 Mar 2015 Posts: 1003 Re: Lorna invests$6,000, some at a 6% annual interest rate and some at 11  [#permalink]

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26 Jan 2017, 09:31
1
Bunuel wrote:
Lorna invests $6,000, some at a 6% annual interest rate and some at 11% annual rate. If she receives a total of$580 at the end of a year from these investments, how much was invested at the 6% interest rate?

A. $160 B.$1,600
C. $2,200 D.$4,400
E. $5,840 Let amount X be invested for 6% then 6000-x is invested in 11% rate thus X*6/100*1 + (6000-x)*11/100*1= 580 6X + 66000 - 11X =58000 5x = 8000 X =$1600

Ans B
Intern
Joined: 01 Jan 2016
Posts: 21
Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11 [#permalink] ### Show Tags 08 May 2018, 23:01 Bunuel wrote: Lorna invests$6,000, some at a 6% annual interest rate and some at 11% annual rate. If she receives a total of $580 at the end of a year from these investments, how much was invested at the 6% interest rate? A.$160
B. $1,600 C.$2,200
D. $4,400 E.$5,840

Does the following set up work: 6000[(x/100)*0.06 + ((1-x)/100)*0.11)] = 580. Solve for x.
Intern
Joined: 01 Jan 2016
Posts: 21
Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11 [#permalink] ### Show Tags 08 May 2018, 23:07 rohit8865 wrote: Bunuel wrote: Lorna invests$6,000, some at a 6% annual interest rate and some at 11% annual rate. If she receives a total of $580 at the end of a year from these investments, how much was invested at the 6% interest rate? A.$160
B. $1,600 C.$2,200
D. $4,400 E.$5,840

Let amount X be invested for 6%
then 6000-x is invested in 11% rate

thus
X*6/100*1 + (6000-x)*11/100*1= 580
6X + 66000 - 11X =58000
5x = 8000
X = $1600 Ans B Can you further elaborate on this explanation? The setup is not clear to me. Thanks. Senior SC Moderator Joined: 22 May 2016 Posts: 2223 Lorna invests$6,000, some at a 6% annual interest rate and some at 11  [#permalink]

### Show Tags

09 May 2018, 10:57
1
Bunuel wrote:
Lorna invests $6,000, some at a 6% annual interest rate and some at 11% annual rate. If she receives a total of$580 at the end of a year from these investments, how much was invested at the 6% interest rate?

A. $160 B.$1,600
C. $2,200 D.$4,400
E. $5,840 Weighted average The amounts invested at different rates sum to $$6,000$$ Let $$x$$ = amount invested at $$6$$% $$(6,000-x)$$ = amount invested at $$11$$% The dollar amounts of interest earned by each portion at each rate sum to$580. Different rates carry different weights.
Weighted average:
$$.06(x) +.11(6,000-x)=580$$
$$.06x + 660 -.11x = 580$$
$$80=.05x$$
$$x=\frac{80}{.05}=\frac{8000}{5}=1,600$$

Start with C) $2200 Amt invested at 6%: $$2200$$ Amt invested at 11%: $$(6000-2200)=3,800$$ Interest earned at 6%:$$(2200*.06)=132$$ Interest earned at 11%:$$(3800*.11)=418$$ Total interest earned?$$(132+418)=550$$ Too low. The 6% rate has too great a portion of the$6,000. If 11% has a greater portion of the $6,000, total interest earned will increase. Eliminate Answers D and E. Their portion for the 6% rate is higher than C, which is already too high. We need a number lower than C, but not by much. Try B)$1600

Amt invested at 6%: $$1600$$
Amt invested at 11%:$$(6000-1600)=4400$$

Interest earned at 6%: $$(1600*.06)=96$$
Interest earned at 11%: $$(4400*.11)=484$$

Total interest earned:
$$(96+484)=580$$
That's correct

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Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11 [#permalink] ### Show Tags 05 Jul 2018, 15:48 Bunuel wrote: Lorna invests$6,000, some at a 6% annual interest rate and some at 11% annual rate. If she receives a total of $580 at the end of a year from these investments, how much was invested at the 6% interest rate? A.$160
B. $1,600 C.$2,200
D. $4,400 E.$5,840

We can let n = the amount invested at 6% and (6000 - n) = the amount invested at 11%; thus:

0.06n + 0.11(6,000 - n) = 580

0.06n + 660 - 0.11n = 580

80 = 0.05n

1,600 = n

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