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Lorna invests $6,000, some at a 6% annual interest rate and some at 11

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Lorna invests $6,000, some at a 6% annual interest rate and some at 11  [#permalink]

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New post 26 Jan 2017, 10:24
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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

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Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11  [#permalink]

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New post 26 Jan 2017, 10:31
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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
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Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11  [#permalink]

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New post 09 May 2018, 00:01
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.
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Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11  [#permalink]

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New post 09 May 2018, 00:07
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.
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Lorna invests $6,000, some at a 6% annual interest rate and some at 11  [#permalink]

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New post 09 May 2018, 11:57
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:
\(.06(x) +.11(6,000-x)=580\)
\(.06x + 660 -.11x = 580\)
\(80=.05x\)
\(x=\frac{80}{.05}=\frac{8000}{5}=1,600\)


ANSWER B

Answer Choices

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

ANSWER B
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Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11  [#permalink]

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New post 05 Jul 2018, 16:48
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

Answer: B
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Re: Lorna invests $6,000, some at a 6% annual interest rate and some at 11 &nbs [#permalink] 05 Jul 2018, 16:48
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