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Lorna invests $6000, with some at 6% annual interest and som

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Director
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Lorna invests $6000, with some at 6% annual interest and som  [#permalink]

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New post 20 Nov 2010, 05:20
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Question Stats:

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Lorna invests $6000, with some at 6% annual interest and some at 11% annual interest. If she receives a total of $580 from these investments at the end of a year, 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: mixtures for Simple interest  [#permalink]

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New post 20 Nov 2010, 05:47
rxs0005 wrote:
Lorna invests $6000, with some at 6% annual interest and some at 11% annual interest. If she receives a total of $580 from these investments at the end of a year, how much was invested at the 6% interest rate?

(A) $160

(B) $1,600

(C) $2,200

(D) $4,400

(E) $5,840


Let the amount invested at 6% be \(x\), then the amount invested at 11% will be \(6000-x\). So we have \(0.06x+0.11(6000-x)=580\) --> \(x=1600\).

Answer: B.
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Re: mixtures for Simple interest  [#permalink]

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New post 12 May 2014, 07:38
Concept of differentials + Approximation gives the formula of success for this problem

We have that 'x' invested at 6%
We also have that '6000-x' invested at 11%

Now then 580/6000 is approx 10%

Therefore, -4x+6000-x=0

5x=5000

x=1200 approximately. Probably higher


Only B matches
Hope it helps
Cheers!
J :)
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Re: mixtures for Simple interest  [#permalink]

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New post 19 May 2014, 04:29
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jlgdr wrote:
Concept of differentials + Approximation gives the formula of success for this problem

We have that 'x' invested at 6%
We also have that '6000-x' invested at 11%

Now then 580/6000 is approx 10%

Therefore, -4x+6000-x=0

5x=5000

x=1200 approximately. Probably higher


Only B matches
Hope it helps
Cheers!
J :)


If he has invested the entire amount at 6%, he would get 360. But now he is getting 580, which is 220 more. This is the result of 5% more interest he is fetching by investing at 11%.

5% * amount invested at 11% = 220

Amount invested at 11% =4400
Amount invested at 6% = 1600


Hope you like this approach!!!

Kudos if you like!!!
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Re: Lorna invests $6000, with some at 6% annual interest and som  [#permalink]

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New post 10 Dec 2015, 05:48
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rxs0005 wrote:
Lorna invests $6000, with some at 6% annual interest and some at 11% annual interest. If she receives a total of $580 from these investments at the end of a year, how much was invested at the 6% interest rate?

(A) $160

(B) $1,600

(C) $2,200

(D) $4,400

(E) $5,840


NOTE: There is no mention of Simple interest or compund interest here. This is because at the end of 1st year, SI = CI
CI is simply SI + interest on interest. Since during the first year, there is no interest accumulated, hence CI = SI


Coming to the question.

Assume amount invested at 6% = $x
Amount invested at 11% = $6000 - x
Period = 1 year.

Total interest = \(\frac{6*x}{100}\) + \(\frac{(6000-x)*11}{100}\) = 580
66000 - 5x = 58000
5x = 8000
x = 1600

Option B
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Re: Lorna invests $6000, with some at 6% annual interest and som  [#permalink]

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New post 07 Jul 2016, 04:23
rxs0005 wrote:
Lorna invests $6000, with some at 6% annual interest and some at 11% annual interest. If she receives a total of $580 from these investments at the end of a year, how much was invested at the 6% interest rate?

(A) $160

(B) $1,600

(C) $2,200

(D) $4,400

(E) $5,840



11+6=17%, therefore 17/100*6000= 1020, now add 1020 + 580= 1600. (B)

Correct me if my technique is wrong. Thank you.
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Re: Lorna invests $6000, with some at 6% annual interest and som   [#permalink] 07 Jul 2016, 04:23

Lorna invests $6000, with some at 6% annual interest and som

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