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20 Nov 2010, 05:20
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Difficulty:

25% (medium)

Question Stats:

80% (02:18) correct 20% (02:13) wrong based on 163 sessions

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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 Math Expert Joined: 02 Sep 2009 Posts: 64249 Re: mixtures for Simple interest [#permalink] ### Show Tags 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 $$0.06x+0.11(6000-x)=580$$ --> $$x=1600$$.

_________________
VP
Joined: 06 Sep 2013
Posts: 1494
Concentration: Finance
Re: mixtures for Simple interest  [#permalink]

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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
Intern
Joined: 16 May 2014
Posts: 35
Re: mixtures for Simple interest  [#permalink]

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19 May 2014, 04:29
1
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!!!
Senior Manager
Joined: 20 Aug 2015
Posts: 379
Location: India
GMAT 1: 760 Q50 V44
Re: Lorna invests $6000, with some at 6% annual interest and som [#permalink] ### Show Tags 10 Dec 2015, 05:48 1 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
Intern
Joined: 18 Jun 2016
Posts: 1
Re: Lorna invests $6000, with some at 6% annual interest and som [#permalink] ### Show Tags 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.