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A person invested a total amount of $1,500,000. A part of it was inves

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Bunuel
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A person invested a total amount of $1,500,000. A part of it was inves [#permalink] New post   03 Apr 2020, 03:17
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A person invested a total amount of $1,500,000. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is $76,000 then what was the amount invested in the fixed deposit?

A. 700,000
B. 800,000
C. 900,000
D. 1,000,000
E. 1,100,000


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A person invested a total amount of $1,500,000. A part of it was inves [#permalink] New post   03 Apr 2020, 03:54
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Bunuel wrote:
A person invested a total amount of $1,500,000. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is $76,000 then what was the amount invested in the fixed deposit?

A. 700,000
B. 800,000
C. 900,000
D. 1,000,000
E. 1,100,000


Let amounts invested at 4% & 3% rate of interest = \(2p\) & \(p\) respectively
--> Amount invested in fixed deposit at 6% rate of interest = \(1,500,000 - 3p\)

Total interest = \(76,000\)
--> \(4\)% of \(2p\) + \(3\)% of \(p\) + \(6\)% of \((1,500,000 - 3p)\) = \(76,000\)

--> \(\frac{4}{100}*2p + \frac{3}{100}*p + \frac{6}{100}*(1,500,000 - 3p) = 76,000\)

--> \(\frac{8p + 3p + 9,000,000 - 18p}{100} = 76,000\)

--> \(9,000,000 - 7p = 7,600,000\)

--> \(7p = 1,400,000\)

--> \(p = 200,000\)

Amount invested in fixed deposit = \(1,500,000 - 3*200,000 = $900,000\)

Option C
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Re: A person invested a total amount of $1,500,000. A part of it was inves [#permalink] New post   03 Apr 2020, 04:24
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Let the amount deposit in fd and other two parts are x, 2y and y. Also removing last three zeros from 1500000 and 76000 to simplify the calculation.
x +3y = 1500
6/100*x + 4/100*2y+3/100*y=76
6x+11y=76
Solving these two equations
X=900
So answer is 900000 that is option c

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A person invested a total amount of $1,500,000. A part of it was inves [#permalink] New post   05 Apr 2020, 00:50
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Let’s say that the remaining amount is x
—> \((1500000 —x)\frac{6}{100} + (\frac{2x}{3})*(\frac{4}{100}) +(\frac{ x}{3})*(\frac{3}{100}) = 76000\)

—> \(90000 —\frac{6x}{100} + \frac{8x}{300} + \frac{3x}{300} = 76000\)
\(14000 = \frac{(18x —8x —3x)}{300}\)
—> \(7x = 14000*300\)
\(x= 600000\)
—> the amount invested in fixed deposit —\(( 1500000 —600000) = 900000 \)

Answer ( C)

hi, gurmukh
I liked your approach how to solve it too.
Kudos for you

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Re: A person invested a total amount of $1,500,000. A part of it was inves [#permalink] New post   05 Apr 2020, 02:09
Thanks lacktutor for complimenting, hopefully I'll solve many people problem in the time to come. This is a good platform and I m finding it nice to give different ways to people to solve a problem. I have been coaching GMAT for 15 years and have lot of innovative to solve problems

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Re: A person invested a total amount of $1,500,000. A part of it was inves [#permalink]
05 Apr 2020, 02:09
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