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by sana.noor » Sat Apr 06, 2013 10:31 pm
a trustee invested a total of $100,000, more than half of which was invested at 6 percent simple interest and the rest of which was invested at 8 percent simple annual interest. what amount was invested at 8 percent?
1) the total annual income on the investment was $6800
2) if twice as much had been invested at 8 percent and the rest at 6 percent, the total annual income on the investment would have been increased by $800

D
i know statement A is sufficient but couldnt solve statement 2. what will be the equation for statement 2?
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by Anju@Gurome » Sat Apr 06, 2013 10:43 pm
sana.noor wrote:a trustee invested a total of $100,000, more than half of which was invested at 6 percent simple interest and the rest of which was invested at 8 percent simple annual interest. what amount was invested at 8 percent?
1) the total annual income on the investment was $6800
2) if twice as much had been invested at 8 percent and the rest at 6 percent, the total annual income on the investment would have been increased by $800
Let us assume, $100x was invested at 6% and $(100,000 - 100x) was invested at 8%.

Statement 1: 6x + 8(100 - x) = 6800
We can easily solve this linear equation in x.

Sufficient

Statement 2: Total annual income on investment = 6x + 8(100 - x)
In given scenario,
  • 2(100,000 - 100x) at 8% and [100,000 - 2(100,000 - 100x)] = (100x - 100,000) at 6%

    Annual income on investment = 16(100 - x) + 6(100 - x)
Hence, 16(100 - x) + 6(100 - x) = 6x + 8(100 - x) + 800
We can easily solve this linear equation in x.

Sufficient

The correct answer is D.
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by rintoo22 » Sat Apr 06, 2013 11:06 pm
a trustee invested a total of $100,000, more than half of which was invested at 6 percent simple interest and the rest of which was invested at 8 percent simple annual interest. what amount was invested at 8 percent?
1) the total annual income on the investment was $6800
2) if twice as much had been invested at 8 percent and the rest at 6 percent, the total annual income on the investment would have been increased by $800
Say x is amount more than half.
Therefore 50000+x was invested in 6% => ((50000+x)*6)/100 .....(1)
And 50000-x was invested in 8% => ((50000-x)*8)/100 ....(2)

Statement 1
((50000+x)*6)/100 + ((50000-x)*8)/100 = 6800
So X can be calculated. So we can easily say that Statement 1 is sufficient. As correctly pointed out by you.

what will be the equation for statement 2?
Statement 2
If twice as much had been invested.
The original amount invested at 8% is ((50000-x)*8)/100 . So twice as much would be
((2(50000-x))*8)/100
And the rest to be invested at 6% would be ((100000 - (2(50000-x)))*6)/100.

((2(50000-x))*8)/100 + ((100000 - (2(50000-x)))*6)/100 = (((50000+x)*6)/100 + ((50000-x)*8)/100) + 800

So its again a linear eq which can be solved fro x
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by bharat.bondalapati » Sun Apr 07, 2013 12:28 am
From Statement 1:

(6x)/100 + (1,00,000-x)8/100 = 6800

Solving this linear equation for x, we have x=60,000
So, the two sums are 60,000 and 40,000

Second equation:
40000*2=80,000 at 8%
80000*8/100 + 20000*6/100 = 7600 which is $ 800 more than the first income.
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