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Bunuel
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Last year Luis invested x dollars for one year, half at 8 percent simp 29 Aug 2017, 09:29
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55% (hard)

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65% (02:16) correct 35% (02:22) wrong based on 468 sessions

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Last year Luis invested x dollars for one year, half at 8 percent simple annual interest and the other half at 12 percent simple annual interest. Now he wants to reinvest the x dollars for one year in the same two types of investments, but the lower rate has decreased. If the higher rate is unchanged, what fraction of the x dollars must he reinvest at the 12 percent rate so that the total interest earned from the x dollars will be the same for both years ?

(1) The lower rate is now 6 percent.
(2) The total amount of interest earned from the two investments last year was $3,000.
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Last year Luis invested x dollars for one year, half at 8 percent simp 29 Aug 2017, 10:40
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last year :
Total SI = x/2 * 8/100* 1 + x/2 * 12/100* 1 = 20x/200 = x/10 ---------------> equation 1

this year :
let p be invested at lower rate 'r' while x-p will be invested at 12%
Total SI = pr/100 + 12(x-p)/100 ---------------------------------------------------> equation 2

We have to find p to know the fraction of the x dollars invested at 12% and total interest earned remains the same.
A.) The lower rate is now 6 percent.
r = 6%
Equating eq 1 and 2 , we can find p easily-
equation 2 becomes => 6p/100 + 12(x-p)/100 => (6p+12x-12p)/100 => 12x-6p/100
12x-6p/100 = x/10
12x-6p = 10x
2x = 6p
x = 3p
therefore p = x/3 invested at 6% while 2x/3 will be invested at 12%
Sufficient

B.) Knowing total interest earned gives us the value of x but still 2 unknowns, p & r.
Not sufficient

Hence, A

Please point out in case I have made a mistake.

Thanks.
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Last year Luis invested x dollars for one year, half at 8 percent simp 30 May 2019, 08:12
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Bunuel, VeritasKarishma,
Cud u plz help us solving this tough question?
Is the following way given by walker correct: https://gmatclub.com/forum/last-year-lu ... 81854.html?

I look forward to hearing from u very soon ??

Posted from my mobile device
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Last year Luis invested x dollars for one year, half at 8 percent simp 30 May 2019, 23:24
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studentsensual
Bunuel, VeritasKarishma,
Cud u plz help us solving this tough question?
Is the following way given by walker correct: https://gmatclub.com/forum/last-year-lu ... 81854.html?

I look forward to hearing from u very soon ??

Posted from my mobile device
Show more

Year 1:
Rate of return for x/2 and x/2 amount are 8% and 12%. This gives us an average rate of return of 10% (since the two amounts are the same, average will just be the simple average)

Year 2:
We need a 10% rate of return. One investment is available at 12%. The other rate of return can be anything from 0% to 8%. The amount invested in the two investments will be adjusted to get 10%.

For example, say the second rate of return is 6%,

w1/w2 = (12 - 10)/(10 - 6) = 1/2
So 1/3 part of x should be invested at 6% and 2/3 part at 12%

Say second rate of return is 4%
w1/w2 = (12 - 10)/(10 - 4) = 1/3
So 1/4 part of x should be invested at 4% and 3/4 part at 12%

Say second rate of return is 0%
w1/w2 = (12 - 10)/(10 - 0) = 1/5
So 1/6 part of x should be invested at 0% and 5/6 part at 12%.

and so on...
We need the fraction that must be invested at 12%. For this, we need to know the second rate of return.

(1) The lower rate is now 6 percent.

Sufficient.

(2) The total amount of interest earned from the two investments last year was $3,000.

Not sufficient. We get the amount x using this information. But the question is still the same - we need total 10% return. What is the second rate of return? Not known.

Not sufficient.

Answer (A)
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Last year Luis invested x dollars for one year, half at 8 percent simp 14 May 2025, 12:27
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Let's analyze the problem. Last year, Luis invested $x$ dollars, half at 8% and half at 12%. The total interest earned last year was:
$$
\frac{x}{2} \cdot 0.08 + \frac{x}{2} \cdot 0.12 = 0.04x + 0.06x = 0.10x
$$
This year, let \(f\) be the fraction of \(x\) dollars invested at 12%, so \($x \cdot f$\) is invested at 12%, and \($x \cdot (1-f)$\) is invested at the lower rate. We want the total interest earned this year to be the same as last year, which is \($0.10x$\).

Statement (1): The lower rate is now 6%. So, we have:
$$
x(1-f) \cdot 0.06 + xf \cdot 0.12 = 0.10x
$$
Dividing by \(x\), we get:
$$
0.06(1-f) + 0.12f = 0.10 \\
0.06 - 0.06f + 0.12f = 0.10 \\
0.06f = 0.04 \\
f = \frac{0.04}{0.06} = \frac{4}{6} = \frac{2}{3}
$$
So, we can find the fraction. Statement (1) is sufficient.

Statement (2): The total amount of interest earned from the two investments last year was $3,000. This means $0.10x = 3000$, so \(x\) = 30000$. However, we don't know the new lower interest rate, so we cannot determine the fraction \(f\). Statement (2) is insufficient.

Therefore, the answer is A.
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