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Russ19
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At the beginning of 2009, equal amounts were invested in two accounts 21 Feb 2022, 06:10
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E
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Difficulty:

85% (hard)

Question Stats:

55% (03:04) correct 45% (02:51) wrong based on 107 sessions

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At the beginning of 2009, equal amounts were invested in two accounts each paying x% annual interest. The first account paid simple interest, and the second paid interest compounded annually. If the interest earned on the second account after two years was 18% greater than the interest earned on the first account, what is the value of x?

A) 9

B) 12

C) 18

D) 24

E) 36
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E
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JerryAtDreamScore
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At the beginning of 2009, equal amounts were invested in two accounts 21 Feb 2022, 17:45
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sjuniv32
At the beginning of 2009, equal amounts were invested in two accounts each paying x% annual interest. The first account paid simple interest, and the second paid interest compounded annually. If the interest earned on the second account after two years was 18% greater than the interest earned on the first account, what is the value of x?

A) 9

B) 12

C) 18

D) 24

E) 36
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Breaking Down the Info:

For simple interest, two years of x% interest is simply \(2x\%\) interest.

Then for compound interest, we need to write the total amount at the end \((1 + x\%)^2 = 1 + 2x\% + (x\%)^2\) first and subtract 1 to get the interest itself is \(2x\% + (x\%)^2\).

The compound interest is 18% greater than the simple interest, so \(2x\% + (x\%)^2 = 1.18 * 2x\%\).

Then \((x\%)^2 = 0.18 * 2x\%\)
\(x^2 = \frac{0.18 * 2x }{ \%} = 36x\)
\(x = 36\)

Answer: E
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Apoorva0406
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At the beginning of 2009, equal amounts were invested in two accounts 23 Jan 2025, 21:58
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why are we subtracting 1 here??
Bunuel can you please explain??
JerryAtDreamScore
sjuniv32
At the beginning of 2009, equal amounts were invested in two accounts each paying x% annual interest. The first account paid simple interest, and the second paid interest compounded annually. If the interest earned on the second account after two years was 18% greater than the interest earned on the first account, what is the value of x?

A) 9

B) 12

C) 18

D) 24

E) 36

Breaking Down the Info:

For simple interest, two years of x% interest is simply \(2x\%\) interest.

Then for compound interest, we need to write the total amount at the end \((1 + x\%)^2 = 1 + 2x\% + (x\%)^2\) first and subtract 1 to get the interest itself is \(2x\% + (x\%)^2\).

The compound interest is 18% greater than the simple interest, so \(2x\% + (x\%)^2 = 1.18 * 2x\%\).

Then \((x\%)^2 = 0.18 * 2x\%\)
\(x^2 = \frac{0.18 * 2x }{ \%} = 36x\)
\(x = 36\)

Answer: E
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hr1212
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At the beginning of 2009, equal amounts were invested in two accounts 29 Jan 2025, 07:38
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Russ19
At the beginning of 2009, equal amounts were invested in two accounts each paying x% annual interest. The first account paid simple interest, and the second paid interest compounded annually. If the interest earned on the second account after two years was 18% greater than the interest earned on the first account, what is the value of x?

A) 9

B) 12

C) 18

D) 24

E) 36
Show more
Simple interest => \(S = P*(1 + rt)\) --- (P is Principal, r is interest rate, t is time)

Compound interest => \(C = P*(1 + \frac{r}{n})^{nt}\) --- (P is Principal, r is interest rate, t is time, n is no. of times it's compounded per year)

In our case we are just interested in calculating interest and not the final amount, which will be,

\(S = P*(1 + rt) - P = P*2x/100\) ----- (r=x/100 and t=2)
\(C = P*(1 + \frac{r}{n})^{nt} - P = P*(1 + \frac{x}{100})^{2} - P = P*(2x/100 + x^2/10^4)\) ----- (r=x/100, n=1 and t=2)

S*1.18 = C
\(P*2x*1.18/100 = P*(2x/100 + x^2/10^4)\)
\(2.36 = 2 + x/100\)
\(x = 36\)

IMO: E
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