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# Exactly 6 years ago, James opened 2 salary accounts, with no

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Exactly 6 years ago, James opened 2 salary accounts, with no [#permalink]

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07 Nov 2012, 04:54
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Exactly 6 years ago, James opened 2 salary accounts, with no further deposits nor withdrawals since then. Each account has earned simple interest, one at 3% per year and the other at 4% per year. Which account has more money now?

(1) When James opened the accounts, the account earning 3% interest contained $1,000 more than the account earning 4%. (2) Last year, the account earning 3% earned exactly$150 in interest.
[Reveal] Spoiler: OA

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Re: Exactly 6 years ago, James opened 2 salary accounts, with no [#permalink]

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07 Nov 2012, 05:22
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Exactly 6 years ago, James opened 2 salary accounts, with no further deposits nor withdrawals since then. Each account has earned simple interest, one at 3% per year and the other at 4% per year. Which account has more money now?

Say the amount invested at 3% was $x and the amount invested at 4% was$y.

(1) When James opened the accounts, the account earning 3% interest contained $1,000 more than the account earning 4%. x=y+1,000. Check extreme cases: if y=0 and x=1,000, then the amount invested at 4% ($0) would still have $0 after 6 years, thus the amount invested at 3% ($1,000) would naturally have more money. But if y=1,000,000 and x=1,001,000, then even after one year the amount invested at 4% would have more money than the amount invested at 3% (1,000,000+40,000=1,040,000 and 1,001,000+30,000+30=1,031,030). Not sufficient.

(2) Last year, the account earning 3% earned exactly $150 in interest. Since accounts earn simple interest, then each account earns the same amount each year --> x*0.03=$150 --> we can find x but still not sufficient.

(1)+(2) Since from (2) we know the value of x, then from x=y+1,000 we can find the value of y, hence we can answer the question. Sufficient.

Hope it's clear.
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Re: Exactly 6 years ago, James opened 2 salary accounts, with no [#permalink]

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07 Nov 2012, 11:05

Meanwhile from "(2) Last year, the account earning 3% earned exactly $150 in interest." Shouldn't it be x(0.03)*5 =$150?
The accounts were opened 6 yrs ago, and last yr would mean the account would have been
active for 5yrs...Other than that the explanations are crystal clear.

Thanks.
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Re: Exactly 6 years ago, James opened 2 salary accounts, with no [#permalink]

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10 Nov 2012, 19:17
gmatbull wrote:

Meanwhile from "(2) Last year, the account earning 3% earned exactly $150 in interest." Shouldn't it be x(0.03)*5 =$150?
The accounts were opened 6 yrs ago, and last yr would mean the account would have been
active for 5yrs...Other than that the explanations are crystal clear.

Thanks.

I was also confused with "simple interest" term. I don't know what exactly GMAT means by that but in real world it means that interest acquires only at the end of the term. Thus it should be x(0.03)^5=150.
But anyhow it doesn't influence our solution. We don't have to calculate this X from the second statement, we just know that X is some definite number, which will help us to put it into first statement. I personally stopped here.
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Re: Exactly 6 years ago, James opened 2 salary accounts, with no [#permalink]

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12 Nov 2012, 09:56
gmatbull wrote:

Meanwhile from "(2) Last year, the account earning 3% earned exactly $150 in interest." Shouldn't it be x(0.03)*5 =$150?
The accounts were opened 6 yrs ago, and last yr would mean the account would have been
active for 5yrs...Other than that the explanations are crystal clear.

Thanks.

The accounts earn simple interest, thus each year the accounts earn the same amount. Thus, if last year (for one, fifth year), the account earning 3% earned exactly $150 in interest, then x*0.03=$150.

Hope it's clear.
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Re: Exactly 6 years ago, James opened 2 salary accounts, with no [#permalink]

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15 Nov 2012, 15:00
Bunuel wrote:
Exactly 6 years ago, James opened 2 salary accounts, with no further deposits nor withdrawals since then. Each account has earned simple interest, one at 3% per year and the other at 4% per year. Which account has more money now?

Say the amount invested at 3% was $x and the amount invested at 4% was$y.

(1) When James opened the accounts, the account earning 3% interest contained $1,000 more than the account earning 4%. x=y+1,000. Check extreme cases: if y=0 and x=1,000, then the amount invested at 4% ($0) would still have $0 after 6 years, thus the amount invested at 3% ($1,000) would naturally have more money. But if y=1,000,000 and x=1,001,000, then even after one year the amount invested at 4% would have more money than the amount invested at 3% (1,000,000+40,000=1,040,000 and 1,001,000+30,000+30=1,031,030). Not sufficient.

(2) Last year, the account earning 3% earned exactly $150 in interest. Since accounts earn simple interest, then each account earns the same amount each year --> x*0.03=$150 --> we can find x but still not sufficient.

(1)+(2) Since from (2) we know the value of x, then from x=y+1,000 we can find the value of y, hence we can answer the question. Sufficient.

Hope it's clear.

Hi there

My question is how do you know when to use the extreme number such as y=1,000,000 and x=1,001,000, whereas I would just use x = 2000 and y = 1000 and would still get the result that by the end of 6 years the 3% account would earn more.
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Re: Exactly 6 years ago, James opened 2 salary accounts, with no [#permalink]

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05 Apr 2014, 06:28
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Re: Exactly 6 years ago, James opened 2 salary accounts, with no   [#permalink] 05 Apr 2014, 06:28
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