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.
Answer: C.
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.