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05 Jun 2012, 05:03
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D
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Difficulty:
5%
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Question Stats:
89% (01:14) correct
11%
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wrong
based on 1235
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On May 1 of last year, Jasmin invested x dollars in a new account at an interest rate of 6 percent per year, compounded monthly. If no other deposits or withdrawals were made in the account and the interest rate did not change, what is the value of x?
(1) As of June 1 of last year, the investment had earned $200 in interest.
(2) As of July 1 of last year, the investment had earned $401 in interest.
ID 01933
(1) As of June 1 of last year, the investment had earned $200 in interest.
(2) As of July 1 of last year, the investment had earned $401 in interest.
ID 01933
fozzzy
on may 1 of last year, jasmin invested X dollars in a new account at an interest rate of 6 percent per year, compounded monthly. If no other deposits or withdrawals were made in the account and the interest rate did not change, what is the value of x?
1) as of june 1 of last year, the investment had earned $200 in interest
2) as of july 1 of lasy year, the investment had earned $401 in interest.
I know its a basic question, you don't really need to solve it but I'm interested in the detailed approach in case I see a problem solving question of this type.
Statement 1-1) as of june 1 of last year, the investment had earned $200 in interest
2) as of july 1 of lasy year, the investment had earned $401 in interest.
I know its a basic question, you don't really need to solve it but I'm interested in the detailed approach in case I see a problem solving question of this type.
Monthly C.I.(Compd. Int.) = 6/12=0.5% per month;
When there is only one period for calculation, then C.I.= S.I.(simple Int.). In this question May to June is one month only.
We know that SI=P*r*t/100 => 200=P*0.5*1/100 => P can be derived. There is no need to calculate exact value in DS question.Just make sure there is a unique value. Sufficient.
Statement 2-
From May to July there are 2 periods, so this is the case of CI rather SI.
We know that CI = P(1+r/100)^n - P => n= # of periods = 2 months => 401=P(1+0.5/100)^2 - P => P can be derived. There is no need to calculate exact value. Sufficient.
Answer D.
-Shalabh Jain
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30 Jun 2012, 23:46
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Principal: $x
Rate (per year): 6% compounded monthly
Rate (per month): \(\frac{6%(per year)}{12(months)} = 0.5%\)
(1) As of June1 of last year, the investment had earned $200 in interest.
In one months time, the interest earned is $200:
\(x * (\frac{0.5}{100}) = 200.\)
From this, we can find out the value of x.
Hence, A is sufficient
(2) As of July 1 of last year, the investment had earned $401 in interest.
In two months, the interest earned on the new principal is $401:
\(x * (1 + \frac{0.5}{100})*(\frac{0.5}{100}) = 401.\)
Above, the expression \(x * (1 + \frac{0.5}{100})\) represents the principal after the end if the first month i.e the initial principal+one month's worth of interest.
This is enough to compute the value of x.
B is also sufficient.
Thus, the answer is D, both statements alone are sufficient.
Rate (per year): 6% compounded monthly
Rate (per month): \(\frac{6%(per year)}{12(months)} = 0.5%\)
(1) As of June1 of last year, the investment had earned $200 in interest.
In one months time, the interest earned is $200:
\(x * (\frac{0.5}{100}) = 200.\)
From this, we can find out the value of x.
Hence, A is sufficient
(2) As of July 1 of last year, the investment had earned $401 in interest.
In two months, the interest earned on the new principal is $401:
\(x * (1 + \frac{0.5}{100})*(\frac{0.5}{100}) = 401.\)
Above, the expression \(x * (1 + \frac{0.5}{100})\) represents the principal after the end if the first month i.e the initial principal+one month's worth of interest.
This is enough to compute the value of x.
B is also sufficient.
Thus, the answer is D, both statements alone are sufficient.
