Hi,
I thought the answer was A) but turned out to be B).... can you explain?
On July 1, 1982, Ms. Fox deposited $10,000 in a new account at the annual interest rate of 12 percent compounded monthly. If no additional deposits or withdrawals were made and if interest was credited on the last day of each month, what was the amount of money in the account on September 1, 1982?
(A) $10,200
(B) $10,201
(C) $11,100
(D) $12,100
(E) $12,544
ps 500 test7 # 10
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- Neo2000
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You know that in Compound Interest, the annual rate will become Rate/Number of Compoundings so rate now becomes 1
Time period "n" will become 12n as we are compounding monthly.
Luckily they asked us only 2 compoundings so you could just as easily have done it the simple interest way
1% of 10000 = 100
Next month
1% of 10100 = 101
Amount at the beginning of the third month
10201
Time period "n" will become 12n as we are compounding monthly.
Luckily they asked us only 2 compoundings so you could just as easily have done it the simple interest way
1% of 10000 = 100
Next month
1% of 10100 = 101
Amount at the beginning of the third month
10201
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On July 1, 1982, Ms. Fox deposited $10,000 in a new account at the annual interest rate of 12 percent compounded monthly. If no additional deposits or withdrawals were made and if interest was credited on the last day of each month, what was the amount of money in the account on September 1, 1982?
(A) $10,200
(B) $10,201
(C) $11,100
(D) $12,100
(E) $12,544
Since the interest is credited on the last day of each month. In the time period mentioned above the interest will be credited twice i.e. on July 31, 1982 and August 31, 1982.
Formula for Amount = P [ 1 + R/100]^n
= 10,000 [ 1 + 12/100*12]^2
= 10,000 * 101/100 * 101/100
= 101 * 101 = 10,201
(A) $10,200
(B) $10,201
(C) $11,100
(D) $12,100
(E) $12,544
Since the interest is credited on the last day of each month. In the time period mentioned above the interest will be credited twice i.e. on July 31, 1982 and August 31, 1982.
Formula for Amount = P [ 1 + R/100]^n
= 10,000 [ 1 + 12/100*12]^2
= 10,000 * 101/100 * 101/100
= 101 * 101 = 10,201
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$$Interest,\ I,\ is\ given\ by\ I=\ \frac{\left[principal\left(p\right)\cdot rate\left(r\right)\cdot Time\left(t\right)\right]}{100}$$
for the first month interval, i.e July 1 to august 1,
p=10000
r=12% annually
t=1/12 years (1 month)
$$therefore,\ I=\ \frac{\left[10000\cdot12\cdot\left(\frac{1}{\left(12\right)}\right)\right]}{100}\ \ =\ 100$$
amount after this period=P+I=10000+100=10100
For the second month interval, i.e august 1 to september 1,
p=10100
r=12% monthly
t= 1/12 years
$$therefore,\ I=\ \frac{\left[10100\cdot12\cdot\left(\frac{1}{\left(12\right)}\right)\right]}{100}\ \ =\ 101$$
the amount after this perios= P+I = 10100+101
=$10201
hence, option B is the correct answer :mrgreen:
for the first month interval, i.e July 1 to august 1,
p=10000
r=12% annually
t=1/12 years (1 month)
$$therefore,\ I=\ \frac{\left[10000\cdot12\cdot\left(\frac{1}{\left(12\right)}\right)\right]}{100}\ \ =\ 100$$
amount after this period=P+I=10000+100=10100
For the second month interval, i.e august 1 to september 1,
p=10100
r=12% monthly
t= 1/12 years
$$therefore,\ I=\ \frac{\left[10100\cdot12\cdot\left(\frac{1}{\left(12\right)}\right)\right]}{100}\ \ =\ 101$$
the amount after this perios= P+I = 10100+101
=$10201
hence, option B is the correct answer :mrgreen:
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- Scott@TargetTestPrep
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We see that the monthly interest rate is 0.12/12 = 0.01. Therefore, on August 1st, the amount of money in the account was:dunkin77 wrote: ↑Fri Apr 06, 2007 1:08 pmHi,
I thought the answer was A) but turned out to be B).... can you explain?
On July 1, 1982, Ms. Fox deposited $10,000 in a new account at the annual interest rate of 12 percent compounded monthly. If no additional deposits or withdrawals were made and if interest was credited on the last day of each month, what was the amount of money in the account on September 1, 1982?
(A) $10,200
(B) $10,201
(C) $11,100
(D) $12,100
(E) $12,544
10,000 x 0.01 + 10,000 = 100 + 10,000 = 10,100
On September 1st, the amount in the account was:
10,100 x 0.01 + 10,100 = 101 + 10,100 = 10,201
Answer: B
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