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ps 500 test7 # 10

by dunkin77 » Fri Apr 06, 2007 1:08 pm

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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
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by Neo2000 » Fri Apr 06, 2007 5:26 pm
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
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by dunkin77 » Fri Apr 06, 2007 8:07 pm
Okay - I got it:)

I was doing 10000 + (1% of 10000) + (1% of 10000)....

we need to multiply 1% of 10000 with the previous year.

Thanks!
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by Cybermusings » Fri Apr 06, 2007 11:54 pm
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
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by BTGmoderatorRO » Sun Oct 29, 2017 9:31 am
$$Interest,\ I,\ is\ given\ by\ I=\ \frac{\left[principal\left(p\right)\cdot rate\left(r\right)\cdot Time\left(t\right)\right]}{100}$$
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:
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Re: ps 500 test7 # 10

by Scott@TargetTestPrep » Sat Jan 18, 2020 4:26 am
dunkin77 wrote: ↑
Fri Apr 06, 2007 1:08 pm
 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
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:

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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