Nathan took out a student loan for 1200$ at 10 percent annual interest

Question

Nathan took out a student loan for 1200$ at 10 percent annual interest, compounded annually. If he did not repay any of the loan or interest during the first 3 years, which of the following is the closest to the amount of interest he owed for the 3 years.

A. 360
B. 390
C. 400
D. 410
E. 420

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A. 360 
B. 390 
C. 400 
D. 410 
E. 420

shashankism · Accepted Answer

Principal amount is : 1200
Total amount after 
1st yr = 1200 x 1.1 = 1320
2nd yr = 1320 x 1,1 = 1452
3rd yr = 1452 x 1.1 = 1597.2

Amount of interest he owed after 3 years = 1597.2 - 1200 = 397.2 ....... The closest value is 400.

Correct Answer C

ab52 · Answer

I did it this way:
1st year's interest: 10% of 1200=120,
2nd year's interest: 10% of 1320=132, ( here, 1320= 1200 +120),
3rd year's interest: 10% of 1452= 145.2 ( 1452= 1320+132)
 hence total interest=  397.2\approx{400}

chetan2u · Answer

a straight Qrequiring direct formulae..
when compounded annualy the final amount=P(1+ r/100)^n..
here P=1200, r=10 and n=3..
1200(1+ 10/100)^3..
1200*1.1^3=1597.2

interest =1597-1200=397, which is nearly 400 in the choices given
ansC

singhabhijeet · Answer

CI =  1200 * (1+\frac{10}{100})^3 
 
 =  1200* (1.10)^3 
 
 = 1200 * 1.331

= 1597.2

Approx amount = 1597.2 - 1200 = 397.8 = 400 (approx)

Explanation

C.I = P *  (1+\frac{r}{100*n})^nt

where, 
P = principal amount
r = rate of Interest
n= number of times interest compounded
t = time period.

So, we have

n = 1200
r = 10
n = 1 (compounded annually)
t = 3 years.

Substitute above values in CI formula.

Ans: Option C

guptarahul · Answer

when compounded annualy the final amount=P(1+ r/100)^n.. here P=1200, r=10 and n=3.. 1200(1+ 10/100)^3.. 1200*1.1^3=1597.2 final amount - P = P(1+ r/100)^n - P = Compound Interest 1597.2-1200 =

aprox (397)= :-D

aprox (400)

StockAnalyst · Answer

Question asked approximation so it is easy
Initial amount 1200 mean at year 1 end==120$(interest)
Now amount 1320 mean at year 2 end==(120+132)
now amount 1452 mean at year 3 end==(120+132+145.2)
total=397.2 close to 400
Option C

Alphatango · Answer

Closer to 400 than 390.
I doubt whether these qstns do come in exam.

Blackishmamba · Answer

Please find the short calculation

Posted from my mobile device

GMATGuruNY · Answer

Since the 10% rate is compounded annually, the amount of interest increases by 10% each year: 
First-year interest = 10% of 1200 = 120. 
Second-year interest = (first-year interest) + 10% = 120 + 12 = 132. 
Third-year interest = (second-year interest) + 10% = 132 + 13.2 = 145.2. 
Total interest = 120 + 132 + 145.2 = 397.2.

C

ScottTargetTestPrep · Answer

We can use the annual compound interest formula: A = P(1 + r)^t. So we have:

A = 1200(1 + 0.1)^3

A = 1200(1.1)^3

A = 1597.2

The amount of $1597.20 is the total amount (principal plus interest) he owes, so the amount of interest he owes is 1597.2 - 1200 = $397.2, or approximately $400.

Answer: C

SchruteDwight · Answer

1200*(1+\frac{1}{10})^3-1200 \implies 1200(1+\frac{2}{10}+\frac{1}{100})(1+\frac{1}{10})-1200 \implies 120+240+24+12+1.2

Hoozan · Answer

For those who do not want to learn any formula can just use the basic understanding of SI and CI and solve

10% of 1200 = 120 (1) ; New amount = 1320.

10% of 1320 = 132 (2) ; New amount = 1450

10% of 1450 = 145 (3)

Add (1) (2) & (3)

Total interest due = $397

(C)

Fisayofalana · Answer

Used estimation.

Year 1: 10% of 1200 = 120 interest
Year 2: 1200+120 = 1320 principal x 10% = 132 interest 
Year 3. 1320+132 = 1452 x 10% = ~145 interest

sum of interest = 120+132+145= 397. 400 is the closest

SOHAL

Nathan took out a student loan for 1200$ at 10 percent annual interest

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