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Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep
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Author:  Kritesh [ 11 May 2017, 10:33 ]
Post subject:  Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

Lynn lent Robert $10,000 at 6 percent simple annual interest to be repaid at the end of 3 years. What was the total amount that Robert had to pay to Lynn?


A. $10,180
B. $10,190
C. $10,600
D. $11,800
E. $11,910

Author:  anugrahs [ 11 May 2017, 10:45 ]
Post subject:  Re: Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

A very easy Simple interest problem.

I = (P*N*R)/100
where
P = principal
N = time (in years)
R = rate

So in this case I = (10000*3*6)/100 = 1800

Thus total amount to be paid = Principal + Interest
= 10000 + 1800 = 11800

Thus D

Regards
Anugrah

Author:  Abhishek009 [ 11 May 2017, 11:31 ]
Post subject:  Re: Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

Kritesh wrote:
Lynn lent Robert $10,000 at 6 percent simple annual interest to be repaid at the end of 3 years. What was the total amount that Robert had to pay to Lynn?


A. $10,180
B. $10,190
C. $10,600
D. $11,800
E. $11,910

_____________________________________________________
If you appreciate the question then please click +1Kudos :)


\(Amount = Principal + Simple \ Interest\)

Or, \(Amount = 10,000 + (\frac{10,000*3*6}{100})\)

Or, \(Amount = 10,000 + 1,800\)

Or, \(Amount = 11,800\)

Thus, the correct answer must be (D) $ 11,800

Author:  sashiim20 [ 12 May 2017, 02:37 ]
Post subject:  Re: Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

Lynn lent Robert $10,000 at 6 percent simple annual interest to be repaid at the end of 3 years. What was the total amount that Robert had to pay to Lynn?

simple interest = \(\frac{10000 * 6 * 3}{100}\)
simple interest = 1800

Total amount = 10000 +1800 = 11800 .... Answer D

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Author:  generis [ 20 Aug 2018, 11:56 ]
Post subject:  Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

Kritesh wrote:
Lynn lent Robert $10,000 at 6 percent simple annual interest to be repaid at the end of 3 years. What was the total amount that Robert had to pay to Lynn?


A. $10,180
B. $10,190
C. $10,600
D. $11,800
E. $11,910

The simple interest AMOUNT owed each year is derived from the original principal alone and does not change.

Six percent of \($10,000 = $600\) each year
\((\frac{6}{100}*$10,000)=$600\)

Start: $10,000 owed
END of Year 1: assessment of $600 owed
(new running total: $10,600 owed)

END of Year 2: assessment of $600 owed
(new running total: $11,200 owed)

END of Year 3: assessment of $600 owed
(new running total: $11,800 owed)

OR: 6% of $10,000 = $600. That amount will be assessed at the end of each of three years.($600 * 3) = $1,800

After 3 years
Principal owed + interest owed = ($10,000 + $1,800) = $11,800

Answer D

Author:  ScottTargetTestPrep [ 23 Aug 2018, 15:37 ]
Post subject:  Re: Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

Kritesh wrote:
Lynn lent Robert $10,000 at 6 percent simple annual interest to be repaid at the end of 3 years. What was the total amount that Robert had to pay to Lynn?


A. $10,180
B. $10,190
C. $10,600
D. $11,800
E. $11,910


Interest owed was:

10,000 x 0.06 x 3 = $1,800

So a total of 10,000 +1,800 = $11,800 was to be paid back.

Answer: D

Author:  Hovkial [ 02 Oct 2020, 07:29 ]
Post subject:  Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

On $100, Robert would pay an interest amount of (6/100)*100 in one year = $6.

On $10,000, Robert would pay an interest amount of $6*(10,000/100) in one year = $600.

On a sum of $10,000, in three years Robert would pay an interest amount of $600*3 = $1800.

Therefore, total amount to be paid by Robert = Initial sum + Total interest amount = $10,000 + $1800 = $11,800.

ANSWER: (D)

Author:  bumpbot [ 15 Jun 2022, 10:11 ]
Post subject:  Re: Lynn lent Robert $10,000 at 6 percent simple annual interest to be rep

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