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- Question
You deposit $600 into a 6 month certificate of deposit. After 6 months the balance is $618. Find the simple annual interest rate.
Options- A. 5%
- B. 4%
- C. 6%
- D. 7%
- Correct Answer
- 6%
ExplanationI=prt
- 1. Determine the exact and ordinary simple interest on a 90-day loan of $8000 at 812 %
Options- A. 150$
- B. $160
- C. $170
- D. $180 Discuss
- 2. If there is a total of 104 weekly instalments and a third deposit, find: the total cost of the computer
Options- A. 3932.55
- B. 4932
- C. 5932
- D. 6932 Discuss
- 3. If there is a total of 104 weekly instalments and a third deposit, find the interest rate
Options- A. 2.82
- B. 3.82
- C. 4.82
- D. 5.82 Discuss
- 4. If there is a total of 104 weekly instalments and a thirddeposit, find the interest charged
Options- A. 237.55
- B. 337
- C. 443
- D. 334 Discuss
- 5. A sapphire ring with a marked price of $1800 is offered to the purchaser on the following terms: $200 deposit and the balance to be paid over 24 equal monthly instalments with interest charged at 11.5% p.a. flat rate. Find the monthly repayments
Options- A. $53
- B. $42
- C. $62
- D. $82 Discuss
- 6. Rondell's parents borrow $6,300 from the bank for a new car. The interest rate is 6% per year. How much simple interest will they pay if they take 2 years to repay the loan?
Options- A. 556
- B. 665
- C. 756
- D. 856 Discuss
- 7. Derrick's dad bought new tires for $900 using a credit card. His card has an interest rate of 19%. If he has no other charges on his card and does not pay off his balance at the end of the month, how much money will he owe after one month?
Options- A. 11.25
- B. 13.25
- C. 14.25
- D. 15.25 Discuss
- 8. An amount of 5,000 is invested at a fixed rate of 8 per cent per annum. What amount will be the value of the investment in five years time, if the interest is compounded every six months?
Options- A. 7401.22
- B. 3456
- C. 4567
- D. 7890 Discuss
- 9. A sum of Rs 1200 amounts to Rs 1740 in 3 years at simple interest. If rate of interest is increased by 3%, then what will be the new amount (in Rs)?
Options- A. 1848
- B. 1946
- C. 1812
- D. 1924 Discuss
- 10. Both S.I. and C.I. is calculated with a similar rate of 10% per annum on a sum of rupees. If C.I. is calculated yearly for two years, then for what period must S.I. be evaluated such that S.I. = C.I.?
Options- A. 4.2 years
- B. 2.1 years
- C. 1.1 years
- D. 1.4 years Discuss
Simple Interest problems
Search Results
Correct Answer: $170
Explanation:
Exact interest, I= prt = $8000 × 0.085 × 90/365 = 167.67
Ordinary Interest, I= Prt = $8000 x 0.085 x 90/360 = 170
Correct Answer: 3932.55
Explanation:
Total cost = deposit + loan + interest
= 1231.67 + 2463.33 + 237.55
= $3932.55
Correct Answer: 4.82
Explanation:
R = (100 x I)/( P x T)
Correct Answer: 237.55
Explanation:
Cash price = $3695
Deposit =1/3 of $3695 = $1231.67
Loan amount = $3695.00 ? $1231.67 = $2463.33
Total cost of loan = $25.97 × 104= $2700.88
Interest charged = total amount ? loan
I = A ? P
= 2700.88 ? 2463.33
= 237.55
Correct Answer: $82
Explanation:
Total repayment amount = $1600 + $368
= $1968
= $82
regular payment= total amount/number of repayments
Correct Answer: 756
Explanation:
I=prt
Correct Answer: 14.25
Explanation:
I=prt
Correct Answer: 7401.22
Explanation:
With slight modifications, the basic formula can be made to deal with compounding at intervals other than annually.
Since the compounding is done at six-monthly intervals, 4 per cent (half of 8 per cent) will be added to the value on each occasion.
Hence we use r = 0.04. Further, there will be ten additions of interest during the five years, and so n = 10. The formula now gives:
V = P(1 + r)10 = 5,000 x (1.04)10 = 7,401.22
Thus the value in this instance will be £7,401.22.
In a case such as this, the 8 per cent is called a nominal annual
rate, and we are actually referring to 4 per cent per six months.
Correct Answer: 1848
Correct Answer: 2.1 years
Explanation:
Let time period of S.I. be T years.
Then for a principal amount, say P,
ATQ, as, S.I. = C.I. for rate =10%p.a. and
time for C.I. = 2(P x 10 x T)/100 = P{[ (100+10)/100 ]2-1}T/10 = [ 110/100 ]2?1 = [(11/10)2?1] = (121-100)/100T/10 = 21/100T = 21/10 = 2.1 years
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