Exercise : Simple Interest - General Questions
✔ Simple Interest -
General Questions
1.
A sum was put at SI at a certain rate for 3 years. Had it been put at a 2% higher rate, it would have fetched $360 more. Find the sum.
View Answer
Answer: Option C
Explanation:
Explanation:
The extra interest percentage over 3 years is \(3 \times 2 = 6\%\). Since \(6\%\) of \(P = 360\), \(P = \frac{360}{0.06} = 6,000\).
2.
A man invested \(\frac{1}{3}\) of his capital at 7%, \(\frac{1}{4}\) at 8%, and the remainder at 10%. If his annual income is $561, find the capital.
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Answer: Option C
Explanation:
Explanation:
Let capital be \(12x\). Interest = \(4x(0.07) + 3x(0.08) + 5x(0.10) = 561\). \(1.02x = 561 \implies x = 550\). Capital = \(12 \times 550 = 6,600\).
3.
What equal annual installment will discharge a debt of $6,450 due in 4 years at 5% SI?
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Answer: Option B
Explanation:
Explanation:
The formula for installment \(x\) is \(Debt = 4x + \frac{x \times 5 \times (3+2+1+0)}{100} = 6450\). \(4.3x = 6450 \implies x = 1,500\).
4.
The difference between SI and CI on a sum for 2 years at 10% is $50. Find the sum.
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Answer: Option B
Explanation:
Explanation:
Difference for 2 years = \(\frac{PR^2}{100^2}\). \(50 = \frac{P \times 100}{10000} \implies P = 5,000\).
5.
Find the Simple Interest (SI) on $1,000 for 2 years at 5% per annum.
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Answer: Option B
Explanation:
Explanation:
Using the formula \(SI = \frac{P \times R \times T}{100}\), we get \(SI = \frac{1000 \times 5 \times 2}{100} = 100\).
6.
At what rate will \(2,000 yield \)400 interest in 4 years?
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Answer: Option B
Explanation:
Explanation:
Using the formula \(R = \frac{SI \times 100}{P \times T}\), we calculate \(R = \frac{400 \times 100}{2000 \times 4} = 5\%\).
7.
A sum of money doubles itself in 10 years at Simple Interest. What is the rate of interest per annum?
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Answer: Option C
Explanation:
Explanation:
If the sum doubles, the Interest (\(SI\)) equals the Principal (\(P\)). Let \(P = 100\), then \(SI = 100\). \(R = \frac{100 \times 100}{100 \times 10} = 10\%\).
8.
Find the Simple Interest on $2,500 at 6% from January 1st to August 8th (a period of 219 days).
View Answer
Answer: Option B
Explanation:
Explanation:
Time \(T = \frac{219}{365} = 0.6\) years. \(SI = \frac{2500 \times 6 \times 0.6}{100} = 90\).
9.
If the Simple Interest is \(\frac{4}{9}\) of the Principal and the Rate (\(R\)) is equal to the Time (\(T\)), find the Rate.
View Answer
Answer: Option B
Explanation:
Explanation:
Given \(\frac{4}{9}P = \frac{P \times R^2}{100}\). Solving for \(R^2 = \frac{400}{9}\), we get \(R = \frac{20}{3} \approx 6.67\%\).
10.
A sum of money at Simple Interest amounts to $815 in 3 years and $854 in 4 years. Find the principal sum.
View Answer
Answer: Option B
Explanation:
Explanation:
Interest for 1 year = \(854 - 815 = 39\). Interest for 3 years = \(39 \times 3 = 117\). Principal \(P = 815 - 117 = 698\).