Q:
The compound interest on a certain sum of money at 11% for 2 years is ₹6963. Its simple interest (in ₹) at the same rate and for the same period is:
| A) ₹6750 | B) ₹6600 |
| C) ₹6000 | D) ₹6500 |
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3 6132
Q:
The compound interest on a certain sum of money at 21% for 2 years is ₹9,282. Its simple interest (in ₹) at the same rate and for the same period is:
| A) 8,500 | B) 8,400 |
| C) 8,000 | D) 8,750 |
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2 4027
Q:
₹4,000 is given at 5% per annum for one year and interest is compounded half yearly. ₹2,000 is given at 40% per annum compounded quarterly for 1 year. The total interest received is nearest to:
| A) ₹1,333.30 | B) ₹1,888.80 |
| C) ₹1,130.70 | D) ₹1,444.40 |
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1 8548
Q:
A sum amounts to ₹18,600 after 3 years and to ₹27,900 after 6 years, at a certain rate percent p.a., when the interest is compounded annually. The sum is:
| A) Rs. 14,600 | B) Rs. 12,400 |
| C) Rs. 11,800 | D) Rs. 14,400 |
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Q:
A sum of ₹x was borrowed and paid back in two equal yearly instalments, each of ₹35,280. If the rate of interest was 5%, compounded annually, then the value of x is:
| A) 64,400 | B) 65,600 |
| C) 64,800 | D) 65,400 |
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4 7737
Q:
What is the compound interest on a sum of ₹8,100 for years at 8% per annum, if the interest is compounded 5-monthly? (Nearest to ₹1)
| A) ₹837 | B) ₹873 |
| C) ₹824 | D) ₹842 |
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2 9704
Q:
Ram deposited an amount of ₹ 8,000 in a bank’s savings account with interest 6.5% compounded monthly. What amount will he get at the end of 18 months?
| A) Rs. 8816.97 | B) Rs. 8788.98 |
| C) Rs. 8907.56 | D) Rs. 8790.54 |
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3 3720
Q:
The difference between the compound interest and simple interest on ₹ x at 9% per annum for 2 years is ₹20.25. What is the value of x ?
| A) 2,500 | B) 2,400 |
| C) 2,800 | D) 2,200 |
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4 2100
Find the difference between the compound interest compounded yearly and half-yearly on Rs. 10000 for 18 months at 10% per annum.
Answer
Verified
Hint: First, find the amount for the compound interest compounded yearly by applying the formula $A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$ for the first year and then the formula $A = P{\left( {1 + \dfrac{r}{{2 \times 100}}} \right)^{t \times 2}}$ for the next 6 months. Then subtract the principal from the amount to get the interest.
Then, find the amount for the compound interest compounded half-yearly by applying the formula $A = P{\left( {1 + \dfrac{r}{{2 \times 100}}}
\right)^{t \times 2}}$. Then subtract the principal from the amount to get the interest. After that subtract the values of the interest to find the difference of the interest.
Complete step by step answer:
The formula for compound interest compounded yearly is,
$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$
The formula for compound interest compounded half-yearly is,
$A = P{\left( {1 + \dfrac{r}{{2 \times 100}}} \right)^{t \times 2}}$
Given: - Principal, P = Rs.
10000
Time, t = 18 months = 1.5 years
Rate, r = 10% p.a.
For the compound interest compounded yearly,
Calculate the amount for 1 year. Then calculate the amount for $\dfrac{1}{2}$ year.
For 1st year,
$P = 10000$
$t = 1$
$r = 10\% $
Then,
$A = 10000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}$
Cancel out common factors and take LCM,
$ \Rightarrow A = 10000\left( {\dfrac{{10 + 1}}{{10}}} \right)$
Add the terms and cancel out the common factor,
$
\Rightarrow A = 1000 \times 11$
Multiply the terms,
$ \Rightarrow A = 11000$
Now, for $\dfrac{1}{2}$ year,
$P = 11000$
$r = 10\% $
$t = \dfrac{1}{2}$
Substitute the values in the formula for compounded half-yearly,
$A = 11000{\left( {1 + \dfrac{{10}}{{2 \times 100}}} \right)^{\dfrac{1}{2} \times 2}}$
Cancel out the common factors,
$ \Rightarrow A = 11000{\left( {1 + \dfrac{1}{{20}}} \right)^1}$
Take LCM,
$ \Rightarrow A = 11000 \times \dfrac{{20 + 1}}{{20}}$
Cancel
out the common factors,
$ \Rightarrow A = 550 \times 21$
Multiply the terms,
$ \Rightarrow A = {\text{Rs}}{\text{. }}11550$
So, the interest is,
$I = A - P$
Substitute the value of amount and principal,
$ \Rightarrow I = 11550 - 10000$
Subtract the term,
$ \Rightarrow I = {\text{Rs}}{\text{. }}1550$.....….. (1)
For the compound interest compounded half-yearly,
$P = 10000$
$r = 10\% $
$t = \dfrac{3}{2}$
Substitute the values in the formula for compounded
half-yearly,
$A = 10000{\left( {1 + \dfrac{{10}}{{2 \times 100}}} \right)^{\dfrac{3}{2} \times 2}}$
Cancel out the common factors,
$ \Rightarrow A = 10000{\left( {1 + \dfrac{1}{{20}}} \right)^3}$
Take LCM,
$ \Rightarrow A = 10000{\left( {\dfrac{{20 + 1}}{{20}}} \right)^3}$
Add the terms,
$ \Rightarrow A = 10000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}}$
Cancel the terms and multiply the remaining terms,
$ \Rightarrow A = {\text{Rs}}{\text{.
}}11576.25$
So, the interest is,
$I = A - P$
Substitute the value of amount and principal,
$ \Rightarrow I = 11576.25 - 10000$
Subtract the term,
$ \Rightarrow I = {\text{Rs}}{\text{. }}1576.25$..........….. (2)
For the difference between two compound interest is,
$\therefore 1576.25 - 1550 = {\text{Rs}}{\text{. 2}}6.25$
Hence, the difference between the compound interest compounded yearly and half-yearly is Rs. 26.25.
Note: The students might
make mistakes in calculating the amount for the 6 months compounded yearly.
Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is different from the simple interest where interest is not added to the principal while calculating the interest during the next period.