1. What is the difference between the compound interests on Rs. 5000 for 1 | |

A. Rs. 2.04 | B. Rs. 4.80 |

C. Rs. 3.06 | D. Rs. 8.30 |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

Amount after 1^{1}⁄_{2} years when interest is compounded yearly

$MF#%= 5000 \times \left(1 + \dfrac{4}{100}\right)^1\times \left(1 + \dfrac{\dfrac{1}{2} \times 4}{100}\right)

= 5000 \times \dfrac{104}{100} \times \left(1 + \dfrac{2}{100}\right) \\\\
= 5000 \times \dfrac{104}{100} \times \dfrac{102}{100} = 50 \times 104 \times \dfrac{51}{50} \\\\
= 104 \times 51 = \text{Rs. }5304$MF#%

Compound Interest for 1 ^{1}⁄_{2} years when interest is compounded yearly

= Rs.(5304 - 5000)

Amount after 1^{1}⁄_{2} years when interest is compounded half-yearly

$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = 5000\left(1 + \dfrac{(4/2)}{100}\right)^{2 \times \frac{3}{2}} = 5000\left(1 + \dfrac{2}{100}\right)^3\\\\
= 5000\left(\dfrac{102}{100}\right)^3 = 5000\left(\dfrac{102}{100}\right)\left(\dfrac{102}{100}\right)\left(\dfrac{102}{100}\right) = 50 \times 102 \times \dfrac{51}{50}\times \dfrac{51}{50} \\\\
= 102 \times 51 \times \dfrac{51}{50} = 51 \times 51 \times \dfrac{51}{25} = \text{Rs. } 5306.04$MF#%

Compound Interest for 1 ^{1}⁄_{2} years when interest is compounded half-yearly

= Rs.(5306.04 - 5000)

Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000) = 5306.04 - 5304 = Rs. 2.04

2. A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is: | |

A. Rs. 120 | B. Rs. 121 |

C. Rs. 123 | D. Rs. 122 |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

Amount after 1 year on Rs. 1600 (deposited on 1st Jan) at 5% when interest calculated half-yearly

$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T}

= 1600\left(1 + \dfrac{\text{(5/2)}}{100}\right)^{2 \times 1}

= 1600\left(1 + \dfrac{1}{40}\right)^2$MF#%

Amount after 1/2 year on Rs. 1600 (deposited on 1st Jul) at 5% when interest calculated half-yearly

$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = 1600\left(1 + \dfrac{\text{(5/2)}}{100}\right)^{2 \times \frac{1}{2}} = 1600\left(1 + \dfrac{1}{40}\right)$MF#%

Total Amount after 1 year

$MF#%=1600\left(1 + \dfrac{1}{40}\right)^2 + 1600\left(1 + \dfrac{1}{40}\right)

= 1600\left( \dfrac{41}{40}\right)^2 + 1600\left(\dfrac{41}{40}\right)

= 1600\left( \dfrac{41}{40}\right)\left[1 + \dfrac{41}{40}\right]\\\\
= 1600\left( \dfrac{41}{40}\right)\left( \dfrac{81}{40}\right) = 41 \times 81 = \text{Rs. }3321$MF#%

Compound Interest = Rs.3321 - Rs.3200 = Rs.121

3. There is 80% increase in an amount in 8 years at simple interest. What will be the compound interest of Rs. 14,000 after 3 years at the same rate? | |

A. Rs.3794 | B. Rs.3714 |

C. Rs.4612 | D. Rs.4634 |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

Let P = Rs.100

Simple Interest = Rs. 80 ( ∵ 80% increase is due to the simple interest)

$MF#%\text{Rate of interest} =\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 80}{100 \times 8} = 10\%\text{ per annum}$MF#%

Now let's find out the compound interest of Rs. 14,000 after 3 years at 10%

P = Rs.14000

T = 3 years

R = 10%

$MF#%\text{Amount after 3 years } = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 14000\left(1 + \dfrac{10}{100}\right)^3 \\\\
= 14000\left(\dfrac{110}{100}\right)^3 = 14000\left(\dfrac{11}{10}\right)^3 = 14 \times 11^3 = 18634$MF#%

Compound Interest = Rs.18634 - Rs.14000 = Rs.4634

4. The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is: | |

A. 1 | B. 2 |

C. 3 | D. 3.5 |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

Let the period be n years

Then, amount after n years = Rs.(30000 + 4347) = Rs. 34347

$MF#%\begin{align}&\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 34347\\\\
&30000\left(1 + \dfrac{7}{100}\right)^\text{n} = 34347\\\\
&30000\left(\dfrac{107}{100}\right)^\text{n} = 34347\\\\
&\left(\dfrac{107}{100}\right)^\text{n} = \dfrac{34347}{30000} = \dfrac{11449}{10000} = \left(\dfrac{107}{100}\right)^2\\\\
&n = 2\text{ years}\end{align}$MF#%

5. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum is: | |

A. Rs.600 | B. Rs.645 |

C. Rs.525 | D. Rs.625 |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

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**Solution 1**

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Let the sum be Rs.x

$MF#%\begin{align}&\text{Amount after 2 years on Rs.x at 4% per annum when interest is compounded annually }\\\\&=\text{x}\left(1 + \dfrac{4}{100}\right)^2 = \text{x}\left(\dfrac{104}{100}\right)^2\\\\&\text{Compound Interest = }\text{x}\left(\dfrac{104}{100}\right)^2 - x \\\\&= x\left[\left(\dfrac{104}{100}\right)^2 - 1\right] = x\left[\left(\dfrac{26}{25}\right)^2 - 1\right] = x\left[\dfrac{676}{625} - 1\right] = x\left[\dfrac{51}{625} \right] = \dfrac{51x}{625}\\\\\\\\
&\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{x \times 4 \times 2}{100} = \dfrac{2x}{25}\end{align}$MF#%

Given that difference between compound interest and simple interest is Rs.1

$MF#%\begin{align}&\Rightarrow \dfrac{51x}{625} - \dfrac{2x}{25} = 1\\\\
&\Rightarrow \dfrac{51x - 50x}{625} = 1\\\\
&\Rightarrow x = 625\end{align}$MF#%

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**Solution 2**

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$MF#%= \text{P}\left(\dfrac{\text{R}}{100}\right)^2$MF#%

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$MF#%\begin{align}&\text{P}\left(\dfrac{\text{R}}{100}\right)^2= 1\\\\
&\text{P}\left(\dfrac{4}{100}\right)^2= 1\\\\
&\text{P}\left(\dfrac{1}{25}\right)^2= 1\\\\
&\text{P}\left(\dfrac{1}{625}\right)= 1\\\\
&\text{P} = 625\end{align}$MF#%

i.e., required sum is Rs.625

R = 30

Let P is the amount

Simple Interest = PRT/100 = P * 30 * (3/2) /100= 9P/20

Compound Interest = P[1 + (R/2)/100]

^{2n}- P

= P[1 + (30/2)/100]

^{(2*3/2)}- P

= P(1.15)

^{3}- P

= P[(1.15)

^{3}-1]

P[(1.15)

^{3}-1] - 9P/20 = 5670

P[(1.15)

^{3}-1- 9/20] = 5670

P = 80000

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