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## Arithmetic: Compound Interest Test -3

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Question 1 |

The simple interest accrued on an amount of Rs.27, 500 at the end of three years is Rs.10, 230. What would be the approximate compound interest accrued on the same amount at the same rate in the same period?

Rs.11, 550 | |

Rs.12, 620 | |

Rs.10, 950 | |

Rs.11, 900 |

Question 1 Explanation:

Principle value = Rs. 27500

Time = 3 years

therefore

$ \displaystyle ~10230=\frac{27500\times r\times 3}{100}$

r=12.4%

Calculating compound interest

Interst=Â Â $ \displaystyle 27500{{(1+12.4/100)}^{3}}-27500=27500\text{ }\!\!\times\!\!\text{ }(1.42-1)=11550$

Time = 3 years

therefore

$ \displaystyle ~10230=\frac{27500\times r\times 3}{100}$

r=12.4%

Calculating compound interest

Interst=Â Â $ \displaystyle 27500{{(1+12.4/100)}^{3}}-27500=27500\text{ }\!\!\times\!\!\text{ }(1.42-1)=11550$

Question 2 |

What would be the compound interest accrued on an amount of Rs.45, 400 at the end of two years at the ratio of 15 p.c.p.a.?

Rs.16411.5 | |

Rs.14461.5 | |

Rs.16461.5 | |

Rs.14641.5 |

Question 2 Explanation:

Principle value = 45,400

We Know that the Compound Interest =$ \displaystyle C.I.=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]$

= (45400 x 23/20 x 23/20) â€“ 45400

= 240166/4 â€“ 45400

= 14641.5

We Know that the Compound Interest =$ \displaystyle C.I.=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]$

= (45400 x 23/20 x 23/20) â€“ 45400

= 240166/4 â€“ 45400

= 14641.5

Question 3 |

The difference between the amount of compound interest and simple interest accured on an amount of Rs.26000 at the end of 3 years is Rs.2994.134. What is the rate of interest p.c.p.a.?

22 | |

17 | |

19 | |

Cannot be determined |

Question 3 Explanation:

Let the Principle be Rs. x =Rs 26000 and the rate of interest be r% per annum.

Amount after 3 years of Simple Interest

$latex =\,\,\frac{x(3r)}{100}$

Amount after 3 years of Compound Interest

=$latex x{{(1+\frac{r}{100})}^{3}}$

$latex Difference\,=\,\,x{{(1+\frac{r}{100})}^{3}}-(x+\frac{3xr}{100})=2994.134$

Solving this we get x=19%

Amount after 3 years of Simple Interest

$latex =\,\,\frac{x(3r)}{100}$

Amount after 3 years of Compound Interest

=$latex x{{(1+\frac{r}{100})}^{3}}$

$latex Difference\,=\,\,x{{(1+\frac{r}{100})}^{3}}-(x+\frac{3xr}{100})=2994.134$

Solving this we get x=19%

Question 4 |

What is the difference between the simple and compound interest on Rs.7300 at the rate of 6 p.c.p.a. in 2 years?

Rs.29.37 | |

Rs.26.28 | |

Rs.31.41 | |

Rs.23.22 |

Question 4 Explanation:

When we are given by the difference between simple Interest and

Compound Interest for two years then the

$ \displaystyle \begin{array}{l}Difference\,=\frac{P\times {{R}^{2}}}{{{\left( 100 \right)}^{2}}}\\=\frac{7300\times 36}{10000}=Rs.\,26.28\end{array}$

Compound Interest for two years then the

$ \displaystyle \begin{array}{l}Difference\,=\frac{P\times {{R}^{2}}}{{{\left( 100 \right)}^{2}}}\\=\frac{7300\times 36}{10000}=Rs.\,26.28\end{array}$

Question 5 |

The simple interest accrued on a sum of certain principal is Rs.6, 400 in four years at the rate of 8 p.c.p.a. What would be the compound interest accrued on that principal at the rate of 2 p.c.p.a. in 2 years?

Rs.800 | |

Rs.808 | |

Rs.704 | |

Rs.700 |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}P=\frac{S.I.\times 100}{Rate\times Time}\\=\frac{6400\times 100}{8\times 4}\\=Rs.\,20000\\CI=P\,\left[ {{\left( 1+\frac{Rate}{100} \right)}^{Time}}-1 \right]\\=20000\,\left[ {{\left( 1+\frac{2}{100} \right)}^{2}}-1 \right]\\=20000\times 0.0404\\=Rs.\,\,808\end{array}$

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