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## Arithmetic: Simple Interest Test -3

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Question 1 |

The simple interest accrued on an amount of Rs.2, 500 at the end of six years is Rs.1, 875. What would be the simple interest accrued on an amount of Rs.6, 875 at the same rate and same period?

Rs.4.556.5 | |

Rs.5.025.25 | |

Rs.4.895.25 | |

none |

Question 1 Explanation:

Simple Interest for 6 years is Rs. 1875

Simple Interest for 1 year is : Rs. 1875/6.

The rate of interest is = $ \frac{1875\times 100}{6\times 2000}=\frac{125}{8}%~$

The SI accrued on Rs. 6,875 @ 125/8 % for 6 years is = Rs. 6445 (approx)

Simple Interest for 1 year is : Rs. 1875/6.

The rate of interest is = $ \frac{1875\times 100}{6\times 2000}=\frac{125}{8}%~$

The SI accrued on Rs. 6,875 @ 125/8 % for 6 years is = Rs. 6445 (approx)

Question 2 |

A sum of money at simple interest amounts to Rs.14,160 in 3 years. If the rate of interest is increased by 25%, the same sum amounts to Rs.14,700 in the same time. The rate of interest is

5% | |

5$ \displaystyle \frac{1}{2}$% | |

6% | |

7% |

Question 2 Explanation:

Let the principal be Rs. x and rate = r %.

Amount in 3 years @ r% is $ \displaystyle x(1+\frac{3r}{100})$ = Rs. 14160

Amount in 3 years @ (25+r)% is x$ \displaystyle \frac{3\left( r+25 \right)}{100}$

=$ \displaystyle \frac{\text{x}\left( 1+\frac{3\left( r+25 \right)}{100} \right)}{\text{x}\left( 1+\frac{3\left( r \right)}{100} \right)}=\frac{14700}{14160}$

Amount in 3 years @ r% is $ \displaystyle x(1+\frac{3r}{100})$ = Rs. 14160

Amount in 3 years @ (25+r)% is x$ \displaystyle \frac{3\left( r+25 \right)}{100}$

=$ \displaystyle \frac{\text{x}\left( 1+\frac{3\left( r+25 \right)}{100} \right)}{\text{x}\left( 1+\frac{3\left( r \right)}{100} \right)}=\frac{14700}{14160}$

Question 3 |

Simple interest on a certain sum at a certain annual rate of interest is 16% of the sum. If the numbers representing rate percent and time in years be equal, then the rate of interest is

4% | |

6% | |

4.5% | |

6.5% |

Question 3 Explanation:

Let the total sum = y

S.I. of the sum = 16/100 of y

we know that

Rate =Time =x

If x is the rate of interest and x years be the time .

then x

Correct option is (a)

S.I. of the sum = 16/100 of y

we know that

Rate =Time =x

If x is the rate of interest and x years be the time .

then x

^{2}=16. x= 4.Correct option is (a)

Question 4 |

A sum of Rs.5000/- amounts to Rs.6, 050/- in two years. What is the rate of interest?

15% p.a. | |

13% p.a. | |

11% p.a. | |

none |

Question 4 Explanation:

S.I. in 2 years is Rs. 1050.

S.I. in 1 year is Rs. 525.

Rate of interest is$ \displaystyle \frac{525}

{5000}X100=\frac{105}{10}=10.5~%~$

S.I. in 1 year is Rs. 525.

Rate of interest is$ \displaystyle \frac{525}

{5000}X100=\frac{105}{10}=10.5~%~$

Question 5 |

Mr. A lends 40% of sum at 15% p.a. 50%of rest sum at 10%p.a. and the rest at 18% p.a. rate of interest. What would be the rate of interest if the interest is calculated on the whole sum?

13.4% p.a. | |

14.33% p.a. | |

14.4% p.a | |

13.33% p.a. |

Question 5 Explanation:

Let the total number is Rs. 100

He lends Rs. 40 @ 15 %

Rs. 30 @ 10% and

Rs. 30 @ 18%.

Or we can say that Interest earned by each at end of 1 year By First Year

$ \displaystyle \Rightarrow \frac{15}{100}\times \frac{40x}{100}=\frac{60}{1000}x$

By second Year

$ \displaystyle \,\Rightarrow \frac{10}{100}\times \frac{30x}{100}=\frac{30}{1000}x$

By Third Year

$ \displaystyle \,\Rightarrow \frac{18}{100}\times \frac{30x}{100}=\frac{54}{1000}x$

Total Interest = $ \displaystyle =\frac{144}{1000}x$

=$ \displaystyle =\frac{144}{1000}x$

Therefore Rate % = $ \displaystyle \frac{\frac{144x}{1000}}{x}\times 100$

$ \displaystyle =14.4%$

He lends Rs. 40 @ 15 %

Rs. 30 @ 10% and

Rs. 30 @ 18%.

Or we can say that Interest earned by each at end of 1 year By First Year

$ \displaystyle \Rightarrow \frac{15}{100}\times \frac{40x}{100}=\frac{60}{1000}x$

By second Year

$ \displaystyle \,\Rightarrow \frac{10}{100}\times \frac{30x}{100}=\frac{30}{1000}x$

By Third Year

$ \displaystyle \,\Rightarrow \frac{18}{100}\times \frac{30x}{100}=\frac{54}{1000}x$

Total Interest = $ \displaystyle =\frac{144}{1000}x$

=$ \displaystyle =\frac{144}{1000}x$

Therefore Rate % = $ \displaystyle \frac{\frac{144x}{1000}}{x}\times 100$

$ \displaystyle =14.4%$

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