# Simple Interest Questions

**FACTS AND FORMULAE FOR SIMPLE INTEREST QUESTIONS**

**1. Principal:** The money borrowed or lent out for a certain period is called the **principal **or the **sum**.

**2. Interest:** Extra money paid for using other's money is called **interest**

**3. Simple Interest (S.I.) : **If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called **simple interest.**

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,

(i) $S.I=\left(\frac{P\times T\times R}{100}\right)$

(ii) $P=\left(\frac{100\times S.I}{R\times T}\right);R=\left(\frac{100\times S.I}{P\times T}\right)andT=\left(\frac{100\times S.I}{P\times R}\right)$

A) 12.5% | B) 13.5% |

C) 11.5% | D) 14.5% |

Explanation:

Let principal = P, Then, S.I.= P and Time = 8 years

We know that S.I. = PTR/100

**Rate** = [(100 x P)/ (P x 8)]% = 12.5% per annum.

A) Rs.2000 | B) Rs.10000 |

C) Rs.15000 | D) Rs.20000 |

Explanation:

Principal = $Rs\left(\frac{100\times 5400}{12\times 3}\right)=Rs.15000$

A) 650 | B) 690 |

C) 698 | D) 700 |

Explanation:

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698

A) Rs.325 | B) Rs.545 |

C) Rs.560 | D) Rs.550 |

Explanation:

Let each instalment be Rs.x .

1st year = [x + (x * 12 * 2)/100]

2nd year = [ x + (x *12 * 1)/100]

3rd year = x

Then, [x + (x * 12 * 2)/100] + [ x + (x *12 * 1)/100] + x =1092

3x + ( 24x/100 ) + ( 12x/100 ) = 1092

336x =109200

Therefore, x = 325

Each instalment = Rs. 325

A) 800, 14% | B) 800, 13% |

C) 800, 12% | D) 800, 19% |

Explanation:

S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156 .

S.I. for 2 years = Rs (156 x $\frac{2}{3}$ x 2)= Rs 208.

Therefore, Principal = Rs (1008 - 208) = Rs 800.

Now, P = 800, T= 2 and S.I. = 208.

Therefore, Rate = (100 x S.I.) / (P x T) = [ (100 x 208)/(800 x 2)]% = 13%

A) 4 years | B) 2.5 years |

C) 3.5 years | D) 2 years |

A) Rs. 112.50 | B) Rs.150.25 |

C) Rs.167.50 | D) Rs.170 |

Explanation:

$gainin2years=Rs.\left[\left(5000*\frac{25}{4}*\frac{2}{100}\right)-\left(\frac{5000*4*2}{100}\right)\right]$

= Rs. (625 - 400)

= Rs. 225

$gainin1year=Rs.\left(\frac{225}{2}\right)=112.50$

A) 1.5 | B) 2.5 |

C) 3.5 | D) 4.5 |

Explanation:

Let the time be 'n' years, Then

$800\times {\left(1+\frac{5}{100}\right)}^{2}n=926.10={\left(1+\frac{5}{100}\right)}^{2}n=\frac{9261}{8000}={\left(\frac{21}{20}\right)}^{2}n={\left(\frac{21}{10}\right)}^{3}$

n = 3/2 or n= 1$\frac{1}{2}$ Years