# 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) 5% | B) 8% |

C) 12% | D) 15% |

Explanation:

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.

S.I. for 5 years = Rs.$\frac{2205}{3}\times 5$= Rs.3675

Principle = Rs.(9800-3675) = Rs.6125

Hence, Rate = $\left[\frac{100\times 3675}{6125\times 5}\right]$ =12%

A) Rs. 100 | B) Rs. 105 |

C) Rs. 115 | D) Rs. 110 |

Explanation:

Amount to be paid = $Rs.\left[100+\frac{200\times 5\times 1}{100}+\frac{100\times 5\times 1}{100}\right]$= Rs. 115**.**

A) 25years | B) 15years |

C) 20years | D) 22years |

Explanation:

From the information provided we know that,

Principal + 8% p.a. interest on principal for n years = 180 …….. (1)

Principal + 4% p.a. interest on principal for n years = 120 ……… (2)

Subtracting equation (2) from equation (1), we get

4% p.a. interest on principal for n years = Rs.60.

Now, we can substitute this value in equation (2),

i.e Principal + 60 = 120

= Principal = Rs.60.

We know that SI = , where p is the principal, n the number of years and r the rate percent of interest.

In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.

Therefore, 60 =(60*n*4)/100

=> n = 100/4 = 25 years.

A) Rs. 500 | B) Rs. 245 |

C) Rs. 1250 | D) Rs. 635 |

Explanation:

(kx5x1)/100 + [(1500 - k)x6x1]/100 = 85

5k/100 + 90 – 6k/100 = 85

k/100 = 5

=> k = 500

A) 30% | B) 25% |

C) 22% | D) 18% |

A) Rs. 580 | B) Rs. 480 |

C) Rs. 550 | D) Rs. 470 |

Explanation:

Let the sum be Rs. P. Then,

$\left[\mathbf{P}{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{25}}{\mathbf{2}\mathbf{x}\mathbf{100}}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{P}\right]$= 510

P[${\left(\frac{9}{8}\right)}^{2}$- 1] = 510.

Sum = Rs. 1920

So, S.I. = (1920 x 25 x 2) / (2 x 100) = Rs. 480

A) Only a is sufficient | B) Neither a nor b is sufficient |

C) Only b is sufficient | D) Both a and b sufficient |

Explanation:

Let the sum be Rs. x

a. gives, S.I = Rs. 9000 and time = 9 years.

b. gives, Sum + S.I for 6 years = 2 x Sum

--> Sum = S.I for 6 years.

Now, S.I for 9 years = Rs. 9000

S.I for 1 year = Rs. 9000/9 = Rs. 1000.

S.I for 6 years = Rs. (1000 x 6)= Rs. 6000.

--> x = Rs. 6000

Thus, both a and b are necessary to answer the question.

A) Rs. 540 | B) Rs. 415 |

C) Rs. 404 | D) Data is not sufficient |

Explanation:

Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

$\left(\frac{P\times T\times (R+2)}{100}\right)-\left(\frac{P\times T\times R}{100}\right)=108\Rightarrow 2PT=10800......\left(1\right)$

And,

$\left(\frac{P\times R\times (T+2)}{100}\right)-\left(\frac{P\times R\times T}{100}\right)=180\Rightarrow 2PR=18000......\left(2\right)$

Clearly, from (1) and (2), we cannot find the value of p

So, the data is not sufficient.