Arithmetic aptitude :: Simple Interest :: General Questions

• Simple Interest - General Questions
• Simple Interest - Data Sufficiency 1
• Simple Interest - Data Sufficiency 2
• Simple Interest - Important Formulas and Facts

Principal :

The amount of money one borrows from a lender for a certain period of time is known as principal amount or the sum.

Interest :

Extra charge paid for the borrowed money is called Interest

Simple Interest (S.I.) :

Simple interest is a method to calculate the interest amount on a loan

Let Pricipal $=P$, Rate $=R\%$ per annum (p.a.) and Time $=N$ years. Then

(i) Simple interest (SI) $=\left(\frac{P\times N\times R}{100}\right)$

(ii) $P$ $=\left(\frac{100\times SI}{R\times N}\right)$ ; $R=\left(\frac{100\times SI}{P\times N}\right)$ and $N=\left(\frac{100\times SI}{P\times R}\right)$.

Rs.1000 is invested at 5% per annum simple interest. If the interest is added to the principal after every 10 years, the amount will becomes Rs2000 after

Answer:

Explanation:

After $10$ years, $SI$ $= \frac{(1000\times 5\times 10)}{100}$ = Rs.500

Principal for $11^{th}$ Year $=1000+500$ = 1500

$SI$ $=2000-1500$ = Rs.500

$T$ $=\frac{(100\times SI)}{P\times R}$

$=\frac{ 500\times 100}{1500\times 5}$

$=6\frac{2}{3}$ Yrs.

Total time $=10+6\frac{2}{3}$ $=16\frac{2}{3}$ Yrs

With rate of simple interest, the ratio of principal and amount for a certain period of time is $4:5$. After 3 years, with the same rate of interest, the ratio of the principal and amount becomes $5:7$. The rate of interest is

Answer:

Explanation:

Case I

$SI$ $=5x-4x$ $=x$

$x$ $=\frac{(4x\times R\times N)}{100}$

$N$ $=\frac{25}{R}$ Years

Case II

$N$ $=\frac{25}{R}+ 3$ $=\frac{(25+3R)}{R}$

$SI$ $=7y-5y$ $=2y$

$2y$ $=\frac{5y\times R\times (25+3R)}{100R}$

R = 5%

Raman invests a sum of Rs. 900 at 8%  simple interest and Gopi invests a sum of Rs. 1200 at $3\frac{1}{2}\%$ simple interest. After how much time will they both have equal amounts in their accounts ?

Answer:

Explanation:

Difference in interest for $1$ year $=\left(\frac{900\times 8\times 1}{100}\right)$ $-\left(\frac{1200\times 3\frac{1}{2}\times 1}{100}\right)$

= Rs. $30$

To cover a difference of $1200-900$ = Rs. $300$

Time investment $=\frac{300}{30}$ = 10 years

A certain sum of money amounts to Rs. 770 in one year and amounts to Rs. 847 in two years at certain rate of interest. Find the sum.

Answer:

Explanation:

Interest for $1$ year $=847-770$ $=77$

Rate of interest $=\frac{77}{770}\times 100$ $=10\%$

Sum $=\frac{(100\times SI)}{N\times R}$

$=\frac{(100\times 77)}{1\times 10}$

= Rs. 770

What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum ?

Answer:

Explanation:

$SI$ $=\frac{P\times N\times R}{100}$ $=\frac{(132\times 5\times 2)}{100}$ $=13.2$

Present worth $=132-13.2$ = Rs. 118.80

• Simple Interest - General Questions
• Simple Interest - Data Sufficiency 1
• Simple Interest - Data Sufficiency 2