Arithmetic aptitude :: Simple Interest :: Data Sufficiency 1

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

A woman lent a total sum of Rs. 24000 from two moneylenders. For one loan, she paid interest at $7\frac{1}{2}\%$ p.a. and for the other $9\%$ p.a. How much money did she borrow at each rate ?

I. The sum of the interests after $1$ year was Rs. $2025$.

II. The interest on one sume was twice that the other.

Answer:

Explanation:

Suppose she borrowed Rs. $x$ at $7\frac{1}{2}\%$ p.a. and Rs. $(24000-x)$ at $9\%$ p.a.

I gives, total interest = Rs. $2025$.

$\therefore$ $\left(x\times 1\times\frac{15}{2}\times\frac{1}{100}\right)$ $+\left\{(24000-x)\times 1\times\frac{9}{100}\right\}$ $=2025$.

This gives $x$.

II gives interest on Rs. $(24000-x)$ $=2\times$ (interest on Rs. $x$).

$\Rightarrow (24000-x)\times\frac{9}{100}\times 1$ $=\left(2\times x\times\frac{15}{2}\times\frac{1}{100}\right)$

This gives $x$.

Thus I as well as II are suffficient to answer the question.

$\therefore$ Correct answer is (C).

Find the percentage of simple interest per annum did Ram pay to Raj ?

I. Ram borrowed Rs. $8000$ from Raj for $4$ years.

II. Ram returned Rs. $8800$ to Raj at the end of $2$ years and settled the loan.

Answer:

Explanation:

Let the rate be $R\%$ p.a.

I gives, $P$ = Rs. $8000$ and $T$ $=4$ years.

II gives, $SI$ = Rs. $(8800-8000)$ = Rs. $800$.

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

$=\left(\frac{100\times 800}{8000\times 4}\right)\%$ $=2\frac{1}{2}\%$ p.a.

Thus, I and II both are needed to get the answer.

$\therefore$ Correct answer is (E).

Calculate the rate of $SI$ ?

I. The total interest earned was Rs. $4000$.

II. The sum was invested for $4$ years.

Answer:

Explanation:

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

Now, I gives, $SI$ = Rs. $4000$.

II gives, $T$ $=4$ years.

But, we cannot calculate $R$ since $P$ is unknown.

So, given data is insufficient to get $R$

$\therefore$ Correct answer is (D).

How much money did $X$ invest ?

I. An increase in the rate of invest from $4\frac{7}{8}\%$ to $5\frac{1}{8}\%$ per annum increase his yearly income by Rs. $25$.

II. The sum invested gets doubled, when invested at $8\%$ p.a for $12\frac{1}{2}$ years.

Answer:

Explanation:

Suppose $X$ invests Rs. $x$.

I gives : $R_{1}$ $=\frac{39}{8}\%$, $R_{2}$ $=\frac{41}{8}\%$.

Increase in $SI$ = Rs. $25$.

$\Rightarrow \left(\frac{x\times 1\times\frac{41}{8}}{100}\right)$ $-\left(\frac{x\times 1\times\frac{39}{8}}{100}\right)$ $=25$

$\Rightarrow (41x-39x)$ $=(25\times 800)$ $\Rightarrow x$ $=\left(\frac{25\times 800}{2}\right)$ $=10000$.

Thus, I only gives the answer.

II gives, $SI$ = Rs. $x$, $R=8\%$ and $T=\frac{25}{2}$ years.

$P$ $=\frac{100\times SI}{R\times T}$ $=\left(\frac{100\times x}{8\times 25}\times 2\right)$

Thus, $P$ is not obtained.

$\therefore$ I alone is sufficient to get the answer and II is not sufficient to get the answer.

$\therefore$ Correct answer is (A).

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