Question Detail

What will the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years.

Answer: Option D

Explanation:

Let the principal be P and rate be R

then

\begin{aligned}
\text{ratio = } [\frac{(\frac{P*R*6}{100})}{(\frac{P*R*9}{100})}] \\

= \frac{6PR}{9PR} = 2:3
\end{aligned}

1. Find the simple interest on the Rs. 2000 at 25/4% per annum for the period from 4th Feb 2005 to 18th April 2005

Rs 25
Rs 30
Rs 35
Rs 40

Answer: Option A

Explanation:

One thing which is tricky in this question is to calculate the number of days.
Always remember that the day on which money is deposited is not counted while the day on which money is withdrawn is counted.
So lets calculate the number of days now,
Time = (24+31+18) days = 73/365 years = 1/5 years
P = 2000
R = 25/4%

\begin{aligned}
\text{ S.I. = } = \frac{2000 \times 25 }{4 \times 5 \times 100} = 25
\end{aligned}

2. Find the rate at Simple interest, at which a sum becomes four times of itself in 15 years.

Answer: Option B

Explanation:

Let sum be x and rate be r%
then, (x*r*15)/100 = 3x [important to note here is that simple interest will be 3x not 4x, beause 3x+x = 4x]

=> r = 20%

3. If a sum of money doubles itself in 8 years at simple interest, the ratepercent per annum is

12
12.5
13
13.5

Answer: Option B

Explanation:

Let sum = x then Simple Interest = x

Rate = (100 * x) / (x * 8) = 12.5

4. We have total amount Rs. 2379, now divide this amount in three parts so that their sum become equal after 2, 3 and 4 years respectively. If rate of interest is 5% per annum then first part will be ?

Answer: Option B

Explanation:

Lets assume that three parts are x, y and z.
Simple Interest, R = 5%

From question we can conclude that, x + interest (on x) for 2 years = y + interest (on y) for 3 years = z + interest (on y) for 4 years

\begin{aligned}
\left( x + \frac{x*5*2}{100} \right) = \left( y + \frac{y*5*3}{100} \right) = \left( z + \frac{z*5*4}{100} \right)\\
\left( x + \frac{x}{10} \right) = \left( y + \frac{3y}{20} \right) = \left( z + \frac{z}{5} \right) \\
=> \frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\

\text{lets assume k = }\frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\
\text{then }x = \frac{10k}{11} \\
y = \frac{20k}{23}\\
z = \frac{5k}{6}\\
\text{we know x+y+z = 2379}\\
=> \frac{10k}{11} + \frac{20k}{23} + \frac{5k}{6} = 2379\\
\text{10k*23*6+20k*11*6+5k*11*23=2379*11*23*6}\\
\text{1380k+1320k+1265k=2379*11*23*6}\\
\text{3965k=2379*11*23*6}\\
k = \frac{2379*11*23*6}{3965}\\
\text{by putting value of k we can get x} \\
x = \frac{10k}{11} \\
=>x = \frac{10}{11}*\frac{2379*11*23*6}{3965}\\
=>x = \frac{10*2379*23*6}{3965}\\
= \frac{2*2379*23*6}{793}\\
= 2 * 3 * 23 * 6 = 828
\end{aligned}

5. Sachin borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends money to Rahul at 25/4% p.a. for 2 years. Find the gain of one year by Sachin.

110.50
111.50
112.50
113.50

Answer: Option C

Explanation:

Two things need to give attention in this question, First we need to calculate gain for 1 year only.
Second, where we take money at some interest and lends at other, then we use to subtract each other to get result in this type of question. Lets solve this Simple Interest question now.

\begin{aligned}
\text{Gain in 2 year = } \\
[(5000 \times \frac{25}{4} \times \frac{2}{100})-(\frac{5000 \times 4 \times 2}{100})] \\
= (625 - 400) = 225 \\
\text{ So gain for 1 year = }\\
\frac{225}{2} = 112.50
\end{aligned}