Question Detail

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%

1. A man took a loan at rate of 12% per annum simple interest. After 3 years he had to pay 5400 interest. The principal amount borrowed by him was.

Rs 14000
Rs 15000
Rs 16000
Rs 17000

Answer: Option B

Explanation:

\begin{aligned}
\text{S.I.} = \frac{P*R*T}{100} \\
=> P = \frac{S.I. * 100}{R*T} \\
=> P = \frac{5400 * 100}{12*3} = Rs 15000
\end{aligned}

2. In how many years Rs 150 will produce the same interest at 8% as Rs. 800 produce in 3 years at 9/2%

Answer: Option B

Explanation:

Clue:
Firstly we need to calculate the SI with prinical 800,Time 3 years and Rate 9/2%, it will be Rs. 108

Then we can get the Time as

Time = (100*108)/(150*8) = 9

3. The simple interest on a certain sum of money at the rate of 5% p.a. for 8 years is Rs. 840. At what rate of intrest the same amount of interest can be received on the same sum after 5 years.

Answer: Option D

Explanation:

Here firstly we need to calculate the principal amount, then we can calculate the new rate.

\begin{aligned}
P = \frac{S.I. * 100}{R*T} \\
P = \frac{840 * 100}{5*8} \\
P = 2100 \\

\text{Required Rate = } \frac{840 * 100}{5*2100} \\
R = 8\%\\

\end{aligned}

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. A sum of money amounts to Rs 9800 after 5 years and Rs 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is

Answer: Option D

Explanation:

We can get SI of 3 years = 12005 - 9800 = 2205

SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]

Principal = 12005 - 3675 = 6125

So Rate = (100*3675)/(6125*5) = 12%