Question Detail

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

1. 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}

2. 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}

3. 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

4. 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}

5. A sum of Rs 12,500 amounts to Rs. 15,500 in the 4 years at the rate of simple interest. Find the rate percent

Answer: Option A

Explanation:

\begin{aligned}
\text{S.I.} = \frac{P*R*T}{100} \\
=> R = \frac{S.I. * 100}{P*T}
\end{aligned}

So, S.I = 15500 - 12500 = 3000.

\begin{aligned}
=> R = \frac{3000 * 100}{12500*4} = 6\%
\end{aligned}