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Question Detail
A lent Rs. 5000 to B for 2 years and Rs 3000 to C for 4 years on simple interest at the same rate of interest and received Rs 2200 in all from both of them as interest. The rate of interest per annum is
- 9%
- 10%
- 11%
- 12%
Answer: Option B
Explanation:
Let R% be the rate of simple interest then,
from question we can conclude that
\begin{aligned}
(\frac{5000*R*2}{100}) + (\frac{3000*R*4}{100}) = 2200 \\
<=> 100R + 120R = 2200 \\
<=> R = 10\%
\end{aligned}
1. In how many years Rs 150 will produce the same interest at 8% as Rs. 800 produce in 3 years at 9/2%
- 8
- 9
- 10
- 11
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
2. 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Always remember that the day on which money is deposited is not counted while the day on which money is withdrawn is counted.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}
3. 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 ?
- 818
- 828
- 838
- 848
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}
4. 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.
- 1:2
- 2:1
- 2:2
- 2:3
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}
5. Find the rate at Simple interest, at which a sum becomes four times of itself in 15 years.
- 10%
- 20%
- 30%
- 40%
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%
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