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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.
- 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}
1. 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
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 ?
- 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}
3. What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum
- 110
- 120
- 130
- 140
Answer: Option B
Explanation:
Let the present worth be Rs.x
Then,S.I.= Rs.(132 - x)
=› (x*5*2/100) = 132 - x
=› 10x = 13200 - 100x
=› 110x = 13200
x= 120
4. 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%
5. If A lends Rs. 3500 to B at 10% p.a. and B lends the same sum to C at 11.5% p.a., then the gain of B (in Rs.) in a period of 3 years is
- Rs. 154.50
- Rs. 155.50
- Rs. 156.50
- Rs. 157.50
Answer: Option D
Explanation:
We need to calculate the profit of B.
It will be,
SI on the rate B lends - SI on the rate B gets
\begin{aligned}
\text{Gain of B}\\ &= \frac{3500\times11.5\times3}{100} - \frac{3500\times10\times3}{100}\\
= 157.50
\end{aligned}
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