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Question Detail
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
1. 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}
2. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs. 800 become in 3 years.
- Rs 1052
- Rs 1152
- Rs 1252
- Rs 1352
Answer: Option A
Explanation:
S.I. = 956 - 800 = Rs 156
\begin{aligned}
R = \frac{156*100}{800*3} \\
R = 6\frac{1}{2}\% \\
\text{ New Rate = }6\frac{1}{2}+4 \\
= \frac{21}{2} \% \\
\text{ New S.I. = }800\times\frac{21}{2}\times{3}{100} \\
= 252
\end{aligned}
Now amount will be 800 + 252 = 1052
3. 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}
4. 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}
5. Reema took a loan of Rs 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest.
- 5%
- 6%
- 7%
- 8%
Answer: Option B
Explanation:
Let rate = R% then Time = R years.
\begin{aligned}
=> \frac{1200*R*R}{100}=432 \\
=> R^2 = 36 \\
=> R = 6\%
\end{aligned}
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