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Question Detail
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}
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. A financier claims to be lending money at simple interest, But he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes.
- 10.25%
- 10%
- 9.25%
- 9%
Answer: Option A
Explanation:
Let the sum is 100.
As financier includes interest every six months., then we will calculate SI for 6 months, then again for six months as below:
SI for first Six Months = (100*10*1)/(100*2) = Rs. 5
Important: now sum will become 100+5 = 105
SI for last Six Months = (105*10*1)/(100*2) = Rs. 5.25
So amount at the end of year will be (100+5+5.25)
= 110.25
Effective rate = 110.25 - 100 = 10.25
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. Sahil took a loan for 6 years at the rate of 5% per annum on Simple Interest, If the total interest paid was Rs. 1230, the principal was
- 4100
- 4200
- 4300
- 4400
Answer: Option A
Explanation:
\begin{aligned}
\text{S.I.} = \frac{P*R*T}{100} \\
=> P = \frac{S.I. * 100}{R*T}
\end{aligned}
By applying above formula we can easily solve this question, as we are already having the simple interest.
\begin{aligned}
=> P = \frac{1230 * 100}{6*5} \\
=> P = 4100
\end{aligned}
5. 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
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