Question 1. Find the simple interest and amount on:

$\displaystyle \text{i) } \text{Rs. } 4500 \text{ for } 2 \frac{1 }{ 2} \text{ years at } 7 \frac{2 }{ 3} \% \text{ per annum }$

$\displaystyle \text{ii) } 6360 \text{ for } 6 \text{ years at } 3 \text{ months at } 8\% \text{ per annum }$

$\displaystyle \text{iii) } \text{Rs. } 19200 \text{ for } 11 \text{ months at } 9 \frac{3 }{ 4} \% \text{ per annum }$

$\displaystyle \text{iv) } 58400 \text{ for } 75 \text{ days at } 6 \frac{1 }{ 2} \% \text{ per annum }$

$\displaystyle \text{i) } \text{Rs. } 4500 \text{ for } 2 \frac{1 }{ 2} \text{ years at } 7 \frac{2 }{ 3} \% \text{ per annum }$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \text{P } = Rs. 45000, \text{R } = 7 \frac{2 }{ 3} \% = \frac{23 }{ 3} \%$

$\displaystyle \text{T } = 2 \frac{1 }{ 2} \text{ year } = \frac{5 }{ 2}$ year

$\displaystyle \text{S.I. } = \frac{45000 \times \frac{23 }{ 3} \times \frac{5 }{ 2 }}{ 100} = 8625 \text{ Rs. }$

$\displaystyle \text{Amount } = \text{P + S.I } = 53625 \text{ Rs. }$

$\displaystyle \text{ii) } 6360 \text{ for } 6 \text{ years at } 3 \text{ months at } 8\% \text{ per annum }$

$\displaystyle \text{P } = 6360 \text{ Rs. }$

$\displaystyle \text{R } = 8\%$

$\displaystyle \text{T } = 6 \frac{1 }{ 4} = \frac{25 }{ 4}$ year

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100} = \frac{6360 \times 8 \times \frac{25 }{ 4} }{ 100} = 3180 \text{ Rs. }$

Amount $\displaystyle \text{Amount } = 6360 + 3180 = 9540 \text{ Rs. }$

$\displaystyle \text{iii) } \text{Rs. } 19200 \text{ for } 11 \text{ months at } 9 \frac{3 }{ 4} \% \text{ per annum }$

$\displaystyle \text{P } = 19200$ \text{ Rs. } \text{R } = 9 \frac{3 }{ 4} \% = \frac{39 }{ 4} \% \$

$\displaystyle \text{T } = 11 \text{ months } = \frac{11 }{ 12} \text{ years }$

S.I. $\displaystyle = \frac{19200 \times \frac{39 }{ 4} \times \frac{11 }{ 12} }{ 100} = 1716 \text{ Rs. }$

Amount $\displaystyle \text{Amount } = P + SI = Rs. 19200 + 1716 = 20916 \text{ Rs. }$

$\displaystyle \text{iv) } 58400 \text{ for } 75 \text{ days at } 6 \frac{1 }{ 2} \% \text{ per annum }$

$\displaystyle \text{P } = 58400 \text{ Rs. } \text{R } = 6 \frac{1 }{ 2} \% = \frac{13 }{ 2} \%$

$\displaystyle \text{T } = 75 \text{ days } = \frac{75 }{ 365} \text{ years }$

$\displaystyle \text{S.I. } = \frac{58400 \times \frac{13 }{ 2} \times \frac{75 }{ 365} }{ 100} = 780 \text{ Rs. }$

Amount $\displaystyle \text{Amount } = 59180 \text{ Rs. }$

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Question 2. Find the simple interest on $\displaystyle \text{Rs. } 8600$ from 18th October, 2006 to 13th march, 2007 at $\displaystyle 8\% \text{ per annum }$. Also find the amount.

$\displaystyle \text{P } = 8600 \text{Rs. } \text{R } = 8\%$

$\displaystyle \text{T } = \text{18 Oct.to 13 March } = 13 + 30 + 31 + 31 + 28 + 13 = 146 \text{ days } \\ = \frac{146 }{ 365} \text{ years }$

$\displaystyle \text{S.I } = \frac{P \times R \times T }{ 100} = \frac{8600 \times 8 \times \frac{146 }{ 365} }{ 100} = 275.20 \text{ Rs. }$

$\displaystyle \text{Amount } = \text{P + S.I } . = 8600 + 275.2 = 8875.2 \text{ Rs. }$

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Question 3. Ashish lent $\displaystyle \text{Rs. } 10500$ to Mark at $\displaystyle 7\% \text{ per annum }$ simple interest. After $\displaystyle 5 \text{ years }$, Mark discharged the debt by giving a watch and $\displaystyle \text{Rs. } 3000$ in cash. What is the value of the watch?

$\displaystyle \text{P } = 10500 \text{ Rs. } , \text{R } = 7\% , T = 5 \text{ years }$

$\displaystyle \text{S.I }= \frac{P \times R \times T }{ 100} = \frac{10500 \times 7 \times 5 }{ 100} = 3675 \text{ Rs. }$

$\displaystyle \text{Amount } = \text{P + S.I } . = 10500 + 3675 = 14175 \text{ Rs. }$

The value of watch $\displaystyle = Amount - 13000 = 14175 - 13000 = 1175 \text{ Rs. }$

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Question 4. In what time will the simple interest on $\displaystyle \text{Rs. } 7560$ be $\displaystyle \text{Rs. } 1102$ at $\displaystyle 6 \frac{1 }{ 4} \% \text{ per annum ? }$

$\displaystyle \text{P } = 7560 \text{ Rs. } , \text{S.I. } = 1102$.5, $\displaystyle \text{R } = 6 \frac{1 }{ 4} \% = \frac{25 }{ 4} \%$

We are requires to compute time in years.

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \Rightarrow T = \frac{S.I. \times 100 }{ P \times R} = \frac{1102.5 \times 100 }{ 7560 \times \frac{25 }{ 4} } = 2.3 \text{ years or 2 years 4 months }$

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Question 5. In how much time will $\displaystyle \text{Rs. } 25600$ amount to $\displaystyle \text{Rs. } 35665$, when money is worth $\displaystyle 9 \frac{1 }{ 4} \% \text{ per annum }$ simple interest?

$\displaystyle \text{Amount } = 35664 \text{ Rs. } , \text{P } = 25600 \text{ Rs. }$

$\displaystyle \text{S.I } = Amount-P = (35664-25600) = 10064 \text{ Rs. }$

$\displaystyle \text{R } = 9 \frac{1 }{ 4} \% = \frac{37 }{ 4} \%$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \Rightarrow T = \frac{S.I. \times 100 }{ P \times R} = \frac{10064 \times 100 }{25600 \times \frac{37 }{ 4}} = 4.25 \text{ years or 4 years 3 months }$

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Question 6. At what rate per cent per annum will $\displaystyle \text{Rs. } 1625$ amount to $\displaystyle \text{Rs. } 2080$in $\displaystyle 3 \frac{1 }{ 2} \text{ years ?}$

$\displaystyle \text{P } = 1625$ Rs, Amount $\displaystyle \text{Amount } = 2080$ Rs, $\displaystyle \text{T } = 3 \frac{1 }{ 2} \text{ years at } = \frac{7 }{ 2} \text{ years }$

We have to compute rate of interest $\displaystyle R$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \Rightarrow \text{R } = \frac{S.I. \times 100 }{ P \times T}$ … … … … … i)

$\displaystyle \text{S.I. = Amount - P} = (2080- 1625) = 455 \text{ Rs. }$

$\displaystyle \text{ Substituting the value of S.I. }, P \& T$ in Equal i) we get

$\displaystyle \text{R } = \frac{455 \times 100 }{ 1625 \times \frac{7 }{ 2}} = 8\%$

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Question 7. At what rate per cent per annum will the simple interest on $\displaystyle \text{Rs. } 6720$ be $\displaystyle \text{Rs. } 1911$ in $\displaystyle 3 \text{ years at } 3$ months?

$\displaystyle \text{P } = 6720 \text{ Rs. } , \text{S.I. } = 1911 \text{ Rs. }$

$\displaystyle \text{Time } = 3 \text{ year } 3 \text{ months } = 3 \frac{1 }{ 4} \text{ years at } = \frac{13 }{ 4} \text{ years }$

We have to compute $\displaystyle R$ .

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \text{R } = \frac{S.I. \times 100 }{ P \times T} = \frac{1911 \times 100 }{ 6720 \times \frac{13 }{ 4}} = 8.75\% = 8 \frac{3 }{ 4} \%$

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Question 8. At what rate per cent of simple interest will a sum of money double itself in $\displaystyle 12 \text{ years }$?

Principle $\displaystyle \text{P } = x$ Amount $\displaystyle \text{Amount } = 2 x$ S.I. $\displaystyle = x$

$\displaystyle \text{Time } = 12 \text{ years }$

We have to compute $\displaystyle R$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \text{R } = \frac{S.I. \times 100 }{ P \times T} = \frac{x \times 100 }{ x \times 12} = \frac{100 }{ 12} = 8 \text{ years } \text{4 months}$

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Question 9. Simple interest will a sum is $\displaystyle \frac{9 }{ 16}$ of the sum. Find the rate per cent and the time if both are numerically equal.

Let Principle $\displaystyle \text{P } = x \text{ Rs. }$

$\displaystyle \text{S.I. } = \frac{9 }{ 16} x$

Numerical $\displaystyle \text{R } = T$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

We have to compute $\displaystyle R$ .

$\displaystyle \text{R } = \frac{S.I. \times 100 }{ P \times T}$

$\displaystyle \text{T } = R$

$\displaystyle R^2 = \frac{S.I. \times 100 }{ P} = \frac{ \frac{9 }{ 16} x \times 100 }{ x} = \frac{9 }{ 16} \times 100$

$\displaystyle \text{R } = \frac{3 }{ 4} \times 10 = 7.5\%$

$\displaystyle \text{Per annum, T } = 7 \frac{1 }{ 2} \text{ years }$

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Question 10. What sum will yield $\displaystyle \text{Rs. } 406$ as simple interest in $\displaystyle 1 \text{ year } 2 \text{ months at } 6 \frac{1 }{ 4} \% \text{ per annum ?}$

$\displaystyle \text{S.I. } = 406 \text{ Rs. }$

$\displaystyle \text{T } = 1 \text{ year } 2 \text{ months } = 1 \frac{2 }{ 12} = 1 \frac{1 }{ 6} = \frac{7 }{ 6} \text{ years }$

$\displaystyle \text{R } = 6 \frac{1 }{ 4} \% = \frac{25 }{4} \%$

We are required to compute $\displaystyle \text{P }$

$\displaystyle \text{S.I. } = \frac{ P \times R \times T }{ 100}$

$\displaystyle \text{P } = \frac{S.I. \times 100 }{ R \times T} = \frac{406 \times 100 }{ \frac{25 }{ 4} \times \frac{7 }{ 6}} = 5568 \text{ Rs. }$

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Question 11. What sum will amount to $\displaystyle 1748 \text{ Rs. }$ in $\displaystyle 2 \frac{1 }{ 2} \text{ years at } 7 \frac{1 }{ 2} \% \text{ per annum }$ simple interest?

$\displaystyle \text{Amount } \text{Rs. } 1748 , \text{T } = 2 \frac{1 }{ 2} \text{ years } , S = 7 \frac{1 }{ 2} \%$

We have to compute $\displaystyle \text{P }$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \text{P } = \frac{S.I. \times 100}{ R \times T}$

$\displaystyle \text{S.I. = Amount - P} = (1748-P)$

$\displaystyle \Rightarrow P = \frac{ (1748 -P) \times 100 }{ 7 \frac{1 }{ 2} \times 2 \frac{1 }{ 2}}$

$\displaystyle \Rightarrow P \times \frac{15 }{ 2} \times \frac{5 }{ 2} = (1748-P)100$

$\displaystyle \Rightarrow P (100 + \frac{75 }{ 4} ) = 1748 \times 100$

$\displaystyle \Rightarrow P ( \frac{475 }{ 4} ) = 1748 \times 100$

$\displaystyle \Rightarrow P = \frac{1748 \times 4 \times 100 }{ 475} = 1472 \text{ Rs. }$

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Question 12. A sum of money becomes $\displaystyle \frac{8 }{ 5}$ of itself in $\displaystyle 5 \text{ years }$ at certain rate of simple interest. Find the rate of interest.

$\displaystyle \text{Let P } = x \text{ Rs. , Amount} = \text{Rs. } \frac{8 }{ 5} x \text{ Time } \text{(T) } = 5 \text{ years }$

We have to compute rate of interest $\displaystyle R$.

$\displaystyle \text{S.I. } = ( \frac{8 }{ 5} x-x) = \frac{3 }{ 5} x$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \Rightarrow \text{R } = \frac{S.I. \times 100 }{ P \times T} = \frac{ \frac{3 }{ 5} x \times 100 }{ x \times 5} = \frac{300 }{ 25} = 12 \% \text{ p.a. }$

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Question 13. What sum of money lent at $\displaystyle 12 \frac{1 }{ 2} \% \text{ per annum }$ will produce the same interest in $\displaystyle 4 \text{ years }$ as $\displaystyle \text{Rs. } 8560$ produces in $\displaystyle 5 \text{ years at } 12 \% \text{ per annum }$?

Consider

$\displaystyle \text{P } = 8560 \text{ Rs. } , \text{R } = 12 \% , T = 5 \text{ years }$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100} = \frac{8560 \times 12 \times 5 }{ 100} = 5136 \text{ Rs. }$

$\displaystyle \text{Now we have to compute P if T } = 4 \text{ years } , \text{R } = 12 \frac{1 }{ 2} \% \text{ and } S.I. = 5316 \text{ Rs. }$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \Rightarrow P = \frac{S.I. \times 100 }{ R \times T} = \frac{5136 \times 100 }{ 12 \frac{1 }{ 2} \times 4} = 10272 \text{ Rs. }$

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Question 14. If $\displaystyle 1250 \text{ Rs. }$ amount to $\displaystyle 1550 \text{ Rs. }$ in $\displaystyle 3 \text{ years }$ at simple interest, what will $\displaystyle 3100 \text{ Rs. }$ amount to in $\displaystyle 4 \text{ years }$ at the same rate?

First compute $\displaystyle R$ for

$\displaystyle \text{P } = 1250 \text{ Rs. } , \text{Amount} = 1550 \text{ Rs. } , \text{T } = 3 \text{ years }$

$\displaystyle \text{S.I. } = \text{Amount - P} = 1550 - 1250 = 300 \text{ Rs. }$

Now for $\displaystyle \text{S.I. } = 300 \text{ Rs. }$

$\displaystyle \text{R } = \frac{S.I. \times 100 }{ P \times T} = \frac{300 \times 100 }{ 1250 \times 3} = 8 \%$ p.a.

Now for $\displaystyle \text{R } = 8 \% , P = 3200 \text{ Rs. } , \text{T } = 4 \text{ years }$

$\displaystyle \text{S.I .} = \frac{P \times R \times T }{ 100} = \frac{3200 \times 8 \times 4 }{ 100} = 1024 \text{ Rs. }$

$\displaystyle \text{Amount} = \text{P + S.I } . = 3200 + 1024 = 4224 \text{ Rs. }$

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Question 15. A sum of money lent at simple interest amounts to $\displaystyle 3224 \text{ Rs. }$ in $\displaystyle 2 \text{ years } \text{ and } 4160 \text{ Rs. }$ in $\displaystyle 5 \text{ years }$. Find the sum and the rate of interest.

Given:

$\displaystyle \text{(1) Amount } = 3224 \text{ Rs. } , \text{T } = 2 \text{ years }$

$\displaystyle \text{(2) Amount } = 4160 \text{ Rs. } , \text{T } = 5 \text{ years }$

$\displaystyle \text{P } \text{ and } R$ are the same in both the above area from the data given in (i)

$\displaystyle \text{S.I. } = (3224-P)$

$\displaystyle \Rightarrow 3224-P = \frac{P \times R \times 2 }{ 100}$ … … … …. … i)

From data given in (ii)

$\displaystyle \text{S.I. } = 4160-P$

$\displaystyle \Rightarrow 4160-P = \frac{P \times R \times 5 }{ 100}$ … … … …. … i)

Divide equation (i) by equation (ii)

$\displaystyle \Rightarrow \frac{3224-P }{ 4160-P} = \frac{ \frac{P \times R \times 2}{100} } {\frac{P \times R \times 5}{100} } = \frac{2}{5}$

$\displaystyle \Rightarrow (3224-P)5 = (4160-P)2$

$\displaystyle \Rightarrow 3224 \times 5-5P = 4160 \times 2-2P$

$\displaystyle \Rightarrow 3P = 3224 \times 5-4160 \times 2$

$\displaystyle \text{P } = \frac{3224 \times 5-4160 \times 2 }{ 3} = 2600$

From data set (i)

$\displaystyle \text{S.I. } = (3224-2600) = 624 \text{ Rs. }$

$\displaystyle \text{S.I. } = \frac{P \times R \times T }{ 100}$

$\displaystyle \Rightarrow \text{R } = \frac{S.I. \times 100 }{ P \times T} = \frac{624 \times 100 }{ 2600 \times 2} = 12 \%$ p.a.

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Question 16. $\displaystyle A$ lends $\displaystyle 2500 \text{ Rs. }$ to $\displaystyle B$ and certain sum to $\displaystyle C$ at the same time at $\displaystyle 7 \% \text{ per annum }$ simple interest. If after $\displaystyle 4 \text{ years }$, a altogether receives $\displaystyle 1120 \text{ Rs. }$ as interest from $\displaystyle B \text{ and } C$ find the sum lent to $\displaystyle C$.

$\displaystyle \text{For A , P} = 2500 \text{ Rs. } , \text{R } = 7 \%$ p.a. and, $\displaystyle \text{T } = 4 \text{ years }$

$\displaystyle \text{S.I. due from B} = \frac{2500 \times 7 \times 4 }{ 100} = 700 \text{ Rs. }$

Total interest received from both $\displaystyle B$ & $\displaystyle C = 1120 \text{ Rs. }$

Total interest received from $\displaystyle C = 1120-700 = 420 \text{ Rs. }$

Now we have to compute sum rent to $\displaystyle \text{P }$

$\displaystyle \text{P } = \frac{S.I. \times 100 }{ R \times T}$

$\displaystyle \text{S.I. } = 420 \text{ Rs. } , \text{T } = 4 \text{ years }$, R $\displaystyle = 7 \%$ p.a.

$\displaystyle \text{Therefore, P } = \frac{420 \times 100 }{ 7 \times 4} = 1500 \text{ Rs. }$

The amount lent by $\displaystyle A$ to $\displaystyle C = 1500 \text{ Rs. }$

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Question 17. The simple interest on a certain sum for $\displaystyle 3 \text{ years at } 8 \% \text{ per annum }$ is $\displaystyle 96 \text{ Rs. }$ more than the simple interest on the same sum for $\displaystyle 2 \text{ years at } 9 \% \text{ per annum }$. Find the sum.

$\displaystyle \text{S.I. }_i$ for a sum $\displaystyle \text{P }$, for; $\displaystyle \text{T } = 3 \text{ years } , \text{R } = 8 \% \text{ p.a. }$

$\displaystyle \Rightarrow \text{S.I. }_1 = \frac{P \times 8 \times 3 }{ 100}$ …………………(i)

$\displaystyle \text{S.I. }_2$ for a sum $\displaystyle \text{P }$, for, $\displaystyle \text{T } = 2 \text{ years } , \text{R } = 9 \%$ p.a.

$\displaystyle \Rightarrow \text{S.I. }_2 = \frac{P \times 9 \times 2 }{ 100}$ ………………………ii)

$\displaystyle \text{S.I. }_1 = S.I._2 + 96$

$\displaystyle \frac{P \times 24 }{ 100} = \frac{P \times 18}{ 100 \times 96}$

$\displaystyle \Rightarrow P( \frac{24 }{ 100} - \frac{18 }{ 100} ) = 96$

$\displaystyle \Rightarrow P \times \frac{6 }{ 100} = 96$

$\displaystyle \Rightarrow P = \frac{96 \times 100 }{ 6} = 1600 \text{ Rs. }$

$\displaystyle \text{The sum } 1600 \text{ Rs. }$

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Question 18. Two equal sums of money were lent at simple interest at $\displaystyle 11 \%$ p.a. for $\displaystyle 3 \frac{1 }{ 2} \text{ years } \text{ and } 4 \frac{1 }{ 2} \text{ years }$ respectively. If the difference for two periods was $\displaystyle 412.50 \text{ Rs. }$, find each sum.

$\displaystyle \text{S.I. }_1$ for sum $\displaystyle = P , R, = 11 \% , T = 3 \frac{1 }{ 2}$ year

$\displaystyle \Rightarrow S.I._1 = \frac{P \times 11 \times 3 \frac{1 }{ 2} }{ 100}$

$\displaystyle \text{S.I. }_2$ in the S.I.for sum $\displaystyle = P , \text{R } = 11 \% , T = 4 \frac{1 }{ 2} \text{ years }$

$\displaystyle \Rightarrow S.I._2 = \frac{P \times 11 \times 4 \frac{1 }{ 2} }{ 100}$

$\displaystyle \text{Given } \text{S.I. }_2 - S.I._1 = 412.5 \text{ Rs. }$

$\displaystyle \Rightarrow \frac{P \times 11 \times 4 \frac{1 }{ 2} }{ 100} - \frac{P \times 11 \times 3 \frac{1 }{ 2} }{ 100} = 412.5$

$\displaystyle \Rightarrow \frac{P \times 11 }{ 100} = 412.5$

$\displaystyle \Rightarrow P = \frac{412.5 \times 100 }{ 11} = 3750 \text{ Rs. }$

The equal sums are $\displaystyle 3750 \text{ Rs. }$ each

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Question 19. Divide $\displaystyle 6000 \text{ Rs. }$ into two parts so that the simple interest on the first pat for $\displaystyle 9 \text{ months at } 12 \% \text{ per annum }$ is equal to the simple interest on the second part for $\displaystyle 1\frac{1}{2} \text{ years at } 10 \% \text{ per annum }$.

Let $\displaystyle x \text{ Rs. be one part of the sum, then} (6000-x) \text{ be other part of sum.}$

$\displaystyle \text{S.I. }_1 \text{is the S.I. on P } = x \text{ Rs. } \text{ , for T } = 9 \text{ months } = \frac{3}{ 4 } \text{ years , at R } = 12 \%$

$\displaystyle \Rightarrow \text{S.I. }_1 = \frac{x \times 12 \times \frac{3 }{ 4} }{ 100}$

$\displaystyle \text{S.I. }_2$ is $\displaystyle \text{S.I. }$ on $\displaystyle ( 6000-x) \text{ for } \text{R } = 10 \% , T = 1 \frac{1 }{ 2} \text{ years }$

$\displaystyle \Rightarrow \text{S.I. }_2 = \frac{(6000-x) \times 10 \times 1 \frac{1 }{ 2} }{ 100}$

$\displaystyle \text{S.I. }_1 = \text{S.I. }_2$

$\displaystyle \frac{x \times 12 \times \frac{3 }{ 4} }{ 100} = \frac{(6000-x) \times 10 \times 1 \frac{1 }{ 2} }{ 100}$

$\displaystyle \Rightarrow x \times 9 = (6000-x) \times 15$

$\displaystyle \Rightarrow 9x + 15x = 6000 \times 15$

$\displaystyle \Rightarrow 24x = 6000 \times 15$

$\displaystyle \Rightarrow x = \frac{6000 \times 15 }{ 24} = 3750 \text{ Rs. }$

The Two Parts are: $\displaystyle 3750 \text{ Rs. and } 2250 \text{ Rs. }$

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Question 20. Divide a sum of $\displaystyle 13500 \text{ Rs. }$ into two parts such that if one part be lent at $\displaystyle 8\frac{1 }{ 3} \% \text{ per annum }$ for $\displaystyle 2 \text{ years at } 9$ months and the other at $\displaystyle 7\frac{1 }{ 2} \% \text{ per annum }$ for $\displaystyle 1 \text{ years at } 8$ months, the total interest received is $\displaystyle 2375 \text{ Rs. }$

Let $\displaystyle x \text{ Rs. }$ be the first part of the sum.

Then $\displaystyle (13500-x)$ becomes the second part of the sum.

$\displaystyle \text{S.I. }_1 \text{ is the interest of P } = x , \text{R } = 8 \frac{1 }{ 3} \% , T = 2 \text{ years at } 9 \text{ months } = 2 \frac{3 }{ 4} \text{ years }$

$\displaystyle \text{S.I. }_1 = \frac{x \times 8 \frac{1 }{ 3} \times 2 \frac{3 }{ 4} }{ 100} = \frac{ \frac{25 }{ 3} \times \frac{11 }{ 4} }{ 100} x = \frac{275 }{ 1200} x$

$\displaystyle \text{S.I. }_2 \text{ is the interest on P } = 13500-x , \text{R } = 7 \frac{1 }{ 2} \% \text{p.a. T } = 1 \text{ year } 8 \text{ months } = 1 \frac{1 }{ 3} \text{ years }$

$\displaystyle \text{S.I. }_2 = \frac{(13500-x) \times 7 \frac{1 }{ 2 }\times 1 \frac{2 }{ 3} }{ 100} = (13500-x) \frac{75 }{ 600}$

$\displaystyle \text{Given } \text{S.I. }_1 + S.I._2 = 2375$

$\displaystyle \frac{275x }{ 1200} + (13500-x) \frac{75 }{ 600} = 2375$

$\displaystyle \Rightarrow 275x + 150(13500-x) = 2375 \times 1200$

$\displaystyle \Rightarrow 275x-150x = 2375 \times 1200-13500 \times 150$

$\displaystyle \Rightarrow 125x = 285000-2025000 = 825000$

$\displaystyle \Rightarrow x = 6600$

The one part of sum is $\displaystyle 6600 \text{ Rs. }$ and second part $\displaystyle 6900 \text{ Rs. }$

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Question 21. In what time will a sum of money lent at $\displaystyle 8 \frac{1 }{ 3} \%$ simple interest become $\displaystyle 4$ times of itself.

Let $\displaystyle \text{P }$ be the sum lent. $\displaystyle \text{Amount } = 4P , \text{R } = 8 \frac{1 }{ 3} \%$

$\displaystyle \text{We have to compute T } = S.I. = 4P-P = 3P$

$\displaystyle 3P = \frac{P \times 8 \frac{1 }{ 3} \times T }{ 100}$

$\displaystyle \Rightarrow 3 = \frac{ 8 \frac{1 }{ 3} \times T }{ 100}$

$\displaystyle \Rightarrow T = \frac{300 }{ 8 \frac{1 }{ 3}} = \frac{300 \times 3 }{ 25} = \frac{900 }{ 5} = 36 \text{ years }$

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Question 22. A certain sum of money lent out at $\displaystyle 6 \frac{2 }{ 3} \%$ p.a. produced the same simple interest in $\displaystyle 6 \text{ years }$ as $\displaystyle \text{Rs. } 3200$ lent out at $\displaystyle 8 \frac{2 }{ 5} \%$ p.a. for $\displaystyle 7 \text{ years }$. Find the sum.

Let $\displaystyle \text{P }$ be the sum lent. $\displaystyle \text{R } = 6 \frac{2 }{ 3} \% , T = 6 \text{ years }$

$\displaystyle \text{S.I. }_1$ be the simple interest earned

$\displaystyle \text{S.I. }_1 = \frac{P \times 6 \frac{2 }{ 3} \times 6 }{ 100} = \frac{40 }{ 100} P = \frac{4 }{ 10} P$

Le $\displaystyle \text{S.I. }_2$ be the

S.I. for $\displaystyle \text{P } = 3200 \text{ Rs. } , \text{R } = 8 \frac{2 }{ 5} \% , T = 7 \%$ p.a.

$\displaystyle \text{S.I. }_2 = \frac{3200 \times 8 \frac{2 }{ 5} \times 7 }{ 100} = 1881.6$

$\displaystyle \text{Given } \text{S.I. }_1 = S.I._2$

$\displaystyle \frac{4 }{ 10} P = 1881.6$

$\displaystyle \Rightarrow P = \frac{1881.6 \times 10 }{ 4} = 4704 \text{ Rs. }$

The sum lent is $\displaystyle = 4704 \text{ Rs. }$

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Question 23. Naveen and Praveen borrowed $\displaystyle 42000 \text{ Rs. and } 55000 \text{ Rs. }$ respectively for $\displaystyle 3 \frac{1 }{ 2} \text{ years }$ at the same rate of interest. If Praveen has to pay $\displaystyle 3640 \text{ Rs. }$ more than Naveen, find the rate of interest.

Let $\displaystyle R$ be the rate interest

$\displaystyle \text{S.I. }_1$  $\displaystyle \text{ be the S.I.paid by Naveen,for P } = 42000 \text{ Rs. } , \text{T } = 3 \frac{1 }{ 3} \text{ years at } = \frac{7 }{ 5} \text{ years }$,

$\displaystyle \text{S.I. }_1 = \frac{42000 \times R \times \frac{7 }{ 2} }{ 100} = 210 \times 7R$

$\displaystyle \text{S.I. }_2$ be the S.I.paid by Praveen for $\displaystyle \text{P } = 55000 \text{ Rs. } , \text{T } = 3 \frac{1 }{ 2} \text{ years at } = \frac{7 }{ 5} \text{ years }$,

$\displaystyle \text{Given } \text{S.I. }_2-S.I._2 = 3640$

$\displaystyle \text{ i.e. } 275 \times 7R-210 \times 7 \text{R } = 3640$

$\displaystyle \Rightarrow (275-210) \times 7 \text{R } = 3640$

$\displaystyle \Rightarrow 65 \times 7 \text{R } = 3640$

$\displaystyle \Rightarrow \text{R } = \frac{3640}{65 \times 7} = 8 \%$ p.a.

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Question 24. A sum of money was put at simple interest at a certain rate for $\displaystyle 2 \text{ years }$. If this sum had been put at $\displaystyle 3 \%$ higher rate, it would have earned $\displaystyle 720 \text{ Rs. }$ more as interest. Find the sum.

Let $\displaystyle \text{P }$ be the sum lent

$\displaystyle \text{S.I. }_1$ is interest for sum $\displaystyle = P$, Rate interest $\displaystyle R \text{ and } T = 2 \text{ years }$

$\displaystyle \text{S.I. }_1 = \frac{P \times R \times 2 }{ 100} = \frac{P \times R }{ 50}$

Let $\displaystyle \text{S.I. }_2$ is the interest for sum $\displaystyle \text{P }$ for interest rate $\displaystyle (R + 3)$

$\displaystyle \text{S.I. }_2 = \frac{P \times (R + 3) \times 2 }{ 100}$

Given,

$\displaystyle \text{S.I. }_2 - S.I._1 = 720$

$\displaystyle \frac{P \times (R + 3) \times 2 }{ 100} - \frac{P \times R \times 2 }{ 100} = 720$

$\displaystyle \frac{P \times R \times 2 }{ 100} + \frac{P \times 3 \times 2 }{ 100} - \frac{P \times R \times 2 }{ 100} = 720$

$\displaystyle \Rightarrow \frac{6P }{ 100} = 720$

$\displaystyle \Rightarrow P = \frac{720 \times 100 }{ 6} = 12000$

Sum lent is $\displaystyle 12000 \text{ Rs. }$

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Question 25. A sum of money invested at $\displaystyle 6 \%$ p.a. simple interest for a certain period of time yield $\displaystyle 960 \text{ Rs. }$ as interest. If this sum had been invested for $\displaystyle 5 \text{ years }$ more, it would have yielded $\displaystyle 2160$ as interest. Find the sum.

Let sum lent is $\displaystyle \text{P } , \text{R } = 6 \% , T \text{ years }$

$\displaystyle \text{S.I. }_1 = \frac{P \times 6 \times T }{ 100}$

$\displaystyle \text{Given } \text{S.I. }_1 = 960 \text{ Rs. }$

$\displaystyle \frac{P \times 6 \times T}{ 100} = 960$

$\displaystyle \Rightarrow P \times T = 16000$ … … … … … (i)

For sum $\displaystyle = P , \text{R } = 6 \% , T = 5 \text{ years }$,

$\displaystyle S.I_1 \text{ is the S.I.}$

$\displaystyle \text{S.I. }_2 = \frac{P \times 6 \times (T + 5) }{ 100} = 2160$

$\displaystyle \Rightarrow \frac{P \times 6 \times T }{ 100} + \frac{P \times 6 \times 5 }{ 100} = 2160$

From (i) , $\displaystyle \text{P } \times T = 16000$

$\displaystyle \Rightarrow \frac{16000 \times 6 }{ 100} + \frac{P \times 30 }{ 100} = 2160$

$\displaystyle \Rightarrow \frac{P \times 30 }{ 100} = 2160-960 = 1200$

$\displaystyle \Rightarrow P = \frac{1200 \times 100 }{ 30} = 4000$

$\displaystyle \text{The sum } = 4000 \text{ Rs. }$