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 }

Answer:

\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.

Answer:

\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?

Answer:

\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 ? }

Answer:

\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?

Answer:

\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  ?}

Answer:

\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?

Answer:

\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 } ?

Answer:

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.

Answer:

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 ?}

Answer:

\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?

Answer:

\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.

Answer:

\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 } ?

Answer:

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?

Answer:

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.

Answer:

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 .

Answer:

\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.

Answer:

\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.

Answer:

\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 } .

Answer:

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. }

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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. }