Question 1. Find the simple interest and amount on:

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

ii)    $6360$ for $6$ years $3$ months at $8\%$ per annum

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

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

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

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

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

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

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

Amount $= P+S.I = 53625$ Rs.

ii)    $6360$ for $6$ years $3$ months at $8\%$ per annum

$P = 6360$ Rs.

$R = 8\%$

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

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

Amount $= 6360+3180=9540$ Rs.

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

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

$T= 11$ Months $=$ $\frac{11 }{ 12}$ years

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

Amount $= P+SI = Rs. 19200+1716 = 20916$ Rs.

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

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

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

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

Amount $= 59180$  Rs.

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

$P= 8600$ Rs.     $R=8\%$

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

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

Amount $= P+S.I.= 8600+275.2 = 8875.2$ Rs.

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

$P = 10500$ Rs.,      $R=7\%$ ,      $T=5$ years

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

Amount $= P+S.I. = 10500+3675 =14175$ Rs.

The value of watch $= Amount - 13000 = 14175 - 13000 = 1175$ Rs.

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

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

We are requires to compute time in years.

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

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

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

Amount $= 35664$  Rs.,     $P=25600$  Rs.

$S.I = Amount-P = (35664-25600) =10064$  Rs

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

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

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

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

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

We have to compute rate of interest $R$

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

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

$S.I.= Amount - P = (2080- 1625) =455$ Rs.

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

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

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

$P=6720$   Rs.,    $S.I.= 1911$ Rs.

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

We have to compute $R$ .

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

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

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

Principle $P=x$      Amount $= 2 x$       S.I. $=x$

Time $= 12$ years

We have to compute $R$

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

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

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

Let Principle $P = x$ Rs.

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

Numerical $R = T$

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

We have to compute $R$ .

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

$T = R$

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

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

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

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

$S.I.= 406$ Rs.

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

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

We are required to compute $P$

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

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

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

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

We have to compute $P$

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

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

$S.I.= Amount - P = (1748-P)$

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

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

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

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

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

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

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

We have to compute rate of interest $R$.

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

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

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

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

Consider

$P= 8560$  Rs.,      $R = 12 \%$ ,      $T = 5$ years

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

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

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

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

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

First compute $R$ for

$P=1250$  Rs.,    $Amount = 1550$  Rs.,    $T = 3$ Years

$S.I.= Amount - P = 1550 - 1250 = 300$ Rs.

Now for $S.I. = 300$ Rs.

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

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

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

$Amount = P+S.I. = 3200+1024 = 4224$  Rs.

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

Given:

(1) $Amount =3224$  Rs.,      $T=2$ years

(2) $Amount = 4160$  Rs.,    $T=5$ years

$P$ and $R$ are the same in both the above area from the data given in (i)

$S.I.= (3224-P)$

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

From data given in (ii)

$S.I.= 4160-P$

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

Divide equation (i) by equation (ii)

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

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

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

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

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

From data set (i)

$S.I.= (3224-2600) = 624$  Rs.

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

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

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

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

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

Total interest received from both $B$ & $C$ $=1120$  Rs.

Total interest received from $C$ $= 1120-700=420$  Rs.

Now we have to compute sum rent to $P$

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

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

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

The amount lent by $A$ to $C$ $= 1500$  Rs.

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

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

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

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

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

$S.I._1 = S.I._2+96$

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

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

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

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

The sum $1600$  Rs.

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

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

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

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

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

Given $S.I._2 - S.I._1 = 412.5$  Rs.

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

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

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

The equal sums are $3750$  Rs. each

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

Let $x$  Rs. be one part of the sum, then $(6000-x)$ be other part of sum.

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

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

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

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

$S.I._1=S.I._2$

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

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

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

$\Rightarrow 24x=6000 \times 15$

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

The Two Parts are: $3750$ Rs. and $2250$ Rs.

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

Let $x$  Rs. be the first part of the sum.

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

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

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

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

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

Given,  $S.I._1+S.I._2=2375$

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

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

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

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

$\Rightarrow x=6600$

The one part of sum is $6600$ Rs. and second part $6900$ Rs.

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

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

We have to compute $T = S.I.=4P-P =3P$

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

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

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

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

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

$S.I._1$ be the simple interest earned

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

Le $S.I._2$ be the

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

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

Given,  $S.I._1 = S.I._2$

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

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

The sum lent is $= 4704$ Rs.

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

Let $R$ be the rate interest

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

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

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

Given $S.I._2-S.I._2=3640$

i.e. $275 \times 7R-210 \times 7R=3640$

$\Rightarrow (275-210) \times 7R=3640$

$\Rightarrow 65 \times 7R = 3640$

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

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

Let $P$ be the sum lent

$S.I._1$ is interest for sum $=P$,     Rate interest $R$ and $T = 2$ years

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

Let $S.I._2$ is the interest for sum $P$ for interest rate $(R+3)$

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

Given,

$S.I._2 - S.I._1=720$

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

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

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

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

Sum lent is $12000$  Rs.

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

Let sum lent is $P$,     $R=6 \%$ ,     $T$ years

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

Given, $S.I._1=960$ Rs.

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

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

For sum $= P$,      $R=6 \%$ ,    $T=5$ years,

$S.I._2$ is the S.I.

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

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

From (i) , $P \times T=16000$

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

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

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

The sum $=4000$  Rs.