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

Answer:

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.

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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.

Answer:

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?

Answer:

 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?

Answer:

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.

Answer:

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?

Answer:

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?

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

 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.

Answer:

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.