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Class 8 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 14: Compound interest

Compound interest Exercise Ex. 14.1

Solution 1

 

(1)

P = Rs. 2000, R = 5 p.c.p.a. N = 2 years

A = Rs. 2205

 

CI = A - P

= 2205 - 2000

= Rs. 205

 

(2)

P = Rs. 5000, R = 8 p.c.p.a., N = 3 years

∴ A= 6298.56

 

CI = A - P

= 6298.56 - 5000

= Rs. 1298.56

 

(3)

P = Rs. 4000, R = 7.5 p.c.p.a., N = 2 years

∴ A = Rs. 4622.5

 

CI = A - P

= 4622.5 - 4000

= Rs. 622.5

Solution 2

Sameerrao has taken a loan of Rs. 12500 i.e. P = Rs. 12,500

Rate of 12 p.c.p.a i.e. R = 12 p.c.p.a.

For 3 years i.e. N = 3 years

∴ A = Rs. 17561.6

Sameerrao should pay Rs. 17561.6 to clear his loan amount.

Solution 3

Shalaka has taken a loan of Rs.8000 i.e. P = Rs. 8000

at a rate of 10 p.c.p.a. i.e. R = 10.5 p.c.p.a

After two years i.e. N = 2 years

∴ A = Rs. 9768.2

CI = A - P

= 9768.2 - 8000

= Rs. 1768.2

Shalaka will have to pay Rs. 1768.2 as compound interest.

Compound interest Exercise Ex. 14.2

Solution 1

P = initial number of workers = 320,

R = rate of interest = 25 p.c.p.a.,

N = number of years = 2 years

  

  

 

  

 

  

 

  

 

  

∴ A= 20 × 25

∴ A = 500

The number of workers after 2 years is 500.

Solution 2

P = Number of sheep initially = 200

N = 3 years

R = 8% every year

  

  

  

 

  

 

  

  

  

∴ A = 251.9424

As number of sheep must be an integer, it should be 252.

Solution 3

P = number of trees = 40,000

N = 3 years

R = 5% per year

  

 

  

 

  

 

  

 

  

 

  

 

  

 

∴ A = 46305

The expected number of trees after 3 years at the rate of 25% every year is 46,305.

Solution 4

P = cost of machine = Rs. 2,50,000

R = 10% per year

N = 2 years

The rate of depreciation means the rate is decreasing of a machine.

  

  

 

  

 

  

 

  

 

  

 

∴ A = 202500

 

Depreciation = P - A

= 2,50,000 - 2,02, 500

= 47,500

The cost of machine after two years will be Rs. 47,500.

Solution 5

P = ?

N = 2 years

A = Rs. 4036.80

R = 16 p.c.p.a.

  

  

 

  

 

∴ P =   

∴ P = 3000

CI = A - P

= 4036.80 - 3000

= Rs. 1036.80

The compound interest is Rs. 1036.80.

Solution 6

P = Rs. 15,000

R = 12 p.c.p.a.

N = 3 years

  

  

 

  

 

  

 

∴ A =   

 

∴ A = 21073.92

The amount to settle the loan after 3 years is Rs. 21073.92

Solution 7

A = Rs. 13924

N = 2 years

R = 18 p.c.p.a.

  

  

 

  

 

  

 

∴ P = 13924       

∴ P = 10000

The principal is Rs. 10,000.

Solution 8

P = 16,000

N = 2 years

R =?

A = 17640

  

  

 

  

 

  

 

   Taking square root on both sides

 

  

 

  =   

 

∴ R = 5%

Rate of interest of population is 5% every year.

Solution 9

P = Rs. 700

A = Rs. 847

R = 10 p.c.p.a.

N =?

0  

 

  

 

  

 

  

 

  

 

We know that 112 = 121 and 102 = 100

Hence, 700 amounts to 847 in 2 years.

Solution 10

P = Rs. 20,000

R = 8 p.c.p.a.

N = 2 years

Simple interest = P × N ×   

= P× 2 ×   

=   ….(i)

 

CI = A - P

  

  

  

  

  ……(ii)

 

CI - SI

 

=       

 

=   

 

=   

 

= 2 × 8 × 8

= 128

The difference between simple interest and compound interest is Rs. 128.