NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3

 NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3:–

Question 1.

Calculate the amount and compound interest on

(a) ₹ 10,800 for 3 years at 12–1/2 % per annum compounded annually.

Solution:

 Given:

P = ₹ 10,800, n = 3 years,

R = 12-1/2 % = 25/2 % pa 

A = P ( 1+ R/100)^n

A = 10800(1+25/2×100)³

A = 10800( 9/8)³

A = 10800×9/8×9/8×9/8

A = 492075/32

A = ₹ 15377.34

CI = A – P = ₹ 15,377.35 – ₹ 10,800 = ₹ 4,577.35

Hence amount = ₹ 15,377.34 and CI = ₹ 4,577.34

(b) ₹ 18,000 for 212 years at 10% per annum compounded annually.

Solution :

 Given: P = ₹ 18,000, n = 212 years = 52years

R = 10% p.a.

The amount for 212 years, i.e., 2 years and 6 months can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years has to be calculated

A = 18000( 1+ 10/100)²

A = 18000(1+ 1/10)²

A = 18000( 11/10)²

A = 18000×11/10 ×11/10

A = ₹21780

.: interest after two years = A–P

= 21780– 18000 = ₹ 3780 

By taking ₹21780 as a principal amount. The SI for next 1/2 year will be calculated 

SI for 1/2 year = P×R×n /100 

                          = 21780 × 10 ×1/ 100×2

                          = ₹ 1089 

Total CI = ₹ 3780 + ₹ 1089 = ₹ 4,869

Amount = P + I = ₹ 21,780 + ₹ 1,089 = ₹ 22,869

Hence, the amount = ₹ 22,869

and CI = ₹ 4,869

(c) ₹ 62,500 for 112 years at 8% per annum compounded half yearly.

Solution :

 Given: P = ₹ 62,500, n = 112 years = 32 years per annum compounded half yearly

= 32 × 2 years = 3 half years

R = 8% = 82 % = 4% half yearly

A = P (1 + R /100)^n

A = 62500 ( 1+ 4/100)³

A = 62500 ( 1+ 1/25 )³

A = 62500 ( 26/25 )³

A = 62500 × 26/25 × 26/25 × 26/25

A = ₹ 70304 

CI = A – P = ₹ 70,304 – ₹ 62,500 = ₹ 7,804

Hence, amount = ₹ 70304 and CI = ₹ 7804

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

Solution :

Given: P = ₹ 8,000, n = 1 years R = 9% per annum compounded half yearly

Since, the interest is compounded half yearly n = 1 × 2 = 2 half years

R = 9/2 % per half year

A = P ( 1+R/100)^n

A = 8000( 1+ 9/ 2×100)²

A = 8000( 1+9/200)²

A = 8000( 209/200)²

A = 8000× 209/200 ×209/200

A = ₹ 8736.20

CI = A – P = ₹ 8,736.20 – ₹ 8,000 = ₹ 736.20

Hence, the amount = ₹ 8736.20 and CI = ₹ 736.20

(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.

Solution :

 Given: P = ₹ 10,000, n = 1 year and R = 8% pa compounded half yearly

Since the interest is compounded half yearly n = 1 × 2 = 2 half years

R = 8× 1/2 = 4% per half year

A = P ( 1+R/100)^n

A = 10000(1+ 4/100)²

A = 10000(1+1/25)²

A = 10000( 26/25)²

A = 10000 × 26/25 × 26/25

A = ₹ 10816

CI = A – P = ₹ 10,816 – ₹ 10,000 = ₹ 816

Hence the amount = ₹ 10,816 and Cl = ₹ 816

Question 2.

Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for 412 years).

Solution:

Given:

P = ₹ 26,400

R = 15% p.a. compounded yearly

n = 2 years and 4 months

Amount for 2 years = P ( 1+R/100)^n

= 26400( 1+ 15/100)²

= 26400( 23/20)²

= 26400 × 23/20 × 23/20

= ₹ 34914

Principal for months i.e.,4 /12 years = ₹34914 

.: SI for 4 months = P×R×n/100

= 34914 × 15 × 4/ 100× 12

= ₹ 1745.70

Amount after 2 years and 4 months = ₹ 34,914 + ₹ 1745.70 = ₹ 36,659.70

Hence, the amount to be paid by Kamla = ₹ 36,659.70

Question 3.

Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Solution:

For Fabina: P = ₹ 12,500, R = 12% p.a. and n = 3 years

.: SI = P×R×n/100

        = 12500 ×12×3/400

        = ₹ 4500

For radha : P = ₹12500, R = 10%p.a. and n = 3 years

.: CI = A–P

        = P ( 1+R/100)^n – P

        = 12500( 1+ 10/100)³ – 12500

        = 12500( 1+ 1/10)³ – 12500

        = 12500( 11/10)³ – 12500

        = 12500× 11/10 ×11/10 ×11/10 –12500

        = 12500 × 1331/1000 – 12500 

        = 12500( 1331/1000 – 1)

        = 12500( 1331 – 1000 / 1000)

        = 12500 × 331/1000

        = ₹ 4137.50

Difference between the two interests = ₹ 4500 – ₹ 4137.50 = ₹ 362.50

Hence, Fabina pays more interest by ₹ 362.50.

Question 4.

I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Solution:

Given: P = ₹ 12,000, R = 6% p.a., n = 2 years

.: SI = P×R×n/100

       = 12000 ×6 ×2/100

       = ₹ 1440

CI = P ( 1+ R/100)^n – P

    = 12000( 1+ 6/100)² – 12000

    = 12000( 53/50)² –12000

    = 12000( 53 × 53 / 50× 50 – 1)

    = 12000 ( 2809 /2500 –1)

    = 12000 ( 2809 – 2500 / 2500)

    = 12000 ( 309 / 2500)

    = ₹ 1483.20

Difference between two interests = ₹ 1483.20 – ₹ 1440 = ₹ 43.20

Hence, the extra amount to be paid = ₹ 43.20

Question 5.

Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months?

(ii) after 1 year?

Solution:

(i) Given: P = ₹ 60,000, R = 12% p.a. compounded half yearly

n = 6 months = 6/12 year = 1/2 year 

Simple interest = P×R×n/100

= 60000×12×1/100×2

= ₹ 3600

.: Amount = SI + P

                   = ₹ 3600 + ₹ 60,000

                   = ₹ 63600

Hence, the required amount = ₹ 63600

(ii) given P = 60000, R = 12/2% = 6%

n = 1× 2 = 2 half year

.: Amount = P ( 1+ R/100)^n

                   = 60000(1+6/100)²

                   = 60000(53/50)²

                   = 60000×53/50×53/50

                   = ₹ 67416

Hence, the required amount = ₹ 67416

Question 6.

Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1–1/2 years if the interest is

(i) compounded annually.

(ii) compounded half yearly.

Solution:

(i) Given: P = ₹ 80,000

R = 10% p.a.

n = 1–1/2 years

Since the interest is compounded annually

.: SI for 1 year = P×R×n/100

                          = 80000×10×1/100

                          = ₹ 8000

Principal for the second year 

= P + SI

= 80000 +8000

= 88000

Now interest for 1/2 year

= 88000× 10×1/ 100 ×2

= ₹ 4400

.: Amount = ₹ 88000 + 4400

                    = ₹ 92400

(ii) if the interest is compounded half yearly 

R = 10/2% = 5% half yearly 

n = 1-1/2

   = 3/2 ×2 = 3 half years 

A = P ( 1+ R/100)^n

   = 80000( 1+5/100)³

   = 80000( 21/20)³

   = 80000× 21/20 × 21/20 × 21/20

   = ₹ 92610

Difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210

Question 7.

Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(i) The amount credited against her name at the end of the second year.

(ii) The interest for the third year.

Solution:

(i) Given: P = ₹ 8,000, R = 5% p.a.

and n = 2 years

A = P ( 1+ R/100)^n

   = 8000(1+5/100)²

   = 8000( 21/20)²

   = 8000 × 21/20 × 21/20

   = ₹ 8820

Thus, the amount credited againsts Maria's name at the end of 2nd year = ₹ 8820

(ii) interest for the 3rd year 

= amount after third year – amount after 2 year

= P ( 1+ R/100)^n – ₹ 8820

= 8000( 1+ 5/100)³ – ₹ 8820

= 8000(1+1/20)³ – ₹ 8820

= 8000(21/20)³ – ₹ 8820

= ₹ 9261 – ₹ 8820

= ₹ 441

Hence, interest for the third year = ₹ 441

Question 8.

Find the amount and the compound interest on ₹ 10,000 for 112 years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Solution:

Given: P = ₹ 10,000, n = 112 years

R = 10% per annum

Since the interest is compounded half yearly

.: n = 1–1/2

      = 3/2 × 2 = 3 half year

R = 10/2 % = 5%

A = P ( 1+ R/100)^n

   = 10000( 1+5/100)³

   = 10000( 1+1/20)³

   = 10000( 21/20)³

   = 11576.25

Compound interest = A – P

= 11576.25 – 10000 

= 1576.25

If the interest is compound annually, then 

n = 1–1/2 , R = 10%

.: SI = P × R × n /100

        = 10000 × 10 ×1 /100

        = ₹ 1000

Principal of the second year 

= ₹ 10000 + ₹ 1000

= ₹ 11000

.: interest for the 1/2 year 

= 11000 × 10 ×1 /100 ×2

= ₹ 550

Total interest = ₹ 1,000 + ₹ 550 = ₹ 1,550

Difference between the two interests = ₹ 1,576.25 – ₹ 1,550 = ₹ 26.25

Hence, the interest will be ₹ 26.25 more when compounded half yearly than the interest when compounded annually.

 Question 9.

Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at 1212 per annum, interest being compounded half yearly.

Solution:

Given: P = ₹ 4,096, R = 1212 % pa, n = 18 months

R = 12–1/2% pa

   = 25/2 ×1/2

   = 25/4% half yearly 

n = 18 months = 18/12 year 

= 18/12 ×2 = 3 half year 

A = P ( 1+ R/100)^n

A = 4096 (1 + 25/4×100)³

A = 4096 ( 1+1/16)³

A = 4096 ( 17/16)³

A = 4096 × 17/16×17/16×17/16

A = ₹ 4913

Hence, the required amount = ₹ 4913

 Question 10.

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(i) Find the population in 2001.

(ii) What would be its population in 2005?

Solution:

(i) Given: Population in 2003 = 54,000

Rate = 5% pa

Time = 2003 – 2001 = 2 years

Population in 2003 = Population in 2001 ×(1+R/100)^n

54000 = Population in 2001 × (1+5/100)²

54000 = Population in 2001 × (1+1/20)²

54000 = Population in 2001 × (21/20)²

54000 = Population in 2001 × 441 /400

.: Population in 2001 = 54000 × 400/441

= 21600000/ 441

= 48979.59 = 48980

(ii) Population in 2005 = Population in 2003 × (1+R/100)^n

= 54000( 1+5/100)²

= 54000( 1+1/20)²

= 54000(21/20)²

= 54000×21/20×21/20

= 59535

Question 11.

In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Solution:

Given: Initial count of bacteria = 5,06,000

Rate = 2.5% per hour

n = 2 hours

Number of bacteria at the end of 2 hours = Number of count of bacteria initially ×(1+R/100)^n

= 506000 (1+ 2.5/100)²

= 506000( 1+1/40)²

= 506000(41/40)²

= 506000×41/40×41/40

= 531616.25

Thus, the number of bacteria after two hours = 5,31,616 (approx).

Question 12.

A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution:

Given: Cost price of the scooter = ₹ 42,000

Rate of depreciation = 8% p.a.

Time = 1 year

Final value of the scooter

= present value ×(1– R/100)^n

= 42000(1– 8 /100)

= 42000(23/25)

= 38640

Hence, the value of scooter after 1 year = ₹ 38,640.