NCERT Solutions Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

NCERT Solutions Class 8 Mathematics 
Chapter – 8 (Comparing Quantities) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 8 Comparing Quantities Exercise 8.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter 8: Comparing Quantities

Exercise – 8.3 

1. Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12\frac{1}{2} % per annum compounded annually

(b) ₹ 18,000 for 2\frac{1}{2} years at 10% per annum compounded annually

(c) ₹ 62,500 for 1\frac{1}{2} years at 8% per annum compounded half yearly
(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)
(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly

Solution –
(a) ₹ 10,800 for 3 years at 12\frac{1}{2} % per annum compounded annually
Principal (P) = ₹ 10,800
Rate (R) = 12\frac{1}{2}% = \frac{25}{2} % (annual)
Number of years (n) = 3
Amount (A) = P\left ( 1+\frac{R}{100} \right )^n

= 10800 \left ( 1+\frac{25}{2\times 100} \right )^3

= 10800 \left ( 1+\frac{25}{200} \right )^3

= 10800 \left ( 1+\frac{25}{200} \right )^3

= 10800 \left (\frac{225}{200} \right )^3

= 10800 \left ( \frac{9}{8} \right )^3

= 10800 \frac{9}{8}\times \frac{9}{8}\times \frac{9}{8}  

= 15377.34375
= ₹ 15377.34 (approximately)
C.I. = A – P = ₹ (15377.34 – 10800)
= ₹ 4,577.34

(b) ₹ 18000 for 2\frac{1}{2}  years at 10% per annum compounded annually.
Principal (P) = ₹ 18,000
Rate (R) = 10% annual
Number of years (n) = 2\frac{1}{2}
Since ‘n’ is 2\frac{1}{2} years, amount can be calculated for 2 years and having amount as principal Simple Interest (S.I.) can be calculated for \frac{1}{2} years because C.I. is only annually

First, the amount for 2 years has to be calculated
Amount, A = P\left ( 1+\frac{R}{100} \right )^n
= 18000 \left ( 1+\frac{10}{100} \right )^2

= 18000 \left ( 1+\frac{1}{10} \right )^2

= 18000 \frac{11}{10}\times \frac{11}{10}

= ₹ 21780

By taking ₹ 21780 as principal, the S.I. for the next \frac{1}{2} year will be calculated
S.I. =  \frac{1}{2} x 21780 x \frac{10}{100} = ₹ 1089
Hence, the interest for the first 2 years = ₹ (21780 – 18000) = ₹ 3780
And, interest for the next ½ year = ₹ 1089
Total C.I. = ₹ 3780 + ₹ 1089 = ₹ 4,869
Therefore,
Amount, A = P + C.I.
= ₹ 18000 + ₹ 4869
= ₹ 22,869

(c) ₹ 62500 for 1\frac{1}{2} years at 8% per annum compounded half yearly.

Solution – Principal (P) = ₹ 62,500
Rate = 8% per annum or 4% per half-year
Number of years = 1\frac{1}{2}
There will be 3 half-years in 1\frac{1}{2} years

Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 62500\left ( 1+\frac{4}{100} \right )^3

= 62500\left ( 1+\frac{1}{25} \right )^3

= 62500\left (\frac{26}{25} \right )^3  = ₹ 70304
C.I. = A – P
= ₹ 70304 – ₹ 62500
= ₹ 7,804

(d) ₹ 8000 for 1 year at 9% per annum compound half yearly. (You can use the year-by-year calculation using S.I. formula to verify)

Solution – Principal (P) = ₹ 8000
Rate of interest = 9% per annum or \frac{9}{2}% per half-year
Number of years = 1 year
There will be 2 half-years in 1 year

Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 8000\left ( 1+\frac{9}{2\times 100} \right )^2

= 8000\left (\frac{209}{200} \right )^2

= 8736.20
C.I. = A – P
= ₹ 8736.20 – ₹ 8000
= ₹ 736.20

(e) ₹ 10000 for 1 year at 8% per annum compounded half yearly.

Solution – Principal (P) = ₹ 10,000
Rate = 8% per annum or 4% per half-year
Number of years = 1 year
There are 2 half-years in 1 year
Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 10000\left ( 1+\frac{4}{100} \right )^2

= 10000\left ( 1+\frac{1}{25} \right )^3

= 10000\left (\frac{26}{25} \right )^2
= ₹ 10816

C.I. = A – P
= ₹ 10816 – ₹ 10000
= ₹ 816

2. Kamala borrowed ₹ 26400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find A for 2 years with interest compounded yearly and then find S.I. on the 2nd year amount for \mathbf{\frac{4}{12}} years.)

Solution – Principal (P) = ₹ 26,400
Rate (R) = 15% per annum
Number of years (n) = 2\frac{4}{12} 

First, the amount for 2 years has to be calculated
Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 26400\left ( 1+\frac{15}{100} \right )^2

= 26400\left ( 1+\frac{3}{20} \right )^2

= 26400 \left (\frac{23}{20} \right )^2
= ₹ 34914
By taking ₹ 34,914 as principal, the S.I. for the next \frac{4}{12} = \frac{1}{3} years will be calculated
S.I. = 34914 × \frac{1}{3} x  \frac{15}{100} = ₹ 1745.70
Interest for the first two years = ₹ (34914 – 26400) = ₹ 8,514
And interest for the next 1/3 year = ₹ 1,745.70
Total C.I. = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70
Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.70

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 – Interest paid by Fabina = \frac{(P\times R\times T)}{100}

= \frac{12500\times 12\times 3}{100} = 4500

Amount paid by Radha at the end of 3 years = A = P\left ( 1+\frac{R}{100} \right )^n
A = 12500\left ( 1+\frac{10}{100} \right )^3

= 12500\left (\frac{11}{10} \right )^3
= ₹ 16637.50
C.I. = A – P
= ₹ 16637.50 – ₹ 12500
= ₹ 4,137.50
The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50
Thus, Fabina pays more interest = ₹ 4500 − ₹ 4137.50 = ₹ 362.50
Hence, Fabina will have to pay ₹ 362.50 more.

4. I borrowed ₹ 12000 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 – P = ₹ 12000
R = 6% per annum
T = 2 years
S.I. = \frac{(P\times R\times T)}{100}

= \frac{12000 \times 6\times 2}{100} = ₹ 1440
To find the compound interest, the amount (A) has to be calculated

Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 12000\left ( 1+\frac{6}{100} \right )^2

= 12000\left ( 1+\frac{3}{50} \right )^2

= 12000\left ( \frac{53}{50} \right )^2 = ₹ 13483.20

∴ C.I. = A − P
= ₹ 13483.20 − ₹ 12000
= ₹ 1,483.20
C.I. − S.I.
= ₹ 1,483.20 − ₹ 1,440
= ₹ 43.20
Therefore, the extra amount to be paid is ₹ 43.20.

5. Vasudevan invested ₹ 60000 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) after 6 months?
P = ₹ 60,000
Rate = 12% per annum = 6% per half-year
n = 6 months = 1 half-year
Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 60000\left ( 1+\frac{6}{100} \right )^1

= 60000\left ( 1+\frac{3}{50} \right )

= 60000 \frac{53}{50}

= ₹ 63600

(ii) There are 2 half-years in 1 year
So, n = 2
Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 60000\left ( 1+\frac{6}{100} \right )^2

= 60000\left ( 1+\frac{3}{50} \right )^2

= 60000 \left ( \frac{53}{50} \right )^2
= ₹ 67416

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 \mathbf{1\frac{1}{2}} years if the interest is
(i) Compounded annually
(ii) Compounded half yearly

Solution –
(i) Compounded annually
P = ₹ 80,000
R = 10% per annum
n = 1\frac{1}{2} years
Since ‘n’ is 1\frac{1}{2} years, the amount can be calculated for 1 year, and having that amount as principal, S.I. can be calculated for the remaining 1/2 year because C.I. is always calculated annually.

First, the amount for 1 year has to be calculated
Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 80000\left ( 1+\frac{10}{100} \right )^1

= 80000 x \frac{11}{10} = ₹ 88000

By taking ₹ 88,000 as principal, the S.I. for the next ½ year will be calculated as

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

= \frac{88000\times 10\times 1}{2\times 100}

= ₹ 4400

Interest for the first year = ₹ 88000 – ₹ 80000 = ₹ 8000
And interest for the next ½ year = ₹ 4,400
Total C.I. = ₹ 8,000 + ₹ 4,400 = ₹ 12,400
A = P + C.I.= ₹ (80000 + 12400)
= ₹ 92,400

(ii) The interest is compounded half yearly
Rate = 10% per annum = 5% per half-year
There will be three half-years in 1\frac{1}{2} years
Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 80000\left ( 1+\frac{5}{100} \right )^3

= 80000\left ( 1+\frac{1}{20} \right )^3

= 80000 \left ( \frac{21}{20} \right )^3

= ₹ 92610
Thus, the difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210

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 3rd year

Solution –

(i) The amount credited against her name at the end of the second year
P = ₹ 8,000
R = 5% per annum
n = 2 years
Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 8000 \left ( 1+\frac{5}{100} \right )^2

= 8000 \left (\frac{21}{20} \right )^2

= ₹ 8820

(ii) The interest for the 3rd year
P = ₹ 8,000
R = 5% per annum
T = 1
S.I. = \frac{(P\times R\times T)}{100}

= \frac{8820 \times 5\times 1}{100} = ₹ 441

8. Find the amount and the compound interest on ₹ 10,000 for \mathbf{1\frac{1}{2}} 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 – P = ₹ 10,000
Rate = 10% per annum = 5% per half-year
n = 1\frac{1}{2}  years
There will 3 half-years in 1\frac{1}{2} years

Amount, A = P\left ( 1+\frac{R}{100} \right )^n

= 10000 \left ( 1+\frac{5}{100} \right )^3

= 10000\left ( 1+\frac{1}{20} \right )^3

= 10000 \left ( \frac{21}{20} \right )^3

= ₹ 11576.25

Interest earned at 10% p.a. compounded half-yearly = A – P
= ₹ 11576.25 – ₹ 10000 = ₹ 1576.25

Now, let’s find the interest when compounded annually at the same rate of interest.
Hence, for 1 year
R = 10%
n = 1

Amount, A = 10000 \left ( 1+\frac{10}{100} \right )^1

= 10000 \left ( \frac{11}{10} \right )
= ₹ 11000
By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as

S.I. = (P x R x T)/100
= (11000 x 10 x ½)/100
= ₹ 550
So, the interest for the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000
Hence, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550
So the difference between two interests = 1576.25 – 1550 = 26.25
Therefore, the interest will be less when compounded annually at the same rate.

9. Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at \mathbf{12\frac{1}{2}} per annum, interest being compounded half-yearly.

Solution – P = ₹ 4,096
R = 12\frac{1}{2} per annum = \frac{25}{2} per annum = \frac{25}{4} per half-year
n = 18 months
There will be 3 half-years in 18 months
Therefore, amount A = P\left ( 1+\frac{R}{100} \right )^n

= 4096 \left ( 1+\frac{25}{4\times 100} \right )^3

= 4096 \left ( 1+\frac{1}{16} \right )^3

= 4096 \left ( \frac{17}{16} \right )^3

= ₹ 4913
Therefore, the required amount is ₹ 4,913.

10. The population of a place increased to 54000 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) find the population in 2001
Let the population in the year 2001 be ‘P’ and the population in 2003 is ‘A’ = 54000
R = 5%,
n = 2
A = P\left ( 1+\frac{R}{100} \right )^n

54000 = P\left ( 1+\frac{5}{100} \right )^2

54000 = P\left ( 1+\frac{1}{20} \right )^2

54000 = P × \left ( \frac{21}{10} \right )^2

P = 54000 × \frac{400}{441}
P = 48979.6
The population in 2001 = 48980 (approx.)

(ii) Population in the year 2005
Now, the population in 2003 is considered as ‘P’ = 540000 and the population in 2005 is ‘A’
R = 5%,
n = 2
A = P\left ( 1+\frac{R}{100} \right )^n

A = 54000\left (1+\frac{5}{100} \right )^2

= 54000 \left ( \frac{21}{20} \right )^2
= 59535
Therefore, the population in the year 2005 would be 59,535.

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 –
The initial count of bacteria is given as 5,06,000
Bacteria at the end of 2 hours = 506000 \left ( 1+\frac{2.5}{100} \right )^2

= 506000 \left ( 1+\frac{1}{40} \right )^2

= 506000 \left ( \frac{41}{40} \right )^2

= 531616.25

Therefore, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).

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 – Principal = Cost price of the scooter = ₹ 42,000
Depreciation = 8% of ₹ 42,000 per year
= \frac{(P\times R\times T)}{100}

= \frac{42000\times 8\times 1}{100}  = ₹ 3360

Thus, the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.

 

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