Exercise 8.3
Q 1 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 1
Calculate the amount and compound interest on
(a) Rs 10800 for 3 years at
per annum compounded annually.Sol :
(a) Principal (P) = Rs 10, 800
Rate (R) =
= 
% (annual)Number of years (n) = 3
Amount, A =
C.I. = A − P = Rs (15377.34 − 10800) = Rs 4,577.34
(b) Rs 18000 for
years at 10% per annum compounded annually.Sol :
(b) Principal (P) = Rs 18,000
Rate (R) = 10% annual
Number of years (n) =
The amount for 2 years and 6 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years.
Firstly, the amount for 2 years has to be calculated.
By taking Rs 21780 as principal, the S.I. for the next
year will be calculated.
∴ Interest for the first 2 years = Rs (21780 − 18000) = Rs 3780
And interest for the next
year = Rs 1089∴ Total C.I. = Rs 3780 + Rs 1089 = Rs 4,869
A = P + C.I. = Rs 18000 + Rs 4869 = Rs 22,869
(c) Rs 62500 for
years at 8% per annum compounded half yearly.Sol :
(c) Principal (P) = Rs 62,500
Rate = 8% per annum or 4% per half year
Number of years =
There will be 3 half years in
years.
C.I. = A − P = Rs 70304 − Rs 62500 = Rs 7,804
(d) Rs 8000 for 1 year at 9% per annum compound half yearly.
(You could use the year by year calculation using SI formula to verify)
Sol :
(d) Principal (P) = Rs 8000
Rate of interest = 9% per annum or
% per half yearNumber of years = 1 year
There will be 2 half years in 1 year.
C.I. = A − P = Rs 8736.20 − Rs 8000 = Rs 736.20
(e) Rs 10000 for 1 year at 8% per annum compounded half yearly.
Sol :
(e) Principal (P) = Rs 10,000
Rate = 8% per annum or 4% per half year
Number of years = 1 year
There are 2 half years in 1 year.
C.I. = A − P = Rs 10816 − Rs 10000 = Rs 816
Q 2 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 2
Kamala borrowed Rs 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 is compounded yearly and then find SI on the 2nd year amount for
years.)Sol :
Principal (P) = Rs 26,400
Rate (R) = 15% per annum
Number of years (n) =
The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years.
Firstly, the amount for 2 years has to be calculated.
By taking Rs 34,914 as principal, the S.I. for the next
will be calculated.
Interest for the first two years = Rs (34914 − 26400) = Rs 8,514
And interest for the next
year = Rs 1,745.70Total C.I. = Rs (8514 + Rs 1745.70) = Rs 10,259.70
Amount = P + C.I. = Rs 26400 + Rs 10259.70 = Rs 36,659.70
Q 3 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Fabina borrows Rs 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?
Sol :
Interest paid by Fabina =
Amount paid by Radha at the end of 3 years = A =
C.I. = A − P = Rs 16637.50 − Rs 12500 = Rs 4,137.50
The interest paid by Fabina is Rs 4,500 and by Radha is Rs 4,137.50.
Thus, Fabina pays more interest.
Rs 4500 − Rs 4137.50 = Rs 362.50
Hence, Fabina will have to pay Rs 362.50 more.
Q 4 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 4
I borrowed Rs 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?
Sol :
P = Rs 12000
R = 6% per annum
T = 2 years
To find the compound interest, the amount (A) has to be calculated.
∴ C.I. = A − P = Rs 13483.20 − Rs 12000 = Rs 1,483.20
C.I. − S.I. = Rs 1,483.20 − Rs 1,440 = Rs 43.20
Thus, the extra amount to be paid is Rs 43.20.
Q 5 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 5
Vasudevan invested Rs 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?
Sol :
(i) P = Rs 60,000
Rate = 12% per annum = 6% per half year
n = 6 months = 1 half year
(ii) There are 2 half years in 1 year.
n = 2
Q 6 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 6
Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after
years if the interest is(i) Compounded annually
(ii) Compounded half yearly
Sol :
(i) P = Rs 80,000
R = 10% per annum
n =
yearsThe amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.
Firstly, the amount for 1 year has to be calculated.
By taking Rs 88,000 as principal, the SI for the next
year will be calculated.
Interest for the first year = Rs 88000 − Rs 80000 = Rs 8,000
And interest for the next
year = Rs 4,400Total C.I. = Rs 8000 + Rs 4,400 = Rs 1,2400
A = P + C.I. = Rs (80000 + 12400) = Rs 92,400
(ii) The interest is compounded half yearly.
Rate = 10% per annum = 5% per half year
There will be three half years in
years.
Difference between the amounts = Rs 92,610 − Rs 92,400 = Rs 210
Q 7 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 7
Maria invested Rs 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.
Sol :
(i) P = Rs 8,000
R = 5% per annum
n = 2 years
(ii) The interest for the next one year, i.e. the third year, has to be calculated.
By taking Rs 8,820 as principal, the S.I. for the next year will be calculated.
Q 8 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 8
Find the amount and the compound interest on Rs 10,000 for
years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?Sol :
P = Rs 10,000
Rate = 10% per annum = 5% per half year
n =
yearsThere will be 3 half years in
years.
C.I. = A − P
= Rs 11576.25 − Rs 10000 = Rs 1,576.25
The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.
The amount for the first year has to be calculated first.
By taking Rs 11,000 as the principal, the S.I. for the next
year will be calculated.
∴ Interest for the first year = Rs 11000 − Rs 10000 = Rs 1,000
∴ Total compound interest = Rs 1000 + Rs 550 = Rs 1,550
Therefore, the interest would be more when compounded half yearly than the interest when compounded annually.
Q 9 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 9
Find the amount which Ram will get on Rs 4,096, he gave it for 18 months at
per annum, interest being compounded half yearly.Sol :
P = Rs 4,096
R =
per annum = 
per half yearn = 18 months
There will be 3 half years in 18 months.
Therefore,
Thus, the required amount is Rs 4,913.
Q 10 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 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?
Sol :
(i) It is given that, population in the year 2003 = 54,000
Therefore,
54000 = (Population in 2001)
Population in 2001
= 48979.59Thus, the population in the year 2001 was approximately 48,980.
(ii) Population in 2005 =
Thus, the population in the year 2005 would be 59,535.
Q 11 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
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.
Sol :
The initial count of bacteria is given as 5,06,000.
Bacteria at the end of 2 hours =
Thus, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).
Q 12 - Ex 8.3 - Comparing Quantities - NCERT Maths Class 8th - Chapter 8
Question 12
A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Sol :
Principal = Cost price of the scooter = Rs 42,000
Depreciation = 8% of Rs 42,000 per year
Value after 1 year = Rs 42000 − Rs 3360 = Rs 38,640
Right answer
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