HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Haryana State Board HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Calculate the amount and compound interest on :
(a) Rs. 10,800 for 3 years at 12\(\frac{1}{2}\)% per annum compounded annually.
(b) Rs. 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.
(c) Rs. 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.
(d) Rs. 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using S.I. formula to verify).
(e) Rs. 10,000 for 1 year at 8% per annum compounded half yearly.
Solution:
(a) A = ?, C.I. = ?, P = Rs. 10,800,
T = 3 years, R = 12 \(\frac{1}{2}\)% per annum
A = P(1 + \(\frac{R}{100}\))T
= 1080o(1 + \(\frac{25}{2}\) × \(\frac{1}{100}\))3
= 10800(\(\frac{17}{16}\))3
= 10800 × \(\frac{17}{16}\) × \(\frac{17}{16}\) × \(\frac{17}{16}\)
= Rs. 12954.20
C.I. = A – P
= Rs. 1295.4.20 – Rs. 10800
= Rs. 2154.20.

(b) P = Rs. 18000, T = 2\(\frac{1}{2}\) years, R = 10% per annum compounded annually, A = ?, C.I. = ?
A = P(1 + \(\frac{R}{100}\))T
= 18000(1 + \(\frac{10}{100}\))2\(\frac{1}{2}\)
So amount for 2 years is given by
A = Rs. 18000(1 + \(\frac{10}{100}\))2
= 18000(1 + \(\frac{1}{10}\))2
= 18000 × \(\frac{11}{10}\) × \(\frac{11}{10}\)
= 180 × 121 = Rs. 21780
Rs. 21780 would act as principal for next \(\frac{1}{2}\) year. We find S.I. on Rs. 21780 for \(\frac{1}{2}\) year
S.I = Rs. \(\frac{21780 \times \frac{1}{2} \times 10}{100}\)
= Rs. 1089
Interest for two years
= Rs. 21780 – Rs. 18000
= Rs. 3780
Interest for next \(\frac{1}{2}\) year = Rs. 1089
Therefore, total compound interest
= Rs. 3780 + Rs. 1089
= Rs. 4869
A = P + C.I.
= Rs. 18000 + Rs. 4869
= Rs. 22869.

(c) P = Rs. 62500, T = 1\(\frac{1}{2}\) years = 3 half years, R = 8% per annum = 4% per half yearly, A = ?, C.I. = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 1
= Rs 70304
C.I. = A – P
= Rs. 70304 – Rs. 62500
= Rs. 7804.

(d) P = Rs. 8000, T = 1 year = 2 half years, R = 9% per annum = \(\frac{9}{2}\) % per half year, A = ?, C.I. = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 2
= Rs. 8736.20
C.I. = A – P
= Rs. 8736.20 – Rs. 8000
= Rs. 736.20.

(e) P = Rs. 10000, T = 1 year = 2 half years, R = 8% per annum = 4% per half year, A = ?, C.I. = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 3
= Rs. 10816
C.I. = A – P
= Rs. 10816 – Rs. 10000
= Rs. 816.

HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 2.
Kamla borrowed Rs. 26,400 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 S.I. on the 2nd year amount for 4/12 years]
Solution:
P = Rs. 26400, R = 15% per annum, T = 2 years and 4 months, A = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 4
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 5
= 66 × 23 × 23
= Rs. 34914
Rs. 34914 would act as principal for next 4 months i.e. \(\frac{1}{3}\) year
So S.I. = P × R × T
= 34914 × \(\frac{15}{100}\) × \(\frac{1}{3}\)
= Rs. 1745.70
Amount paid by Kamla after 2 years
= Rs. 34914 + Rs. 1745.70
= Rs. 36659.70.

Question 3.
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 ?
Solution:
In case of Fabina :
P = Rs. 12500,
R = 12% per annum, T = 3 years
S.I = P × R × T
= 12500 × \(\frac{12}{100}\) × 3 = RS. 4500.
In case of Radha :
P = Rs. 12500,
R = 10% per annum,
T = 3 years, C.I. = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 6
= Rs. 16637.50
C.I. = A – P
= Rs. 16637.50 – Rs. 12500
= Rs. 4137.50
It is obvious that Fabina would pay more interest by Rs. 4500 – Rs. 4137.50 = Rs. 362.50 So,
Fabina paid Rs. 362.50 more than Radha.

HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 4.
I borrowed Rs. 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:
P = Rs. 12000, R = 6% per annum, T = 2 years, S.I. = ?, C.I. = ?
In case of simple interest:
S.I = P × R × T
= 12000 × \(\frac{6}{100}\) × 2
= Rs. 1440
In case of compound interest:
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 7
= Rs. 13483.20
So, Amount = Rs. 13483.20
∴ C.I. = A – P
= Rs. 13483.20 – Rs. 12000
= Rs. 1483.20
It is obvious that C.I. > S.I
So extra amount paid in case of C.I. is
= Rs. 1483.20 – Rs. 1440
= Rs. 43.20.

Question 5.
Vasudevan invested Rs. 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:
P = Rs. 60000,
R = 12% per annum compounded half yearly
= 6% half yearly
A6 = ?
A12 = ?
T6 = 6 months = 1 half year
T12 = 1 year = 2 half years

(i) Amount after 6 months
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 8
= Rs. 67416.
∴ A12 = Rs. 67416.

HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

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 1\(\frac{1}{2}\) years if the interest is
(i) Compounded annually.
(ii) Compounded half yearly.
Solution:
P = Rs. 80000,
R = 10% per annum,
T = 1\(\frac{1}{2}\) years
= 5% half yearly = 3 half years
In 1st Case : When amount calculated annual compounding
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 9
Rs. 88000 would be principal for half year to calculate S.I. for \(\frac{1}{2}\) year
S.I. = P × R × T
= 88000 × \(\frac{10}{100}\) × \(\frac{1}{2}\)
= Rs. 4400
∴ C.I. for 1\(\frac{1}{2}\) on given principal
= (Rs. 88000 – Rs. 80000) + Rs. 4400
= Rs. 12400
So, A = P + C.I.
= Rs. 80000 + Rs. 12400
= Rs. 92400
In 2nd Case : When amount calculated half yearly compounding
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 10
= Rs. 92610
Difference in amount in both the cases
= Rs. 92610 – Rs. 92400
= Rs. 210.

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.
Solution:
(i) P = Rs. 8000, R = 5% per annum, T = 2 year, T = 3 year, A2 = ?, A3 = ?
Amount after 2 years,
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 11
Amount credited after 2 years = Rs. 8820.
Amount after 3 years,
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 12
= Rs. 9261
∴ Interest for 3rd year = A3 – A2
= Rs. 9261 – Rs. 8820
= Rs. 441.

HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 8.
Find the amount and the compound interest on Rs. 10,000 for 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 = Rs. 10000
T = 1\(\frac{1}{2}\) years = 3 half years
R = 10% per annum = 5% half yearly
Ah =?, Aa = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 13
= Rs. 11576.25
So, Amount = Rs. 11576.25
∴ C.I. = A – P
= Rs. 11576.25 – Rs. 10000
= Rs. 1576.25
If n = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) years
Aa = P(1 + \(\frac{R}{100}\))T
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 14
= 1000 × 1.05 = Rs. 11550
∴ C.I. = Aa – P = 11550 – 10000
= Rs. 1550.
or, Rs. 11000 would be principal for \(\frac{1}{2}\) year to calculate S.I. for \(\frac{1}{2}\) year.
S.I. = P × R × T
= 11000 × \(\frac{10}{100}\) × \(\frac{1}{2}\) = Rs. 550
∴ C.I. = (Rs. 11000 – Rs. 10000) + Rs. 550 = Rs. 1550
It is obvious that C.I. calculated half yearly is greater than that when calculated yearly.
Extra amount of interest = Rs. 1576.25 – Rs. 1500 = Rs. 76.25.

Question 9.
Find the amount which Ram will get on Rs. 4096, if have gave it for 18 months at 12\(\frac{1}{2}\) % per annum, interest being compounded half yearly.
Solution:
P = Rs. 4096,
T = 18 months = 3 half years,
R = 12\(\frac{1}{2}\) % per annum
= \(\frac{25}{4}\) % per half year
A = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 15
= Rs. 4913

HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

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) P = 54000, R = 5% per annum, A = ?, T = 2 years
Population before 2 years
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 16

(ii) For population in 2005 i.e. 2 years after 2003
P = 54000, T = 2 years, R = 5% per annum, A = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 17
∴ Population in 2005 be 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:
Initial count of bacteria P = 506000, T = 2 hour
Rate of increasing bacteria R = 2.5% per hour, A = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 18
∴ Count of bacteria after 2 hour = 531616.

HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

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.
Solution:
P = Rs. 42000, R = 8% per annum, T = 1 year, A = ?
HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 19

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