NCERT Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3

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 1/2% per annum compounded annually.
b) Rs 18,000 for 2 1/2 years at 10% per annum compounded annually.
c) Rs 62,500 for 1 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 SI formula to verify).
e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.
Answer:

a)
Principal = ₹10,800
Time period = 3 years
Rate of interest = 12 1/2% = 25/2%
Amount = ?
= 10800(1 + 25/200)³
= 10800(1 + 1/8)³
= 10800(8/8 + 1/8)³
= 10800(9/8)³
= 10800 × 9/8 × 9/8 × 9/8
= ₹15377.34
Compound interest = Amount - Principal
= ₹15377.34 - ₹10800
= ₹4577.34

b)
Principal = ₹18,000
Time period = 2 1/2 years
Rate of interest = 10%
Amount = ?
(Calculating amount for 2 years)
= 18000(1 + 10/100)²
= 18000(1 + 1/10)²
= 18000(10/10 + 1/10)²
= 18000(11/10)²
= 18000 × 11/10 × 11/10
= ₹21780
(Calculating amount for 1/2 year)
= 21780 × 10 × 1/100 × 2
= ₹1089
Amount = ₹21780 + ₹1089 = ₹22869
Compound Interest = Amount - Principal
= ₹22869 - ₹18000
= ₹4869

c)
Principal = ₹62,500
Time period = 1 1/2 years = 3 half years
Rate of interest = 8% = 8/2 = 4%
Amount = ?
= 62500(1 + 4/100)³
= 62500(100/100 + 4/100)³
= 62500(104/100)³
= 62500 × 104/100 × 104/100 × 104/100
= ₹70304
Compound interest = Amount – Principal
= ₹70304 - ₹62500
= ₹7804

d)
Principal = ₹8,000
Time period = 1 year = 2 half years
Rate of interest = 9% = 9/2 = 4.5%
Amount = ?
= 8000(1 + 9/200)²
= 8000(200/200 + 9/200)²
= 8000(209/200)²
= 8000 × 209/200 × 209/200
= ₹8736.2
Compound interest = Amount - Principal
= ₹8736.2 - ₹8000
= ₹736.2

e)
Principal = ₹10,000
Time period = 1 year = 2 half years
Rate of interest = 8% = 8/2 = 4%
Amount = ?
= 10000(1 + 4/100)²
= 10000(100/100 + 4/100)²
= 10000(104/100)²
= 10000 × 104/100 × 104/100
= ₹10816
Compound interest = Amount - Principal
= ₹10816 - ₹10000
= ₹816

Question 2: Kamala 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 SI on the 2nd year amount for 4/12 years).
Answer:

Principal = ₹26,400
Time period = 2 years 4 months
Rate of interest = 15%
(Calculating amount for 2 years)
= 26400(1 + 15/100)²
= 26400(1 + 3/20)²
= 26400(20/20 + 3/20)²
= 26400(23/20)²
= 26400 × 23/20 × 23/20
= ₹34914
(Calculating amount for 4 months)
4/12 years = 1/3 years
= 34914 × 1 × 15/100 × 3
= ₹1745.7
Total amount = ₹34914 + ₹1745.7 = ₹36659.7
Therefore, Kamala has to pay ₹36659.7 at the end of 2 years and 4 months to clear the loan.

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?
Answer:

Fabina
Principal = ₹12,500
Time period = 3 years
Rate of interest = 12%
Amount = ?
= 12500 × 3 × 12/100
= ₹4500

Radha
Principal = ₹12,500
Time period = 3 years
Rate of interest = 10%
Amount = ?
= 12500(1 + 10/100)³
= 12500(1 + 1/10)³
= 12500(10/10 + 1/10)³
= 12500(11/10)³
= 12500 × 11/10 × 11/10 × 11/10
= ₹16637.5
Compound interest = Amount - Principal
= ₹16637.5 - ₹12500
= ₹4137.5

Comparing
₹4500 > ₹4137.5
Fabina paid more interest.

₹4500 - ₹4137.5 = ₹362.5
Therefore, Fabina paid more by ₹362.5.

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?
Answer:

Simple Interest
Principal = ₹12,000
Rate of interest = 6%
Time period = 2 years
Amount = ?
= P × T × R/100
= 12000 × 2 × 6/100
= ₹1440

Compound interest
Principal = ₹12,000
Rate of interest = 6%
Time period = 2 years
Amount = ?
= P(1 + R/100)ⁿ
= 12000(1 + 6/100)²
= 12000(100/100 + 6/100)²
= 12000(106/100)²
= 12000 × 106/100 × 106/100
= ₹13483.2
Compound interest = Amount - Principal
= ₹13483.2 - ₹12000
= ₹1483.2

The extra amount that I had to pay = ₹1483.2 - ₹1440 = ₹43.2
Therefore, I had to pay ₹43.2 extra if I had borrowed it.

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?
Answer:

i)
Principal = ₹60,000
Time period = 6 months
Rate of interest = 12% = 12/2 = 6%
Amount = ?
= P(1 + R/100)ⁿ
= 60000(1 + 6/100)¹
= 60000(100/100 + 6/100)
= 60000(106/100)
= 60000 × 106/100
= ₹63600
Therefore, Vasudevan would get ₹63600 after 6 months.

ii)
Principal = ₹60,000
Time period = 1 year = 2 half years
Rate of interest = 12% = 12/2 = 6%
Amount = ?
= P(1 + R/100)ⁿ
= 60000(1 + 6/100)²
= 60000(100/100 + 6/100)²
= 60000(106/100)²
= 60000 × 106/100 × 106/100
= ₹67416
Therefore, Vasudevan would get ₹67416 after 1 years.

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 1/2 years if the interest is
i) compounded annually.
ii) compounded half yearly.
Answer:

i)
Principal = ₹80000
Time period = 1 1/2 years
Rate of Interest = 10%
Amount = ?
(Calculating amount for 1 years)
= P(1 + R/100)ⁿ
= 80000(1 + 10/100)¹
= 80000(1 + 1/10)¹
= 80000(10/10 + 1/10)¹
= 80000(11/10)¹
= 80000 × 11/10
= ₹88000
(Calculating amount for 1/2 years)
= P × T × R/100
= 88000 × 1 × 10/100 × 2
= ₹4400
Total amount = ₹88000 + ₹4400 = ₹92400
Therefore, Arif has to pay ₹92400 if he took loan from a bank at the rate of 10% for 1 1/2 years compounded annually.

ii)
Principal = ₹80000
Time period = 1 1/2 years = 3 half years
Rate of Interest = 10% = 10/2 = 5%
Amount = ?
= P(1 + R/100)ⁿ
= 80000(1 + 5/100)³
= 80000(100/100 + 5/100)³
= 80000(105/100)³
= 80000 × 105/100 × 105/100 × 105/100
= ₹92610
Therefore, Arif has to pay ₹92610 if he took loan from a bank at the rate of 10% for 1 1/2 years compounded half yearly.

= ₹92610 - ₹92400
= ₹210
Therefore, Arif has to pay ₹210 more if he took loan from a bank at the rate of 10% for 1 1/2 years compounded half yearly.

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 second year.
ii) The interest for the 3rd year.
Answer:

i)
Principal = ₹8000
Time period = 2 years
Rate of interest = 5%
Amount = ?
= P(1 + R/100)ⁿ
= 8000(1 + 5/100)²
= 8000(100/100 + 5/100)²
= 8000(105/100)²
= 8000 × 105/100 × 105/100
= ₹8820
Therefore, the amount credited against her name at the end of second year is ₹8820.

ii)
Principal = ₹8000
Time period = 3 years
Rate of interest = 5%
Amount = ?
= P(1 + R/100)ⁿ
= 8000(1 + 5/100)³
= 8000(100/100 + 5/100)³
= 8000(105/100)³
= 8000 × 105/100 × 105/100 × 105/100
= ₹9261
Interest for 3rd year = Amount of 3rd year - Amount for 2nd year
= ₹9261 - ₹8820
= ₹441
Therefore, the interest for 3rd year was ₹441.

Question 8: Find the amount and the compound interest on Rs 10,000 for 1 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?
Answer:

→ Compounded half yearly
Principal = ₹10000
Time period = 1 1/2 years = 3 half years
Rate of interest = 10% = 10/2 = 5%
Amount = ?
= P(1 + R/100)ⁿ
= 10000(1 + 5/100)³
= 10000(100/100 + 5/100)³
= 10000(105/100)³
= 10000 × 105/100 × 105/100 × 105/100
= ₹11576.25
Compound interest = Amount - Principal
= ₹11576.25 - ₹10000
= ₹1576.25
Therefore, the amount and compound interest for ₹10000 for 1 1/2 years at 10% per annum, compounded half yearly will be ₹11576.25 and ₹1576.25 respectively.

→ Compounded yearly
Principal = ₹10000
Time period = 1 1/2 years
Rate of interest = 10%
Amount = ?
(Calculating for 1 year)
= P(1 + R/100)ⁿ
= 10000(1 + 10/100)¹
= 10000(1 + 1/10)¹
= 10000(10/10 + 1/10)¹
= 10000 × 11/10
= ₹11000
Compound interest = Amount – Principal
= ₹11000 - ₹10000
= ₹1000

Comparing
₹1576.25 > ₹1000
Yes, the interest be more than the interest he would get if it was compounded annually.

Question 9: Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 1/2% per annum, interest being compounded half yearly.
Answer:

Principal = ₹4096
Time period = 18 months = 3 half years
Rate of interest = 12 1/2% = 25/4%
Amount = ?
= P(1 + R/100)ⁿ
= 4096(1 + 25/4 × 100)³
= 4096(1 + 25/400)³
= 4096(400/400 + 25/400)³
= 4096(425/400)³
= 4096(17/16)³
= 4096 × 17/16 × 17/16 × 17/16
= ₹4913
Therefore, the amount which Ram will get on ₹4096, if he gave it for 18 months at 12 1/2% per annum, interest being compounded half yearly will be ₹4913.

Question 10: The population of a place increased to 54,000 in 2003 at the rate of 5% per annum
i) find the population in 2001.
ii) what would be its population in 2005?
Answer:

Population in 2003 = 54000
Rate = 5%
Time = 2003 - 2001 = 2 years
Population in 2003 = Population in 2001 × (1 + R/100)ⁿ
Let the population in 2001 be a.
54000 = a × (1 + 5/100)²
54000 = a × (1 + 1/20)²
54000 = a × (20/20 + 1/20)²
54000 = a × (21/20)²
54000 = a × 21/20 × 21/20
54000 = a × 441/400
54000 ÷ 441/400 = a
a = 48979.59
a = 48980 (approximately)
Therefore, the population of that place in 2001 is 48980.

ii)
Population in 2003 = 54000
Rate = 5%
Time = 2005 - 2003 = 2 years
Population in 2005 = Population in 2003 × (1 + R/100)ⁿ
Let the population in 2005 be p.
p = 54000 × (1 + 5/100)²
p = 54000 × (1 + 1/20)²
p = 54000 × (20/20 + 1/20)²
p = 54000 × (21/20)²
p = 54000 × 21/20 × 21/20
p = 54000 × 441/400
p = 59535
Therefore, the population of that place in 2005 is 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.
Answer:

Initial count of bacteria = 506000
Rate = 2.5% per hour
Time = 2 hours
Number of bacteria at the end of 2 hours = number of count of bacterial initially × (1 + R/100)ⁿ
Let the number of bacteria at the end of 2 hours be m.
= m = 506000 × (1 + 25/1000)²
= m = 506000 × (1 + 1/40)²
= m = 506000 × (40/40 + 1/40)²
= m = 506000 × (41/40)²
= m = 506000 × 41/40 × 41/40
= m = 531616.25
= m = 531616 (approximately)
Therefore, the number of bacteria at the end of 2 hours will be 531616.

Question 12: A scooter was brought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Answer:

Cost Price of scooter = ₹42000
Rate of depreciated = 5%
Time period = 1 year
Value after 1 year = present value × (1 - R/100)ⁿ
Let the value of scooter after 1 year be h.
= h = 42000 × (1 - 5/100)¹
= h = 42000 × (1 - 1/20)¹
= h = 42000 × (20/20 - 1/20)¹
= h = 42000 × (19/20)¹
= h = 42000 × 19/20
= h = ₹39900
Therefore, the cost of scooter after 1 year will be ₹39900.

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