## 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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