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NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Exercise: 7.3

Q1: 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 is 2005?

Sol: Population in 2003 is P = 54000
(i) Let the population in 2001 (i.e. 2 years ago) = P
Since rate of increment in population = 5% p.a.
∴ Present population NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Thus, the population in 2001 was about 48980.

(ii) Initial population (in 2003), i.e. P = 54000
Rate of increment in population = 5% p.a.
Time = 2 years ⇒n = 2
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= 59535

Thus, the population in 2005 = 59,535.


Q2: 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: Initial count of bacteria (P) = 5,06,000
Increasing rate (R) = 2.5% per hour
Time (T) = 2 hour. ⇒ n = 2
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= 531616.25 or 531616 (approx.)
Thus, the count of bacteria after 2 hours will be 531616 (approx.).


Q3: A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Sol: Initial cost (value) of the scooter (P) = Rs 42000
Depreciation rate = 8% p.a.
Time = 1 year ⇒ n = 1
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= ₹ 420 × 92 = ₹ 38,640

Thus, the value of the scooter after 1 year will be ₹ 38,640.

Try These

Q1: Machinery worth Rs 10,500 depreciated by 5%. Find its value after one year.

Ans: Here, P = Rs 10,500, R = –5% p.a., T = 1 year, n = 1
∵  NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) [∵ Depreciation is there, ∴ r = –5%]
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)
= Rs10500 * 19/20 = Rs 525 * 19 = Rs 9975
Thus, machinery value after 1 year = Rs 9975


Q2: Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%

Ans: Present population, P = 12 lakh
Rate of increase, R = 4% p.a.
Time, T = 2 years
∴ Population after 2 years = NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)
Thus, the population of the town will be 12,97,920 after 2 years.
Note: For depreciation, we use the formula as
NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Old NCERT Questions

Try These

Q1:  Find CI on a sum of Rs 8000 for 2 years at 5% per annum compounded annually. Solution: We have P = Rs 8000, R = 5% p.a., T = 2 years

Sol: 

∵  NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

∴  NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs (20 * 21 * 21) = Rs 8820

Now, compound interest = A – P

= Rs 8820 – Rs 8000

= Rs 820

Remember

  1. The time period after which the interest is added each time to form a new principal is called the conversion period.
  2. If the interest is compounded half yearly, the time period becomes twice and rate becomes half of the annual rate.
  3. If the interest compounded quarterly, the time period becomes 4 times and rate become one-fourth of the annual rate.
  4. If the conversion period is not specified, then it is taken as one year and the interest is compounded annually.

Q2: Find the time period and rate for each. 

  1. A sum taken for NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) years at 8% per annum is compounded half yearly.
  2. A sum taken for 2 years at 4% per annum compounded half yearly. 

Sol: 

1. We have an interest rate 8% per annum for NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) year.

∴ Time period   NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) half years

Rate (R) = 1/2 (8%) = 4% per half year.

2. We have interest rate 4% per annum for 2 years.

∴ Time period (n) = 2(2) = 4 half years

Rate (R) = 1/2 (4%)= 2% per half year.

Q3: Find the amount to be paid

  1. At the end of 2 years on Rs 2,400 at 5% per annum compounded annually.
  2. At the end of 1 year on Rs 1,800 at 8% per annum compounded quarterly.

Sol:

1. We have: P = Rs 2400, R = 5% p.a., T = 2 years

∵ Interest is compounded annually i.e. n = 2

∴  NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

2. Here, interest compounded quarterly.

∴ R = 8% p.a. = 8/4% i.e. 2% per quarter

T = 1 year = 4 * 1, i.e. 4 quarters or n = 4

Now

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)


Q4: Calculate the amount and compound interest on   

(a) Rs 10,800 for 3 years at NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) per annum compounded annually.

(b) Rs 18,000 for NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) years at 10% per annum compounded annually.

(c) Rs 62,500 for NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) 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.  

Sol: (a ) Here P = Rs 10800, T = 3 years, R = NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

We have:

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

[∵ Interest compounded annually,∴ n = 3]

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

∴ Amount = Rs 15377.34Rs 15377.34

Now, compound interest = Rs 15377.34 – Rs 10800

= Rs 4577.34

(b) Here P = Rs 18000, T = NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) years, R = 10% p.a.

∵ Interest is compounded annually,

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 22869

∴ Amount = Rs 22869

CI = Rs 22869 – Rs 18000 = Rs 4869

(c) Here P = Rs 62500, T = NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) r = 8% p.a.

Compounding half yearly,

R = 8% p.a. = 4% per half year

T = NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)  year → n = 3 half years.

∴ Amount 

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) 

= Rs 4 * 26 * 26 * 26 = Rs 70304

Amount = RS 70304

CI = Rs 70304 – Rs 62500 = Rs 7804

(d) Here P = Rs 8000, T = 1 year, R = 9% p.a.

Interest is compounded half yearly,

∴ T = 1 year = 2 half years

 R = 9% p.a = 9/2% half yearly

∴ Amount =

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

CI = Rs 8736.20 – Rs 8000

= Rs 736.20

(e) Here P = Rs 10000, T = 1 year

R = 8% p.a. compounded half yearly.

∴ R = 8% p.a. = 4% per half yearly

T = 1 year → n = 2 * 1 = 2

Now, amount

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 16 * 26 * 26 = Rs 10816

CI = Rs 10816 – Rs 10000 = Rs 816

Q5: 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 an the 2nd year for 4/12 years.)

Sol:

Note: Here, we shall calculate the amount for 2 years using the CI formula. Then this amount will become the principal for next 4 months, i.e. 4/12 years.

Here, P = Rs 26400, T = 2 years, R = 15% p.a

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Again, P = 34914, T = 4 months = 4/12 years, R = 15% p.a.

∴ Using   NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) we have

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 1745.70

Amount = S.I. + P

= Rs (1745.70 + 34914) = Rs 36659.70

Thus, the required amount to be paid to the bank after 2 years 4 month = Rs 36659.70

Q6: 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: 

For Fabina

P = Rs 12500 

T = 3 years   

R = 12% p.a.

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 125 X 3 X 12 

For Radha

P = Rs 12500

T = 3 years

R = 10% p.a. (Compounded annually)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 16637.5

∴ CI = 16637.5 – RS 12500 = 4137.50

 Difference = Rs 4500 – Rs 4137.50 = Rs 362.50

Thus, Fabina pays Rs 362.50 more.

Q7: 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?

Sol:

For SI  

Principal = Rs 12000

Time = 2 years   

Rate = 6% p.a. 

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

 For CI

  Principal = Rs 12000

Time = 2 years

 Rate = 6% p.a.

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

 

Thus, excess amount = Rs 1483.20 – Rs 1440

= Rs 43.20.

Q8: 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?

Sol:

(i) Amount after 6 months

P = Rs 60000

T = 1/2 year , n = 1           [∵ Interest is compounded half yearly.]

R = 12% p.a. = 6% per half year

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

(ii) Amount after 1 year

P = Rs 60000

T = 1 year; n = 2

R = 12% p.a. = 6% per half year

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Thus, amount after 6 months = Rs 63600

and amount after 1 year = Rs 67416

Q9: 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 NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) years if the interest is (i) Compounded annually. (ii) Compounded half yearly.

Sol: 

(i) Compounded annually

P = Rs 80000

R = 10% p.a.

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Amount for 1st year.

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 440 *10 = Rs 4400

∴ Amount = Rs 88000 + Rs 44000

= Rs 92400

(ii) Compounded half yearly 

P = Rs 80000

R = 10% p.a. = 5% per half year

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 10 * 21 * 21 * 21

= Rs 92610

Thus, the difference between the two amounts = Rs 92610 – Rs 92400

= Rs 210

Q10: 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: Principal = Rs 8000

Rate = 5% p.a. compounded annually

(i) Time = 2 years ⇒ n = 2

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

∴ Amount credited against her name at the end of two years = Rs 8820

(ii) Time = 3 years ⇒ n = 3

∴    NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs (21 * 21 * 21) = Rs 9261

∵ Interest paid during 3rd year

= [Amount at the end of 3rd year] – [Amount at the end of 2nd year]

= Rs 9261 – Rs 8820 = Rs 441

Q11: Find the amount and the compound interest on Rs 10,000 for NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) 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:

Principal = Rs 10,000

Time = NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) %

Rate = 10% p.a.

Case I. Interest on compounded half yearly

We have r = 10 p.a. = 5% per half yearly

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

∴  NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

∴ Amount = Rs 11576.25

Now CI = Amount – Principal

= Rs 11576.25 – Rs 10,000 = Rs 1576.25

Case II. Interest on compounded annually

We have R = 10% p.a.

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Amount for 1 year =   NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) 

∴ Interest for 1st year = Rs 11000 – Rs 10,000 = Rs 1000

Interest for next 1/2 year on Rs 11000  NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

= Rs 55 * 10 = Rs 550

∴ Total interest = Rs 1000 + Rs 550

= Rs 1550

Since Rs 1576.25 > Rs 1550

∴ Interest would be more in case of it is compounded half yearly.

Q12: Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at  NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) per annum, interest being compounded half yearly.

Sol:

We have P = Rs 4096

T = 18 months

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

∵ Interest is compounded half yearly.

T = 18 months ⇒ n = 18/6 = 3 six months

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Now,

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

Thus, the required amount = Rs 4913

The document NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3) is a part of the Class 8 Course NCERT Textbooks & Solutions for Class 8.
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FAQs on NCERT Solutions for Class 8 Maths - Comparing Quantities - 3 (Exercise: 7.3)

1. What are the key concepts covered in Exercise 7.3 of Comparing Quantities in Old NCERT?
Ans.Exercise 7.3 covers concepts such as percentage increase and decrease, simple interest, and the application of ratios in real-life problems. It provides various problems that help in understanding how to compare quantities effectively.
2. How can I solve problems related to percentage increase and decrease in Exercise 7.3?
Ans.To solve percentage increase and decrease problems, identify the initial value and the final value. Use the formula: Percentage Change = [(Final Value - Initial Value) / Initial Value] × 100. This will help you determine the percentage change.
3. Are there any specific formulas to remember for solving questions in Comparing Quantities?
Ans.Yes, some key formulas include: 1. Simple Interest (SI) = (Principal × Rate × Time) / 100. 2. Percentage = (Part / Whole) × 100. 3. Percentage Increase = [(New Value - Original Value) / Original Value] × 100.
4. How can I practice effectively for the Comparing Quantities chapter?
Ans.To practice effectively, solve the exercises provided in the NCERT textbook, refer to previous years' question papers, and use additional worksheets. Understanding the concepts through practical applications will enhance your skills.
5. What are the common mistakes to avoid while solving Exercise 7.3 questions?
Ans.Common mistakes include miscalculating percentages, misunderstanding the problem statement, and neglecting to convert units when necessary. Always double-check your calculations and ensure you interpret the questions correctly.
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