The document NCERT Solutions(Part- 4)- Comparing Quantities Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by Full Circle.

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**Question 1.** **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**

∵

∴

= Rs (20 * 21 * 21) = Rs 8820

Now, compound interest = A – P

= Rs 8820 – Rs 8000

= Rs 820

**Remember**

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

**Question:** **Find the time period and rate for each. **

**A sum taken for years at 8% per annum is compounded half yearly.****A sum taken for 2 years at 4% per annum compounded half yearly.**

**Solution: **

**1.** We have interest rate 8% per annum for year.

∴ Time period 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.

**Question:** **Find the amount to be paid**

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

**Solution:**

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

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

∴

**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

**Question 1. ****A machinery worth Rs 10,500 depreciated by 5%. Find its value after one year.**

**Solution:** Here, P = Rs 10,500, R = –5% p.a., T = 1 year, n = 1

∵ [∵ Depreciation is there, ∴ r = –5%]

∴

= Rs10500 * 19/20 = Rs 525 * 19 = Rs 9975

Thus, machinery value after 1 year = Rs 9975

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

**Solution:**

Present population, P = 12 lakh

Rate of increase, R = 4% p.a.

Time, T = 2 years

∴ Population after 2 years =

Thus, the population of the town will be 12,97,920 after 2 years.

Note: For depreciation, we use the formula as

**Exercise 8.3**

**Question 1.** **Calculate the amount and compound interest on **

**(a) Rs 10,800 for 3 years at per annum compounded annually.**

**(b) Rs 18,000 for years at 10% per annum compounded annually.**

**(c) Rs 62,500 for 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. **

**Solution:** (a ) Here P = Rs 10800, T = 3 years, R =

We have:

[∵ Interest compounded annually,∴ n = 3]

∴ Amount = Rs 15377.34Rs 15377.34

Now, compound interest = Rs 15377.34 – Rs 10800

= Rs 4577.34

**(b)** Here P = Rs 18000, T = years, R = 10% p.a.

∵ Interest is compounded annually,

= Rs 22869

∴ Amount = Rs 22869

CI = Rs 22869 – Rs 18000 = Rs 4869

**(c)** Here P = Rs 62500, T = r = 8% p.a.

Compounding half yearly,

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

T = year → n = 3 half years.

∴ Amount

= 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 =

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

= Rs 16 * 26 * 26 = Rs 10816

CI = Rs 10816 – Rs 10000 = Rs 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 an the 2nd year for 4/12 years.)**

**Solution:**

**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

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

∴ Using we have

= 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

**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: **

P = Rs 12500 T = 3 years R = 12% p.a. = Rs 125 X 3 X 12 |
P = Rs 12500 T = 3 years R = 10% p.a. (Compounded annually) = 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.

**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:**

Principal = Rs 12000 Time = 2 years Rate = 6% p.a. |
Principal = Rs 12000 Time = 2 years Rate = 6% p.a. |

Thus, excess amount = 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:**

**(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

**(ii) **Amount after 1 year

P = Rs 60000

T = 1 year; n = 2

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

Thus, amount after 6 months = Rs 63600

and amount after 1 year = Rs 67416

**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.**

**Solution: **

**(i)** Compounded annually

P = Rs 80000

R = 10% p.a.

Amount for 1st year.

= 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

= Rs 10 * 21 * 21 * 21

= Rs 92610

Thus, the difference between the two amounts = 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:** Principal = Rs 8000

Rate = 5% p.a. compounded annually

**(i)** Time = 2 years ⇒ n = 2

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

**(ii)** Time = 3 years ⇒ n = 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

**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?**

**Solution:**

Principal = Rs 10,000

Time = %

Rate = 10% p.a.

**Case I. **Interest on compounded half yearly

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

∴

∴ 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.

Amount for 1 year =

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

Interest for next 1/2 year on Rs 11000

= 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.

**Question 9.** **Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at per annum, interest being compounded half yearly.**

**Solution:**

We have P = Rs 4096

T = 18 months

∵ Interest is compounded half yearly.

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

Now,

Thus, the required amount = Rs 4913

**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 is 2005?**

**Solution:** 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

or

or

or

= 48980 (approx.)

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

∴

= 135 * 21 * 21 = 59535

Thus, the population in 2005 = 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) = 5,06,000

Increasing rate (R) = 2.5% per hour

Time (T) = 2 hour. ⇒ n = 2

∵

= 531616.25 or 531616 (approx.)

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

**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:** Initial cost (value) of the scooter (P) = Rs 42000

Depreciation rate = 8% p.a.

Time = 1 year ⇒ n = 1

= Rs 420 * 92 = Rs 38640

Thus, the value of the scooter after 1 year will be Rs 38640.

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