Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

1. Calculate the amount and compound interest on:

Question (a)

₹ 10,800 for 3 years at 12\(\frac {1}{2}\) per annum compounded annually.

Solution:

Here, P = ₹ 10,800;

R = 12\(\frac {1}{2}\) % = \(\frac {25}{2}\) %;

T = 3 years; ∴ n = 3

Amount = ₹ 15,377.34

Compound’interest = Amount – Principal

= ₹ (15377.34 – 10800)

= ₹ 4577.34

Question (b)

₹ 18,000 for 2\(\frac {1}{2}\) years at 10% per annum compounded annually.

Solution:

Here, P = ₹ 18,000; R = 10 %;

T = 2\(\frac {1}{2}\) years; ∴ n = 2 + \(\frac {1}{2}\)

Amount = ₹ 22,869

Compoimd interest = Amount – Principal

= ₹ (22869 – 18000)

= ₹ 4869

Question (c)

₹ 62,500 for 1\(\frac {1}{2}\) years at 8% per annum compounded half yearly.

Solution:

Here, the interest is compounded half-yearly.

Here, P = ₹ 62,500; R = \(\frac {8}{2}\) = 4 %

T = 1\(\frac {1}{2}\) years ∴ n = \(\frac {3}{2}\) × 2 = 3

Amount = ₹ 70,304

Compound interest = Amount – Principal

= ₹ (70304 – 62500)

= ₹ 7804

Question (d)

₹ 8000 for 1 year at 9 % per annum compounded half-yearly.

(You could use the year-by-year calculation using SI formula to verify.)

Solution:

Here, the interest is compounded half-yearly.

Here, P = ₹ 8000; R = \(\frac {9}{2}\) %;

T = 1 year ∴ n = 2

Amount = ₹ 8736.20

Compound interest = Amount – Principal

= ₹ (8736.20 – 8000)

= ₹ 736.20

[Note : By finding simple interest also we can calculate.)

SI = \(\frac {PRT}{100}\)

= \(\frac{8000 \times 9 \times 1}{2 \times 100}\)

= ₹ 376.20

Thus, total interest of 1 year

= ₹ (360 + 376.20)

= ₹ 736.20

Question (e)

₹ 10,000 for 1 year 8% per annum compounded half yearly.

Solution:

Here, the interest is compounded half-yearly.

Here, P = ₹ 10,000; R = \(\frac {8}{2}\) = 4 %;

T = 1 year ∴ n = 1 × 2 = 2

Amount = ₹ 10,816

Compound interest = Amount – Principal

= ₹ (10816 – 10000)

= ₹ 816

2. Kamala borrowed ₹ 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 \(\frac {4}{12}\) years)

Solution:

[Note: Here, find amount after 2 years by compound interest. This amount is principal for \(\frac {4}{12}\) year. For this \(\frac {4}{12}\) year, find simple interest.)

Here, P = ₹ 26,400; R = 15%;

T = 2 years ∴ n = 2

Now, ₹ 34,914 will be principal to find interest of 4 months.

SI = \(\frac {PRT}{100}\)

= \(\frac{34914 \times 15 \times 4}{100 \times 12}\)

= \(\frac{174570}{100}\)

= ₹ 174570

= ₹ 1745.70

Amount = ₹ (34914 + 1745.70)

= ₹ 36,659.70

Thus, Kamala Mil have to pay ₹ 36,659.70 to clear the loan.

3. Fabina borrows ₹ 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:

For Fabina:

Here, P = ₹ 12,500; R = 12%; T = 3 years

SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{12500 \times 12 \times 3}{100}\)

= 125 × 12 × 3

= ₹ 4500

Simple interest = ₹ 4500

For Radha:

Here, P = ₹ 12,500; R = 10%;

T = 3 years ∴ n = 3

Amount = ₹ 16,637.50

Compound interest = Amount – Principal

= ₹ (16637.50 – 12500)

= ₹ 4137.50

Fabina has to pay ₹ 4500 as interest and Radha has to pay ₹ 4137.50 as interest.

∴ Fabina has to pay more interest.

Difference in interest = ₹ (4500 – 4137.50)

= ₹ 362.50

Thus, Fabina has to pay ₹ 362.50 more than Radha as interest.

4. I borrowed ₹ 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:

For Simple Interest:

Here, P = ₹ 12,000; R = 6 %; T = 2 years

SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{12000 \times 6 \times 2}{100}\)

= 120 × 6 × 2

= ₹ 1440

SI = ₹ 1440

For Compound Interest:

Here, P = ₹ 12,000, R = 6 %

T = 2 years ∴ n = 2

= ₹ 13,483.20

Amount = ₹ 13,483.20

CI = A – P

= ₹ (13483.20 – 12000)

= ₹ 1483.20

∴ Extra amount to be paid

= ₹ (1483.20 – 1440)

= ₹ 43.20

Thus, I have to pay ₹ 43.20 as extra amount.

5. Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get.

Question (i)

after 6 months?

Solution:

Interest after 6 months:

Here, P = ₹ 60,000; R = \(\frac {12}{2}\) = 6%;

T = 6 months ∴ n = 1

∴ Amount = ₹ 63,600

Question (ii)

after 1 year?

Solution:

After 1 year:

∴ Amount = ₹ 67,416

Here, P = ₹ 60,000; R = \(\frac {12}{2}\) = 6 %;

T = 1 year ∴ n = 2

∴ Amount = ₹ 67, 146

Thus, Vasudevan will get ₹ 63,600 after 6 months and ₹ 67,416 after 1 year.

6. Arif took a loan of ₹ 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

Question (i)

compounded annually.

Solution:

Calculation of CI annually:

Here, P = ₹ 80,000; R = 10 %;

T = 1\(\frac {1}{2}\) year

For 1st year:

R = 10% and n = 1

∴ Amount of CI after 1 year = ₹ 88,000

Now, calculate simple interest of ₹ 88,000 for 6 months.

P = ₹ 88,000; R = 10 %;

T = 6 months = \(\frac {1}{2}\) year

∴ Interest = \(\frac{P \times R \times T}{100}\)

= \(\frac{88000 \times 10 \times 1}{100 \times 2}\)

= 4400

∴ Interest of 6 months = ₹ 4400

Thus, A = P + I

= ₹ (88000 + 4400)

= ₹ 92,400

Thus, according to CI, Arif has to pay ₹ 92,400

Question (ii)

compounded half yearly.

Solution:

If interest is compounded half yearly

Here, P = ₹ 88000, R = \(\frac {10}{2}\) = 5%

T = 1\(\frac {1}{2}\) years ∴ n = 3

∴ Amount = ₹ 92,610

Thus, according to half-yearly CI, Arif has to pay ₹ 92,610

∴ Difference = ₹ (92,610 – 92,400)

= ₹ 210

7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

Question (i)

The amount credited against her name at the end of the second year,

Solution:

(i) Here, P = ₹ 8000, R = 5 %,

T = 2 years ∴ n = 2

Thus, amount credited against Marlas name at the end of second year is ₹ 8820.

Question (ii)

The interest for the 3rd year.

Solution:

To find the interest for the 3rd year:

P = ₹ 8820, R = 5%, T = 1 years

SI = \(\frac{\mathrm{PRT}}{100}=\frac{8820 \times 5 \times 1}{100}\) = 441

The interest for 3rd year is ₹ 441

OR

The interest for the 3rd year:

Here, P = ₹ 8000, R = 5 %, T = 3 years ∴ n = 3

∴ At the end of 3rd year ₹ 9261 will be credited against Maria’s name.

∴ Interest of the 3rd year = Amount of 3 years – Amount of 2 years

= ₹ (9261 – 8820)

= ₹ 441

8. Find the amount and the compound interest on ₹ 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:

(i) Here, interest is compounded half-yearly.

Here, P = ₹ 10,000; R = \(\frac {10}{2}\) = 5 %;

T = 1\(\frac {1}{2}\) years ∴ n = 3

Amount = ₹ 11,576.25

CI = A – P

= ₹ (11576.25 – 10000)

= ₹ 1576.25

(ii) Here, interest is compounded yearly. CI for 1 year:

Here, P = ₹ 10,000; R = 10%;

T = 1 year ∴ n = 1

Amount at the end of 1 year = ₹ 11,000

∴ CI = A – P

= ₹ (11000 – 10000)

= ₹ 1000

Now, calculate SI for 6 months.

Here, P = ₹ 11,000; R = 10%; T = \(\frac {1}{2}\) year

SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{11000 \times 10 \times 1}{100 \times 2}\)

= 550

∴ Total interest of 1\(\frac {1}{2}\) years = ₹ (1000 + 550)

= ₹ 1550

After comparing (i) and (ii), we can conclude ₹ 1576.25 > ₹ 1550

∴ Yes, the interest is more if compounded half-yearly.

9. Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12\(\frac {1}{2}\) % per annum, interest being compounded half-yearly.

Solution:

Here, interest is compounded half-yearly.

Here, P = ₹ 4096, R = 12\(\frac{1}{2} \times \frac{1}{2}=\frac{25}{4}\)

T = 18 months = 1\(\frac {1}{2}\) years ∴ n = 3

Amount = ₹ 4913

Thus, Ram will get ₹ 4913 at the end of period.

10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

Question (i)

find the population in 2001.

Solution:

[Note: In 1st case, we have to find P as population of 2003 is given. From that we have to find population of 2001.]

Population in 2003 = 54,000

Here, A = 54,000; R = 5%; T = 2 years ∴ n = 2

∴ P = 48979.59 (approx)

P = 48980 (approx)

Thus, the population in 2001 is 48,980.

Question (ii)

What would be its population in 2005?

Solution:

Here, P = 54,000; R = 5 %;

T = 2 years ∴ n = 2

Thus, the population in 2005 is 59,535.

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 = 5,06,000

Rate of increasing = 2.5% per hour

Bacteria count after 2 hours

Here, P = 5,06,000, R = 2.5 % = \(\frac {5}{2}\)%;

T = 2 hours ∴ n = 2

A = 531616 (approx)

Thus, the number of bacteria count after 2 hours will be 5,31,616 (approx).

12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8 % per annum. Find its value after one year.

Solution:

CP of a scooter = ₹ 42,000

Here, P = ₹ 42,000; R = 8 %; T = 1 ∴ n = 1

R = – 8 % (as depreciation)

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