The DAV Class 8 Maths Book Solutions Pdf and **DAV Class 8 Maths Chapter 6 Worksheet 1** Solutions of Compound Interest offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 6 WS 1 Solutions

Question 1.

Find the compound interest on ₹ 25000 at the rate of 12% p.a. for 3 years.

Solution:

Here P = ₹ 25000,

R = 12% p.a.,

T = 3 years

∴ Interest for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{25000 \times 12 \times 1}{100}\)

= ₹ 3000

Principal for the second year = ₹ 25000 + ₹ 3000

= ₹ 28000

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{28000 \times 12 \times 1}{100}\)

= ₹ 3360

Principal for the third year = ₹ (28000 + 3360) = ₹ 31360

Interest for the third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{31360 \times 12 \times 1}{100}\)

= ₹ 3763.20

∴ Amount at the end of third year = ₹ (31360 + 3763.20)

= ₹ 35123.20

∴ Compound interest = A – P

= ₹ 35123.20 – ₹ 25000

= ₹ 10123.20

Hence, the required C.I. = ₹ 10123.20.

Question 2.

Find the compound interest on ₹ 6500 for 2 years at 9% p.a.

Solution:

Here P = ₹ 6500,

T = 2 years,

R = 9% p.a.

∴ Interest for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{6500 \times 9 \times 1}{100}\)

= ₹ 585

Principal for the second year = ₹ (6500 + 585) = ₹ 7085

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{7085 \times 9 \times 1}{100}\)

= ₹ 637.65

Amount at the end of second year = ₹ (7085 + 637.65)

= ₹ 7722.65

∴ C.I. = A – P

= ₹ 7722.65 – ₹ 6500

= ₹ 1222.65

Hence, the compound interest = ₹ 1222.65.

Question 3.

Find the amount and compound interest on a sum of ₹ 8,000 at 5% per annum for 3 years compounded annually.

Solution:

Here P = ₹ 8000,

R = 5%,

T = 3 years

Interest for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{8000 \times 5 \times 1}{100}\)

= ₹ 400

Principal for the second year = ₹ 8000 + ₹ 400 = ₹ 8400

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{8400 \times 5 \times 1}{100}\)

= ₹ 420

Principal for the third year = ₹ 8400 + ₹ 420 = ₹ 8820

Interest for the third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{8820 \times 5 \times 1}{100}\)

= ₹ 441

Amount after 3 years = ₹ 8820 + ₹ 441

= ₹ 9261

Compound interest = A – P

= ₹ 9261 – ₹ 8000

= ₹ 1261

Hence, the amount = ₹ 9261

and C.I. = ₹ 1261.

Question 4.

Harvinder deposited ₹ 40,000 in a post office for a period of 3 years. The post office credits the interest yearly in his account at 7% p.a. compounded annually. Find the balance in his account after 3 years.

Solution:

Here P = ₹ 40,000,

R = 7% p.a.,

T = 3 years

Interest for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\) = \(\frac{40000 \times 7 \times 1}{100}\)

= ₹ 2800

Principal for the second year = ₹ 40000 + ₹ 2800 = ₹ 42800

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{42800 \times 7 \times 1}{100}\)

= ₹ 2996

Principal for the third year = ₹ 42800 + ₹ 2996 = ₹ 45796

Interest for the third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{45796 \times 7 \times 1}{100}\)

= ₹ 3205.72

Amount after 3 years = ₹ 45796 + ₹ 3205.72

= ₹ 49001.72

C.I. after 3 years = ₹ 49001.72 – ₹ 40000

= ₹ 9001.72

Hence, the balance in his account after 3 years = ₹ 49001.72

and C.I. = ₹ 9001.72.

Question 5.

Monika borrowed ₹ 4096 from Shalini for 3 years at 6\(\frac{1}{4}\) % p.a. Find the amount and the compound interest paid by her to Shalini after 3 years if the interest is compounded annually.

Solution:

Here P = ₹ 4096,

R = 6 \(\frac{1}{4}\) %, p.a. or \(\frac{25}{4}\)%

T = 3 years

Interest for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\) = \(\frac{4096 \times 25 \times 1}{100 \times 4}\)

= ₹ 256

Principal for the second year = ₹ 4096 + ₹ 256

= ₹ 4352

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{4352 \times 25 \times 1}{100 \times 4}\)

= ₹ 272

Principal for the third year = ₹ 4352 + ₹ 272

= ₹ 4624

Interest for the third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{4624 \times 25 \times 1}{100 \times 4}\)

= ₹ 289

∴ Amount after 3 years = ₹ 4624 + ₹ 289

= ₹ 4913

∴ C.I = A – P

= ₹ 4913 – ₹ 4096

= ₹ 817

Hence, A = ₹ 4913

and C.I. = ₹ 817.

Question 6.

Ravi purchased a house from DDA on credit. If the cost of the house is ₹ 7,50,000 and DDA charges interest at 6% p.a. compounded annually, find the interest paid by Ravi, if he makes payment to DDA after 3 years.

Solution:

Here P = ₹ 7,50,000,

R = 6%, p.a.,

T = 3 years

Interest for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{750000 \times 6 \times 1}{100}\)

= ₹ 45000

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{795000 \times 6 \times 1}{100}\)

= ₹ 47700

Interest for the third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{842700 \times 6 \times 1}{100}\)

= ₹ 50562

Amount after 3 years = ₹ 842700 + ₹ 50562

= ₹ 893262

C.I. for 3 years = A – P

= ₹ 893262 – ₹ 750000

= ₹ 1,43,262

Hence, the interest to be paid by Ravi = ₹ 1,43,262.

### DAV Class 8 Maths Chapter 6 Worksheet 1 Notes

The interest of first year calculated on the principal is added to the principal and the amount so obtained becomes the principal for second year. The amount at the end of second year becomes the principal for the third year and so on.

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

where P = Principal,

R = Rate of interest and

T = Time.

Compound interest, C.I. = A – P

= P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\) – P

= P [\(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\) – 1]

Note:

- If the interest is compounded half yearly, then the time is 2T and the rate is \(\frac{R}{2}\)%.
- If the interest is compounded quarterly, then the time is 4T and the rate is \(\frac{R}{4}\)%.
- If the interest is compounded monthly, then the time is 12T and the rate is \(\frac{R}{12}\)%.

**To Calculate the Compound Interest without Using the Formula**

Example 1.

Find the amount and the compound interest on 2500 for 2 years at 11% per annum.

Solution:

Here, P = 2500,

T = 2 years and

R = 11% p.a.

S.I. for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{2500 \times 11 \times 1}{100}\)

= ₹ 275

Principal for the second year = P + S.I.

= ₹ 2500 + ₹ 275

= ₹ 2775

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{2775 \times 11 \times 1}{100}\)

= ₹ 305.25

Amount after 2 years = ₹ 2775 + ₹ 305.25 = ₹ 3080.25

and the compound interest = ₹ 3080.25 – ₹ 2500 = ₹ 580.25.

Example 2.

Find the amount and compound interest on 12000 for 3 years at 20% per annum.

Solution:

Here P = ₹ 12000,

R = 20% p.a. and

T = 3 years.

S.I. for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{12000 \times 20 \times 1}{100}\)

= ₹ 2400

Principal for the second year = ₹ 12000 + ₹ 2400 = ₹ 14400

Interest for second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{14400 \times 20 \times 1}{100}\)

= ₹ 2880

Principal for the third year = ₹ 14400 + ₹ 2880 = ₹ 17280

Interest for third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{1728 \times 20 \times 1}{100}\)

= ₹ 3456

Amount at the end of third year = ₹ 17280 + ₹ 3456 = ₹ 20736

∴ Compound interest for 3 years = Amount – Principal

= ₹ 20736 – ₹ 12000 = ₹ 8736.

Example 3.

Find the amount and compound interest on Z 20,000 for 3 years at 9% per annum.

Solution:

Here P = ₹ 20,000,

R = 9% p.a.,

Time = 3 years

S.I. for the first year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{20,000 \times 9 \times 1}{100}\)

= ₹ 1800

Amount for the second year = ₹ 20,000 + ₹ 1800 = ₹ 21800

Interest for the second year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{21800 \times 9 \times 1}{100}\)

= ₹ 1962

Amount for the third year = ₹ 21800 + ₹ 1962 = ₹ 23762

Interest for third year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{23762 \times 9 \times 1}{100}\)

= ₹ 2138.58

Amount at the end of third year = ₹ 23762 + ₹ 2138.58 = ₹ 25900.58

∴ C.I. = A – P

= ₹ 25900.58 – ₹ 20,000 = ₹ 5900.58.

Example 4.

Find the compound interest on ₹ 5000 for 1\(\frac{1}{2}\) years at 16% per annum compounded half yearly.

Solution:

As the interest is compounded half yearly,

∴ Rate of interest, R = \(\frac{16}{2}\) = 8% and

time, T = 2 × 1\(\frac{1}{2}\) = 3 half years.

∴ Interest for the first half year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{5000 \times 8 \times 1}{100}\)

= ₹ 400

Principal for the second half year = ₹ 5000 + ₹ 400 = ₹ 5400

∴ Interest for the second half year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{5400 \times 8 \times 1}{100}\)

= ₹ 432

Principal for the third half year = ₹ 5400 + ₹ 432 = ₹ 5832

∴ Interest for the third half year = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{5832 \times 8 \times 1}{100}\)

= ₹ 466.56

Amount at the end of third half year = ₹ 5832 + ₹ 466.56 = ₹ 6298.56

∴ Compound Interest = ₹ 6298.56 – ₹ 5000 = ₹ 1298.56.

Example 5.

Find the compound interest on ₹ 15625 at 16% p.a. for 9 months when compounded quarterly.

Solution:

As the interest is compounded quarterly,

∴ T = \(\frac{19}{2}\) × 2 = 3 quarters and

R = \(\frac{16}{4}\) = 4%

∴ Interest for the first quarter = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{15625 \times 4 \times 1}{100}\)

= ₹ 625

Principal for the second quarter = ₹ 15625 + ₹ 625 = ₹ 16250

∴ Interest for the second quarter = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{16250 \times 4 \times 1}{100}\)

= ₹ 650

Principal for the third quarter = ₹ 16250 + ₹ 650 = ₹ 16900

∴ Interest for the third quarter = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{16900 \times 4 \times 1}{100}\)

= ₹ 676

Amount at the end of third quarter = ₹ 16900 + ₹ 676 = ₹ 17576

C.I. = A – P

= ₹ 17576 – ₹ 15625

= ₹ 1951.

**To Calculate the Compound Interest Using Formula**

Example 6.

Calculate the amount and the compound interest on a sum of ₹ 7500 at the end of 3 years at the rate of 4% p.a. compounded annually.

Solution:

Here P = ₹ 7500,

R = 4% p.a.,

T = ₹ 7500

∴ A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 7500 \(\left[1+\frac{4}{100}\right]^3\)

= 7500 \(\left[\frac{26}{25}\right]^3\)

= 8436.48

∴ C.I. = A – P

= ₹ 8436.48 – ₹ 7500 = ₹ 936.48

Hence amount = ₹ 8436.48 and

CI. = ₹ 936.48.

Example 7.

Find the amount and the compound interest on ₹ 3200 is 2 years, if the rate is 5% for the first year and 6% for the second year.

Solution:

Here P = ₹ 3200,

R_{1} = 5% for the first year and

R_{2} = 6 \(\frac{1}{4}\)% or \(\frac{25}{4}\)% for the second year

∴ A = P [1 + \(\frac{\mathrm{R}_1}{100}\)] [1 + \(\frac{\mathrm{R}_2}{100}\)]

= 3200 [1 + \(\frac{5}{100}\)] [1 + \(\frac{25}{4 \times 100}\)]

= 3200 × \(\frac{21}{20} \times \frac{17}{16}\)

= ₹ 3570

∴ C.I. = A – P

= ₹ 3570 – ₹ 3200 = ₹ 370

Hence, A = ₹ 3570 and C.I. = ₹ 370.

Example 8.

Find the difference between the compound interest and simple interest on a sum of ₹ 3375 for 2 years at the rate of 6% p.a.

Solution:

Here P = ₹ 3375,

R = 6\(\frac{2}{3}\) % or \(\frac{20}{3}\)% p.a.,

T = 2 years

∴ S.I.= \(\frac{P \times R \times T}{100}\)

= \(=\frac{3375 \times 20 \times 2}{100 \times 3}\)

= ₹ 450

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 3375 \(\left[1+\frac{20}{3 \times 100}\right]^2\)

= 3375 × \(\frac{16}{15} \times \frac{16}{15}\)

∴ C.I. = A – P

= ₹ 3840 – ₹ 3375

= ₹ 465

C.I. – S.I. = ₹ 465 – ₹ 450 = ₹ 15

Hence, the difference between C.I. and S.I. = ₹ 15.

Example 9.

What sum of money will amount to ₹ 3630 in 2 years at 10% p.a. compound interest?

Solution:

Here A = ₹ 3620,

T = 2 years,

R = 10% p.a.

∴ A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

3630 = P \(\left(1+\frac{10}{100}\right)^2\)

3630 = P \(\left(\frac{11}{10}\right)^2\)

3630 = P × \(\frac{11}{10} \times \frac{11}{10}\)

P = \(\frac{3630 \times 10 \times 10}{11 \times 11}\)

= ₹ 3000

Hence, the required principal = ₹ 3000.

Example 10.

At what rate per cent per annum compound interest will ₹ 6250 amount to ₹ 7290 in 2 years?

Solution:

Here P = ₹ 6250,

A = ₹ 7290,

T = 2 years

A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

⇒ 7290 = 6250 \(\left[1+\frac{\mathrm{R}}{100}\right]^2\)

⇒ \(\frac{7290}{6250}=\left[1+\frac{\mathrm{R}}{100}\right]^2\)

⇒ \(\left(\frac{27}{25}\right)^2=\left[1+\frac{\mathrm{R}}{100}\right]^2\)

⇒ \(\frac{27}{25}\) = 1 + \(\frac{R}{100}\)

⇒ \(\frac{R}{100}\) = \(\frac{27}{25}\) – 1

⇒ \(\frac{R}{100}=\frac{27}{25}\)

⇒ R = \(\frac{2 \times 100}{25}\)

⇒ R = 8%

Hence, the rate of interest is 8%.

Example 11.

The compound interest on ₹ 30000 at 7% p.a. for a certain period is ₹ 4347. Find the period for which the sum is invested.

Solution:

Here P = ₹ 30000,

C.I. = ₹ 4347,

R = 7% p.a.

∴ C.I. = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\) – P

⇒ 4347 = 300oo \(\left[1+\frac{7}{100}\right]^{\mathrm{T}}\) – P

⇒ 4347 = P \(\left[\left(1+\frac{7}{100}\right)^{\mathrm{T}}-1\right]\)

⇒ 4347 = 30000 \(\left[\left(\frac{107}{100}\right)^{\mathrm{T}}-1\right]\)

⇒ \(\frac{4347}{30000}=\left(\frac{107}{100}\right)^{\mathrm{T}}-1\)

⇒ \(\frac{4347}{30000}+1=\left(\frac{107}{100}\right)^{\mathrm{T}}\)

⇒ \(\frac{4347+30000}{30000}=\left(\frac{107}{100}\right)^{\mathrm{T}}\)

⇒ \(\frac{34347}{30000}=\left(\frac{107}{100}\right)^{\mathrm{T}}\)

⇒ \(\frac{11449}{10000}=\left(\frac{107}{100}\right)^{\mathrm{T}}\)

⇒ \(\left(\frac{107}{100}\right)^3=\left(\frac{107}{100}\right)^{\mathrm{T}}\)

∴ T = 3 years

Hence, the required time is 3 years.