**ML Aggarwal CBSE Solutions Class 8 Math 7th Chapter Compound Interest Exercise 7.1**

**(1) **A = P (1+ R/100)^{n}

Given P = 6000/-

R = 10%

N = 2 years

= 6000 (1+ 10/100)^{2}

= 6000 X (100+10/100)^{2}

= 6000 X (110/100)^{2}

= 6000 X 110/100 X 110/100

= 7260

∴ C.I. = A – P = 7260 – 6000

= 1260 /- (Ans)

**(4) **(i) P_{1} for 1 year = 46875 /-

R = 4%

T_{1} = 1 year

I for 1 year = P_{1}RT_{1}/100

= 46875 X 4 X 1/100

= 1875 /- (Ans)

(ii) P_{2} for 2 year = 46875 + 1875

= 48750 /-

R = 4 %

T_{1} = 2 years

I for 2^{nd} year = P_{2} R T_{2}/100

= 48750 X 4 X 1/100

= 1950

Amount after 2^{nd} year = 48750 + 1950

= 50700

(iii) P_{3} for 3^{rd} year = 50700 /-

R = 4%

T_{3} = 3 years

I for 3 years = P_{3} R T_{3}/100

= 50700 X 4 X 1/100 = 2028 /- (Ans)

**(5) **For 1^{st} year =>

P_{1} = 6000, R = 10% T_{1} = 1

∴ I, P_{1 }R T_{1}/100

= 6000 X 10 X 1/100 = 600/-

∴ A_{1} at the end of 1^{st} year = 6000 + 600

= 6600 /-

For 2^{nd} year à

P_{2} = 6600, R = 10%, T_{2} = 1

∴ I_{2} = P_{2}R T_{2}/100 = 6600 X 10 X 1/100

= 660/- (Ans)

A_{2} at the end of 2^{nd} year = 6600 + 660

= 7260/-

For, 3^{rd} year ->

P_{3} = 7260, R = 10% , T_{2} = 1

∴ I_{3} = P_{3} R T_{3 }/100

= 7260 X 10 X 1/100

= 726

∴ A_{3} at the end of 3^{rd} year = (7260+726)

= 7986 /- (Ans)

**(6) **For 1 year

P_{1} = 5000 /-

R_{1} = 6%

T_{1} = 1

∴ I_{2} = P_{1} R_{1} T_{1}/100 = 5000 X 6 X 1/100

= 300 /-

End of 1 year the amount will be =>

∴ A_{1} = 5000 + 300 = 5300 /-

For 2^{nd} year =

P_{2} = 5300 /-

R_{2} = 8%

T_{2} = 1 year

∴ I_{2} = P_{2} R_{2} T_{2}/100 = 5300 X 8 X 1/100

= 424 /-

End of 2 year the amount will be ->

∴ A_{2} = 5300 + 424

= 5724 /- (Ans)

∴ C.I = 5724 – 5000 = 724/- (Ans)

**(7) **For, simple interest ->

P = 20000/-, T = 2 years, R = 8%

∴ I = PRT/100 = 20000 X 2 X 8/100 = 3200/-

For compound interest ->

For 1 year –

P_{1} = 2000/-, R = 8%, T = 1 year

∴ I = PRT/100 = 20000 X 8 X 1/100 = 1600/-

∴ A_{1} after 1 year = 20000 + 1600 = 21600/-

For, 2^{nd} years –

P_{2} = 21600/- R = 8% T = 1 year

∴ I = PRT/100

= 26100 X 8 X 1/100

= 2088/-

A_{2} at after 2^{nd} year = 26100 + 2088

= 28188/-

∴ C.I. = 28188 – 20000 = 8188/-

∴ Difference between C.I & S.I

= 8188 – 3200 = 4988/- (Ans)