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) P1 for 1 year = 46875 /-
R = 4%
T1 = 1 year
I for 1 year = P1RT1/100
= 46875 X 4 X 1/100
= 1875 /- (Ans)
(ii) P2 for 2 year = 46875 + 1875
= 48750 /-
R = 4 %
T1 = 2 years
I for 2nd year = P2 R T2/100
= 48750 X 4 X 1/100
= 1950
Amount after 2nd year = 48750 + 1950
= 50700
(iii) P3 for 3rd year = 50700 /-
R = 4%
T3 = 3 years
I for 3 years = P3 R T3/100
= 50700 X 4 X 1/100 = 2028 /- (Ans)
(5) For 1st year =>
P1 = 6000, R = 10% T1 = 1
∴ I, P1 R T1/100
= 6000 X 10 X 1/100 = 600/-
∴ A1 at the end of 1st year = 6000 + 600
= 6600 /-
For 2nd year à
P2 = 6600, R = 10%, T2 = 1
∴ I2 = P2R T2/100 = 6600 X 10 X 1/100
= 660/- (Ans)
A2 at the end of 2nd year = 6600 + 660
= 7260/-
For, 3rd year ->
P3 = 7260, R = 10% , T2 = 1
∴ I3 = P3 R T3 /100
= 7260 X 10 X 1/100
= 726
∴ A3 at the end of 3rd year = (7260+726)
= 7986 /- (Ans)
(6) For 1 year
P1 = 5000 /-
R1 = 6%
T1 = 1
∴ I2 = P1 R1 T1/100 = 5000 X 6 X 1/100
= 300 /-
End of 1 year the amount will be =>
∴ A1 = 5000 + 300 = 5300 /-
For 2nd year =
P2 = 5300 /-
R2 = 8%
T2 = 1 year
∴ I2 = P2 R2 T2/100 = 5300 X 8 X 1/100
= 424 /-
End of 2 year the amount will be ->
∴ A2 = 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 –
P1 = 2000/-, R = 8%, T = 1 year
∴ I = PRT/100 = 20000 X 8 X 1/100 = 1600/-
∴ A1 after 1 year = 20000 + 1600 = 21600/-
For, 2nd years –
P2 = 21600/- R = 8% T = 1 year
∴ I = PRT/100
= 26100 X 8 X 1/100
= 2088/-
A2 at after 2nd year = 26100 + 2088
= 28188/-
∴ C.I. = 28188 – 20000 = 8188/-
∴ Difference between C.I & S.I
= 8188 – 3200 = 4988/- (Ans)
