ML Aggarwal CBSE Solutions Class 8 Math 7th Chapter Compound Interest Exercise 7.2
(2) For SI –
I – PRT/100 = 2500 X 4 X 2/100 = 200/-
For CI –
P = 2500, R = 4/2 = 2% [∵ semi annually]
Years = 2 = 24 month.
∴ n = 24/6 = 4
∴ A = p [1+ r/100]n
= 2500 [1+ 2/100]4
= 2500 [100+2/100]4
= 2500 X (102/100)4
= 2500 X 102/100 X 102/100 X 102/100 X 102/100
= 255X 102 X 102 X 102/100000
= 2706.08
∴ CI = A – P = 2706.08 – 2500
= 206.08 /-
∴ Difference between = CI & SI
= 206.08 – 200
= 60.8 /- (Ans)
(3) P = 3125 /-
R1 = 4%
R2 = 5%
R3 = 6%
∴ A = P (1+ R1/100) (1+ R2/100) (1+ R3/100)
= 3125 (1+ 4/100) (1+ 5/100) (1+ 6/100)
= 3125 (100+4/100) (100+5/100) (100+6/100)
= 3125 X 104/100 X 105/100 X 106/100
= 26 X 105 X 106/16
= 18,086.25 /-
∴ CI = A – P = (18086.25 – 3125.00) /-
= 14961.25 /-
(5) For simple interest –
P = 18000/-
R = 8%
T = 2 years
∴ SI = PRT/100
= 18000 X 8 X 2/100
∴ A = 18000 + 2880
= 20880 /-
For compound interest –
P = 18000 /-
R = 8%
N = 2 year
∴ A = P (1+ r/100)n
= 18000 (1+ 8/100)2
= 18000 (100+8/100)2
= 18000 X 108/100 X 108/100
= 20995.2 /-
∴ Amount difference = 20995.2 – 20880
= 115.20 /-
(6) Rohit borrows from Arun –
∴ P = 86000/-
R = 5%
T = 2 years
∴ SI = 86000 X 5 X 2/100
= 8600/-
Now, Rohit lend out in compoundedly
P = 86000/-
R = 5%
N = 2
∴ A = P (1+ r/100)n
= 86000 (1+ 5/100)2 = 86000 (100+5/100)2
= 86000 X (1005/100)2 = 86000 X 105/100 X 105/100
= 94815 /-
∴ CI = 94815 – 86000
= 8815 /-
∴ 8815 > 8600
∴ Profit of Rohit = 8815 – 8600
= 215 /-
(7) (i) For compounded annually –
P = 75000/-
R = 12%
N = 1 years
∴ A = P (1 + r/100)n
= 75000 (1+ 12/100)1
= 75000 (100+12/100)1
= 75000 (112/100)
= 75000 X 112/100
= 84000 /-
Now, next 1/2 years will be calculated
∴ P = 84000 r = 12% n = 1/2 year
∴ I = PRT/100 = 84000 X 12 X (1/2)/100 = 5040 /-
∴ A after 1 1/2 years = 84000 + 5040
= 89040 /- (Ans)
(7) (ii) compounded half yearly –
∴ P = 75000/-
R = 12/2 % = 6%
After 1 1/2 years (3/2 X 2)
∵ n = 3
∴ A = P [1+ r/100]n
= 75000 (1+ 6/100)3
= 75000 (100+6/100)3
= 75000 (106/100)3
= 75000 X 106/100 X 106/100 X 106/100
= 36,570 /- (Ans)
(8) (i) For 2 years –
P = 10000 /-
R = 7%
N = 2
∴ A = P (1+ r/100)n
= 10000 (1+ 7/100)2
= 10000 (100+7/100)2
= 10000 X 107/100 X 107/100
= 11449/- (Ans)
(ii) For 3 years
P = 10000/-
R = 7%
N = 3
∴ A = P (1+ r/100)n
= 10000 (1+ 7/100)3
= 10000 (100+7/100)3
= 10000 X (107/100)3
= 10000 X 107/100 X 107/100 X 107/100
= 12250.43/- (Ans)
(10) P = 2000/-
A = 2315.25/-
N = 3
R =?
According to qne –
A = P (1+ r/100)n
Or, 2315.25 = 2000 (100+r/100)3
Or, 231525/100 = 2000 X (100+r)3/100 X 100 X 100
Or, (100 + r)3 = 231525 X 100 X 100/2000
Or, (100 + r)3 = 1157625
Or, 100 + r = 3 √1157625
Or, 100 + r = 105
Or r = 105 – 100 = 5
∴ r = 5% (Ans)
(12) P = 152625 /-
A = 17576 /-
R = 4%
N = ?
According to qne –
A = P (1+ r/100)n
Or, 17576 = 15625 (1+ 4/100)n
Or, 17576/15625 = (100+4/100)n
Or, (26/25)3 = (104/100)n
Or, (26/25)3 = (26/25)n
Or, n = 3
∴ Time will be 3 years (Ans)
(11) P = 40000/-
A = 46305/-
N = 1 1/2 years
∵ half yearly compounded –
∴ n will be = 3/2 X 2 = 3 years
R =?
According to question –
A = P (1+ r/100)n
Or, 46305 = 40000 (100+r/100)3
Or, 46305/40000 = (100+r/100)3
Or, (21/20)3 = (100+r/100)3
Or, 100+r/100 = 21/200
Or, 2000+20r = 2100
Or, 20r = 2100 – 2000 = 100
Or, r = 100/20
Or, r = 5
∴ Interest rate = 5% (Ans)
(13) A = 18522
P = 16000
R = 10% = 10%/2 = 5%
N =?
According to qne –
A = P (1+ r/100)n
Or, 18522 = 16000 (100+5/100)n
Or, 18522/16000 = (105/100)n
Or, 1.157625 = (1.05)n
Or, (1.05)3 = (1.05)n
Or, n = 3
∵ Semi annually period will be completed
∴Time will be = 3/2
= 1.5 years (Ans)
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