ML Aggarwal CBSE Solutions Class 8 Math Seventh Chapter Compound Interest Exercise 7.2

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)

Updated: May 29, 2021 — 5:42 pm