Exercise 8B

Solution 1:

Amount=P(1+R/100)^n
=625(1+4/100)^2
=625*26/25*26/25
=676

CI=Amount-Principal
=676-625
=51

Solution 4:
Reena
Simple interest = principle × rate × time/100

Rate = 10%

Principle = 60000

Time = 3 years

Simple interest = 60000 × 3 × 0.1 = 18000

Rs 18000
Ruchira
Compound interest = Accumulated amount less the principle

Accumulated amount = P(1 + i)ⁿ

P = 50000

i = 10%

n = 3

A = 50000(1.1)³ = 66550

Compound interest = 66550 - 50000 = 16550

Rs 16550
Reena paid more interest by :
18000 - 16550 = 1450
Rs 1450

Solution 5:

P is 40960

R is 12.5% p.a semi annually

T is 1 1/2 years =3/2 years

A=p(1+r/200)^2n

=49130=amount

CI=A-P

=49130-40960

=8170Ans.

Solution 6:
Sudhir's Case

Principal = 2000
Rate = 10% p.a.
A = P ( 1 + r)n
                 100
A=  2000[ 1 + 10/100]1   =  2000 * 110/100
   =  2200

Prashant's Case
Principal = 2000
Rate = 10% p.a.  or  5% half - yearly
A = 2000 (1 + 5/100)2-no. of half years
   = 2000 * 105/100 * 105//100
   =2205
The Difference
2205 - 2200 = 5 rupees .
Hope this will help you .

Solution 7:

Case 1
Given : P = Rs. 16000, R = 5 %, T = 3/2 years
Simple Interest = (P*R*T)/100
= (16000*5*3)/(2*100)
Simple Interest = Rs. 1200
Case 2
Compound Interest
Given : P = Rs. 16000, R = 5 % per annum and compounded half yearly so, rate of interest = 2.5 %
T = 3/2 years = 1 year and one half year = 3 half years
A = P (1 +r/100)ⁿ
= 16000 (1 + 2.5/100)³
= 16000 × 102.5/100 × 102.5/100 × 102.5/100
A = Rs. 17230.25
So, compound interest = 17230.25 - 16000 = Rs. 1230.25
Difference between compound interest and simple interest = 1230.25 - 1200 = Rs 30.25
Answer.

Solution 8:
Principal (P)=10000.
Rate of compound interest (r)=12%.
No. Of years(n)=3years.

So Amount(A)=P[(1+r/100)^n].

=10000[1+12/100)^3].
= 10000[(112/100)^3].
= 10000*(112/100)*(112/100)*(112/100).
Upon solving this we get
A =14049.28
Interest = A- P.
=14049.28 - 10000.
= 4049.28Rs