Question

Compound interest of 2 year is rs 156 and 3 year is rs 254. what is rate of interest?

Answer 1

C.I = P [(1+R/100)^n -1]

P[(1 + R/100)³ - 1] = 254 --- (I)
P[(1 + R/100)² - 1] = 156 --- (II)

Dividing eq I by II and Let q =( 1 + r/100)

(q³ - 1)/(q² - 1) = 254/156

Formula used
[ a³-b³= (a-b)(a²+b²+ab) & a²-b²= (a+b)(a-b)]

[(q - 1)(q² + q + 1)]/(q - 1)(q + 1) = 127/78

(q²+ q + 1)/(q + 1) = 127/78

78(q² + q + 1) = 127(q+1)

78q²+78q+78= 127q+127

78q²+ 78q-127q+78-127=0

78q²-49q-49=0

78q²-91q+42q-49=0

13q(6q-7)+7(6q-7)=0

(13q+7)(6q-7)=0


6q-7=0

q=7/6

q=(1+r/100)=7/6

r/100=7/6-1

r/100=( (7-6)/6)

r/100= 1/6

6r= 100

3r= 50

r= 50/3

r=16⅔%

Hence, the rate of interest (R) = 16⅔%

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Hope this will help you...