find the difference between ci and si on 12000 an in 3÷2 at 10 % compounded half yearly
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Jahanvi004:
weong answer
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Heya ☺
Given that
P = 12,000
R = 10 % p.a.
T = 3/2 years
Solution
When the interest is compounded half yearly
A = P(1+R/100)^n
= 12,000(1+10/200)^3/2
= 12,000(200+10/200)^3/2 × 2
= 12,000(210/200)^3
= 12,000(21/20)^3
= 12,000 × 9,261/8,8000
= ₹13,891.50
C.I. = A - P
= ₹(13,891.50 - 12,000)
= ₹1,891.50
S.I. = P×R×T/100
= 12,000×10×3/100×2
= ₹1,800
Difference = C.I. - S.I.
= ₹(1,891.50 - 1,800)
= ₹91.50
Thanks
Given that
P = 12,000
R = 10 % p.a.
T = 3/2 years
Solution
When the interest is compounded half yearly
A = P(1+R/100)^n
= 12,000(1+10/200)^3/2
= 12,000(200+10/200)^3/2 × 2
= 12,000(210/200)^3
= 12,000(21/20)^3
= 12,000 × 9,261/8,8000
= ₹13,891.50
C.I. = A - P
= ₹(13,891.50 - 12,000)
= ₹1,891.50
S.I. = P×R×T/100
= 12,000×10×3/100×2
= ₹1,800
Difference = C.I. - S.I.
= ₹(1,891.50 - 1,800)
= ₹91.50
Thanks
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