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06 Aug 2024, 15:45
Kudos
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B
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Difficulty:
95%
(hard)
Question Stats:
41% (02:41) correct
59%
(02:33)
wrong
based on 64
sessions
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The ratio of the simple interest and compound interest, compounded annually, on a certain sum in 2 years is 12 : 13. If the ratio of the compound interest and the simple interest in three years is m : n, where m and n are co-primes, what is the value of (m + n)?
A. 220
B. 235
C. 255
D. 260
E. 270
A. 220
B. 235
C. 255
D. 260
E. 270
06 Aug 2024, 16:41
Kudos
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a = capital
r = interest rate (percentage/100)
In two years:
\(\frac{a + 2ar - a}{a(1+r)^2-a}=\frac{12}{13}\)
\(\frac{2ar}{a(1+2r+r^2)-a}=\frac{12}{13}\)
\(\frac{2r}{2r+r^2}=\frac{12}{13}\)
\(\frac{2}{2+r}=\frac{12}{13}\)
\(26=24+12r\)
\(r=\frac{1}{6}\)
In three years:
\(\frac{a + 3ar - a}{a(1+r)^3-a}=\frac{3r}{(1+r)^3-1}=\frac{3*\frac{1}{6}}{(1+\frac{1}{6})^3-1}=\frac{\frac{1}{2}}{(\frac{7}{6})^3-1}=\frac{\frac{1}{2}}{\frac{343}{216}-1}=\frac{\frac{1}{2}}{\frac{127}{216}}=\frac{108}{127}\)
\(m+n=108+127=235\)
IMO B
r = interest rate (percentage/100)
In two years:
\(\frac{a + 2ar - a}{a(1+r)^2-a}=\frac{12}{13}\)
\(\frac{2ar}{a(1+2r+r^2)-a}=\frac{12}{13}\)
\(\frac{2r}{2r+r^2}=\frac{12}{13}\)
\(\frac{2}{2+r}=\frac{12}{13}\)
\(26=24+12r\)
\(r=\frac{1}{6}\)
In three years:
\(\frac{a + 3ar - a}{a(1+r)^3-a}=\frac{3r}{(1+r)^3-1}=\frac{3*\frac{1}{6}}{(1+\frac{1}{6})^3-1}=\frac{\frac{1}{2}}{(\frac{7}{6})^3-1}=\frac{\frac{1}{2}}{\frac{343}{216}-1}=\frac{\frac{1}{2}}{\frac{127}{216}}=\frac{108}{127}\)
\(m+n=108+127=235\)
IMO B
02 Sep 2025, 01:40
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