Weight of Manoj is \(\frac{5}{7}\) times of weight of Sanju. By what fraction weight of Sanju is more than the weight of Manoj?
\(\frac{2}{5}\)
Let the weight of Sanju be S. Then the weight of Manoj is \(\frac{5}{7}\)S.
Sanju is heavier by S − \(\frac{5}{7}\)S = \(\frac{2}{7}\)S.
The question asks for this excess as a fraction of Manoj's weight, so divide by \(\frac{5}{7}\)S: (\(\frac{2}{7}\)S) ÷ (\(\frac{5}{7}\)S) = \(\frac{2}{5}\).
Hence, the answer is \(\frac{2}{5}\).
Similar Questions
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Important Questions from Fractions
5 \(\frac{3}{4}\) + x + 2 \(\frac{1}{2}\) = 10 \(\frac{1}{8}\) Find the value of x.
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Number 0.232323 can be written in rational form as:
Solve: \(\frac{1}{2}\) [{-2(2 + 3)*20}/2]
Match the following.
Column I
Column II
a.
Equivalent fraction of \(\frac{7}{12}\) is
i.
Proper fraction
b.
Equivalent fraction of \(\frac{9}{15}\) is
ii.
Improper fraction
c.
\(\frac{7}{11}\) is
iii.
\(\frac{21}{36}\)
d.
\(\frac{19}{5}\) is
iv.
\(\frac{3}{5}\)
