Sharing this great and fun math and critical thinking puzzle with students is the best way to get them thinking mathematically.
Challenge:
There are 4 fractions in the figure above with the same difference. What is the sum of all fractions?
A-\(\frac{1}{126}\)
B- \(\frac{7}{33}\)
C- \(\frac{13}{42}\)
D- \(\frac{10}{66}\)
E- \(\frac{13}{21}\)
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The correct answer is E.
There are four fractions with the same difference. First, we need to find the missing fractions.
The difference of \(\frac{1}{7}\) and \(\frac{1}{6}\) is:
\(\frac{1}{6} – \frac{1}{7}/7 = \frac{7-6}{42} = \frac{1}{42}\)
The difference of \(\frac{1}{7}\) and the first fraction is:
\(\frac{1}{42} ÷3 = \frac{1}{126}\)
The first missing fraction is:
\(\frac{1}{7} + \frac{1}{126}= \frac{18+1}{126} = \frac{19}{126}\)
The second missing fraction is:
\(\frac{19}{126} + \frac{1}{126}= \frac{20}{126}\)
The sum of all fractions equals:
\(\frac{1}{7} + \frac{19}{126}+ \frac{20}{126} + \frac{1}{6} = \frac{18}{126}+ \frac{19}{126} + \frac{20}{126} + \frac{21}{126} = \frac{78}{126} = \frac{13}{21}\)
