Question : $(1-\frac{1}{5})(1-\frac{1}{6})(1-\frac{1}{7}).......(1-\frac{1}{100})$ is equal to:
Option 1: $0$
Option 2: $\frac{1}{25}$
Option 3: $\frac{1}{100}$
Option 4: $\frac{1}{50}$

Question

Question : $(1-\frac{1}{5})(1-\frac{1}{6})(1-\frac{1}{7}).......(1-\frac{1}{100})$ is equal to:

Option 1: $0$

Option 2: $\frac{1}{25}$

Option 3: $\frac{1}{100}$

Option 4: $\frac{1}{50}$

Team Careers360 · Accepted Answer

Correct Answer: $\frac{1}{25}$

Solution : $(1-\frac{1}{5})(1-\frac{1}{6})(1-\frac{1}{7}).......(1-\frac{1}{100})$
= $(\frac{4}{5})(\frac{5}{6})(\frac{6}{7}).......(\frac{99}{100})$
Notice that the denominator of the first term cancels the numerator of the second term.
So, $(\frac{4}{5})(\frac{5}{6})(\frac{6}{7}).......(\frac{99}{100})$ = $\frac{4}{100}$ = $\frac{1}{25}$
Hence, the correct answer is $\frac{1}{25}$.

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Question : $(1 - \frac{1}{5}) (1 - \frac{1}{6}) (1 - \frac{1}{7}) . . . . . . . (1 - \frac{1}{100})$ is equal to:
Option 1: $0$
Option 2: $\frac{1}{25}$
Option 3: $\frac{1}{100}$
Option 4: $\frac{1}{50}$