Question : What is the digit in the unit's place in the number $\frac{15!}{100}$?
Option 1: 5
Option 2: 7
Option 3: 3
Option 4: 0
Correct Answer: 0
Solution : Number of trailing zeroes in $n$! = [$\frac{n}{5}$] + [$\frac{n}{25}$] + [$\frac{n}{125}$] + ....... where [ ] denotes greatest integer function. So, the number of trailing zeroes in 15! = [$\frac{15}{5}$] + [$\frac{15}{25}$] + [$\frac{15}{125}$] + ....... = 3 + 0 + 0 + ..... = 3 ⇒ Number of zeroes in the product = 3 ⇒ Unit's digit in $\frac{15!}{100}$ = 0 Hence, the correct answer is 0.
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Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
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 : What should be the value in place of (?) in $7 \frac{5}{8}+\frac{5}{8}$ of $184 × 15 ÷ 5 - (?) = 0$.
Option 1: $-352 \frac{5}{8}$
Option 2: $152 \frac{1}{8}$
Option 3: $-152 \frac{1}{8}$
Option 4: $352 \frac{5}{8}$
Question : The value of $\frac{\frac{5}{2}-\frac{3}{7} \times 1 \frac{4}{5} \div 3 \frac{6}{7}}{\frac{3}{2}+1 \frac{2}{5} \div 3 \frac{1}{2} \times 1 \frac{1}{4}}$ is:
Option 1: $2 \frac{3}{20}$
Option 2: $1\frac{2}{20}$
Option 3: $1 \frac{3}{20}$
Option 4: $1 \frac{7}{20}$
Question : The greatest fraction among $\frac{2}{3}, \frac{5}{6}, \frac{11}{15} \text{ and } \frac{7}{8} \text{ is:}$
Option 1: $\frac{7}{8}$
Option 2: $\frac{11}{15}$
Option 3: $\frac{5}{6}$
Option 4: $\frac{2}{3}$
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