Question : Raju ate $\frac{3}{8}$th part of a pizza and Adam ate $\frac{3}{10}$th part of the remaining pizza. Then Renu ate $\frac{4}{7}$th part of the pizza that was left. What fraction of the pizza is still left?
Option 1: $\frac{5}{12}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{8}$
Option 4: $\frac{3}{16}$
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Correct Answer: $\frac{3}{16}$
Solution : Let the total pizza be 1 part. Raju ate $\frac{3}{8}$th part of it. Remaining Part of the pizza $=(1 - \frac{3}{8})=\frac{5}{8}$ part Adam ate $\frac{3}{10}$th part of the remaining pizza. So, Adam ate $(\frac{5}{8} × \frac{3}{10}) = \frac{15}{80} = \frac{3}{16}$ part of the total pizza Total pizza is eaten $(\frac{3}{8} + \frac{3}{16})=\frac{9}{16}$ part Remaining pizza $=(1 - \frac{9}{16})=\frac{7}{16}$ part Then Renu ate $\frac{4}{7}$th part of the pizza. Renu ate $(\frac{7}{16} × \frac{4}{7}) = \frac{1}{4}$ part of the pizza Together they ate $(\frac{3}{8} + \frac{3}{16} + \frac{1}{4})=\frac{13}{16}$ part Final remaining pizza $=(1 - \frac{13}{16})=\frac{3}{16}$ part Hence, the correct answer is $\frac{3}{16}$.
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Question : The value of $15 \div 8-\frac{5}{4}$ of $\left(\frac{8}{3} \times \frac{9}{16}\right)+\left(\frac{9}{8} \times \frac{3}{4}\right)-\left(\frac{5}{32} \div \frac{5}{7}\right)+\frac{3}{8}$ is:
Question : Simplify the following expression. $\frac{7}{10} \div \frac{3}{7}$ of $\left(2 \frac{3}{10}+2 \frac{3}{5}\right)+\frac{1}{5} \div 1 \frac{2}{5}-\frac{2}{7}$
Question : The sum of three fractions is $2\frac{11}{24}$. On dividing the largest faction by the smallest fraction, $\frac{7}{6}$ is obtained which is $\frac{1}{3}$ greater than the middle fraction. The smallest fraction is:
Question : The value of $3 \frac{1}{5} \div 4 \frac{1}{2}$ of $5 \frac{1}{3}+\frac{1}{8} \div \frac{1}{2}$ of $\frac{1}{4}-\frac{1}{4}\left(\frac{1}{2} \div \frac{1}{8} \times \frac{1}{4}\right)$ is:
Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
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