Simplify: $\left(\frac{3}{4} + \frac{1}{6}\right) \div \left(\frac{5}{8} - \frac{1}{4}\right)$

First evaluate the brackets. Left bracket: $\frac{3}{4} + \frac{1}{6}$. The LCM of $4$ and $6$ is $12$. So $\frac{3}{4} = \frac{9}{12}$ and $\frac{1}{6} = \frac{2}{12}$. Hence $\frac{9}{12} + \frac{2}{12} = \frac{11}{12}$. Right bracket: $\frac{5}{8} - \frac{1}{4}$. The LCM of $8$ and $4$ is $8$. So $\frac{1}{4} = \frac{2}{8}$. Hence $\frac{5}{8} - \frac{2}{8} = \frac{3}{8}$. Now divide: $\frac{11}{12} \div \frac{3}{8} = \frac{11}{12} \times \frac{8}{3}$. Cancel: $8$ and $12$ have a common factor $4$, giving $\frac{11}{3} \times \frac{2}{3} = \frac{22}{9}$. Convert to a mixed number: $\frac{22}{9} = 2\frac{4}{9}$. Answer: $2\frac{4}{9}$.

A shopkeeper has $360$ oranges. He sells $\frac{2}{5}$ of them in the morning and $\frac{1}{3}$ of the remainder in the afternoon. How many oranges are left unsold?

Total oranges $= 360$. Morning sales: $\frac{2}{5} \times 360 = \frac{720}{5} = 144$ oranges. Remainder after morning: $360 - 144 = 216$ oranges. Afternoon sales: $\frac{1}{3}$ of the remainder $= \frac{1}{3} \times 216 = 72$ oranges. Oranges left unsold: $216 - 72 = 144$. Answer: $144$ oranges.

Kamala spends $\frac{1}{4}$ of her salary on rent and $\frac{1}{3}$ of the remainder on food. If she has Rs. $9000$ left, what is her monthly salary?

Let the salary be $x$. Rent: $\frac{1}{4}x$. Remainder after rent: $x - \frac{1}{4}x = \frac{3}{4}x$. Food: $\frac{1}{3}$ of the remainder $= \frac{1}{3} \times \frac{3}{4}x = \frac{1}{4}x$. Total spent: $\frac{1}{4}x + \frac{1}{4}x = \frac{1}{2}x$. Money left: $x - \frac{1}{2}x = \frac{1}{2}x$. Given $\frac{1}{2}x = 9000$, so $x = 9000 \times 2 = 18000$. Answer: Rs. $18000$.

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විසඳුම් සහිතව භාග පිළිබඳ සා/පෙළ විභාග ප්‍රශ්න පුහුණු වන්න

Simplify: $\left(\frac{3}{4} + \frac{1}{6}\right) \div \left(\frac{5}{8} - \frac{1}{4}\right)$

විසඳුම බලන්න

First evaluate the brackets. Left bracket: $\frac{3}{4} + \frac{1}{6}$. The LCM of $4$ and $6$ is $12$. So $\frac{3}{4} = \frac{9}{12}$ and $\frac{1}{6} = \frac{2}{12}$. Hence $\frac{9}{12} + \frac{2}{12} = \frac{11}{12}$. Right bracket: $\frac{5}{8} - \frac{1}{4}$. The LCM of $8$ and $4$ is $8$. So $\frac{1}{4} = \frac{2}{8}$. Hence $\frac{5}{8} - \frac{2}{8} = \frac{3}{8}$. Now divide: $\frac{11}{12} \div \frac{3}{8} = \frac{11}{12} \times \frac{8}{3}$. Cancel: $8$ and $12$ have a common factor $4$, giving $\frac{11}{3} \times \frac{2}{3} = \frac{22}{9}$. Convert to a mixed number: $\frac{22}{9} = 2\frac{4}{9}$. Answer: $2\frac{4}{9}$.

ආදර්ශයට ගත්තේ: modelled on the MATHEMATICS OL_10 exam pattern
A shopkeeper has $360$ oranges. He sells $\frac{2}{5}$ of them in the morning and $\frac{1}{3}$ of the remainder in the afternoon. How many oranges are left unsold?

විසඳුම බලන්න

Total oranges $= 360$. Morning sales: $\frac{2}{5} \times 360 = \frac{720}{5} = 144$ oranges. Remainder after morning: $360 - 144 = 216$ oranges. Afternoon sales: $\frac{1}{3}$ of the remainder $= \frac{1}{3} \times 216 = 72$ oranges. Oranges left unsold: $216 - 72 = 144$. Answer: $144$ oranges.

ආදර්ශයට ගත්තේ: modelled on the MATHEMATICS OL_10 exam pattern
Kamala spends $\frac{1}{4}$ of her salary on rent and $\frac{1}{3}$ of the remainder on food. If she has Rs. $9000$ left, what is her monthly salary?

විසඳුම බලන්න

Let the salary be $x$. Rent: $\frac{1}{4}x$. Remainder after rent: $x - \frac{1}{4}x = \frac{3}{4}x$. Food: $\frac{1}{3}$ of the remainder $= \frac{1}{3} \times \frac{3}{4}x = \frac{1}{4}x$. Total spent: $\frac{1}{4}x + \frac{1}{4}x = \frac{1}{2}x$. Money left: $x - \frac{1}{2}x = \frac{1}{2}x$. Given $\frac{1}{2}x = 9000$, so $x = 9000 \times 2 = 18000$. Answer: Rs. $18000$.

ආදර්ශයට ගත්තේ: modelled on the MATHEMATICS OL_10 exam pattern

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