Working with Fractions: questions with answers

Each “of” means multiplication. Multiply $\frac{2}{5}$, $\frac{1}{2}$, and $\frac{3}{4}$ to get $\frac{3}{20}$ of the original square. Adding gives $\frac{33}{20}$, which does not describe successive parts. Using $\frac{3}{4}$ alone omits two factors.

After the road takes $\frac{2}{7}$, the remaining land is $\frac{5}{7}$ of the original area. The daughter receives half of this remainder, which is $\frac{5}{14}$ of the original area. Taking half of the original whole would ignore the land already used for the road.

Multiplying the numerator and the denominator by the same number makes an equivalent fraction, so $\frac{14}{21}$ is $\frac{2}{3}$ written another way. Its value has not changed at all. To get the product $\frac{2}{3} \times 7$, this fraction must still be multiplied by $7$, which gives $\frac{14}{3}$.

Write $1\frac{3}{4}$ hours as $\frac{7}{4}$ hours. Multiplying $\frac{7}{4}$ by $8$ gives a cost of $14$ rupees. Both the whole and fractional parts of the time must be multiplied by the hourly cost. Leaving the fractional part unchanged gives $8\frac{3}{4}$ rupees, which charges too little for the extra part of an hour.

Each of the $9$ rows contains $4$ cells. Multiplying the number of rows by the number of cells in each row gives $36$ equal parts. Adding the row and column counts gives $13$, which does not count the cells.

Changing the order leaves the product of the numerators unchanged. It also leaves the product of the denominators unchanged. Therefore, $\frac{3}{8} \times \frac{5}{6}$ equals $\frac{5}{16}$.

Dividing each numerator by 4 divides their product by 16, since $4 \times 4 = 16$. The denominators stay unchanged. Multiply the new product by 16 to get the original product.

15 is a numerator and 9 is a denominator, so a factor shared by them may be cancelled against each other. The largest number that divides both 15 and 9 is 3. Divide both by 3 and then multiply; the product is unchanged. The two numbers do not have to be equal, as in the chapter's own step where 12 and 24 are divided by 12 to leave 1 and 2.

The starting number $4$ is positive. The multiplier $\frac{4}{3}$ is greater than 1, so the product is greater than $4$. Being written as a fraction does not mean the multiplier is less than 1.

Both factors are between 0 and 1, so their product is smaller than either factor. With a common denominator, the smallest numerator gives the smallest fraction. Since $12 < 15 < 32$, the product's numerator is $12$.

The reciprocal of $5$ is $\frac{1}{5}$. Multiply the dividend $\frac{3}{8}$ by this reciprocal to obtain $\frac{3}{40}$. The positive dividend becomes smaller because the divisor is greater than 1.

Multiply by the divisor’s reciprocal: $\frac{1}{3} \times \frac{5}{2}$. The result is $\frac{5}{6}$. Since the dividend is positive and the divisor is between 0 and 1, this quotient is larger than $\frac{1}{3}$.

Dividing by a fraction means multiplying by its reciprocal. $9 \div \frac{3}{5} = 9 \times \frac{5}{3} = 15$. Reversing the division gives $\frac{3}{5} \div 9 = \frac{3}{5} \times \frac{1}{ 9 } = \frac{1}{15}$. Since $\frac{1}{15} < 1 < 15$, $\frac{3}{5} \div 9 < 9 \div \frac{3}{5}$.

Interchange the numerator and denominator of $\frac{2}{7}$ to obtain $\frac{7}{2}$. Their product is 1, which checks the reciprocal. By contrast, $\frac{5}{7}$ adds to 1 with the given fraction; that is not the reciprocal.

To divide, multiply the dividend by the reciprocal of the divisor. The divisor $\frac{1}{2}$ is between 0 and 1, so its reciprocal is greater than 1. Multiplying the positive dividend $\frac{3}{8}$ by this reciprocal gives $\frac{3}{4}$, which is greater than the dividend.