Working with Fractions

Aaron's pet tortoise walks at a much slower pace. It can walk only $\dfrac{1}{4}$ kilometre in 1 hour. How far can it walk in 3 hours?

Answer

Given:

Distance covered by the tortoise in 1 hour = $\dfrac{1}{4}$ km

Distance covered in 3 hours = ?

Number of hours = 3

Distance covered in 3 hours = (Distance covered in 1 hour) × (Number of hours)

Substituting the values, we get:

$= \Big(3 \times \dfrac{1}{4}\Big)\text{ km} \\[1em] = \dfrac{3 \times 1}{4}\text{ km} \\[1em] = \dfrac{3}{4}\text{ km}$

∴ The tortoise can walk $\mathbf {\dfrac{3}{4}}$ km in 3 hours.

(i) We saw that Aaron can walk 3 kilometres in 1 hour. How far can he walk in $\dfrac{1}{5}$ hours?

(ii) How far can Aaron walk in $\dfrac{2}{5}$ hours?

Answer

Given:

Distance covered by Aaron in 1 hour = 3 km

Distance covered in $\dfrac{1}{5}$ hours = ?

Distance covered in $\dfrac{1}{5}$ hours = (Distance covered in 1 hour) × (Number of hours)

Substituting the values, we get:

$= \Big(3 \times \dfrac{1}{5}\Big)\text{ km} \\[1em] = \dfrac{3 \times 1}{5}\text{ km} \\[1em] = \dfrac{3}{5}\text{ km}$

∴ Aaron can walk $\mathbf {\dfrac{3}{5}}$ km in $\mathbf {\dfrac{1}{5}}$ hours.

(ii)

Given:

Distance covered by Aaron in 1 hour = 3 km

Distance covered in $\dfrac{2}{5}$ hours = ?

Distance covered in $\dfrac{2}{5}$ hours = (Distance covered in 1 hour) × (Number of hours)

Substituting the values, we get:

$= \Big(3 \times \dfrac{2}{5}\Big)\text{ km} \\[1em] = \dfrac{3 \times 2}{5}\text{ km} \\[1em] = \dfrac{6}{5}\text{ km} \\[1em] = 1\dfrac{1}{5}\text{ km}$

∴ Aaron can walk $\mathbf{\dfrac{6}{5}}$ km, that is $\mathbf {1\dfrac{1}{5}}$ km, in $\mathbf {\dfrac{2}{5}}$ hours.

Tenzin drinks $\dfrac{1}{2}$ glass of milk every day. How many glasses of milk does he drink in a week? How many glasses of milk did he drink in the month of January?

Answer

Given:

Milk Tenzin drinks each day = $\dfrac{1}{2}$ glass

First, we find the milk he drinks in a week.

Number of days in a week = 7

Milk in a week = (Milk each day) × (Number of days)

$= \Big(\dfrac{1}{2} \times 7\Big)\text{ glasses} \\[1em] = \dfrac{7}{2}\text{ glasses}$

Next, we find the milk he drinks in the month of January.

Number of days in January = 31

Milk in January = (Milk each day) × (Number of days)

$= \Big(\dfrac{1}{2} \times 31\Big)\text{ glasses} \\[1em] = \dfrac{31}{2}\text{ glasses}$

∴ Tenzin drinks $\mathbf {\dfrac{7}{2}}$ glasses of milk in a week and $\mathbf {\dfrac{31}{2}}$ glasses of milk in the month of January.

A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make .......... km of the water canal. If they work 5 days a week, they can make .......... km of the water canal in a week.

Answer

Given:

Length of canal made in 8 days = 1 km

First, we find the length of canal made in one day.

Canal made in 1 day = (Total length) ÷ (Number of days)

= $\dfrac{1}{8}\text{ km}$

So, in one day, the team can make $\dfrac{1}{8}$ km of the water canal.

Next, we find the length made in a week of 5 working days.

Canal made in a week = (Canal made in 1 day) × (Number of working days)

$= \Big(\dfrac{1}{8} \times 5 \Big)\text{ km} \\[1em] = \dfrac{5}{8}\text{ km}$

∴ In one day the team can make $\mathbf {\dfrac{1}{8}}$ km, and in a week (5 days) they can make $\mathbf {\dfrac{5}{8}}$ km of the water canal.

Manju and two of her neighbours buy 5 litres of oil every week and share it equally among the 3 families. How much oil does each family get in a week? How much oil will one family get in 4 weeks?

Answer

Given:

Oil bought every week = 5 litres

Number of families sharing equally = 3

First, we find the oil each family gets in a week.

Oil per family in a week = (Total oil) ÷ (Number of families)

= $\dfrac{5}{3}\text{ litres}$

Next, we find the oil one family gets in 4 weeks.

Oil in 4 weeks = (Oil per family in a week) × (Number of weeks)

$= \Big(\dfrac{5}{3} \times 4\Big)\text{ litres} \\[1em] = \dfrac{5 \times 4}{3}\text{ litres} \\[1em] = \dfrac{20}{3}\text{ litres}$

∴ Each family gets $\mathbf {\dfrac{5}{3}}$ litres of oil in a week, and one family gets $\mathbf {\dfrac{20}{3}}$ litres in 4 weeks.

Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets $\dfrac{5}{6}$ hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?

Answer

Given:

Each day the Moon sets $\dfrac{5}{6}$ hour later than the previous day.

The Moon was seen setting on Monday at 10 pm.

From Monday to Thursday = 3 days.

So the Moon sets $\dfrac{5}{6}$ hour later on each of the 3 successive days.

Total delay by Thursday = (Daily delay) × (Number of days)

$= \Big(\dfrac{5}{6} \times 3\Big)\text{ hours} \\[1em] = \dfrac{5 \times \overset{1}{\cancel{3}}}{\underset{2}{\cancel{6}}}\text{ hours} \\[1em] = \dfrac{5 \times 1}{2}\text{ hours} \\[1em] = \dfrac{5}{2}\text{ hours} \\[1em] = 2.5 \text{ hours}$

Thus, the Moon will set $2\dfrac{1}{2}$ hours after 10 pm, which is 12:30 am.

∴ The Moon will set $\mathbf {2\dfrac{1}{2}}$ hours after 10 pm, i.e., at 12:30 am.

Multiply and then convert it into a mixed fraction:

(a) $7 \times \dfrac{3}{5}$

(b) $4 \times \dfrac{1}{3}$

(c) $\dfrac{9}{7} \times 6$

(d) $\dfrac{13}{11} \times 6$

Answer

(a)

We have:

$\phantom{=} 7 \times \dfrac{3}{5} \\[1em] = \dfrac{7 \times 3}{5} \\[1em] = \dfrac{21}{5} \\[1em] = 4\dfrac{1}{5}$

∴ The answer is $\mathbf {4\dfrac{1}{5}}$.

(b)

We have:

$\phantom{=} 4 \times \dfrac{1}{3} \\[1em] = \dfrac{4 \times 1}{3} \\[1em] = \dfrac{4}{3} \\[1em] = 1\dfrac{1}{3}$

∴ The answer is $\mathbf{1\dfrac{1}{3}}$.

(c)

We have:

$\phantom{=} \dfrac{9}{7} \times 6 \\[1em] = \dfrac{9 \times 6}{7} \\[1em] = \dfrac{54}{7} \\[1em] = 7\dfrac{5}{7}$

∴ The answer is $\mathbf {7\dfrac{5}{7}}$.

(d)

We have:

$\phantom{=} \dfrac{13}{11} \times 6 \\[1em] = \dfrac{13 \times 6}{11} \\[1em] = \dfrac{78}{11} \\[1em] = 7\dfrac{1}{11}$

∴ The answer is $\mathbf {7\dfrac{1}{11}}$.

We know, that Aaron's pet tortoise can walk only $\dfrac{1}{4}$ km in 1 hour. How far can it walk in half an hour?

Answer

Given:

Distance covered by the tortoise in 1 hour = $\dfrac{1}{4}$ km

Distance covered in $\dfrac{1}{2}$ hour = ?

Distance covered in $\dfrac{1}{2}$ hour = (Distance covered in 1 hour) × (Number of hours)

Substituting the values, we get:

$= \Big(\dfrac{1}{4} \times \dfrac{1}{2}\Big)\text{ km} \\[1em] = \dfrac{1 \times 1}{4 \times 2}\text{ km} \\[1em] = \dfrac{1}{8}\text{ km}$

∴ The tortoise can walk $\mathbf {\dfrac{1}{8}}$ km in half an hour.

If the tortoise walks faster and it can cover $\dfrac{2}{5}$ km in 1 hour, how far will it walk in $\dfrac{3}{4}$ of an hour?

Answer

Given:

Distance covered by the tortoise in 1 hour = $\dfrac{2}{5}$ km

Distance covered in $\dfrac{3}{4}$ of an hour = ?

Distance covered in $\dfrac{3}{4}$ of an hour = (Distance covered in 1 hour) × (Number of hours)

Substituting the values, we get:

$= \Big(\dfrac{2}{5} \times \dfrac{3}{4}\Big)\text{ km} \\[1em] = \dfrac{2 \times 3}{5 \times 4}\text{ km} \\[1em] = \dfrac{\overset{1}{\cancel{2}} \times 3}{5 \times \underset{2}{\cancel{4}}}\text{ km} \\[1em] = \dfrac{1 \times 3}{5 \times 2}\text{ km} \\[1em] = \dfrac{3}{10}\text{ km}$

∴ The tortoise will walk $\mathbf {\dfrac{3}{10}}$ km in $\mathbf {\dfrac{3}{4}}$ of an hour.

Multiply $\dfrac{5}{4} \times \dfrac{3}{2}$.

Answer

We have:

$\dfrac{5}{4} \times \dfrac{3}{2}$

Taking the unit square as the whole, we first represent the multiplicand $\dfrac{3}{2}$ and divide it into 4 equal parts (the denominator of the multiplier). We then multiply the result by 5 (its numerator):

$\phantom{=} \dfrac{5}{4} \times \dfrac{3}{2} \\[1em] = 5 \times \dfrac{3}{2 \times 4} \quad \text{[Dividing } \dfrac{3}{2} \text{ into 4 equal parts]} \\[1em] = 5 \times \dfrac{3}{8} \\[1em] = \dfrac{5 \times 3}{8} \\[1em] = \dfrac{15}{8} \\[1em]$

∴ The answer is $\mathbf {\dfrac{15}{8}}$

Length and breadth of a rectangle are $\dfrac{1}{2}$ unit and $\dfrac{1}{4}$ unit respectively. Its area is $\dfrac{1}{8}$ sq units.
Do you see any relation between the area and the product of length and breadth?

Answer

Given:

Length = $\dfrac{1}{2}$ unit, Breadth = $\dfrac{1}{4}$ unit and Area = $\dfrac{1}{8}$ sq units

Yes. When a rectangle is drawn inside the unit square with its length and breadth taken as two fractions, the area of that rectangle is exactly equal to the product of those two fractions.

For example, the shaded rectangle formed with length $\dfrac{1}{2}$ unit and breadth $\dfrac{1}{4}$ unit has an area of $\dfrac{1}{8}$ square unit, which is the same as $\dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}$.

∴ The area of a rectangle with fractional sides equals the product of its sides. So, to find the product of two fractions, we can find the area of the rectangle formed with the two fractions as its sides.

Find the following products. Use a unit square as a whole for representing the fractions:

(a) $\dfrac{1}{3} \times \dfrac{1}{5}$

(b) $\dfrac{1}{4} \times \dfrac{1}{3}$

(c) $\dfrac{1}{5} \times \dfrac{1}{2}$

(d) $\dfrac{1}{6} \times \dfrac{1}{5}$

Now, find $\dfrac{1}{12} \times \dfrac{1}{18}$.

Answer

Taking the unit square as the whole and dividing it into rows and columns equal to the two denominators, the whole is split into (product of the denominators) equal parts, of which exactly one part is shaded. This gives the rule:

$\dfrac{1}{b} \times \dfrac{1}{d} = \dfrac{1}{b \times d}$

(a) $\dfrac{1}{3} \times \dfrac{1}{5}$

The unit square is divided into $3 \times 5 = 15$ equal parts, of which 1 part is shaded.

$\dfrac{1}{3} \times \dfrac{1}{5} = \dfrac{1}{3 \times 5} = \dfrac{1}{15}$

∴ The answer is $\mathbf {\dfrac{1}{15}}$

(b) $\dfrac{1}{4} \times \dfrac{1}{3}$

The unit square is divided into 4 × 3 = 12 equal parts, of which 1 part is shaded.

$\dfrac{1}{4} \times \dfrac{1}{3} = \dfrac{1}{4 \times 3} = \dfrac{1}{12}$

∴ The answer is $\mathbf {\dfrac{1}{12}}$.

(c) $\dfrac{1}{5} \times \dfrac{1}{2}$

The unit square is divided into 5 × 2 = 10 equal parts, of which 1 part is shaded.

$\dfrac{1}{5} \times \dfrac{1}{2} = \dfrac{1}{5 \times 2} = \dfrac{1}{10}$

∴ The answer is $\mathbf {\dfrac{1}{10}}$.

(d) $\dfrac{1}{6} \times \dfrac{1}{5}$

The unit square is divided into 6 × 5 = 30 equal parts, of which 1 part is shaded.

$\dfrac{1}{6} \times \dfrac{1}{5} = \dfrac{1}{6 \times 5} = \dfrac{1}{30}$

∴ The answer is $\mathbf {\dfrac{1}{30}}$.

Now, drawing the unit square for $\dfrac{1}{12} \times \dfrac{1}{18}$ is cumbersome, so we apply the same rule directly. The whole would be divided into 12 × 18 equal parts, with 1 part shaded:

$\dfrac{1}{12} \times \dfrac{1}{18} = \dfrac{1}{12 \times 18} = \dfrac{1}{216}$

∴ The answer is $\mathbf {\dfrac{1}{216}}$.

Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations.

(a) $\dfrac{2}{3} \times \dfrac{4}{5}$

(b) $\dfrac{1}{4} \times \dfrac{2}{3}$

(c) $\dfrac{3}{5} \times \dfrac{1}{2}$

(d) $\dfrac{4}{6} \times \dfrac{3}{5}$

Answer

Taking the unit square as the whole, we divide it into rows equal to the denominator of the multiplicand and columns equal to the denominator of the multiplier. The number of shaded parts is the product of the numerators. This gives the general rule:

$\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$

(a) $\dfrac{2}{3} \times \dfrac{4}{5}$

The whole is divided into 5 rows and 3 columns, creating 5 × 3 = 15 equal parts, of which 4 × 2 = 8 parts are shaded.

$\phantom{=} \dfrac{2}{3} \times \dfrac{4}{5} \\[1em] = \dfrac{2 \times 4}{3 \times 5} \\[1em] = \dfrac{8}{15}$

∴ The answer is $\mathbf {\dfrac{8}{15}}$.

(b) $\dfrac{1}{4} \times \dfrac{2}{3}$

The whole is divided into 3 rows and 4 columns, creating 3 × 4 = 12 equal parts, of which 2 × 1 = 2 parts are shaded.

$\phantom{=} \dfrac{1}{4} \times \dfrac{2}{3} \\[1em] = \dfrac{1 \times 2}{4 \times 3} \\[1em] = \dfrac{1 \times \overset{1}{\cancel{2}}}{\underset{2}{\cancel{4}} \times 3} \\[1em] = \dfrac{1 \times 1}{2 \times 3} \\[1em] = \dfrac{1}{6}$

∴ The answer is $\mathbf {\dfrac{2}{12} = \dfrac{1}{6}}$.

(c) $\dfrac{3}{5} \times \dfrac{1}{2}$

The whole is divided into 2 rows and 5 columns, creating 2 × 5 = 10 equal parts, of which 1 × 3 = 3 parts are shaded.

$\phantom{=} \dfrac{3}{5} \times \dfrac{1}{2} \\[1em] = \dfrac{3 \times 1}{5 \times 2} \\[1em] = \dfrac{3}{10}$

∴ The answer is $\mathbf {\dfrac{3}{10}}$.

(d) $\dfrac{4}{6} \times \dfrac{3}{5}$

The whole is divided into 5 rows and 6 columns, creating 5 × 6 = 30 equal parts, of which 3 × 4 = 12 parts are shaded.

$\phantom{=} \dfrac{4}{6} \times \dfrac{3}{5} \\[1em] = \dfrac{4 \times 3}{6 \times 5} \\[1em] = \dfrac{4 \times \overset{1}{\cancel{3}}}{\underset{2}{\cancel{6}} \times 5} \\[1em] = \dfrac{\overset{2}{\cancel{4}} \times 1}{\cancel{2} \times 5} \\[1em] = \dfrac{2 \times 1}{1 \times 5} \\[1em] = \dfrac{2}{5}$

∴ The answer is $\mathbf {\dfrac{12}{30} = \dfrac{2}{5}}$.

Multiply the following fractions and express the product in its lowest form:

$\dfrac{12}{7} \times \dfrac{5}{24}$

Answer

We have:

$\phantom{=} \dfrac{12}{7} \times \dfrac{5}{24} \\[1em] = \dfrac{12 \times 5}{7 \times 24} \\[1em] = \dfrac{\overset{1}{\cancel{12}} \times 5}{7 \times \underset{2}{\cancel{24}}} \\[1em] = \dfrac{1 \times 5}{7 \times 2} \\[1em] = \dfrac{5}{14}$

∴ The answer is $\mathbf {\dfrac{5}{14}}$.

A water tank is filled from a tap. If the tap is open for 1 hour, $\dfrac{7}{10}$ of the tank gets filled. How much of the tank is filled if the tap is open for

(a) $\dfrac{1}{3}$ hour ...............

(b) $\dfrac{2}{3}$ hour ...............

(c) $\dfrac{3}{4}$ hour ...............

(d) $\dfrac{7}{10}$ hour ...............

(e) For the tank to be full, how long should the tap be running?

Answer

(a) In $\dfrac{1}{3}$ hour:

Given:

Part of the tank filled in 1 hour = $\dfrac{7}{10}$

Part of the tank filled in $\dfrac{1}{3}$ hour = (Part filled in 1 hour) × (Number of hours)

$= \dfrac{7}{10} \times \dfrac{1}{3} \\[1em] = \dfrac{7 \times 1}{10 \times 3} \\[1em] = \dfrac{7}{30}$

∴ $\mathbf {\dfrac{7}{30}}$ of the tank is filled.

(b) In $\dfrac{2}{3}$ hour:

Given:

Part of the tank filled in 1 hour = $\dfrac{7}{10}$

Part of the tank filled in $\dfrac{2}{3}$ hour = (Part filled in 1 hour) × (Number of hours)

$= \dfrac{7}{10} \times \dfrac{2}{3} \\[1em] = \dfrac{7 \times 2}{10 \times 3} \\[1em] = \dfrac{7 \times \overset{1}{\cancel{2}}}{\underset{5}{\cancel{10}} \times 3} \\[1em] = \dfrac{7 \times 1}{5 \times 3} \\[1em] = \dfrac{7}{15}$

∴ $\mathbf {\dfrac{7}{15}}$ of the tank is filled.

(c) In $\dfrac{3}{4}$ hour:

Given:

Part of the tank filled in 1 hour = $\dfrac{7}{10}$

Part of the tank filled in $\dfrac{3}{4}$ hour = (Part filled in 1 hour) × (Number of hours)

$= \dfrac{7}{10} \times \dfrac{3}{4} \\[1em] = \dfrac{7 \times 3}{10 \times 4} \\[1em] = \dfrac{21}{40}$

∴ $\mathbf {\dfrac{21}{40}}$ of the tank is filled.

(d) In $\dfrac{7}{10}$ hour:

Given:

Part of the tank filled in 1 hour = $\dfrac{7}{10}$

Part of the tank filled in $\dfrac{7}{10}$ hour = (Part filled in 1 hour) × (Number of hours)

$\phantom{=} \dfrac{7}{10} \times \dfrac{7}{10} \\[1em] = \dfrac{7 \times 7}{10 \times 10} \\[1em] = \dfrac{49}{100}$

∴ $\mathbf {\dfrac{49}{100}}$ of the tank is filled.

(e) For the tank to be full, 1 whole tank must be filled. Since $\dfrac{7}{10}$ of the tank is filled in 1 hour, the time needed to fill the whole tank is:

$\phantom{=} 1 \div \dfrac{7}{10} \\[1em] = 1 \times \dfrac{10}{7} \\[1em] \text{[Dividing by a fraction = multiplying by its reciprocal]} \\[1em] = \dfrac{10}{7} \\[1em]$

∴ The tap should run for $\mathbf {\dfrac{10}{7}}$ hours for the tank to be full.

The government has taken $\dfrac{1}{6}$ of Somu's land to build a road. What part of the land remains with Somu now? She gives half of the remaining part of the land to her daughter Krishna and $\dfrac{1}{3}$ of it to her son Bora. After giving them their shares, she keeps the remaining land for herself.

(a) What part of the original land did Krishna get?

(b) What part of the original land did Bora get?

(c) What part of the original land did Somu keep for herself?

Answer

Given:

Part of the land taken for the road = $\dfrac{1}{6}$

Taking the whole land as 1, the part remaining with Somu is:

$\phantom{=} 1 - \dfrac{1}{6} \\[1em] = \dfrac{6}{6} - \dfrac{1}{6} \\[1em] = \dfrac{5}{6}$

∴ $\mathbf {\dfrac{5}{6}}$ of the original land remains with Somu.

(a) Krishna gets half of the remaining part:

$\phantom{=} \dfrac{1}{2} \times \dfrac{5}{6} \\[1em] = \dfrac{1 \times 5}{2 \times 6} \\[1em] = \dfrac{5}{12}$

∴ Krishna got $\mathbf {\dfrac{5}{12}}$ of the original land.

(b) Bora gets $\dfrac{1}{3}$ of the remaining part:

$\phantom{=} \dfrac{1}{3} \times \dfrac{5}{6} \\[1em] = \dfrac{1 \times 5}{3 \times 6} \\[1em] = \dfrac{5}{18}$

∴ Bora got $\mathbf {\dfrac{5}{18}}$ of the original land.

(c) The part Somu keeps for herself is the remaining land after Krishna's and Bora's shares:

$\phantom{=} \dfrac{5}{6} - \dfrac{5}{12} - \dfrac{5}{18} \\[1em] = \dfrac{30}{36} - \dfrac{15}{36} - \dfrac{10}{36} \\[1em] \text{[Taking LCM of 6, 12 and 18 as 36]} \\[1em] = \dfrac{30 - 15 - 10}{36} \\[1em] = \dfrac{5}{36}$

∴ Somu kept $\mathbf {\dfrac{5}{36}}$ of the original land for herself.

Find the area of a rectangle of sides $3\dfrac{3}{4}$ ft and $9\dfrac{3}{5}$ ft.

Answer

Given:

Length of the rectangle = $9\dfrac{3}{5}$ ft = $\dfrac{48}{5}$ ft

Breadth of the rectangle = $3\dfrac{3}{4}$ ft = $\dfrac{15}{4}$ ft

Area of a rectangle = Length × Breadth

Substituting the values, we get:

$= \Big(\dfrac{15}{4} \times \dfrac{48}{5}\Big)\text{ sq ft} \\[1em] = \Big(\dfrac{15 \times 48}{4 \times 5}\Big)\text{ sq ft} \\[1em] = \Big(\dfrac{\overset{3}{\cancel{15}} \times 48}{4 \times \cancel{5}}\Big) \\[1em] = \Big(\dfrac{3 \times \overset{12}{\cancel{48}}}{\cancel{4} \times 1}\Big) \\[1em] = \Big(\dfrac{3 \times 12}{1 \times 1}\Big)\text{ sq ft} \\[1em] = 36\text{ sq ft}$

∴ The area of the rectangle = 36 sq ft.

Tsewang plants four saplings in a row in his garden. The distance between two saplings is $\dfrac{3}{4}$ m. Find the distance between the first and last sapling. [Hint: Draw a rough diagram with four saplings with distance between two saplings as $\dfrac{3}{4}$ m]

Answer

With 4 saplings in a row, there are 3 equal gaps between the first and the last sapling. Since each gap measures $\dfrac{3}{4}$ m, the total distance is:

$\phantom{=} 3 \times \dfrac{3}{4} \\[1em] = \dfrac{3 \times 3}{4} \\[1em] = \dfrac{9}{4} \\[1em] = 2\dfrac{1}{4}\text{ m}$

∴ The distance between the first and the last sapling = $\mathbf {2\dfrac{1}{4}}$ m.

Which is heavier: $\dfrac{12}{15}$ of 500 grams or $\dfrac{3}{20}$ of 4 kg?

Answer

We find the value of each quantity and then compare them.

First quantity — $\dfrac{12}{15}$ of 500 grams:

$\phantom{=} \dfrac{12}{15} \times 500 \text{ g} \\[1em] = \dfrac{12 \times 500}{15} \text{ g} \\[1em] = 12 \times \dfrac{\overset{100}{\cancel{500}}}{\underset{3}{\cancel{15}}} \text{ g} \\[1em] = 12 \times \dfrac{100}{3} \text{ g} \\[1em] = \dfrac{1200}{3} \text{ g} \\[1em] = 400 \text{ g}$

Second quantity — $\dfrac{3}{20}$ of 4 kg.

Converting 4 kg to grams, 4 kg = 4000 g:

$\phantom{=} \dfrac{3}{20} \times 4000 \text{ g} \\[1em] = \dfrac{3 \times \overset{200}{\cancel{4000}}}{\cancel{20}} \text{ g} \\[1em] = 3 \times 200 \text{ g} \\[1em] = 600 \text{ g}$

Comparing the two: 600 g > 400 g.

∴ $\mathbf {\dfrac{3}{20}}$ of 4 kg (600 g) is heavier than $\mathbf {\dfrac{12}{15}}$ of 500 grams (400 g).

What can you conclude about the relationship between the numbers multiplied and the product? Fill in the blanks:

• When one of the numbers being multiplied is between 0 and 1, the product is ............... (greater/less) than the other number.
• When one of the numbers being multiplied is greater than 1, the product is ............... (greater/less) than the other number.

Answer

• When one of the numbers being multiplied is between 0 and 1, the product is less than the other number.
• When one of the numbers being multiplied is greater than 1, the product is greater than the other number.

In each of the figures given below, find the fraction of the big square that the shaded region occupies.

Answer

First figure.

Take the area of the big square to be 1. The diagonal from the top-left corner to the bottom-right corner divides the square into two equal triangles. Therefore, the area of one triangular half is $\dfrac{1}{2}$.

In this half, the small unshaded triangle at the bottom-left corner has base and height equal to half the side of the big square. Hence, its area is:

$\phantom{=} \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{1}{8}$

Therefore, the shaded strip occupies:

$\phantom{=} \dfrac{1}{2} - \dfrac{1}{8} \\[1em] = \dfrac{4}{8} - \dfrac{1}{8} \\[1em] = \dfrac{3}{8}$

∴ In the first figure the shaded region occupies $\mathbf {\dfrac{3}{8}}$ of the big square.

Second figure.

Again take the area of the big square to be 1. The square is divided into four equal quadrants, so each quadrant is $\dfrac{1}{4}$ of the big square. All of the shading lies inside the top‑left quadrant, where a tilted square (a diamond joining the mid‑points of the quadrant) has been drawn.

The tilted square occupies half of that quadrant, and the four corner triangles left over make up the other half — so the four corner triangles together are $\dfrac{1}{2}$ of the quadrant, and each corner triangle is $\dfrac{1}{8}$ of the quadrant. The two shaded corner triangles (on the right) therefore make up:

$\phantom{=} 2 \times \dfrac{1}{8} \text{ of the quadrant} \\[1em] = \dfrac{1}{4} \text{ of the quadrant} \\[1em] = \dfrac{1}{4} \times \dfrac{1}{4} \\[1em] \text{[the quadrant is } \dfrac{1}{4} \text{ of the big square]} \\[1em] = \dfrac{1}{16}$

∴ In the second figure the shaded region occupies $\mathbf {\dfrac{1}{16}}$ of the big square.

If we assume 1 gold dinar = 12 silver drammas, 1 silver dramma = 4 copper panas, 1 copper pana = 6 mashakas, and 1 pana = 30 cowrie shells, and $1 \text{ copper pana} = \dfrac{1}{48} \text{ gold dinar} \left(\dfrac{1}{12} \times \dfrac{1}{4}\right)$, fill in the blanks:

1 cowrie shell = .......... copper panas

1 cowrie shell = .......... gold dinar

Answer

We are told that 1 copper pana = 30 cowrie shells.

Turning this around, one cowrie shell is one of those 30 equal parts of a copper pana:

1 cowrie shell

= $\dfrac{1}{30} \text{ copper panas}$

To change this into gold dinars, we use 1 copper pana = $\dfrac{1}{48}$ gold dinar:

1 cowrie shell

$= \dfrac{1}{30} \text{ copper panas} \\[1em] = \dfrac{1}{30} \times \dfrac{1}{48} \text{ gold dinar} \\[1em] \quad \text{[each copper pana } = \dfrac{1}{48} \text{ gold dinar]} \\[1em] = \dfrac{1}{1440} \text{ gold dinar}$

This agrees with counting directly: 1 gold dinar = 12 × 4 × 30 = 1440 cowrie shells.

∴ 1 cowrie shell = $\mathbf {\dfrac{1}{30}}$ copper panas, and 1 cowrie shell = $\mathbf {\dfrac{1}{1440}}$ gold dinar.

Evaluate the following:

Answer

To divide by a fraction we multiply by its reciprocal.

$\phantom{=} 3 \div \dfrac{7}{9} \\[1em] = 3 \times \dfrac{9}{7} \quad \text{[reciprocal of } \dfrac{7}{9} \text{ ⇒ } \dfrac{9}{7}\text{]} \\[1em] = \dfrac{27}{7}$

∴ $\mathbf {3 \div \dfrac{7}{9} = \dfrac{27}{7}}$

$\phantom{=} \dfrac{14}{4} \div 2 \\[1em] = \dfrac{14}{4} \times \dfrac{1}{2} \quad \text{[reciprocal of } 2 \text{ ⇒ } \dfrac{1}{2}\text{]} \\[1em] = \dfrac{14}{8} \\[1em] = \dfrac{7}{4} \\[1em]$

∴ $\mathbf {\dfrac{14}{4} \div 2 = \dfrac{7}{4}}$

$\phantom{=} \dfrac{2}{3} \div \dfrac{2}{3} \\[1em] = \dfrac{2}{3} \times \dfrac{3}{2} \quad \text{[reciprocal of } \dfrac{2}{3} \text{ ⇒ } \dfrac{3}{2}\text{]} \\[1em] = \dfrac{6}{6} \\[1em] = 1$

∴ $\mathbf {\dfrac{2}{3} \div \dfrac{2}{3} = 1}$

$\phantom{=} \dfrac{14}{6} \div \dfrac{7}{3} \\[1em] = \dfrac{14}{6} \times \dfrac{3}{7} \quad \text{[reciprocal of } \dfrac{7}{3} \text{ ⇒ } \dfrac{3}{7}\text{]} \\[1em] = \dfrac{\overset{2}{\cancel{14}}}{6} \times \dfrac{3}{\cancel{7}} \\[1em] = \dfrac{2}{\underset{2}{\cancel{6}}} \times \dfrac{\overset{1}{\cancel{3}}}{1} \\[1em] = \dfrac{2}{2} \times \dfrac{1}{1} \\[1em] = 1$

∴ $\mathbf {\dfrac{14}{6} \div \dfrac{7}{3} = 1}$

$\phantom{=} \dfrac{4}{3} \div \dfrac{3}{4} \\[1em] = \dfrac{4}{3} \times \dfrac{4}{3} \quad \text{[reciprocal of } \dfrac{3}{4} \text{ ⇒ } \dfrac{4}{3}\text{]} \\[1em] = \dfrac{16}{9} \\[1em]$

∴ $\mathbf {\dfrac{4}{3} \div \dfrac{3}{4} = \dfrac{16}{9}}$

$\phantom{=} \dfrac{7}{4} \div \dfrac{1}{7} \\[1em] = \dfrac{7}{4} \times \dfrac{7}{1} \quad \text{[reciprocal of } \dfrac{1}{7} \text{ ⇒ } 7\text{]} \\[1em] = \dfrac{49}{4}$

∴ $\mathbf {\dfrac{7}{4} \div \dfrac{1}{7} = \dfrac{49}{4}}$

$\phantom{=} \dfrac{8}{2} \div \dfrac{4}{15} \\[1em] = \dfrac{8}{2} \times \dfrac{15}{4} \quad \text{[reciprocal of } \dfrac{4}{15} \text{ ⇒ } \dfrac{15}{4}\text{]} \\[1em] = 4 \times \dfrac{15}{4} \quad \text{[} \dfrac{8}{2} = 4 \text{]} \\[1em] = 15$

∴ $\mathbf {\dfrac{8}{2} \div \dfrac{4}{15} = 15}$

$\phantom{=} \dfrac{1}{5} \div \dfrac{1}{9} \\[1em] = \dfrac{1}{5} \times \dfrac{9}{1} \quad \text{[reciprocal of } \dfrac{1}{9} \text{ ⇒ } 9\text{]} \\[1em] = \dfrac{9}{5}$

∴ $\mathbf {\dfrac{1}{5} \div \dfrac{1}{9} = \dfrac{9}{5}}$

$\phantom{=} \dfrac{1}{6} \div \dfrac{11}{12} \\[1em] = \dfrac{1}{6} \times \dfrac{12}{11} \quad \text{[reciprocal of } \dfrac{11}{12} \text{ ⇒ } \dfrac{12}{11}\text{]} \\[1em] = \dfrac{1}{\cancel{6}} \times \dfrac{\overset{2}{\cancel{12}}}{11} \\[1em] = \dfrac{1}{1} \times \dfrac{2}{11} \\[1em] = \dfrac{2}{11}$

∴ $\mathbf {\dfrac{1}{6} \div \dfrac{11}{12} = \dfrac{2}{11}}$

$\phantom{=} 3\dfrac{2}{3} \div 1\dfrac{3}{8} \\[1em] = \dfrac{11}{3} \div \dfrac{11}{8} \\[1em] = \dfrac{11}{3} \times \dfrac{8}{11} \quad \text{[reciprocal of } \dfrac{11}{8} \text{ ⇒ } \dfrac{8}{11}\text{]} \\[1em] = \dfrac{1}{3} \times \dfrac{8}{1} \\[1em] = \dfrac{8}{3} \\[1em] = 2\dfrac{2}{3}$

∴ $\mathbf {3\dfrac{2}{3} \div 1\dfrac{3}{8} = 2\dfrac{2}{3}}$

For each of the questions below, choose the expression that describes the solution. Then simplify it.

(a) Maria bought 8 m of lace to decorate the bags she made for school. She used $\dfrac{1}{4}$ m for each bag and finished the lace. How many bags did she decorate?

1. $8 \times \dfrac{1}{4}$

2. $\dfrac{1}{8} \times \dfrac{1}{4}$

3. $8 \div \dfrac{1}{4}$

4. $\dfrac{1}{4} \div 8$

Answer

Given:

Length of lace Maria bought = 8 m

Length of lace used for each bag = $\dfrac{1}{4}$ m

To find the number of bags Maria decorated, we need to find how many pieces of length $\dfrac{1}{4}$ m can be cut from 8 m of lace, which is a division.

Therefore, the correct expression is $8 \div \dfrac{1}{4}$.

$\phantom{=} 8 \div \dfrac{1}{4} \\[1em] = 8 \times \dfrac{4}{1} \quad \text{[reciprocal of } \dfrac{1}{4} \text{ ⇒ } 4\text{]} \\[1em] = 32$

∴ The correct expression is $\mathbf {8 \div \dfrac{1}{4}}$, and Maria decorated 32 bags.

(b) $\dfrac{1}{2}$ meter of ribbon is used to make 8 badges. What is the length of the ribbon used for each badge?

1. $8 \times \dfrac{1}{2}$

2. $\dfrac{1}{2} \div \dfrac{1}{8}$

3. $8 \div \dfrac{1}{2}$

4. $\dfrac{1}{2} \div 8$

Answer

Given:

Length of ribbon used to make 8 badges = $\dfrac{1}{2}$ m

Number of badges = 8

To find the length of ribbon used for each badge, we divide the total length of ribbon by the number of badges.

Therefore, the correct expression is $\dfrac{1}{2} \div 8$

$\phantom{=} \dfrac{1}{2} \div 8 \\[1em] = \dfrac{1}{2} \times \dfrac{1}{8} \quad \text{[reciprocal of } 8 \text{ ⇒ } \dfrac{1}{8}\text{]} \\[1em] = \dfrac{1}{16}$

∴ The correct expression is $\mathbf {\dfrac{1}{2} \div 8}$, and each badge uses $\mathbf {\dfrac{1}{16}}$ m of ribbon.

(c) A baker needs $\dfrac{1}{6}$ kg of flour to make one loaf of bread. He has 5 kg of flour. How many loaves of bread can he make?

1. $5 \times \dfrac{1}{6}$

2. $\dfrac{1}{6} \div 5$

3. $5 \div \dfrac{1}{6}$

4. $5 \times 6$

Answer

Given:

Flour needed to make one loaf of bread = $\dfrac{1}{6}$ kg

Total flour available = 5 kg

To find the number of loaves the baker can make, we need to find how many portions of $\dfrac{1}{6}$ kg are contained in 5 kg.

Therefore, the correct expression is $5 \div \dfrac{1}{6}$

$\phantom{=} 5 \div \dfrac{1}{6} \\[1em] = 5 \times \dfrac{6}{1} \quad \text{[reciprocal of } \dfrac{1}{6} \text{ ⇒ } 6\text{]} \\[1em] = 30$

∴ The correct expression is $\mathbf {5 \div \dfrac{1}{6}}$, and the baker can make 30 loaves of bread.

If $\dfrac{1}{4}$ kg of flour is used to make 12 rotis, how much flour is used to make 6 rotis?

Answer

Since 6 rotis are $\dfrac{6}{12}$ of 12 rotis, the flour needed is $\dfrac{6}{12}$ of $\dfrac{1}{4}$ kg.

$\phantom{=} \dfrac{6}{12} \times \dfrac{1}{4} \\[1em] = \dfrac{\overset{1}{\cancel{6}}}{\underset{2}{\cancel{12}}} \times \dfrac{1}{4} \\[1em] = \dfrac{1}{2} \times \dfrac{1}{4} \\[1em] = \dfrac{1}{8}$

∴ $\mathbf {\dfrac{1}{8}}$ kg of flour is used to make 6 rotis.

Pāṭīgaṇita, a book written by Sridharacharya in the 9th century CE, mentions this problem: "Friend, after thinking, what sum will be obtained by adding together $1 \div \dfrac{1}{6}$, $1 \div \dfrac{1}{10}$, $1 \div \dfrac{1}{13}$, $1 \div \dfrac{1}{9}$, and $1 \div \dfrac{1}{2}$". What should the friend say?

Answer

Dividing 1 by a unit fraction gives its denominator, since $1 \div \dfrac{1}{n} = 1 \times n = n$.

$\phantom{=} 1 \div \dfrac{1}{6} + 1 \div \dfrac{1}{10} + 1 \div \dfrac{1}{13} + 1 \div \dfrac{1}{9} + 1 \div \dfrac{1}{2} \\[1em] = 6 + 10 + 13 + 9 + 2 \quad \text{[each } 1 \div \dfrac{1}{n} = n\text{]} \\[1em] = 40$

∴ The friend should say the sum is 40.

Mira is reading a novel that has 400 pages. She read $\dfrac{1}{5}$ of the pages yesterday and $\dfrac{3}{10}$ of the pages today. How many more pages does she need to read to finish the novel?

Answer

First find the fraction of the book already read.

$\phantom{=} \dfrac{1}{5} + \dfrac{3}{10} \\[1em] = \dfrac{2}{10} + \dfrac{3}{10} \\[1em] = \dfrac{5}{10} \\[1em] = \dfrac{1}{2}$

So the fraction still to be read is $1 - \dfrac{1}{2} = \dfrac{1}{2}$.

In pages:

$\phantom{=} \dfrac{1}{2} \times 400 \\[1em] = 200$

∴ Mira needs to read 200 more pages.

A car runs 16 km using 1 litre of petrol. How far will it go using $2\dfrac{3}{4}$ litres of petrol?

Answer

Given:

Distance travelled using 1 litre of petrol = 16 km

Petrol available = $2\dfrac{3}{4}$ litres

To find the distance travelled using $2\dfrac{3}{4}$ litres of petrol, we multiply 16 by $2\dfrac{3}{4}$.

$\phantom{=} 16 \times 2\dfrac{3}{4} \\[1em] = \overset{4}{\cancel{16}} \times \dfrac{11}{\cancel{4}} \quad \text{[writing } 2\dfrac{3}{4} = \dfrac{11}{4}\text{]} \\[1em] = 4 \times 11 \\[1em] = 44$

∴ The car will go 44 km.

Amritpal decides on a destination for his vacation. If he takes a train, it will take him $5\dfrac{1}{6}$ hours to get there. If he takes a plane, it will take him $\dfrac{1}{2}$ hour. How many hours does the plane save?

Answer

Given:

Time taken by train = $5\dfrac{1}{6}$ hours

Time taken by plane = $\dfrac{1}{2}$ hour

The time saved by travelling by plane = Train time − Plane time

$\phantom{=} 5\dfrac{1}{6} - \dfrac{1}{2} \\[1em] = \dfrac{31}{6} - \dfrac{1}{2} \quad \text{[writing } 5\dfrac{1}{6} = \dfrac{31}{6}\text{]} \\[1em] = \dfrac{31}{6} - \dfrac{3}{6} \\[1em] = \dfrac{28}{6} \\[1em] = \dfrac{14}{3} \\[1em] = 4\dfrac{2}{3}$

∴ The plane saves $\mathbf {4\dfrac{2}{3}}$ hours.

Mariam's grandmother baked a cake. Mariam and her cousins finished $\dfrac{4}{5}$ of the cake. The remaining cake was shared equally by Mariam's three friends. How much of the cake did each friend get?

Answer

Given:

Part of the cake eaten by Mariam and her cousins = $\dfrac{4}{5}$

Total cake = 1

Therefore, the remaining part of the cake is

$1 - \dfrac{4}{5} = \dfrac{1}{5}$

This is shared equally among 3 friends.

So, the share of each friend is

$\phantom{=} \dfrac{1}{5} \div 3 \\[1em] = \dfrac{1}{5} \times \dfrac{1}{3} \quad \text{[reciprocal of } 3 \text{ ⇒ } \dfrac{1}{3}\text{]} \\[1em] = \dfrac{1}{15}$

∴ Each friend got $\mathbf {\dfrac{1}{15}}$ of the cake.

Choose the option(s) describing the product of $\left(\dfrac{565}{465} \times \dfrac{707}{676}\right)$:

(a) $\gt \dfrac{565}{465}$

(b) $\lt \dfrac{565}{465}$

(c) $\gt \dfrac{707}{676}$

(d) $\lt \dfrac{707}{676}$

(e) $\gt 1$

(f) $\lt 1$

Answer

We observe that

565 > 465 ⇒ $\dfrac{565}{465} \gt 1$

and

707 > 676 ⇒ $\dfrac{707}{676} \gt 1$

Both factors are greater than 1.

When two numbers greater than 1 are multiplied, the product is greater than each of the numbers being multiplied.

$\dfrac{565}{465} \times \dfrac{707}{676} \gt \dfrac{565}{465}$

and

$\dfrac{565}{465} \times \dfrac{707}{676} \gt \dfrac{707}{676}$

Also, since both factors are greater than 1, their product is greater than 1.

∴ The correct options are (a), (c) and (e).

What fraction of the whole square is shaded?

Answer

Take the side of the whole square as 1 unit, so its area is 1 square unit. The square is divided into four equal quadrants. Therefore, the bottom-right quadrant has area $\dfrac{1}{4}$ square unit.

The shaded region lies in the left half of this quadrant. This left half is a rectangle of width $\dfrac{1}{4}$ unit and height $\dfrac{1}{2}$ unit, so its area is:

$\phantom{=} \dfrac{1}{4} \times \dfrac{1}{2} \\[1em] = \dfrac{1}{8}$

From this rectangle, the small unshaded triangle at the top has base $\dfrac{1}{4}$ unit and height $\dfrac{1}{4}$ unit. Its area is:

$\phantom{=} \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{4} \\[1em] = \dfrac{1}{32}$

Therefore, the shaded area is:

$\phantom{=} \dfrac{1}{8} - \dfrac{1}{32} \\[1em] = \dfrac{4}{32} - \dfrac{1}{32} \\[1em] = \dfrac{3}{32}$

∴ $\mathbf {\dfrac{3}{32}}$ of the whole square is shaded.

A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the Fig. 8.7) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?

Answer

Let the original group of ants be taken as 1 (the whole). At every splitting point the group divides equally, so each outgoing path carries an equal share of the ants arriving at that point.

At the 1^st point (bottom), the whole group splits into 2 equal parts:

$\text{Each path gets } = 1 \times \dfrac{1}{2} = \dfrac{1}{2}$

One of these paths leads straight to the mango tree, and the other continues upwards.

At the 2^nd point, the group of $\dfrac{1}{2}$ splits into 2 equal parts:

$\text{Each path gets } = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

One of these paths leads to the mango tree, and the other continues upwards.

At the 3^rd point, the group of $\dfrac{1}{4}$ splits into 4 equal parts:

Each path gets

$= \dfrac{1}{4} \times \dfrac{1}{4} \\[1em] = \dfrac{1 \times 1}{4 \times 4} \\[1em] = \dfrac{1}{16}$

Of these 4 paths, two lead to the mango tree, one leads to the sugarcane field, and one continues upwards.

At the 4^th point, the group of $\dfrac{1}{16}$ splits into 2 equal parts:

$\phantom{=} \text{Each path gets } \\[1em] = \dfrac{1}{16} \times \dfrac{1}{2} \\[1em] = \dfrac{1 \times 1}{16 \times 2} \\[1em] = \dfrac{1}{32}$

One of these paths leads to the mango tree and the other to the sugarcane field.

Fraction of the group that reached the mango tree

Five paths reach the mango tree, carrying $\dfrac{1}{2}$, $\dfrac{1}{4}$, $\dfrac{1}{16}$, $\dfrac{1}{16}$ and $\dfrac{1}{32}$ of the group.

$\phantom{=} \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{16} + \dfrac{1}{16} + \dfrac{1}{32} \\[1em] = \dfrac{16}{32} + \dfrac{8}{32} + \dfrac{2}{32} + \dfrac{2}{32} + \dfrac{1}{32} \\[1em] \text{[Taking LCM of 2, 4, 16 and 32 as 32]} \\[1em] = \dfrac{16 + 8 + 2 + 2 + 1}{32} \\[1em] = \dfrac{29}{32}$

Fraction of the group that reached the sugarcane field

Two paths reach the sugarcane field, carrying $\dfrac{1}{16}$ and $\dfrac{1}{32}$ of the group.

$\phantom{=} \dfrac{1}{16} + \dfrac{1}{32} \\[1em] = \dfrac{2}{32} + \dfrac{1}{32} \\[1em] \text{[Taking LCM of 16 and 32 as 32]} \\[1em] = \dfrac{2 + 1}{32} \\[1em] = \dfrac{3}{32}$

Check: $\dfrac{29}{32} + \dfrac{3}{32} = \dfrac{32}{32} = 1$, i.e. the whole colony is accounted for.

∴ $\mathbf {\dfrac{29}{32}}$ of the original group reached the mango tree and $\mathbf {\dfrac{3}{32}}$ of the original group reached the sugarcane field.

What is $1 - \dfrac{1}{2}$?

$\left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right)$?

$\left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right) \times \left(1 - \dfrac{1}{4}\right) \times \left(1 - \dfrac{1}{5}\right)$?

$\left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right) \times \left(1 - \dfrac{1}{4}\right) \times \left(1 - \dfrac{1}{5}\right) \times \left(1 - \dfrac{1}{6}\right) \times \left(1 - \dfrac{1}{7}\right) \times \left(1 - \dfrac{1}{8}\right) \times \left(1 - \dfrac{1}{9}\right) \times \left(1 - \dfrac{1}{10}\right)$?

Make a general statement and explain.

Answer

$\phantom{=} 1 - \dfrac{1}{2} \\[1em] = \dfrac{2 - 1}{2} \\[1em] = \dfrac{1}{2}$

Hence, the answer is $\mathbf {\dfrac{1}{2}}$

$\phantom{=} \left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right) \\[1em] = \left(\dfrac{2 - 1}{2}\right) \times \left(\dfrac{3 - 1}{3}\right) \\[1em] = \dfrac{1}{2} \times \dfrac{2}{3} \\[1em] = \dfrac{1}{3}$

Hence, the answer is $\mathbf {\dfrac{1}{3}}$

$\phantom{=} \left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right) \times \left(1 - \dfrac{1}{4}\right) \times \left(1 - \dfrac{1}{5}\right) \\[1em] = \left(\dfrac{2 - 1}{2}\right) \times \left(\dfrac{3 - 1}{3}\right) \times \left(\dfrac{4 - 1}{4}\right) \times \left(\dfrac{5 - 1}{5}\right) \\[1em] = \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} \\[1em] = \dfrac{1 \times 2 \times 3 \times 4}{2 \times 3 \times 4 \times 5} \\[1em] = \dfrac{1}{5}$

Hence, the answer is $\mathbf {\dfrac{1}{5}}$

$\phantom{=} \left(1 - \dfrac{1}{2}\right) \times \left(1 - \dfrac{1}{3}\right) \times \left(1 - \dfrac{1}{4}\right) \times \left(1 - \dfrac{1}{5}\right) \times \left(1 - \dfrac{1}{6}\right) \times \left(1 - \dfrac{1}{7}\right) \times \left(1 - \dfrac{1}{8}\right) \times \left(1 - \dfrac{1}{9}\right) \times \left(1 - \dfrac{1}{10}\right) \\[1em] = \left(\dfrac{2 - 1}{2}\right) \times \left(\dfrac{3 - 1}{3}\right) \times \left(\dfrac{4 - 1}{4}\right) \times \left(\dfrac{5 - 1}{5}\right) \times \left(\dfrac{6 - 1}{6}\right) \times \left(\dfrac{7 - 1}{7}\right) \times \left(\dfrac{8 - 1}{8}\right) \times \left(\dfrac{9 - 1}{9}\right) \times \left(\dfrac{10 - 1}{10}\right) \\[1em] = \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} \times \dfrac{5}{6} \times \dfrac{6}{7} \times \dfrac{7}{8} \times \dfrac{8}{9} \times \dfrac{9}{10} \\[1em] = \dfrac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9}{2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10} \\[1em] = \dfrac{1}{10}$

Hence, the answer is $\mathbf {\dfrac{1}{10}}$

General statement. For any whole number $n \geq 2$,

$\begin{array}{ll} \phantom{=} \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\cdots\left(1 - \dfrac{1}{n}\right) \\\\ = \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \times \cdots \times \dfrac{n-1}{n} \\\\ = \dfrac{1}{n} \end{array}$