Q5: Multiply and then convert it into a mixed fraction: (a) 7 × 3/5 (b) 4 × 1/3 (c) 9/7 × 6 (d) 13/11 × 6

To Find: The product of the given numbers and express it as a mixed fraction. Solution: (a) 7 × 3/5 7 × 3/5 = 7 × 3/5 = 21/5 To convert to a mixed fraction, we divide 21 by 5. Quotient = 4, Remainder = 1. So, 21/5 = 41/5. (b) 4 × 1/3 4 × 1/3 = 4 × 1/3 = 4/3 To convert to a mixed fraction, we divide…

Q3: Find the area of a rectangle of sides 33/4 ft and 93/5 ft.

Given: Length of the rectangle = 93/5 ft Breadth of the rectangle = 33/4 ft To Find: Area of the rectangle. Formula: Area of a rectangle = Length × Breadth Solution: First, convert the mixed fractions to improper fractions. Length = 93/5 = (9 × 5) + 3/5 = 45 + 3/5 = 48/5 ft Breadth = 33/4 = (3 × 4)…

Q1: Evaluate the following: 3 ÷ 7/9 14/4 ÷ 2 2/3 ÷ 2/3 14/6 ÷ 7/3 --- --- --- --- 4/3 ÷ 3/4 7/4 ÷ 1/7 8/2 ÷ 4/15 1/5 ÷ 1/9 1/6 ÷ 11/12 32/3 ÷ 13/8

To Find: The value of the given division expressions. Formula: Division of fractions: a/b ÷ c/d = a/b × d/c Solution: 3 ÷ 7/9 = 3 × 9/7 = 27/7 14/4 ÷ 2 = 14/4 × 1/2 = 14/8 = 7/4 2/3 ÷ 2/3 = 2/3 × 3/2 = 1 14/6 ÷ 7/3 = 14/6 × 3/7 = 2 × 7 × 3/2 × 3 × 7 = 1 4/3 ÷ 3/4 = 4/3 × 4/3 = 16/9 7/4 ÷ 1/7 = 7/4…

Working with Fractions Class 7 Mathematics NCERT Solutions

Q1Figure it Out 1

Tenzin drinks $\frac{1}{2}$ 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?

Solution

Given: Milk Tenzin drinks per day =

\frac{1}{2}

glass

To Find:

Total glasses of milk he drinks in a week.
Total glasses of milk he drank in the month of January.

Solution:

Milk in a week: Number of days in a week = 7 Total milk in a week = Number of days $\times$ Milk per day $= 7 \times \frac{1}{2} = \frac{7}{2} = 3\frac{1}{2} \text{ glasses}$
Milk in January: Number of days in January = 31 Total milk in January = Number of days $\times$ Milk per day $= 31 \times \frac{1}{2} = \frac{31}{2} = 15\frac{1}{2} \text{ glasses}$

Final Answer: Tenzin drinks

3\frac{1}{2}

glasses of milk in a week and

15\frac{1}{2}

glasses of milk in the month of January.

Q2Figure it Out 1

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.

Solution

Given: Work done in 8 days = 1 km of water canal. Working days in a week = 5 days.

To Find:

The length of the canal made in one day.
The length of the canal made in one week (5 working days).

Solution:

Canal made in one day: If 1 km is made in 8 days, then in one day the team makes: $1 \div 8 = \frac{1}{8} \text{ km}$
Canal made in a week: The team works 5 days a week. Length of canal made in a week = Work per day $\times$ Number of working days $= \frac{1}{8} \times 5 = \frac{5}{8} \text{ km}$

Final Answer: In one day, the team can make $\frac{1}{8}$ km of the water canal. If they work 5 days a week, they can make $\frac{5}{8}$ km of the water canal in a week.

Q3Figure it Out 1

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?

Solution

Given: Total oil bought per week = 5 litres Number of families sharing the oil = 3 (Manju + two neighbours)

To Find:

The amount of oil each family gets in a week.
The amount of oil one family gets in 4 weeks.

Solution:

Oil per family per week: Total oil is shared equally among 3 families. Oil per family = Total oil $\div$ Number of families $= 5 \div 3 = \frac{5}{3} = 1\frac{2}{3} \text{ litres}$
Oil for one family in 4 weeks: Total oil for one family in 4 weeks = Oil per week $\times$ Number of weeks $= \frac{5}{3} \times 4 = \frac{20}{3} = 6\frac{2}{3} \text{ litres}$

Final Answer: Each family gets

1\frac{2}{3}

litres of oil in a week. One family will get

6\frac{2}{3}

litres of oil in 4 weeks.

Q4Figure it Out 1

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

Solution

Given: Moon setting time on Monday = 10 pm Daily delay in moon set =

\frac{5}{6}

hour

To Find: The total delay in moon setting time on Thursday compared to Monday at 10 pm.

Solution: The number of days from Monday to Thursday is 3 (Tuesday, Wednesday, Thursday). The moon set time will be later each day. Total delay = Number of days

\times

Daily delay

= 3 \times \frac{5}{6} \text{ hours}

= \frac{15}{6} \text{ hours}

Simplifying the fraction:

= \frac{5}{2} = 2\frac{1}{2} \text{ hours}

Final Answer: The moon will set

2\frac{1}{2}

hours after 10 pm on Thursday.

Q5Figure it Out 1

Multiply and then convert it into a mixed fraction:

(a)

7 \times \frac{3}{5}

(b)

4 \times \frac{1}{3}

(c)

\frac{9}{7} \times 6

(d)

\frac{13}{11} \times 6

Solution

To Find: The product of the given numbers and express it as a mixed fraction.

Solution: (a) $7 \times \frac{3}{5}$

7 \times \frac{3}{5} = \frac{7 \times 3}{5} = \frac{21}{5}

To convert to a mixed fraction, we divide 21 by 5. Quotient = 4, Remainder = 1. So,

\frac{21}{5} = 4\frac{1}{5}

.

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

4 \times \frac{1}{3} = \frac{4 \times 1}{3} = \frac{4}{3}

To convert to a mixed fraction, we divide 4 by 3. Quotient = 1, Remainder = 1. So,

\frac{4}{3} = 1\frac{1}{3}

.

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

\frac{9}{7} \times 6 = \frac{9 \times 6}{7} = \frac{54}{7}

To convert to a mixed fraction, we divide 54 by 7. Quotient = 7, Remainder = 5. So,

\frac{54}{7} = 7\frac{5}{7}

.

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

\frac{13}{11} \times 6 = \frac{13 \times 6}{11} = \frac{78}{11}

To convert to a mixed fraction, we divide 78 by 11. Quotient = 7, Remainder = 1. So,

\frac{78}{11} = 7\frac{1}{11}

.

Final Answer: (a)

4\frac{1}{5}

(b)

1\frac{1}{3}

(c)

7\frac{5}{7}

(d)

7\frac{1}{11}

Q1Figure it Out 2

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

(a)

\frac{1}{3} \times \frac{1}{5}

(b)

\frac{1}{4} \times \frac{1}{3}

(c)

\frac{1}{5} \times \frac{1}{2}

(d)

\frac{1}{6} \times \frac{1}{5}

Now, find

\frac{1}{12} \times \frac{1}{18}

.

Solution

To Find: The product of the given fractions.

Concept: To multiply two fractions like

\frac{1}{a} \times \frac{1}{b}

using a unit square, we can divide the square into 'b' columns and 'a' rows. This creates

a \times b

smaller equal rectangles. The product represents the area of one of these small rectangles, which is

\frac{1}{a \times b}

of the whole square.

Solution: (a)

\frac{1}{3} \times \frac{1}{5}

We multiply the numerators and the denominators.

\frac{1}{3} \times \frac{1}{5} = \frac{1 \times 1}{3 \times 5} = \frac{1}{15}

(b)

\frac{1}{4} \times \frac{1}{3}

\frac{1}{4} \times \frac{1}{3} = \frac{1 \times 1}{4 \times 3} = \frac{1}{12}

(c)

\frac{1}{5} \times \frac{1}{2}

\frac{1}{5} \times \frac{1}{2} = \frac{1 \times 1}{5 \times 2} = \frac{1}{10}

(d)

\frac{1}{6} \times \frac{1}{5}

\frac{1}{6} \times \frac{1}{5} = \frac{1 \times 1}{6 \times 5} = \frac{1}{30}

Now, find

\frac{1}{12} \times \frac{1}{18}

. Using the same rule, we multiply the denominators.

\frac{1}{12} \times \frac{1}{18} = \frac{1}{12 \times 18}

Calculation:

12 \times 18 = 12 \times (10 + 8) = 120 + 96 = 216

. So,

\frac{1}{12} \times \frac{1}{18} = \frac{1}{216}

Final Answer: (a)

\frac{1}{15}

(b)

\frac{1}{12}

(c)

\frac{1}{10}

(d)

\frac{1}{30}

\frac{1}{12} \times \frac{1}{18} = \frac{1}{216}

Q2Figure it Out 2

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

(a)

\frac{2}{3} \times \frac{4}{5}

(b)

\frac{1}{4} \times \frac{2}{3}

(c)

\frac{3}{5} \times \frac{1}{2}

(d)

\frac{4}{6} \times \frac{3}{5}

Solution

To Find: The product of the given fractions.

Concept: To multiply two fractions like

\frac{a}{b} \times \frac{c}{d}

using a unit square, we divide the square into 'd' rows and 'b' columns, creating

b \times d

small rectangles. The fraction

\frac{c}{d}

represents 'c' of these rows, and

\frac{a}{b}

of that area is found by taking 'a' columns of that shaded region. This results in an area of

a \times c

small rectangles out of a total of

b \times d

rectangles, which is

\frac{a \times c}{b \times d}

.

Formula:

\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}

Solution: (a)

\frac{2}{3} \times \frac{4}{5}

\frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \frac{8}{15}

(b)

\frac{1}{4} \times \frac{2}{3}

\frac{1}{4} \times \frac{2}{3} = \frac{1 \times 2}{4 \times 3} = \frac{2}{12}

Simplifying the fraction by dividing the numerator and denominator by their common factor, 2:

\frac{2}{12} = \frac{1}{6}

(c)

\frac{3}{5} \times \frac{1}{2}

\frac{3}{5} \times \frac{1}{2} = \frac{3 \times 1}{5 \times 2} = \frac{3}{10}

(d)

\frac{4}{6} \times \frac{3}{5}

\frac{4}{6} \times \frac{3}{5} = \frac{4 \times 3}{6 \times 5} = \frac{12}{30}

Simplifying the fraction by dividing the numerator and denominator by their greatest common factor, 6:

\frac{12}{30} = \frac{2}{5}

Alternatively, we could simplify before multiplying:

\frac{4}{6} \times \frac{3}{5} = \frac{2}{3} \times \frac{3}{5} = \frac{2 \times 3}{3 \times 5} = \frac{2}{5}

Final Answer: (a)

\frac{8}{15}

(b)

\frac{1}{6}

(c)

\frac{3}{10}

(d)

\frac{2}{5}

Q1Figure it Out 3

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

(a)

\frac{1}{3}

hour _____

(b)

\frac{2}{3}

hour _____

(c)

\frac{3}{4}

hour _____

(d)

\frac{7}{10}

hour _____

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

Solution

Given: Fraction of tank filled in 1 hour =

\frac{7}{10}

To Find: Fraction of tank filled for different durations and the total time to fill the tank.

Solution: Fraction filled = Rate of filling

\times

Time Rate of filling =

\frac{7}{10}

tank per hour.

(a) For

\frac{1}{3}

hour: Fraction filled =

\frac{7}{10} \times \frac{1}{3} = \frac{7}{30}

of the tank.

(b) For

\frac{2}{3}

hour: Fraction filled =

\frac{7}{10} \times \frac{2}{3} = \frac{14}{30} = \frac{7}{15}

of the tank.

(c) For

\frac{3}{4}

hour: Fraction filled =

\frac{7}{10} \times \frac{3}{4} = \frac{21}{40}

of the tank.

(d) For

\frac{7}{10}

hour: Fraction filled =

\frac{7}{10} \times \frac{7}{10} = \frac{49}{100}

of the tank.

(e) Time for the tank to be full: Let T be the total time in hours. Rate of filling

\times

T = 1 (for a full tank)

\frac{7}{10} \times T = 1

T = 1 \div \frac{7}{10} = 1 \times \frac{10}{7} = \frac{10}{7} \text{ hours}

As a mixed fraction,

T = 1\frac{3}{7}

hours.

Final Answer: (a)

\frac{7}{30}

(b)

\frac{7}{15}

(c)

\frac{21}{40}

(d)

\frac{49}{100}

(e) The tap should be running for

1\frac{3}{7}

hours for the tank to be full.

Q2Figure it Out 3

The government has taken $\frac{1}{6}$ 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 $\frac{1}{3}$ 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?

Solution

Given: Let the original land be 1 unit. Part of land taken by government =

\frac{1}{6}

Krishna's share =

\frac{1}{2}

of the remaining land. Bora's share =

\frac{1}{3}

of the remaining land.

Solution: First, find the remaining land after the government took its share. Remaining land =

1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}

of the original land.

(a) Krishna's share: Krishna got

\frac{1}{2}

of the remaining

\frac{5}{6}

of the land. Krishna's share =

\frac{1}{2} \times \frac{5}{6} = \frac{5}{12}

of the original land.

(b) Bora's share: Bora got

\frac{1}{3}

of the remaining

\frac{5}{6}

of the land. Bora's share =

\frac{1}{3} \times \frac{5}{6} = \frac{5}{18}

of the original land.

(c) Somu's share: Somu's share is the part of the remaining land left after giving shares to Krishna and Bora. Total given away from the remaining part = Krishna's share + Bora's share Fraction of remaining land given away =

\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

of the remaining land. Fraction of remaining land Somu keeps =

1 - \frac{5}{6} = \frac{1}{6}

of the remaining land. Somu's share of original land =

\frac{1}{6} \times (\text{Remaining land}) = \frac{1}{6} \times \frac{5}{6} = \frac{5}{36}

of the original land.

Final Answer: (a) Krishna got

\frac{5}{12}

of the original land. (b) Bora got

\frac{5}{18}

of the original land. (c) Somu kept

\frac{5}{36}

of the original land for herself.

Q3Figure it Out 3

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

Solution

Given: Length of the rectangle =

9\frac{3}{5}

ft Breadth of the rectangle =

3\frac{3}{4}

ft

To Find: Area of the rectangle.

Formula: Area of a rectangle = Length

\times

Breadth

Solution: First, convert the mixed fractions to improper fractions. Length =

9\frac{3}{5} = \frac{(9 \times 5) + 3}{5} = \frac{45 + 3}{5} = \frac{48}{5}

ft Breadth =

3\frac{3}{4} = \frac{(3 \times 4) + 3}{4} = \frac{12 + 3}{4} = \frac{15}{4}

ft

Now, calculate the area: Area =

\frac{48}{5} \times \frac{15}{4}

We can simplify before multiplying by cancelling common factors. 48 and 4 have a common factor of 4 (

48 \div 4 = 12

). 15 and 5 have a common factor of 5 (

15 \div 5 = 3

). Area =

\frac{12}{1} \times \frac{3}{1} = 12 \times 3 = 36

Final Answer: The area of the rectangle is 36 square feet.

Q4Figure it Out 3

Tsewang plants four saplings in a row in his garden. The distance between two saplings is $\frac{3}{4}$ m. Find the distance between the first and last sapling.

Solution

Given: Number of saplings = 4 Distance between two adjacent saplings =

\frac{3}{4}

m

To Find: The distance between the first and the last sapling.

Solution: When four saplings are planted in a row, there are three gaps between them. Let S1, S2, S3, S4 be the four saplings. The gaps are between (S1, S2), (S2, S3), and (S3, S4). Total number of gaps = Number of saplings - 1 =

4 - 1 = 3

.

Each gap has a distance of

\frac{3}{4}

m. Total distance = Number of gaps

\times

Distance per gap

= 3 \times \frac{3}{4} = \frac{9}{4} \text{ m}

As a mixed fraction, this is

2\frac{1}{4}

m.

Final Answer: The distance between the first and last sapling is

2\frac{1}{4}

m.

Q5Figure it Out 3

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

Solution

Given: Two quantities to compare:

$\frac{12}{15}$ of 500 grams
$\frac{3}{20}$ of 4 kg

To Find: Which quantity is heavier.

Solution: First, we need to express both quantities in the same unit. Let's use grams (g). We know that 1 kg = 1000 g. So, 4 kg =

4 \times 1000 = 4000

g.

Now, calculate the value of each quantity.

First quantity: $\frac{12}{15} \text{ of } 500 \text{ g} = \frac{12}{15} \times 500$ Simplify the fraction $\frac{12}{15} = \frac{4}{5}$ . $= \frac{4}{5} \times 500 = 4 \times \frac{500}{5} = 4 \times 100 = 400 \text{ g}$
Second quantity: $\frac{3}{20} \text{ of } 4 \text{ kg} = \frac{3}{20} \text{ of } 4000 \text{ g}$ $= \frac{3}{20} \times 4000 = 3 \times \frac{4000}{20} = 3 \times 200 = 600 \text{ g}$

Comparison: Comparing the two values: 400 g and 600 g. Since

600 > 400

, the second quantity is heavier.

Final Answer:

\frac{3}{20}

of 4 kg is heavier.

Q1Figure it Out 4

Evaluate the following:
$3 \div \frac{7}{9}$ $\frac{14}{4} \div 2$ $\frac{2}{3} \div \frac{2}{3}$ $\frac{14}{6} \div \frac{7}{3}$
$\frac{4}{3} \div \frac{3}{4}$ $\frac{7}{4} \div \frac{1}{7}$ $\frac{8}{2} \div \frac{4}{15}$
$\frac{1}{5} \div \frac{1}{9}$ $\frac{1}{6} \div \frac{11}{12}$ $3\frac{2}{3} \div 1\frac{3}{8}$

$3 \div \frac{7}{9}$	$\frac{14}{4} \div 2$	$\frac{2}{3} \div \frac{2}{3}$	$\frac{14}{6} \div \frac{7}{3}$
$\frac{4}{3} \div \frac{3}{4}$	$\frac{7}{4} \div \frac{1}{7}$	$\frac{8}{2} \div \frac{4}{15}$
$\frac{1}{5} \div \frac{1}{9}$	$\frac{1}{6} \div \frac{11}{12}$	$3\frac{2}{3} \div 1\frac{3}{8}$

Solution

To Find: The value of the given division expressions.

Formula: Division of fractions:

\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

Solution:

$3 \div \frac{7}{9} = 3 \times \frac{9}{7} = \frac{27}{7}$
$\frac{14}{4} \div 2 = \frac{14}{4} \times \frac{1}{2} = \frac{14}{8} = \frac{7}{4}$
$\frac{2}{3} \div \frac{2}{3} = \frac{2}{3} \times \frac{3}{2} = 1$
$\frac{14}{6} \div \frac{7}{3} = \frac{14}{6} \times \frac{3}{7} = \frac{2 \times 7 \times 3}{2 \times 3 \times 7} = 1$
$\frac{4}{3} \div \frac{3}{4} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$
$\frac{7}{4} \div \frac{1}{7} = \frac{7}{4} \times 7 = \frac{49}{4}$
$\frac{8}{2} \div \frac{4}{15} = 4 \div \frac{4}{15} = 4 \times \frac{15}{4} = 15$
$\frac{1}{5} \div \frac{1}{9} = \frac{1}{5} \times 9 = \frac{9}{5}$
$\frac{1}{6} \div \frac{11}{12} = \frac{1}{6} \times \frac{12}{11} = \frac{12}{66} = \frac{2}{11}$
$3\frac{2}{3} \div 1\frac{3}{8} = \frac{11}{3} \div \frac{11}{8} = \frac{11}{3} \times \frac{8}{11} = \frac{8}{3}$

Final Answer:

$3 \div \frac{7}{9} = \frac{27}{7}$
$\frac{14}{4} \div 2 = \frac{7}{4}$
$\frac{2}{3} \div \frac{2}{3} = 1$
$\frac{14}{6} \div \frac{7}{3} = 1$
$\frac{4}{3} \div \frac{3}{4} = \frac{16}{9}$
$\frac{7}{4} \div \frac{1}{7} = \frac{49}{4}$
$\frac{8}{2} \div \frac{4}{15} = 15$
$\frac{1}{5} \div \frac{1}{9} = \frac{9}{5}$
$\frac{1}{6} \div \frac{11}{12} = \frac{2}{11}$
$3\frac{2}{3} \div 1\frac{3}{8} = \frac{8}{3}$

Q2Figure it Out 4

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 $\frac{1}{4}$ m for each bag and finished the lace. How many bags did she decorate?

(i)

$8 \times \frac{1}{4}$

(ii)

$\frac{1}{8} \times \frac{1}{4}$

(iii)

$8 \div \frac{1}{4}$

(iv)

$\frac{1}{4} \div 8$

(b)

\frac{1}{2}

meter of ribbon is used to make 8 badges. What is the length of the ribbon used for each badge?

(i)

$8 \times \frac{1}{2}$

(ii)

$\frac{1}{2} \div \frac{1}{8}$

(iii)

$8 \div \frac{1}{2}$

(iv)

$\frac{1}{2} \div 8$

(c)

A baker needs

\frac{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?

(i)

$5 \times \frac{1}{6}$

(ii)

$\frac{1}{6} \div 5$

(iii)

$5 \div \frac{1}{6}$

(iv)

$5 \times 6$

Solution

Solution:

(a) Problem: Maria has 8 m of lace and uses

\frac{1}{4}

m for each bag. Find the number of bags. Logic: The total length of lace is divided by the length used per bag. Expression: Total lace

\div

Lace per bag =

8 \div \frac{1}{4}

. Correct Option: (iii) Calculation:

8 \div \frac{1}{4} = 8 \times 4 = 32

Maria decorated 32 bags.

(b) Problem:

\frac{1}{2}

m of ribbon is used for 8 badges. Find the length per badge. Logic: The total length of ribbon is divided by the number of badges. Expression: Total ribbon

\div

Number of badges =

\frac{1}{2} \div 8

. Correct Option: (iv) Calculation:

\frac{1}{2} \div 8 = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}

The length of ribbon for each badge is

\frac{1}{16}

m.

(c) Problem: A baker has 5 kg of flour and needs

\frac{1}{6}

kg per loaf. Find the number of loaves. Logic: The total amount of flour is divided by the amount needed per loaf. Expression: Total flour

\div

Flour per loaf =

5 \div \frac{1}{6}

. Correct Option: (iii) (Note: Option (iv)

5 \times 6

gives the same result, as

5 \div \frac{1}{6} = 5 \times 6

.) Calculation:

5 \div \frac{1}{6} = 5 \times 6 = 30

He can make 30 loaves of bread.

Final Answer: (a) Correct expression is (iii)

8 \div \frac{1}{4}

. She decorated 32 bags. (b) Correct expression is (iv)

\frac{1}{2} \div 8

. The length used for each badge is

\frac{1}{16}

m. (c) Correct expression is (iii)

5 \div \frac{1}{6}

. He can make 30 loaves.

Q3Figure it Out 4

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

Solution

Given: Flour for 12 rotis =

\frac{1}{4}

kg.

To Find: Flour needed for 6 rotis.

Solution: Method 1: Unitary Method First, find the flour needed for 1 roti. Flour per roti = Total flour

\div

Number of rotis

= \frac{1}{4} \div 12 = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48} \text{ kg}

Now, find the flour needed for 6 rotis. Flour for 6 rotis = Flour per roti

\times

6

= \frac{1}{48} \times 6 = \frac{6}{48} = \frac{1}{8} \text{ kg}

Method 2: Proportional Method 6 rotis is half the number of 12 rotis (

6 = \frac{1}{2} \times 12

). So, the amount of flour needed will also be half. Flour for 6 rotis =

\frac{1}{2} \times

(Flour for 12 rotis)

= \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \text{ kg}

Final Answer:

\frac{1}{8}

kg of flour is used to make 6 rotis.

Q4Figure it Out 4

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

Solution

Given: The expression to evaluate is the sum of several division problems:

(1 \div \frac{1}{6}) + (1 \div \frac{1}{10}) + (1 \div \frac{1}{13}) + (1 \div \frac{1}{9}) + (1 \div \frac{1}{2})

To Find: The sum of these values.

Solution: First, evaluate each division. Dividing 1 by a fraction is the same as finding the reciprocal of that fraction.

$1 \div \frac{1}{6} = 1 \times 6 = 6$
$1 \div \frac{1}{10} = 1 \times 10 = 10$
$1 \div \frac{1}{13} = 1 \times 13 = 13$
$1 \div \frac{1}{9} = 1 \times 9 = 9$
$1 \div \frac{1}{2} = 1 \times 2 = 2$

Now, add these results together: Sum =

6 + 10 + 13 + 9 + 2

Sum =

16 + 13 + 9 + 2

Sum =

29 + 9 + 2

Sum =

38 + 2 = 40

Final Answer: The friend should say the sum is 40.

Q5Figure it Out 4

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

Solution

Given: Total pages in the novel = 400 Fraction of pages read yesterday =

\frac{1}{5}

Fraction of pages read today =

\frac{3}{10}

To Find: The number of pages remaining to be read.

Solution: Step 1: Calculate pages read each day. Pages read yesterday =

\frac{1}{5} \times 400 = \frac{400}{5} = 80

pages. Pages read today =

\frac{3}{10} \times 400 = 3 \times \frac{400}{10} = 3 \times 40 = 120

pages.

Step 2: Calculate total pages read. Total pages read = Pages yesterday + Pages today =

80 + 120 = 200

pages.

Step 3: Calculate remaining pages. Remaining pages = Total pages - Total pages read Remaining pages =

400 - 200 = 200

pages.

Alternative Method: Step 1: Find the total fraction of pages read. Total fraction read =

\frac{1}{5} + \frac{3}{10} = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2}

.

Step 2: Find the fraction of pages remaining. Fraction remaining =

1 - \frac{1}{2} = \frac{1}{2}

.

Step 3: Calculate the number of remaining pages. Remaining pages = Fraction remaining

\times

Total pages Remaining pages =

\frac{1}{2} \times 400 = 200

pages.

Final Answer: Mira needs to read 200 more pages to finish the novel.

Q6Figure it Out 4

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

Solution

Given: Distance covered per litre of petrol = 16 km Amount of petrol used =

2\frac{3}{4}

litres

To Find: Total distance the car will go.

Solution: First, convert the mixed fraction for petrol into an improper fraction. Petrol used =

2\frac{3}{4} = \frac{(2 \times 4) + 3}{4} = \frac{8+3}{4} = \frac{11}{4}

litres.

Total distance = Distance per litre

\times

Total litres of petrol

= 16 \times \frac{11}{4}

= \frac{16}{4} \times 11

= 4 \times 11 = 44 \text{ km}

Final Answer: The car will go 44 km.

Q7Figure it Out 4

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

Solution

Given: Time taken by train =

5\frac{1}{6}

hours Time taken by plane =

\frac{1}{2}

hour

To Find: The time saved by taking the plane.

Solution: Time saved = Time by train - Time by plane

= 5\frac{1}{6} - \frac{1}{2}

First, convert the mixed fraction to an improper fraction.

5\frac{1}{6} = \frac{(5 \times 6) + 1}{6} = \frac{31}{6}

Now, perform the subtraction. Find a common denominator, which is 6.

\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}

Time saved =

\frac{31}{6} - \frac{3}{6} = \frac{31 - 3}{6} = \frac{28}{6}

Simplify the fraction:

\frac{28}{6} = \frac{14}{3}

Convert to a mixed fraction:

\frac{14}{3} = 4\frac{2}{3}

hours.

Final Answer: The plane saves

4\frac{2}{3}

hours.

Q8Figure it Out 4

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

Solution

Given: Fraction of cake finished by Mariam and cousins =

\frac{4}{5}

Number of friends sharing the remaining cake = 3

To Find: The fraction of the original cake each friend got.

Solution: Step 1: Find the fraction of the remaining cake. Remaining cake =

1 - (\text{Fraction eaten})

= 1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}

So,

\frac{1}{5}

of the cake remains.

Step 2: Divide the remaining cake among the three friends. This remaining

\frac{1}{5}

of the cake is shared equally by 3 friends. Share per friend = (Remaining cake)

\div

(Number of friends)

= \frac{1}{5} \div 3

= \frac{1}{5} \times \frac{1}{3} = \frac{1}{15}

Final Answer: Each friend got

\frac{1}{15}

of the cake.

Q9Figure it Out 4

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

(a)

> \frac{565}{465}

(b)

< \frac{565}{465}

(c)

> \frac{707}{676}

(d)

< \frac{707}{676}

(e)

> 1

(f)

< 1

Solution

Given: The product is

(\frac{565}{465} \times \frac{707}{676})

.

To Find: The relationship of the product with its factors and with 1.

Analysis: First, let's analyze the two fractions being multiplied.

First fraction: $\frac{565}{465}$ . Since the numerator (565) is greater than the denominator (465), this fraction is greater than 1. ( $565 > 465 \implies \frac{565}{465} > 1$ ).
Second fraction: $\frac{707}{676}$ . Since the numerator (707) is greater than the denominator (676), this fraction is also greater than 1. ( $707 > 676 \implies \frac{707}{676} > 1$ ).

Rules of Multiplication:

When a number is multiplied by a value greater than 1, the product is greater than the original number.
The product of two numbers, both greater than 1, will also be greater than 1.

Applying the Rules: Let

P = \frac{565}{465} \times \frac{707}{676}

.

Since we are multiplying $\frac{565}{465}$ by $\frac{707}{676}$ (which is > 1), the product $P$ will be greater than $\frac{565}{465}$ . So, (a) is correct.
Since we are multiplying $\frac{707}{676}$ by $\frac{565}{465}$ (which is > 1), the product $P$ will be greater than $\frac{707}{676}$ . So, (c) is correct.
Since both fractions are greater than 1, their product will also be greater than 1. So, (e) is correct.

Options (b), (d), and (f) are incorrect based on this analysis.

Final Answer: The correct options are (a), (c), and (e).

Q10Figure it Out 4

What fraction of the whole square is shaded?

Solution

Given: A large square is shown with some regions shaded. The square is divided into four equal smaller squares (quadrants).

The top-left quadrant is fully shaded.
The bottom-right quadrant is further divided into four equal tiny squares, and one of these tiny squares is shaded.

To Find: The total fraction of the whole square that is shaded.

Solution: Let the area of the whole square be 1 unit.

Step 1: Calculate the area of the first shaded region. The whole square is divided into 4 equal quadrants. The top-left quadrant is shaded. Area of the top-left shaded quadrant =

\frac{1}{4}

of the whole square.

Step 2: Calculate the area of the second shaded region. The bottom-right quadrant has an area of

\frac{1}{4}

of the whole square. This quadrant is further divided into 4 equal parts. One of these parts is shaded. Area of the small shaded part =

\frac{1}{4}

of the area of the bottom-right quadrant. Area =

\frac{1}{4} \times (\text{Area of quadrant}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

of the whole square.

Step 3: Calculate the total shaded area. Total shaded area = Area of first shaded region + Area of second shaded region

= \frac{1}{4} + \frac{1}{16}

To add these fractions, find a common denominator, which is 16.

= \frac{4}{16} + \frac{1}{16} = \frac{5}{16}

Final Answer:

\frac{5}{16}

of the whole square is shaded.

Q11Figure it Out 4

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?

Solution

Given: A colony of ants starts at one point. The path branches out, and at each branch point (junction), the group of ants splits equally into two groups.

To reach the mango tree, the ants pass through 3 such junctions.
To reach the sugarcane field, the ants also pass through 3 such junctions.

To Find: The fraction of the original group that reached the mango tree and the sugarcane field.

Solution: Let the original group of ants be represented by 1. At each junction, the group splits into two equal halves. This means the number of ants taking any path is multiplied by

\frac{1}{2}

.

Fraction reaching the mango tree: The path to the mango tree involves 3 splits.

After the first split, $\frac{1}{2}$ of the group continues on the path.
After the second split, $\frac{1}{2}$ of the remaining group continues, which is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the original group.
After the third split, $\frac{1}{2}$ of that group continues, which is $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ of the original group. So, the fraction of ants reaching the mango tree is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ .

Fraction reaching the sugarcane field: The path to the sugarcane field also involves 3 splits. Similar to the path to the mango tree, the fraction of the original group is reduced by half at each junction. Fraction reaching the sugarcane field =

\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}

.

Final Answer:

\frac{1}{8}

of the original group reached the mango tree, and

\frac{1}{8}

of the original group reached the sugarcane field.

Q12Figure it Out 4

What is $1-\frac{1}{2}$ ? $(1-\frac{1}{2}) \times (1-\frac{1}{3})$ ? $(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{4}) \times (1-\frac{1}{5})$ ? $(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{4}) \times (1-\frac{1}{5}) \times (1-\frac{1}{6}) \times (1-\frac{1}{7}) \times (1-\frac{1}{8}) \times (1-\frac{1}{9}) \times (1-\frac{1}{10})$ ? Make a general statement and explain.

Solution

To Find: The value of the given products and a general rule.

Solution: Let's evaluate each expression step-by-step.

$1-\frac{1}{2}$ $1-\frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}$
$(1-\frac{1}{2}) \times (1-\frac{1}{3})$ $= \left(\frac{1}{2}\right) \times \left(1-\frac{1}{3}\right) = \frac{1}{2} \times \left(\frac{3}{3}-\frac{1}{3}\right) = \frac{1}{2} \times \frac{2}{3} = \frac{1 \times 2}{2 \times 3} = \frac{2}{6} = \frac{1}{3}$ Notice that the numerator of the first fraction (after simplification) cancels with the denominator of the second.
$(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{4}) \times (1-\frac{1}{5})$ $= \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{4}{5}\right)$ $= \frac{1 \times 2 \times 3 \times 4}{2 \times 3 \times 4 \times 5}$ By cancelling terms, we are left with $\frac{1}{5}$ .
$(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times ... \times (1-\frac{1}{10})$ The product is: $= \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{4}{5}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{6}{7}\right) \times \left(\frac{7}{8}\right) \times \left(\frac{8}{9}\right) \times \left(\frac{9}{10}\right)$ The numerator of each fraction cancels the denominator of the next fraction. $= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \frac{\cancel{4}}{\cancel{5}} \times \frac{\cancel{5}}{\cancel{6}} \times \frac{\cancel{6}}{\cancel{7}} \times \frac{\cancel{7}}{\cancel{8}} \times \frac{\cancel{8}}{\cancel{9}} \times \frac{\cancel{9}}{10} = \frac{1}{10}$

General Statement and Explanation:

General Statement: The product of the series

(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{4}) \times ... \times (1-\frac{1}{n})

is equal to

\frac{1}{n}

for any integer

n \geq 2

.

Explanation: Each term

(1 - \frac{1}{k})

in the series can be simplified to

\frac{k-1}{k}

. So the product can be written as:

\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times ... \times \frac{n-1}{n}

In this product, the numerator of each term (except the last) is the same as the denominator of the preceding term. This creates a telescoping product where intermediate terms cancel out:

= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times ... \times \frac{\cancel{n-1}}{n}

Only the numerator of the first term (1) and the denominator of the last term (n) remain. Therefore, the final product is

\frac{1}{n}

.

Final Answer:

$1-\frac{1}{2} = \frac{1}{2}$
$(1-\frac{1}{2}) \times (1-\frac{1}{3}) = \frac{1}{3}$
$(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{4}) \times (1-\frac{1}{5}) = \frac{1}{5}$
$(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times ... \times (1-\frac{1}{10}) = \frac{1}{10}$
The general statement is that the product up to the term $(1-\frac{1}{n})$ is $\frac{1}{n}$ .

Q1Figure it Out (Applications of Fraction Multiplication)

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

(a)

\frac{1}{3}

hour _____

(b)

\frac{2}{3}

hour _____

(c)

\frac{3}{4}

hour _____

(d)

\frac{7}{10}

hour _____

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

Solution

Given: Rate of filling =

\frac{7}{10}

of the tank per hour.

To Find: Fraction of the tank filled for different durations and the total time to fill the tank.

Formula: Fraction filled = Rate of filling

\times

Time

Solution:

(a) For $\frac{1}{3}$ hour: Fraction filled =

\frac{7}{10} \times \frac{1}{3} = \frac{7 \times 1}{10 \times 3} = \frac{7}{30}

of the tank.

(b) For $\frac{2}{3}$ hour: Fraction filled =

\frac{7}{10} \times \frac{2}{3} = \frac{14}{30} = \frac{7}{15}

of the tank.

(c) For $\frac{3}{4}$ hour: Fraction filled =

\frac{7}{10} \times \frac{3}{4} = \frac{21}{40}

of the tank.

(d) For $\frac{7}{10}$ hour: Fraction filled =

\frac{7}{10} \times \frac{7}{10} = \frac{49}{100}

of the tank.

(e) Time for the tank to be full: Let T be the time in hours. A full tank is represented by 1. Rate

\times

Time = 1

\frac{7}{10} \times T = 1

To find T, we divide 1 by the rate:

T = 1 \div \frac{7}{10} = 1 \times \frac{10}{7} = \frac{10}{7} \text{ hours}

Converting to a mixed fraction:

10 \div 7 = 1

with a remainder of 3.

T = 1\frac{3}{7} \text{ hours}

Final Answer: (a)

\frac{7}{30}

of the tank is filled. (b)

\frac{7}{15}

of the tank is filled. (c)

\frac{21}{40}

of the tank is filled. (d)

\frac{49}{100}

of the tank is filled. (e) The tap should be running for

1\frac{3}{7}

hours for the tank to be full.

Q2Figure it Out (Applications of Fraction Multiplication)

The government has taken $\frac{1}{6}$ 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 $\frac{1}{3}$ 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?

Solution

Given: Fraction of land taken by government =

\frac{1}{6}

. Krishna's share =

\frac{1}{2}

of the remaining land. Bora's share =

\frac{1}{3}

of the remaining land.

Solution:

Step 1: Find the remaining land Let the original land be 1 unit. Remaining land = Original land - Land taken by government

= 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}

So,

\frac{5}{6}

of the original land remains.

(a) Krishna's share of the original land Krishna gets

\frac{1}{2}

of the remaining land. Krishna's share =

\frac{1}{2} \times \frac{5}{6} = \frac{1 \times 5}{2 \times 6} = \frac{5}{12}

of the original land.

(b) Bora's share of the original land Bora gets

\frac{1}{3}

of the remaining land. Bora's share =

\frac{1}{3} \times \frac{5}{6} = \frac{1 \times 5}{3 \times 6} = \frac{5}{18}

of the original land.

(c) Somu's share of the original land Somu's share is the part of the remaining land left after giving shares to Krishna and Bora. Total given away from the remaining part = Krishna's share + Bora's share

= \frac{5}{12} + \frac{5}{18}

To add these fractions, find a common denominator (LCM of 12 and 18 is 36).

= \frac{5 \times 3}{12 \times 3} + \frac{5 \times 2}{18 \times 2} = \frac{15}{36} + \frac{10}{36} = \frac{25}{36}

This is the fraction of the original land given to the children. Somu's share = Remaining land - Land given to children

= \frac{5}{6} - \frac{25}{36}

Convert

\frac{5}{6}

to have a denominator of 36:

= \frac{5 \times 6}{6 \times 6} - \frac{25}{36} = \frac{30}{36} - \frac{25}{36} = \frac{5}{36}

So, Somu keeps

\frac{5}{36}

of the original land for herself.

Final Answer: (a) Krishna got

\frac{5}{12}

of the original land. (b) Bora got

\frac{5}{18}

of the original land. (c) Somu kept

\frac{5}{36}

of the original land for herself.

Q3Figure it Out (Applications of Fraction Multiplication)

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

Solution

Given: Length of the rectangle =

9\frac{3}{5} \text{ ft}

Breadth of the rectangle =

3\frac{3}{4} \text{ ft}

To Find: Area of the rectangle.

Formula: Area of a rectangle = Length

\times

Breadth

Solution: Step 1: Convert mixed fractions to improper fractions. Length =

9\frac{3}{5} = \frac{(9 \times 5) + 3}{5} = \frac{45 + 3}{5} = \frac{48}{5} \text{ ft}

Breadth =

3\frac{3}{4} = \frac{(3 \times 4) + 3}{4} = \frac{12 + 3}{4} = \frac{15}{4} \text{ ft}

Step 2: Multiply the fractions to find the area. Area =

\frac{48}{5} \times \frac{15}{4}

We can simplify before multiplying by cancelling common factors.

= \frac{48 \times 15}{5 \times 4}

48 is divisible by 4 (

48 \div 4 = 12

) and 15 is divisible by 5 (

15 \div 5 = 3

).

= \frac{12 \times 3}{1 \times 1} = 36

The unit for area will be square feet (

\text{ft}^2

).

Final Answer: The area of the rectangle is

36 \text{ ft}^2

.

Q4Figure it Out (Applications of Fraction Multiplication)

Tsewang plants four saplings in a row in his garden. The distance between two saplings is $\frac{3}{4} \text{ m}$ . Find the distance between the first and last sapling.

Solution

Given: Number of saplings = 4 Distance between two adjacent saplings =

\frac{3}{4} \text{ m}

To Find: The distance between the first and the last sapling.

Solution: When 4 saplings are planted in a row, there are 3 gaps or intervals between them. Sapling 1 --- Gap 1 --- Sapling 2 --- Gap 2 --- Sapling 3 --- Gap 3 --- Sapling 4

The total distance between the first and the last sapling is the sum of the lengths of these 3 gaps.

Total distance = (Number of gaps)

\times

(Distance of one gap)

= 3 \times \frac{3}{4}

= \frac{3 \times 3}{4} = \frac{9}{4} \text{ m}

To express this as a mixed fraction, we divide 9 by 4.

9 \div 4 = 2

with a remainder of 1. So, the distance is

2\frac{1}{4} \text{ m}

.

Final Answer: The distance between the first and last sapling is

\frac{9}{4} \text{ m}

or

2\frac{1}{4} \text{ m}

.

Q5Figure it Out (Applications of Fraction Multiplication)

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

Solution

Given: Weight 1 =

\frac{12}{15}

of 500 grams Weight 2 =

\frac{3}{20}

of 4 kg

To Find: Which of the two weights is heavier.

Solution: To compare the weights, we must express them in the same unit. Let's convert both to grams. We know that

1 \text{ kg} = 1000 \text{ grams}

.

Step 1: Calculate Weight 1 in grams. Weight 1 =

\frac{12}{15} \times 500 \text{ g}

First, simplify the fraction

\frac{12}{15}

by dividing numerator and denominator by 3.

\frac{12}{15} = \frac{4}{5}

Weight 1 =

\frac{4}{5} \times 500 = 4 \times \frac{500}{5} = 4 \times 100 = 400 \text{ g}

.

Step 2: Calculate Weight 2 in grams. First, convert 4 kg to grams:

4 \text{ kg} = 4 \times 1000 = 4000 \text{ g}

. Weight 2 =

\frac{3}{20} \times 4000 \text{ g}

= 3 \times \frac{4000}{20} = 3 \times 200 = 600 \text{ g}

Step 3: Compare the two weights. Weight 1 = 400 g Weight 2 = 600 g Since

600 > 400

, Weight 2 is heavier than Weight 1.

Final Answer:

\frac{3}{20}

of 4 kg is heavier.

Q1Figure it Out (Division and Mixed Problems)

Evaluate the following:
$3 \div \frac{7}{9}$ $\frac{14}{4} \div 2$ $\frac{2}{3} \div \frac{2}{3}$ $\frac{14}{6} \div \frac{7}{3}$
$\frac{4}{3} \div \frac{3}{4}$ $\frac{7}{4} \div \frac{1}{7}$ $\frac{8}{2} \div \frac{4}{15}$
$\frac{1}{5} \div \frac{1}{9}$ $\frac{1}{6} \div \frac{11}{12}$ $3\frac{2}{3} \div 1\frac{3}{8}$

$3 \div \frac{7}{9}$	$\frac{14}{4} \div 2$	$\frac{2}{3} \div \frac{2}{3}$	$\frac{14}{6} \div \frac{7}{3}$
$\frac{4}{3} \div \frac{3}{4}$	$\frac{7}{4} \div \frac{1}{7}$	$\frac{8}{2} \div \frac{4}{15}$
$\frac{1}{5} \div \frac{1}{9}$	$\frac{1}{6} \div \frac{11}{12}$	$3\frac{2}{3} \div 1\frac{3}{8}$

Solution

To Find: The value of each division expression.

Formula: To divide by a fraction, we multiply by its reciprocal.

\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

Solution:

$3 \div \frac{7}{9} = 3 \times \frac{9}{7} = \frac{27}{7}$
$\frac{14}{4} \div 2 = \frac{14}{4} \times \frac{1}{2} = \frac{14}{8} = \frac{7}{4}$
$\frac{2}{3} \div \frac{2}{3} = \frac{2}{3} \times \frac{3}{2} = \frac{6}{6} = 1$
$\frac{14}{6} \div \frac{7}{3} = \frac{14}{6} \times \frac{3}{7} = \frac{14 \times 3}{6 \times 7} = \frac{42}{42} = 1$
$\frac{4}{3} \div \frac{3}{4} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$
$\frac{7}{4} \div \frac{1}{7} = \frac{7}{4} \times 7 = \frac{49}{4}$
$\frac{8}{2} \div \frac{4}{15} = 4 \div \frac{4}{15} = 4 \times \frac{15}{4} = 15$
$\frac{1}{5} \div \frac{1}{9} = \frac{1}{5} \times 9 = \frac{9}{5}$
$\frac{1}{6} \div \frac{11}{12} = \frac{1}{6} \times \frac{12}{11} = \frac{12}{66} = \frac{2}{11}$
$3\frac{2}{3} \div 1\frac{3}{8} = \frac{11}{3} \div \frac{11}{8} = \frac{11}{3} \times \frac{8}{11} = \frac{8}{3}$

Final Answer:

Expression	Result
$3 \div \frac{7}{9}$	$\frac{27}{7}$
$\frac{14}{4} \div 2$	$\frac{7}{4}$
$\frac{2}{3} \div \frac{2}{3}$	$1$
$\frac{14}{6} \div \frac{7}{3}$	$1$
$\frac{4}{3} \div \frac{3}{4}$	$\frac{16}{9}$
$\frac{7}{4} \div \frac{1}{7}$	$\frac{49}{4}$
$\frac{8}{2} \div \frac{4}{15}$	$15$
$\frac{1}{5} \div \frac{1}{9}$	$\frac{9}{5}$
$\frac{1}{6} \div \frac{11}{12}$	$\frac{2}{11}$
$3\frac{2}{3} \div 1\frac{3}{8}$	$\frac{8}{3}$

Q2Figure it Out (Division and Mixed Problems)

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 $\frac{1}{4}$ m for each bag and finished the lace. How many bags did she decorate?

(i)

$8 \times \frac{1}{4}$

(ii)

$\frac{1}{8} \times \frac{1}{4}$

(iii)

$8 \div \frac{1}{4}$

(iv)

$\frac{1}{4} \div 8$

(b)

\frac{1}{2}

meter of ribbon is used to make 8 badges. What is the length of the ribbon used for each badge?

(i)

$8 \times \frac{1}{2}$

(ii)

$\frac{1}{2} \div \frac{1}{8}$

(iii)

$8 \div \frac{1}{2}$

(iv)

$\frac{1}{2} \div 8$

(c)

A baker needs

\frac{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?

(i)

$5 \times \frac{1}{6}$

(ii)

$\frac{1}{6} \div 5$

(iii)

$5 \div \frac{1}{6}$

(iv)

$5 \times 6$

Solution

Part (a) Problem: Maria has 8 m of lace and uses

\frac{1}{4}

m for each bag. We need to find the number of bags. Logic: We are dividing the total amount of lace by the amount used per bag. Number of bags = (Total lace)

\div

(Lace per bag) =

8 \div \frac{1}{4}

. Correct Expression: (iii)

8 \div \frac{1}{4}

Solution:

8 \div \frac{1}{4} = 8 \times 4 = 32

Maria decorated 32 bags.

Part (b) Problem:

\frac{1}{2}

m of ribbon is used for 8 badges. We need to find the length per badge. Logic: We are dividing the total length of ribbon by the number of badges. Length per badge = (Total ribbon)

\div

(Number of badges) =

\frac{1}{2} \div 8

. Correct Expression: (iv)

\frac{1}{2} \div 8

Solution:

\frac{1}{2} \div 8 = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}

Each badge uses

\frac{1}{16}

m of ribbon.

Part (c) Problem: A baker has 5 kg of flour and needs

\frac{1}{6}

kg for one loaf. We need to find the number of loaves he can make. Logic: We are dividing the total amount of flour by the amount needed per loaf. Number of loaves = (Total flour)

\div

(Flour per loaf) =

5 \div \frac{1}{6}

. Correct Expression: (iii)

5 \div \frac{1}{6}

Solution:

5 \div \frac{1}{6} = 5 \times 6 = 30

He can make 30 loaves of bread. (Note: Option (iv)

5 \times 6

gives the same result and is arithmetically equivalent to option (iii)).

Final Answer: (a) Correct expression is (iii) $8 \div \frac{1}{4}$ . The simplified answer is 32 bags. (b) Correct expression is (iv) $\frac{1}{2} \div 8$ . The simplified answer is $\frac{1}{16}$ m. (c) Correct expression is (iii) $5 \div \frac{1}{6}$ . The simplified answer is 30 loaves.

Q3Figure it Out (Division and Mixed Problems)

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

Solution

Given: Flour for 12 rotis =

\frac{1}{4}

kg.

To Find: Amount of flour needed for 6 rotis.

Method 1: Unitary Method Step 1: Find the flour needed for 1 roti. Flour per roti = (Total flour)

\div

(Number of rotis)

= \frac{1}{4} \div 12

= \frac{1}{4} \times \frac{1}{12} = \frac{1}{48} \text{ kg}

Step 2: Find the flour needed for 6 rotis. Flour for 6 rotis = (Flour per roti)

\times

6

= \frac{1}{48} \times 6 = \frac{6}{48}

Simplifying the fraction:

= \frac{1}{8} \text{ kg}

Method 2: Proportional Method 6 rotis is half the number of 12 rotis (since

6 = \frac{1}{2} \times 12

). Therefore, the amount of flour needed will also be half of the original amount. Flour for 6 rotis =

\frac{1}{2} \times

(Flour for 12 rotis)

= \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \text{ kg}

Final Answer:

\frac{1}{8}

kg of flour is used to make 6 rotis.

Q4Figure it Out (Division and Mixed Problems)

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

Solution

Given: The expression to evaluate is the sum:

(1 \div \frac{1}{6}) + (1 \div \frac{1}{10}) + (1 \div \frac{1}{13}) + (1 \div \frac{1}{9}) + (1 \div \frac{1}{2})

.

To Find: The value of the sum.

Solution: First, we evaluate each division term. Dividing 1 by a fraction is the same as finding the reciprocal of that fraction.

$1 \div \frac{1}{6} = 1 \times 6 = 6$
$1 \div \frac{1}{10} = 1 \times 10 = 10$
$1 \div \frac{1}{13} = 1 \times 13 = 13$
$1 \div \frac{1}{9} = 1 \times 9 = 9$
$1 \div \frac{1}{2} = 1 \times 2 = 2$

Now, we add these results together: Sum =

6 + 10 + 13 + 9 + 2

Sum =

16 + 13 + 9 + 2

Sum =

29 + 9 + 2

Sum =

38 + 2

Sum =

40

Final Answer: The friend should say the sum is 40.

Q5Figure it Out (Division and Mixed Problems)

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

Solution

Given: Total pages in the novel = 400. Fraction of pages read yesterday =

\frac{1}{5}

. Fraction of pages read today =

\frac{3}{10}

.

To Find: The number of pages remaining to be read.

Solution: Step 1: Calculate the number of pages read each day. Pages read yesterday =

\frac{1}{5} \times 400 = \frac{400}{5} = 80

pages. Pages read today =

\frac{3}{10} \times 400 = 3 \times \frac{400}{10} = 3 \times 40 = 120

pages.

Step 2: Calculate the total number of pages read. Total pages read = Pages read yesterday + Pages read today

= 80 + 120 = 200 \text{ pages}

Step 3: Calculate the number of pages remaining. Pages remaining = Total pages - Total pages read

= 400 - 200 = 200 \text{ pages}

Final Answer: Mira needs to read 200 more pages to finish the novel.

Q6Figure it Out (Division and Mixed Problems)

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

Solution

Given: Distance covered with 1 litre of petrol = 16 km. Total petrol available =

2\frac{3}{4}

litres.

To Find: Total distance the car can go.

Solution: Step 1: Convert the mixed fraction to an improper fraction. Petrol available =

2\frac{3}{4} = \frac{(2 \times 4) + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}

litres.

Step 2: Calculate the total distance. Total distance = (Distance per litre)

\times

(Total litres of petrol)

= 16 \times \frac{11}{4}

= \frac{16 \times 11}{4}

We can simplify by dividing 16 by 4.

= 4 \times 11 = 44 \text{ km}

Final Answer: The car will go 44 km using

2\frac{3}{4}

litres of petrol.

Q7Figure it Out (Division and Mixed Problems)

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

Solution

Given: Time taken by train =

5\frac{1}{6}

hours. Time taken by plane =

\frac{1}{2}

hour.

To Find: The time saved by taking the plane.

Solution: Time saved = Time taken by train - Time taken by plane

Step 1: Convert the mixed fraction to an improper fraction. Time by train =

5\frac{1}{6} = \frac{(5 \times 6) + 1}{6} = \frac{31}{6}

hours.

Step 2: Subtract the times. Time saved =

\frac{31}{6} - \frac{1}{2}

To subtract, we need a common denominator, which is 6. Convert

\frac{1}{2}

to an equivalent fraction with a denominator of 6:

\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}

Now subtract: Time saved =

\frac{31}{6} - \frac{3}{6} = \frac{31 - 3}{6} = \frac{28}{6}

hours.

Step 3: Simplify the result. Divide the numerator and denominator by their greatest common divisor, 2:

\frac{28 \div 2}{6 \div 2} = \frac{14}{3}

hours. To convert to a mixed fraction, divide 14 by 3.

14 \div 3 = 4

with a remainder of 2. So, time saved =

4\frac{2}{3}

hours.

Final Answer: The plane saves

4\frac{2}{3}

hours.

Q8Figure it Out (Division and Mixed Problems)

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

Solution

Given: Fraction of cake eaten by Mariam and cousins =

\frac{4}{5}

. Number of friends sharing the remaining cake = 3.

To Find: The fraction of the whole cake that each friend got.

Solution: Step 1: Find the fraction of the cake remaining. Let the whole cake be 1. Remaining cake =

1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}

. So,

\frac{1}{5}

of the cake was remaining.

Step 2: Divide the remaining cake among the 3 friends. The remaining

\frac{1}{5}

of the cake is shared equally by 3 friends. Share per friend = (Remaining cake)

\div

(Number of friends)

= \frac{1}{5} \div 3

= \frac{1}{5} \times \frac{1}{3} = \frac{1}{15}

Each friend got

\frac{1}{15}

of the whole cake.

Final Answer: Each friend got

\frac{1}{15}

of the cake.

Q9Figure it Out (Division and Mixed Problems)

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

(a)

> \frac{565}{465}

(b)

< \frac{565}{465}

(c)

> \frac{707}{676}

(d)

< \frac{707}{676}

(e)

> 1

(f)

< 1

Solution

Given: The product is

P = \frac{565}{465} \times \frac{707}{676}

.

To Find: The relationship of the product P with the numbers being multiplied and with 1.

Analysis: Step 1: Analyze the first fraction, $\frac{565}{465}$ . Since the numerator (565) is greater than the denominator (465), this fraction is an improper fraction and its value is greater than 1.

\frac{565}{465} > 1

.

Step 2: Analyze the second fraction, $\frac{707}{676}$ . Since the numerator (707) is greater than the denominator (676), this fraction is also an improper fraction and its value is greater than 1.

\frac{707}{676} > 1

.

Step 3: Analyze the product. We are multiplying two numbers, both of which are greater than 1. The product of two numbers that are each greater than 1 will be greater than both of the original numbers.

Therefore:

$P > \frac{565}{465}$ . This means option (a) is correct.
$P > \frac{707}{676}$ . This means option (c) is correct.
Since both fractions are greater than 1, their product must also be greater than 1. This means option (e) is correct.

Options (b), (d), and (f) are incorrect.

Final Answer: The correct options are (a), (c), and (e).

Q10Figure it Out (Division and Mixed Problems)

What fraction of the whole square is shaded?

Solution

Given: A diagram of a square with shaded regions. The pattern is as follows: The main square is divided into four equal smaller squares (quadrants). The top-left quadrant is shaded. The top-right quadrant is itself divided into four smaller squares, and its top-left part is shaded. This process is repeated two more times, creating a nested pattern of shaded squares.

To Find: The total fraction of the whole square that is shaded.

Solution: Let the area of the whole square be 1 unit.

Largest shaded square: This is the top-left quadrant of the whole square. Its area is $\frac{1}{4}$ of the whole. Area 1 = $\frac{1}{4}$ .
Second shaded square: This is the top-left quadrant of the top-right quadrant. Its area is $\frac{1}{4}$ of the area of the top-right quadrant. The top-right quadrant itself has an area of $\frac{1}{4}$ . Area 2 = $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ .
Third shaded square: This is in the top-right quadrant of the second square (which had area 1/16). Its area is $\frac{1}{4}$ of the area of the region it is in. Area 3 = $\frac{1}{4} \times \frac{1}{16} = \frac{1}{64}$ .
Smallest shaded square: This is in the top-right quadrant of the third square (which had area 1/64). Area 4 = $\frac{1}{4} \times \frac{1}{64} = \frac{1}{256}$ .

Total shaded area is the sum of the areas of all shaded parts: Total Area = Area 1 + Area 2 + Area 3 + Area 4

= \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256}

To add these fractions, we find a common denominator, which is 256.

= \frac{1 \times 64}{4 \times 64} + \frac{1 \times 16}{16 \times 16} + \frac{1 \times 4}{64 \times 4} + \frac{1}{256}

= \frac{64}{256} + \frac{16}{256} + \frac{4}{256} + \frac{1}{256}

= \frac{64 + 16 + 4 + 1}{256} = \frac{85}{256}

Final Answer:

\frac{85}{256}

of the whole square is shaded.

Q11Figure it Out (Division and Mixed Problems)

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?

Solution

Given: A colony of ants splits equally at each junction point. The paths lead to two food sources: a mango tree and a sugarcane field. Describing the path diagram:

The initial group splits into a top path and a bottom path.
The top path splits again. One of these new paths leads to the Mango Tree. The other path splits again, with both resulting paths leading to the Sugarcane Field.
The bottom path from the initial split also splits, with both resulting paths leading to the Sugarcane Field.

To Find: The fraction of the original group that reached the mango tree and the fraction that reached the sugarcane field.

Solution: Let the original group of ants be 1.

Path to the Mango Tree:

At the first split, $\frac{1}{2}$ of the ants take the top path.
On the top path, there is a second split. $\frac{1}{2}$ of this group (which is $\frac{1}{2}$ of the original $\frac{1}{2}$ ) goes to the Mango Tree.
Fraction reaching Mango Tree = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ .

Paths to the Sugarcane Field: There are three separate routes leading to the sugarcane field.

Route 1: From the top path. After the second split, the remaining group ( $\frac{1}{2}$ $\frac{1}{2}$ of the top path group, or $\frac{1}{4}$ $\frac{1}{4}$ of the original) reaches a third split. This group splits equally, with both paths going to the sugarcane field.
- Fraction via Route 1a = $\frac{1}{2} \times (\frac{1}{4} \text{ of original}) = \frac{1}{8}$ .
- Fraction via Route 1b = $\frac{1}{2} \times (\frac{1}{4} \text{ of original}) = \frac{1}{8}$ .
Route 2: From the bottom path. At the first split, $\frac{1}{2}$ $\frac{1}{2}$ of the ants take the bottom path. This group reaches a second split, and both resulting paths go to the sugarcane field.
- Fraction via Route 2a = $\frac{1}{2} \times (\frac{1}{2} \text{ of original}) = \frac{1}{4}$ .
- Fraction via Route 2b = $\frac{1}{2} \times (\frac{1}{2} \text{ of original}) = \frac{1}{4}$ .

Total fraction reaching the Sugarcane Field: Add the fractions from all routes leading to the sugarcane field: Total =

\frac{1}{8} + \frac{1}{8} + \frac{1}{4} + \frac{1}{4}

= \frac{2}{8} + \frac{1}{4} + \frac{1}{4}

= \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}

Check: Fraction at Mango Tree + Fraction at Sugarcane Field =

\frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1

. This accounts for the entire colony.

Final Answer:

\frac{1}{4}

of the original group reached the mango tree, and

\frac{3}{4}

of the original group reached the sugarcane field.

Q12Figure it Out (Division and Mixed Problems)

What is $1-\frac{1}{2}$ ? $(1-\frac{1}{2}) \times (1-\frac{1}{3})$ ? $(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{4}) \times (1-\frac{1}{5})$ ? $(1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{4}) \times (1-\frac{1}{5}) \times (1-\frac{1}{6}) \times (1-\frac{1}{7}) \times (1-\frac{1}{8}) \times (1-\frac{1}{9}) \times (1-\frac{1}{10})$ ? Make a general statement and explain.

Solution

To Find: The value of the given expressions and a general statement for the pattern.

Solution: First, let's evaluate each expression by simplifying the terms in the parentheses.

$1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}$
$(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) = (\frac{1}{2}) \times (\frac{3}{3} - \frac{1}{3}) = \frac{1}{2} \times \frac{2}{3} = \frac{1 \times 2}{2 \times 3} = \frac{2}{6} = \frac{1}{3}$
$(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times (1 - \frac{1}{4}) \times (1 - \frac{1}{5})$ $= (\frac{1}{2}) \times (\frac{2}{3}) \times (\frac{3}{4}) \times (\frac{4}{5})$ We can see a pattern of cancellation: the numerator of each fraction cancels with the denominator of the previous fraction. $= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \frac{\cancel{4}}{5} = \frac{1}{5}$
$(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times (1 - \frac{1}{4}) \times (1 - \frac{1}{5}) \times (1 - \frac{1}{6}) \times (1 - \frac{1}{7}) \times (1 - \frac{1}{8}) \times (1 - \frac{1}{9}) \times (1 - \frac{1}{10})$ $= (\frac{1}{2}) \times (\frac{2}{3}) \times (\frac{3}{4}) \times (\frac{4}{5}) \times (\frac{5}{6}) \times (\frac{6}{7}) \times (\frac{7}{8}) \times (\frac{8}{9}) \times (\frac{9}{10})$ Following the cancellation pattern, all intermediate numerators and denominators will cancel out, leaving only the first numerator (1) and the last denominator (10). $= \frac{1}{10}$

General Statement: The product of the series

(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times \dots \times (1 - \frac{1}{n})

for any integer

n \geq 2

is equal to

\frac{1}{n}

.

Explanation: The general term in the product is

(1 - \frac{1}{k}) = \frac{k-1}{k}

. So the product can be written as:

\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \dots \times \frac{n-2}{n-1} \times \frac{n-1}{n}

In this chain of multiplication, the numerator of each fraction (e.g., 2 in

\frac{2}{3}

) is cancelled by the denominator of the preceding fraction (e.g., 2 in

\frac{1}{2}

). This telescoping cancellation continues through the entire series, leaving only the numerator of the very first fraction (1) and the denominator of the very last fraction (n). Therefore, the result is

\frac{1}{n}

.

Final Answer:

$1 - \frac{1}{2} = \frac{1}{2}$
$(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) = \frac{1}{3}$
$(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times (1 - \frac{1}{4}) \times (1 - \frac{1}{5}) = \frac{1}{5}$
The product of the series up to $(1 - \frac{1}{10})$ is $\frac{1}{10}$ .
The general statement is: The product $(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times \dots \times (1 - \frac{1}{n}) = \frac{1}{n}$ .

Q1Figure it Out (Multiplication of Fractions and Whole Numbers)

Tenzin drinks $\frac{1}{2}$ 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?

Solution

Given: Milk Tenzin drinks per day =

\frac{1}{2}

glass

To Find:

Total milk consumed in a week.
Total milk consumed in the month of January.

Solution:

Part 1: Milk consumed in a week Number of days in a week = 7 Total milk consumed in a week = (Milk per day)

\times

(Number of days)

= \frac{1}{2} \times 7

= \frac{7}{2}

Converting to a mixed fraction:

7 \div 2

gives quotient 3 and remainder 1.

= 3\frac{1}{2} \text{ glasses}

Part 2: Milk consumed in January Number of days in January = 31 Total milk consumed in January = (Milk per day)

\times

(Number of days)

= \frac{1}{2} \times 31

= \frac{31}{2}

Converting to a mixed fraction:

31 \div 2

gives quotient 15 and remainder 1.

= 15\frac{1}{2} \text{ glasses}

Final Answer: Tenzin drinks

3\frac{1}{2}

glasses of milk in a week and

15\frac{1}{2}

glasses of milk in the month of January.

Q2Figure it Out (Multiplication of Fractions and Whole Numbers)

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.

Solution

Given: Work done in 8 days = 1 km of water canal. Working days per week = 5 days.

To Find:

Work done in one day.
Work done in one week (5 working days).

Solution:

Part 1: Work done in one day If 1 km is made in 8 days, the work done in one day is:

1 \text{ km} \div 8 \text{ days} = \frac{1}{8} \text{ km/day}

So, in one day, the team can make

\frac{1}{8}

km of the water canal.

Part 2: Work done in one week The team works 5 days a week. Work done in a week = (Work done per day)

\times

(Number of working days)

= \frac{1}{8} \times 5

= \frac{5}{8} \text{ km}

So, they can make

\frac{5}{8}

km of the water canal in a week.

Final Answer: In one day, the team can make

\frac{1}{8}

km of the water canal. If they work 5 days a week, they can make

\frac{5}{8}

km of the water canal in a week.

Q3Figure it Out (Multiplication of Fractions and Whole Numbers)

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?

Solution

Given: Total oil bought per week = 5 litres. Number of families sharing = Manju + 2 neighbours = 3 families.

To Find:

Amount of oil each family gets in a week.
Amount of oil one family gets in 4 weeks.

Solution:

Part 1: Oil per family per week The 5 litres of oil is shared equally among 3 families. Oil per family = Total oil

\div

Number of families

= 5 \div 3

= \frac{5}{3} \text{ litres}

Converting to a mixed fraction:

5 \div 3

gives quotient 1 and remainder 2.

= 1\frac{2}{3} \text{ litres}

Part 2: Oil for one family in 4 weeks Oil for one family in 4 weeks = (Oil per family per week)

\times

(Number of weeks)

= \frac{5}{3} \times 4

= \frac{20}{3} \text{ litres}

Converting to a mixed fraction:

20 \div 3

gives quotient 6 and remainder 2.

= 6\frac{2}{3} \text{ litres}

Final Answer: Each family gets

\frac{5}{3}

(or

1\frac{2}{3}

) litres of oil in a week. One family will get

\frac{20}{3}

(or

6\frac{2}{3}

) litres of oil in 4 weeks.

Q4Figure it Out (Multiplication of Fractions and Whole Numbers)

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

Solution

Given: Moon setting time on Monday = 10 pm. Daily delay in moon set =

\frac{5}{6}

hour.

To Find: How many hours after 10 pm the moon will set on Thursday.

Solution: We need to calculate the total delay from Monday to Thursday. The number of days that have passed from Monday evening to Thursday evening is 3 days (Tuesday, Wednesday, Thursday).

Total delay = (Daily delay)

\times

(Number of days)

= \frac{5}{6} \times 3

= \frac{15}{6}

Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3:

= \frac{15 \div 3}{6 \div 3} = \frac{5}{2}

Converting to a mixed fraction or decimal:

= 2\frac{1}{2} \text{ hours, or } 2.5 \text{ hours}

This means on Thursday, the moon will set

2\frac{1}{2}

hours after 10 pm.

Final Answer: The moon will set

2\frac{1}{2}

hours after 10 pm on Thursday.

Q5Figure it Out (Multiplication of Fractions and Whole Numbers)

Multiply and then convert it into a mixed fraction:

(a)

7 \times \frac{3}{5}

(b)

4 \times \frac{1}{3}

(c)

\frac{9}{7} \times 6

(d)

\frac{13}{11} \times 6

Solution

To Find: The product for each expression and convert it into a mixed fraction.

Solution:

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

7 \times \frac{3}{5} = \frac{7 \times 3}{5} = \frac{21}{5}

To convert to a mixed fraction, we divide 21 by 5.

21 \div 5 = 4

with a remainder of 1. So,

\frac{21}{5} = 4\frac{1}{5}

.

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

4 \times \frac{1}{3} = \frac{4 \times 1}{3} = \frac{4}{3}

To convert to a mixed fraction, we divide 4 by 3.

4 \div 3 = 1

with a remainder of 1. So,

\frac{4}{3} = 1\frac{1}{3}

.

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

\frac{9}{7} \times 6 = \frac{9 \times 6}{7} = \frac{54}{7}

To convert to a mixed fraction, we divide 54 by 7.

54 \div 7 = 7

with a remainder of 5. So,

\frac{54}{7} = 7\frac{5}{7}

.

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

\frac{13}{11} \times 6 = \frac{13 \times 6}{11} = \frac{78}{11}

To convert to a mixed fraction, we divide 78 by 11.

78 \div 11 = 7

with a remainder of 1. So,

\frac{78}{11} = 7\frac{1}{11}

.

Final Answer: (a)

4\frac{1}{5}

(b)

1\frac{1}{3}

(c)

7\frac{5}{7}

(d)

7\frac{1}{11}

Q1Figure it Out (Multiplication of Two Fractions)

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

(a)

\frac{1}{3} \times \frac{1}{5}

(b)

\frac{1}{4} \times \frac{1}{3}

(c)

\frac{1}{5} \times \frac{1}{2}

(d)

\frac{1}{6} \times \frac{1}{5}

Now, find

\frac{1}{12} \times \frac{1}{18}

.

Solution

To Find: The products of the given fractions.

Formula: To multiply two fractions, we multiply their numerators and multiply their denominators.

\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}

Solution:

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

\frac{1 \times 1}{3 \times 5} = \frac{1}{15}

(Conceptually, this is finding

\frac{1}{3}

of a region that is

\frac{1}{5}

of a whole. Dividing a whole into 5 columns and 3 rows creates 15 equal parts, and we are interested in one of them.)

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

\frac{1 \times 1}{4 \times 3} = \frac{1}{12}

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

\frac{1 \times 1}{5 \times 2} = \frac{1}{10}

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

\frac{1 \times 1}{6 \times 5} = \frac{1}{30}

Now, find $\frac{1}{12} \times \frac{1}{18}$ Using the same rule:

\frac{1}{12} \times \frac{1}{18} = \frac{1 \times 1}{12 \times 18}

To calculate

12 \times 18

:

12 \times 10 = 120

12 \times 8 = 96

120 + 96 = 216

So, the product is

\frac{1}{216}

.

Final Answer: (a)

\frac{1}{15}

(b)

\frac{1}{12}

(c)

\frac{1}{10}

(d)

\frac{1}{30}

The product of

\frac{1}{12} \times \frac{1}{18}

is

\frac{1}{216}

.

Q2Figure it Out (Multiplication of Two Fractions)

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

(a)

\frac{2}{3} \times \frac{4}{5}

(b)

\frac{1}{4} \times \frac{2}{3}

(c)

\frac{3}{5} \times \frac{1}{2}

(d)

\frac{4}{6} \times \frac{3}{5}

Solution

To Find: The products of the given fractions.

Formula: To multiply two fractions, we multiply their numerators and multiply their denominators.

\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}

Solution:

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

\frac{2 \times 4}{3 \times 5} = \frac{8}{15}

(Conceptually, a unit square is divided into 5 columns and 3 rows, creating 15 parts. We consider a region that is 4 columns wide and 2 rows high, which covers 8 of the 15 parts.)

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

\frac{1 \times 2}{4 \times 3} = \frac{2}{12}

This fraction can be simplified by dividing the numerator and denominator by their greatest common divisor, 2.

\frac{2 \div 2}{12 \div 2} = \frac{1}{6}

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

\frac{3 \times 1}{5 \times 2} = \frac{3}{10}

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

\frac{4 \times 3}{6 \times 5} = \frac{12}{30}

This fraction can be simplified by dividing the numerator and denominator by their greatest common divisor, 6.

\frac{12 \div 6}{30 \div 6} = \frac{2}{5}

Alternatively, we can simplify before multiplying:

\frac{4}{6} \times \frac{3}{5} = \frac{2}{3} \times \frac{3}{5} = \frac{2 \times 3}{3 \times 5} = \frac{6}{15} = \frac{2}{5}

Final Answer: (a)

\frac{8}{15}

(b)

\frac{1}{6}

(c)

\frac{3}{10}

(d)

\frac{2}{5}

Working with FractionsClass 7 Mathematics NCERT Solutions

Tenzin drinks 12\frac{1}{2}21​ 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?

Solution

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.

Solution

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?

Solution

Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets 56\frac{5}{6}65​ hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?

Solution

Multiply and then convert it into a mixed fraction:

Solution

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

Solution

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

Solution

A water tank is filled from a tap. If the tap is open for 1 hour, 710\frac{7}{10}107​ of the tank gets filled. How much of the tank is filled if the tap is open for

Solution

Solution

Find the area of a rectangle of sides 3343\frac{3}{4}343​ ft and 9359\frac{3}{5}953​ ft.

Solution

Tsewang plants four saplings in a row in his garden. The distance between two saplings is 34\frac{3}{4}43​ m. Find the distance between the first and last sapling.

Solution

Which is heavier: 1215\frac{12}{15}1512​ of 500 grams or 320\frac{3}{20}203​ of 4 kg ?

Solution

Solution

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 14\frac{1}{4}41​ m for each bag and finished the lace. How many bags did she decorate?

8×148 \times \frac{1}{4}8×41​

18×14\frac{1}{8} \times \frac{1}{4}81​×41​

8÷148 \div \frac{1}{4}8÷41​

14÷8\frac{1}{4} \div 841​÷8

8×128 \times \frac{1}{2}8×21​

12÷18\frac{1}{2} \div \frac{1}{8}21​÷81​

8÷128 \div \frac{1}{2}8÷21​

12÷8\frac{1}{2} \div 821​÷8

5×165 \times \frac{1}{6}5×61​

16÷5\frac{1}{6} \div 561​÷5

5÷165 \div \frac{1}{6}5÷61​

5×65 \times 65×6

Solution

If 14\frac{1}{4}41​ kg of flour is used to make 12 rotis, how much flour is used to make 6 rotis?

Solution

Solution

Mira is reading a novel that has 400 pages. She read 15\frac{1}{5}51​ of the pages yesterday and 310\frac{3}{10}103​ of the pages today. How many more pages does she need to read to finish the novel?

Solution

A car runs 16 km using 1 litre of petrol. How far will it go using 2342\frac{3}{4}243​ litres of petrol?

Solution

Amritpal decides on a destination for his vacation. If he takes a train, it will take him 5165\frac{1}{6}561​ hours to get there. If he takes a plane, it will take him 12\frac{1}{2}21​ hour. How many hours does the plane save?

Solution

Mariam's grandmother baked a cake. Mariam and her cousins finished 45\frac{4}{5}54​ of the cake. The remaining cake was shared equally by Mariam's three friends. How much of the cake did each friend get?

Solution

Choose the option(s) describing the product of (565465×707676)(\frac{565}{465} \times \frac{707}{676})(465565​×676707​):

Solution

What fraction of the whole square is shaded?

Solution

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?

Solution

Solution

A water tank is filled from a tap. If the tap is open for 1 hour, 710\frac{7}{10}107​ of the tank gets filled. How much of the tank is filled if the tap is open for

Solution

Solution

Find the area of a rectangle of sides 334 ft3\frac{3}{4} \text{ ft}343​ ft and 935 ft9\frac{3}{5} \text{ ft}953​ ft.

Solution

Tsewang plants four saplings in a row in his garden. The distance between two saplings is 34 m\frac{3}{4} \text{ m}43​ m. Find the distance between the first and last sapling.

Solution

Which is heavier: 1215\frac{12}{15}1512​ of 500 grams or 320\frac{3}{20}203​ of 4 kg ?

Solution

Solution

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 14\frac{1}{4}41​ m for each bag and finished the lace. How many bags did she decorate?

8×148 \times \frac{1}{4}8×41​

18×14\frac{1}{8} \times \frac{1}{4}81​×41​

8÷148 \div \frac{1}{4}8÷41​

14÷8\frac{1}{4} \div 841​÷8

8×128 \times \frac{1}{2}8×21​

12÷18\frac{1}{2} \div \frac{1}{8}21​÷81​

8÷128 \div \frac{1}{2}8÷21​

12÷8\frac{1}{2} \div 821​÷8

5×165 \times \frac{1}{6}5×61​

16÷5\frac{1}{6} \div 561​÷5

5÷165 \div \frac{1}{6}5÷61​

5×65 \times 65×6

Tenzin drinks $\frac{1}{2}$ 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?

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.

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

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

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

Tsewang plants four saplings in a row in his garden. The distance between two saplings is $\frac{3}{4}$ m. Find the distance between the first and last sapling.

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

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 $\frac{1}{4}$ m for each bag and finished the lace. How many bags did she decorate?

$8 \times \frac{1}{4}$

$\frac{1}{8} \times \frac{1}{4}$

$8 \div \frac{1}{4}$

$\frac{1}{4} \div 8$

$8 \times \frac{1}{2}$

$\frac{1}{2} \div \frac{1}{8}$

$8 \div \frac{1}{2}$

$\frac{1}{2} \div 8$

$5 \times \frac{1}{6}$

$\frac{1}{6} \div 5$

$5 \div \frac{1}{6}$

$5 \times 6$

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

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

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

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

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

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

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

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

Tsewang plants four saplings in a row in his garden. The distance between two saplings is $\frac{3}{4} \text{ m}$ . Find the distance between the first and last sapling.

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

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 $\frac{1}{4}$ m for each bag and finished the lace. How many bags did she decorate?

$8 \times \frac{1}{4}$

$\frac{1}{8} \times \frac{1}{4}$

$8 \div \frac{1}{4}$

$\frac{1}{4} \div 8$

$8 \times \frac{1}{2}$

$\frac{1}{2} \div \frac{1}{8}$

$8 \div \frac{1}{2}$

$\frac{1}{2} \div 8$

$5 \times \frac{1}{6}$

$\frac{1}{6} \div 5$

$5 \div \frac{1}{6}$

$5 \times 6$

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

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

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

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

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

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

Tenzin drinks $\frac{1}{2}$ 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?

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.

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