Sharing and Measuring — NCERT Solutions

Given: Garden divided into 7 equal parts.

• Marigold: $\frac{1}{7}$ → 1 part
• Rose: $\frac{3}{7}$ → 3 parts
• Hibiscus: $\frac{3}{7}$ → 3 parts

Check: $\frac{1}{7} + \frac{3}{7} + \frac{3}{7} = \frac{7}{7} = 1$ ✓

Answer: Divide the garden strip into 7 equal parts. Colour/label 1 part as Marigold, 3 parts as Rose, and 3 parts as Hibiscus. (The remaining seeds — Mogra, Jasmine, Lily, Periwinkle — are not planted in this arrangement.)

Given: Garden divided into 7 equal parts.

• Lily: $\frac{3}{7}$ → 3 parts
• Marigold: $\frac{2}{7}$ → 2 parts
• Periwinkle: $\frac{2}{7}$ → 2 parts

Check: $\frac{3}{7} + \frac{2}{7} + \frac{2}{7} = \frac{7}{7} = 1$ ✓

Answer: Divide the garden strip into 7 equal parts. Label 3 parts as Lily, 2 parts as Marigold, and 2 parts as Periwinkle.

Given: Garden divided into 7 equal parts.

• Mogra: $\frac{5}{7}$ → 5 parts
• Hibiscus: $\frac{2}{7}$ → 2 parts

Check: $\frac{5}{7} + \frac{2}{7} = \frac{7}{7} = 1$ ✓

Answer: Divide the garden strip into 7 equal parts. Label 5 parts as Mogra and 2 parts as Hibiscus.

Given: The dosa is divided into parts (refer to textbook image). Based on the image description, the dosa appears to be divided into 4 equal parts with 2 toppings shown.

Note: The exact shading is in the textbook image. Applying the concept that the dosa is divided into equal parts:

If the dosa is divided into 4 equal parts and Chilli paneer covers 2 parts and Classic potato covers 2 parts:

1. Chilli paneer = $\frac{2}{4}$ (or $\frac{1}{2}$)
2. Classic potato = $\frac{2}{4}$ (or $\frac{1}{2}$)

(Students should count the actual shaded parts in their textbook image and write the fraction as: number of parts with that topping ÷ total number of equal parts.)

Given: The dosa is divided into equal parts with 3 toppings (refer to textbook image).

Concept: Fraction = $\frac{\text{number of parts with that topping}}{\text{total equal parts}}$

If the dosa is divided into 4 equal parts with each topping getting some parts:
Example answer (students verify with their image):

1. Tangy tomato = $\frac{1}{4}$
2. Classic potato = $\frac{2}{4}$
3. Spicy onion = $\frac{1}{4}$

Check: $\frac{1}{4} + \frac{2}{4} + \frac{1}{4} = \frac{4}{4} = 1$ ✓

Given: The dosa is divided into equal parts with 2 toppings (refer to textbook image).

Concept: Fraction = $\frac{\text{number of parts with that topping}}{\text{total equal parts}}$

Example answer (students verify with their image):

1. Spicy onion = $\frac{3}{4}$
2. Tangy tomato = $\frac{1}{4}$

Check: $\frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$ ✓

Given: The dosa is divided into equal parts with 3 toppings (refer to textbook image).

Concept: Fraction = $\frac{\text{number of parts with that topping}}{\text{total equal parts}}$

Example answer (students verify with their image):

1. Tangy tomato = $\frac{2}{6}$
2. Classic potato = $\frac{3}{6}$
3. Spicy onion = $\frac{1}{6}$

Check: $\frac{2}{6} + \frac{3}{6} + \frac{1}{6} = \frac{6}{6} = 1$ ✓

Given: Dosa to be divided into 3 equal parts.

• Spicy onion: $\frac{2}{3}$ → 2 parts out of 3
• Classic potato: $\frac{1}{3}$ → 1 part out of 3

Check: $\frac{2}{3} + \frac{1}{3} = \frac{3}{3} = 1$ ✓

Answer: Draw a circle (dosa) and divide it into 3 equal parts. Shade/label 2 parts as Spicy onion and 1 part as Classic potato.

Given: Dosa to be divided into 8 equal parts.

• Classic potato: $\frac{3}{8}$ → 3 parts
• Chilly paneer: $\frac{1}{8}$ → 1 part
• Tangy tomato: $\frac{4}{8}$ → 4 parts

Check: $\frac{3}{8} + \frac{1}{8} + \frac{4}{8} = \frac{8}{8} = 1$ ✓

Answer: Draw a circle (dosa) and divide it into 8 equal parts. Label 3 parts as Classic potato, 1 part as Chilly paneer, and 4 parts as Tangy tomato.

Given: Total cookies = 12.

Concept: When 12 cookies are shared equally among $n$ children, each child gets $\frac{12}{n}$ cookies, and each child's share as a fraction of the whole is $\frac{1}{n}$.

a) 3 children:
Each child gets $\frac{12}{3} = 4$ cookies.
Fraction each child gets = $\frac{1}{3}$ of the total (i.e., 4 out of 12).
Circle groups of 4 cookies each.

b) 6 children:
Each child gets $\frac{12}{6} = 2$ cookies.
Fraction each child gets = $\frac{1}{6}$ of the total (i.e., 2 out of 12).
Circle groups of 2 cookies each.

c) 2 children:
Each child gets $\frac{12}{2} = 6$ cookies.
Fraction each child gets = $\frac{1}{2}$ of the total (i.e., 6 out of 12).
Circle groups of 6 cookies each.

d) 4 children:
Each child gets $\frac{12}{4} = 3$ cookies.
Fraction each child gets = $\frac{1}{4}$ of the total (i.e., 3 out of 12).
Circle groups of 3 cookies each.

Given: A group of friends (refer to textbook image for the number). $\frac{1}{3}$ of them receive hairbands.

Method: Count the total number of friends shown. Divide that number by 3. The result is the number of friends who get hairbands.

Example: If there are 9 friends shown:
$\frac{1}{3}$ of 9 = $9 \div 3 = 3$ friends get hairbands.

Answer: Divide the friends into 3 equal groups. Place hairbands on 1 group (one-third of the total). (Students should count the friends in their textbook image and apply this method.)

Given: A number of pots shown in the textbook image.

Method: Count the total number of pots. Divide by 5. Draw flowers in that many pots.

Example: If there are 10 pots:
$\frac{1}{5}$ of 10 = $10 \div 5 = 2$ pots should have flowers drawn in them.

Answer: Count the pots, divide by 5, and draw flowers in that many pots. (Students should count the actual pots in their textbook image and apply this method.)

Given: Total barfis = 16. Divided into 4 equal parts.

Working:
$$\frac{1}{4} \text{ of } 16 = 16 \div 4 = 4$$

Draw 16 barfis arranged in 4 equal groups of 4.

Answer: Each part contains $\mathbf{4}$ barfis. $\frac{1}{4}$ of 16 = 4.

Given: Mohan's ribbon length is shown in the textbook. Mohan's ribbon = $\frac{1}{4}$ of Rohan's ribbon.

Concept: If Mohan's ribbon = $\frac{1}{4}$ of Rohan's ribbon, then Rohan's ribbon = 4 × Mohan's ribbon.

Method: Measure Mohan's ribbon length. Multiply by 4 to get Rohan's ribbon length.

Example: If Mohan's ribbon = 3 cm, then Rohan's ribbon = $4 \times 3 = 12$ cm.

Answer: Rohan's ribbon is 4 times as long as Mohan's ribbon. Draw Rohan's ribbon as 4 copies of Mohan's ribbon placed end to end.

Given: A rectangle folded into 3 equal parts; 1 part is coloured = $\frac{1}{3}$.

Step 1: Fold into 3 equal parts → coloured part = $\frac{1}{3}$
Step 2: Fold in half (now 6 equal parts) → coloured part = $\frac{2}{6}$
Step 3: Fold in half again (now 12 equal parts) → coloured part = $\frac{4}{12}$
Step 4: Fold in half again (now 24 equal parts) → coloured part = $\frac{8}{24}$
Step 5: Fold in half again (now 48 equal parts) → coloured part = $\frac{16}{48}$

All these fractions are equal to $\frac{1}{3}$ — they are called equivalent fractions.

Answer: $$\frac{1}{3} = \frac{2}{6} = \frac{4}{12} = \frac{8}{24} = \frac{16}{48}$$

Observation: Each time we fold in half, both the numerator and denominator double, but the value of the fraction stays the same.

Given: A rectangle folded into 2 equal parts; 1 part is coloured = $\frac{1}{2}$.

Step 1: Fold into 2 equal parts → coloured part = $\frac{1}{2}$
Step 2: Fold in half (now 4 equal parts) → coloured part = $\frac{2}{4}$
Step 3: Fold in half again (now 8 equal parts) → coloured part = $\frac{4}{8}$
Step 4: Fold in half again (now 16 equal parts) → coloured part = $\frac{8}{16}$
Step 5: Fold in half again (now 32 equal parts) → coloured part = $\frac{16}{32}$

All these are equivalent fractions equal to $\frac{1}{2}$.

Answer: $$\frac{1}{2} = \frac{2}{4} = \frac{4}{8} = \frac{8}{16} = \frac{16}{32}$$

Given: Fraction chart showing $\frac{1}{2}$ and $\frac{1}{4}$ strips.

Working: $\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

Answer: Two pieces of $\frac{1}{4}$ are equal to $\frac{1}{2}$.

Given: Compare $\frac{2}{3}$ and $\frac{1}{2}$.

Method using fraction chart: Place two $\frac{1}{3}$ pieces against one $\frac{1}{2}$ piece.

Working: $\frac{2}{3}$ means 2 out of 3 equal parts. $\frac{1}{2}$ means 1 out of 2 equal parts.
Converting to same denominator: $\frac{2}{3} = \frac{4}{6}$ and $\frac{1}{2} = \frac{3}{6}$.
Since $\frac{4}{6} > \frac{3}{6}$, we have $\frac{2}{3} > \frac{1}{2}$.

Answer: $\frac{2}{3}$ is greater than $\frac{1}{2}$.

Working: $\underbrace{\frac{1}{10} + \frac{1}{10} + \cdots + \frac{1}{10}}_{10 \text{ times}} = \frac{10}{10} = 1$

Answer: True. Ten pieces of $\frac{1}{10}$ make exactly one complete whole, because $\frac{10}{10} = 1$.

Working:
Three pieces of $\frac{1}{6}$: $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

Two pieces of $\frac{1}{8}$: $\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$

Since $\frac{1}{2} \neq \frac{1}{4}$, the statement is false.

Answer: False. Three pieces of $\frac{1}{6}$ equal $\frac{1}{2}$, while two pieces of $\frac{1}{8}$ equal $\frac{1}{4}$. These are not equal.

Working: We need to find $n$ such that $n \times \frac{1}{8} = \frac{1}{4}$.
$$n = \frac{1}{4} \div \frac{1}{8} = \frac{1}{4} \times 8 = 2$$

Verification: $\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$ ✓

Answer: 2 pieces of $\frac{1}{8}$ make $\frac{1}{4}$.

Using the fraction chart, here are some combinations:

• $\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$ (two quarters make a half)
• $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$ (three sixths make a half)
• $\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$ (two eighths make a quarter)
• $\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$
• $\frac{1}{4} + \frac{1}{8} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$

Answer: Many combinations of smaller pieces can be put together to make a bigger piece. The key is that the fractions must add up to equal the bigger piece.

Given: Several squares divided in different ways (refer to textbook images).

Concept: A square is cut into equal parts only when all resulting pieces have the same size and shape.

Method: For each square:

1. Check if all parts are equal in size.
2. If yes, circle it and write the fraction: $\frac{\text{number of shaded parts}}{\text{total equal parts}}$.
3. If parts are unequal, do not circle it.

Answer: Circle only the squares where all parts are equal. For each circled square, count the total parts and the shaded parts, then write the fraction. (Students apply this rule to each square shown in their textbook image.)

Given: Six pictures with shaded parts and children's claims about the fractions.

Concept: A fraction is correct only if (a) the shape is divided into equal parts, and (b) the number of shaded parts matches the numerator while the total equal parts match the denominator.

Method for each picture:

1. Check if all parts are equal.
2. Count total parts (= denominator).
3. Count shaded parts (= numerator).
4. Compare with the child's claim.

General guidance (students verify with their textbook images):

• If the shape is divided into 3 equal parts and 1 is shaded → $\frac{1}{3}$ is correct.
• If parts are unequal, the fraction claim is incorrect regardless of the numbers.
• You can draw additional lines to make parts equal and then verify.

Answer: Circle the claims that match the actual equal division and shading. Cross out claims where either the parts are unequal or the fraction does not match the shading. (Students should examine each image carefully.)

Given: Six pictures with coloured parts (refer to textbook images).

Concept: Fraction = $\frac{\text{number of coloured parts}}{\text{total number of equal parts}}$

Method: For each picture, count the total equal parts and the number of coloured parts, then write the fraction.

Answer: (Students count the parts in each image and write the fraction.)
Example answers based on typical textbook figures:

• Figure 1: $\frac{1}{2}$
• Figure 2: $\frac{2}{4}$ or $\frac{1}{2}$
• Figure 3: $\frac{3}{6}$ or $\frac{1}{2}$
• Figure 4: $\frac{1}{3}$
• Figure 5: $\frac{2}{5}$
• Figure 6: $\frac{3}{8}$

(Actual answers depend on the specific images in the textbook.)

Given: Six pictures with blue parts (refer to textbook images).

Concept: Fraction of blue parts = $\frac{\text{number of blue parts}}{\text{total equal parts}}$

Method: Count the total equal parts in each figure and the number of blue parts. Write the fraction.

Answer: (Students count the blue parts and total parts in each image.)
Example answers:

• Figure 1: $\frac{1}{4}$
• Figure 2: $\frac{2}{6}$ or $\frac{1}{3}$
• Figure 3: $\frac{3}{8}$
• Figure 4: $\frac{2}{5}$
• Figure 5: $\frac{1}{3}$
• Figure 6: $\frac{3}{4}$

(Actual answers depend on the specific images in the textbook.)

Given: Four shapes to be divided and shaded.

a) Shade $\frac{2}{3}$:
Divide the shape into 3 equal parts. Shade any 2 of the 3 parts.

b) Shade $\frac{4}{6}$:
Divide the shape into 6 equal parts. Shade any 4 of the 6 parts.
Note: $\frac{4}{6} = \frac{2}{3}$, so this is the same amount as (a).

c) Shade $\frac{1}{4}$:
Divide the shape into 4 equal parts. Shade 1 part.
To also show $\frac{1}{8}$: Divide the shape into 8 equal parts. Shade 1 part (which is half the size of $\frac{1}{4}$).

d) Shade $\frac{3}{4}$:
Divide the shape into 4 equal parts. Shade 3 of the 4 parts.

Answer: Draw the dividing lines carefully to ensure all parts are equal, then shade the required number of parts as described above.