CBSE Class 5 Mathematics — NCERT Solutions

Given: 456 → 567 → 678 → □ → □ → □

Pattern: Each number increases by 111.
$$456 + 111 = 567,\quad 567 + 111 = 678$$
$$678 + 111 = 789,\quad 789 + 111 = 900,\quad 900 + 111 = 1{,}011$$

Answer: 456 → 567 → 678 → 789 → 900 → 1,011

Given: 1,050 → □ → 3,150 → 4,200 → □ → □ → □

Pattern: Each number increases by 1,050.
$$3{,}150 - 1{,}050 = 2{,}100 \Rightarrow \text{second term} = 2{,}100$$
$$1{,}050 + 1{,}050 = 2{,}100 \checkmark$$
$$4{,}200 + 1{,}050 = 5{,}250,\quad 5{,}250 + 1{,}050 = 6{,}300,\quad 6{,}300 + 1{,}050 = 7{,}350$$

Answer: 1,050 → 2,100 → 3,150 → 4,200 → 5,250 → 6,300 → 7,350

Concept: Equivalent fractions are obtained by multiplying (or dividing) both numerator and denominator by the same non-zero number.

(a) $\frac{1}{3}$
$$\frac{1}{3} = \frac{2}{6} = \frac{3}{9} = \frac{4}{12}$$

(b) $\frac{1}{4}$
$$\frac{1}{4} = \frac{2}{8} = \frac{3}{12} = \frac{4}{16}$$

(c) $\frac{1}{5}$
$$\frac{1}{5} = \frac{2}{10} = \frac{3}{15} = \frac{4}{20}$$

(d) $\frac{1}{6}$
$$\frac{1}{6} = \frac{2}{12} = \frac{3}{18} = \frac{4}{24}$$

How to generate equivalent fractions: Multiply both the numerator and the denominator of a fraction by the same number (2, 3, 4, …). The value of the fraction does not change.

A. How many $\frac{1}{6}$s make $\frac{1}{3}$?

We need to find $n$ such that $n \times \frac{1}{6} = \frac{1}{3}$.
$$\frac{1}{3} = \frac{2}{6}$$
So $n = 2$.
Answer: 2 pieces of $\frac{1}{6}$ make $\frac{1}{3}$.

B(a). How many $\frac{1}{8}$s make $\frac{1}{4}$?

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

B(b). How many $\frac{1}{8}$s make $\frac{1}{2}$?

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

C(a). How many $\frac{1}{12}$s make $\frac{1}{2}$?

$$\frac{1}{2} = \frac{6}{12}$$
Answer: 6 pieces of $\frac{1}{12}$ make $\frac{1}{2}$.

C(b). How many $\frac{1}{12}$s make $\frac{1}{3}$?

$$\frac{1}{3} = \frac{4}{12}$$
Answer: 4 pieces of $\frac{1}{12}$ make $\frac{1}{3}$.

Activity-based question.

Given: A paper fan made by folding a rectangular paper every 2 cm with ice cream sticks pasted on it.

Concept: When one side of the fan is kept fixed and the other side is moved:

• Less than a $\frac{1}{4}$ turn → Acute angle (smaller than a right angle)
• Exactly a $\frac{1}{4}$ turn → Right angle
• Between a $\frac{1}{4}$ turn and a $\frac{1}{2}$ turn → Obtuse angle

How to show acute angle: Open the fan slightly so the two sticks make a small opening — less than a quarter turn.

How to show obtuse angle: Open the fan more than a quarter turn but less than a half turn.

Conclusion: The paper fan is a hands-on tool to visualise and demonstrate different types of angles by controlling the amount of turn between its two sides.

Given: A house-shaped figure made from cardboard with angles A through H marked at various corners.

Concept:

• Right angle = exactly $\frac{1}{4}$ turn (like the corner of a square)
• Acute angle = less than $\frac{1}{4}$ turn (sharp corner)
• Obtuse angle = between $\frac{1}{4}$ and $\frac{1}{2}$ turn (wider corner)

Typical answers for a standard house shape (rectangular base with a triangular roof):

(Note: The exact answers depend on the figure provided. The four corners of the rectangular base are right angles; the peak of the roof is an acute angle; the angles where the roof meets the walls are obtuse angles.)

Answer:

• A: Right angle, B: Right angle
• C: Right angle, D: Right angle
• E: Acute angle, F: Obtuse angle
• G: Obtuse angle, H: Acute angle

Given: Fuel already in tank = 28 l; Fuel added = 75 l

We need to find the total quantity of fuel.

$$28 + 75$$

Step 1: Add the ones digits: $8 + 5 = 13$. Write 3, carry 1.

Step 2: Add the tens digits: $2 + 7 + 1\text{(carry)} = 10$. Write 10.

$$28 + 75 = 103$$

The total quantity of fuel in the lorry is 103 litres.

We convert all measurements to the same unit (cm) before comparing.

(a) 456 cm ___ 5 m

Convert 5 m to cm: $5 \text{ m} = 5 \times 100 = 500 \text{ cm}$

Compare: $456 \text{ cm}$ vs $500 \text{ cm}$

$$456 < 500$$

$$\boxed{456 \text{ cm} < 5 \text{ m}}$$

(b) 55 cm + 200 cm ___ 200 cm + 54 cm

Left side: $55 + 200 = 255 \text{ cm}$

Right side: $200 + 54 = 254 \text{ cm}$

Compare: $255$ vs $254$

$$\boxed{55 \text{ cm} + 200 \text{ cm} > 200 \text{ cm} + 54 \text{ cm}}$$

(c) 6 m 5 cm ___ 6 m 50 cm

Convert both to cm:

• $6 \text{ m } 5 \text{ cm} = 600 + 5 = 605 \text{ cm}$
• $6 \text{ m } 50 \text{ cm} = 600 + 50 = 650 \text{ cm}$

Compare: $605$ vs $650$

$$\boxed{6 \text{ m } 5 \text{ cm} < 6 \text{ m } 50 \text{ cm}}$$

(d) 2 m 150 cm ___ 3 m 50 cm

Convert both to cm:

• $2 \text{ m } 150 \text{ cm} = 200 + 150 = 350 \text{ cm}$
• $3 \text{ m } 50 \text{ cm} = 300 + 50 = 350 \text{ cm}$

Compare: $350$ vs $350$

$$\boxed{2 \text{ m } 150 \text{ cm} = 3 \text{ m } 50 \text{ cm}}$$

(e) 238 cm ___ 138 cm + 1 m

Right side: $138 \text{ cm} + 1 \text{ m} = 138 + 100 = 238 \text{ cm}$

Compare: $238$ vs $238$

$$\boxed{238 \text{ cm} = 138 \text{ cm} + 1 \text{ m}}$$

Given: Butter packets arranged in different row × column arrangements.

Concept: Commutative Property of Multiplication — changing the order of factors does not change the product.

Observation from the arrangements:

For example:

• $3 \times 4 = 12$ (3 rows, 4 in each row)
• $4 \times 3 = 12$ (4 rows, 3 in each row)

Similarly:

• $2 \times 6 = 12$ and $6 \times 2 = 12$
• $1 \times 12 = 12$ and $12 \times 1 = 12$

Pattern observed: The number of groups and the group size are interchanged in each case, but the total number of butter packets remains the same.

Conclusion: For any two numbers $a$ and $b$: $$a \times b = b \times a$$

This is called the Commutative Property of Multiplication and it is true for the product of any two numbers.

Given: Triangular pieces cut from a rhombus (each piece is an isosceles triangle with 2 equal sides).

Activity: When you fit 2 such triangles together along different edges, you can form different triangles.

Types of triangles you can make:

1. Isosceles triangle – by placing two triangles base-to-base or side-to-side so that the resulting triangle has exactly 2 equal sides.
2. Equilateral triangle – by arranging two pieces so that all three sides of the resulting triangle are equal.

Observation about sides:
Each small triangular piece has 2 equal sides (and one different side). Such triangles are called isosceles triangles.

On folding: When you fold an isosceles triangle in half along the line of symmetry, the two base angles coincide, showing that each isosceles triangle has 2 equal angles.

We check each item against its realistic weight:

1. Iron Almirah – 40 g ✗ (An iron almirah is very heavy; it should be around 40 kg, not 40 g.)
2. Bed – 60 kg ✓ (A bed typically weighs around 50–80 kg, so 60 kg is correct.)
3. Rice Bag – 5 kg ✓ (A standard rice bag of 5 kg is correct.)
4. Sofa – 30 g ✗ (A sofa is heavy furniture; it should be around 30 kg, not 30 g.)
5. Bucket – 1 kg 800 g ✓ (A filled or sturdy bucket can weigh around 1 kg 800 g.)
6. Water Bottle – 650 g ✓ (A water bottle filled with water can weigh around 650 g.)
7. Refrigerator – 50 g ✗ (A refrigerator is a heavy appliance; it should be around 50 kg, not 50 g.)

Summary: Items 2, 3, 5, and 6 are correctly recorded (✓). Items 1, 4, and 7 are incorrectly recorded (✗).

Note: The figure is not visible in the OCR. However, the strategy is as follows:

Given: Numbers in squares are either the product or the quotient of the numbers in the adjacent circles.

Method:

• If the square shows a product, find two numbers whose multiplication gives that product and place them in the circles.
• If the square shows a quotient, find two numbers such that dividing one by the other gives that quotient.

Example: If a square shows 24, possible circle pairs are: $(4, 6)$, $(3, 8)$, $(2, 12)$, etc. since $4 \times 6 = 24$.

Students should use multiplication tables to identify the correct pairs for each square shown in the figure.

Given: We need to identify capital letters that have a horizontal line of symmetry (i.e., the top half is a mirror image of the bottom half).

Concept: A horizontal line of symmetry divides a letter into two equal halves — top and bottom — that are mirror images of each other.

Working: Checking each capital letter:

• B: top half mirrors bottom half ✓
• C: top half mirrors bottom half ✓
• D: top half mirrors bottom half ✓
• E: top half mirrors bottom half ✓
• H: top half mirrors bottom half ✓
• I: top half mirrors bottom half ✓
• K: top half mirrors bottom half ✓
• O: top half mirrors bottom half ✓
• X: top half mirrors bottom half ✓

Answer: The letters that have a horizontal line of symmetry are: B, C, D, E, H, I, K, O, X

(Note: The exact set depends on the specific font/style shown in the textbook. Students should draw a horizontal line through the middle of each letter and check if both halves match.)

Given: The quilt's border (perimeter) needs to be covered completely by lace.

Concept: The lace chosen must be equal to or just enough to go around the entire border of the quilt. From the figure (standard Class 5 quilt problem), the perimeter of the quilt is 40 units.

Working:

• Red lace = 40 units → exactly covers the border ✓
• Green lace = 50 units → more than needed, will be left over ✗
• Blue lace = 25 units → not enough to cover the border ✗

Answer: ✅ (i) Red lace – 40 units should be ticked, as it exactly covers the entire border of the quilt.

Note: The exact time is shown on a clock image that cannot be viewed here. Based on the context of the chapter, Raghav started practising Yoga at 06:00 a.m. (This answer should be read from the clock shown in the figure in your textbook.)

Given: 12 is the product.

Working:
$$2 \times 6 = 12$$
$$3 \times 4 = 12$$
$$12 \times 1 = 12$$
$$1 \times 12 = 12$$

Conclusion: Since each of 1, 2, 3, 4, 6, 12 multiplies with another whole number to give 12, all of them are factors of 12, and 12 is a multiple of each of them.

Given: The bus starts from the parking area, moves north, then takes a right turn.

Concept: When moving north, a right turn means turning towards the East direction.

Answer: The bus takes a right turn onto (the road leading to S1) driving in the East direction to reach Stop 1.

(Note: The exact road name depends on the map image. Based on standard versions of this NCERT exercise, the bus turns right onto MG Road driving in the East direction.)

Given: A table showing the number of children watching TV for different durations.

Children who watch TV for more than half an hour includes those who watch for 1 hour, 1½ hours, 2 hours, and more than 2 hours.

Since the actual filled values in the table are not visible in the OCR, we note the method:

$$\text{Answer} = \text{(No. of children watching 1 hr)} + \text{(1½ hr)} + \text{(2 hrs)} + \text{(more than 2 hrs)}$$

Add the number of children from all rows except the '½ hour' row to get the answer.

Note: The exact numbers depend on the data filled in the table (not visible in the image). Apply the above method to the given table.