CBSE Class 6 Mathematics — NCERT Solutions

Given: Various number sequences represented pictorially in Table 2.

Concept: Each sequence follows a specific pattern; to draw the next picture, we identify the rule and extend it.

Working and next pictures for each sequence:

1. Counting Numbers (1, 2, 3, 4, 5, ...): Each picture has one more dot than the previous. The next picture after 5 dots is 6 dots arranged in a row.

2. Odd Numbers (1, 3, 5, 7, 9, ...): Each picture adds 2 more dots. The next picture after 9 dots is 11 dots.

3. Even Numbers (2, 4, 6, 8, 10, ...): Each picture adds 2 more dots. The next picture after 10 dots is 12 dots.

4. Triangular Numbers (1, 3, 6, 10, 15, ...): Each picture forms a triangle by adding one more row. The next triangular number is $15 + 6 = 21$, so the next picture is a triangle with 6 rows (21 dots total).

5. Square Numbers (1, 4, 9, 16, 25, ...): Each picture is a square grid. The next square after $5 \times 5 = 25$ is a $6 \times 6$ grid (36 dots).

6. Cube Numbers (1, 8, 27, 64, 125, ...): Each picture is a cube. The next cube after $5^3 = 125$ is a $6 \times 6 \times 6$ cube (216 dots).

7. Virahānka Numbers (1, 1, 2, 3, 5, 8, 13, ...): Each number is the sum of the two preceding numbers. The next number is $8 + 13 = \mathbf{21}$.

8. Powers of 2 (1, 2, 4, 8, 16, ...): Each picture doubles. The next picture after 16 is 32.

Answer: Draw each sequence as described above, extending by one step following the identified rule.

Triangular Numbers (1, 3, 6, 10, 15, ...):

These are called triangular numbers because the dots representing each number can be arranged in the shape of an equilateral triangle.

• 1 dot → a triangle with 1 row
• 3 dots → a triangle with 2 rows (1 + 2)
• 6 dots → a triangle with 3 rows (1 + 2 + 3)
• 10 dots → a triangle with 4 rows (1 + 2 + 3 + 4)
• 15 dots → a triangle with 5 rows (1 + 2 + 3 + 4 + 5)

In general, the $n$-th triangular number $= 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$.

Square Numbers (1, 4, 9, 16, 25, ...):

These are called square numbers because the dots can be arranged in a perfect square grid.

• $1 = 1 \times 1$, $4 = 2 \times 2$, $9 = 3 \times 3$, $16 = 4 \times 4$, $25 = 5 \times 5$

In general, the $n$-th square number $= n^2$. The shape formed is always a square.

Cube Numbers (1, 8, 27, 64, 125, ...):

These are called cubes because the dots can be arranged to fill a perfect cube (3D box with equal sides).

• $1 = 1^3$, $8 = 2^3$, $27 = 3^3$, $64 = 4^3$, $125 = 5^3$

In general, the $n$-th cube number $= n^3$. The shape formed is always a cube.

Answer: The names come from the geometric shapes that the respective numbers of dots can form — triangles, squares, and cubes.

Rihan's case (one point):
Given: A single point on paper.
Concept: Through a single point, infinitely many lines can be drawn — we can draw lines in any direction through that point.
Answer: Rihan can draw infinitely many (countless) lines through one point.

Sheetal's case (two points):
Given: Two distinct points on paper.
Concept: Through two distinct points, one and only one straight line can be drawn.
Answer: Sheetal can draw exactly one line that passes through both points.

Note: Fig. 2.4 shows points A, B, C, D, E with line segments connecting some of them. Based on the standard version of this figure, the line segments are $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DE}$ (four line segments in a chain).

Line segments named: $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$.

Points on exactly one line segment: A and E — point A is only on $\overline{AB}$, and point E is only on $\overline{DE}$. These are the endpoints at the two ends of the chain.

Points on two line segments: B, C, and D — point B is on $\overline{AB}$ and $\overline{BC}$; point C is on $\overline{BC}$ and $\overline{CD}$; point D is on $\overline{CD}$ and $\overline{DE}$.

Given: A row of numbers. A cell is a supercell if its number is greater than all its adjacent (left and right) neighbours.

Step 1 – List the numbers with their positions:
Position 1: 6828, Position 2: 670, Position 3: 9435, Position 4: 3780, Position 5: 3708, Position 6: 7308, Position 7: 8000, Position 8: 5583, Position 9: 52

Step 2 – Check each cell:

• 6828 (pos 1): only right neighbour is 670. Since $6828 > 670$, it IS a supercell. ✓
• 670 (pos 2): neighbours 6828 and 9435. Since $670 < 6828$, it is NOT a supercell.
• 9435 (pos 3): neighbours 670 and 3780. Since $9435 > 670$ and $9435 > 3780$, it IS a supercell. ✓
• 3780 (pos 4): neighbours 9435 and 3708. Since $3780 < 9435$, it is NOT a supercell.
• 3708 (pos 5): neighbours 3780 and 7308. Since $3708 < 7308$, it is NOT a supercell.
• 7308 (pos 6): neighbours 3708 and 8000. Since $7308 < 8000$, it is NOT a supercell.
• 8000 (pos 7): neighbours 7308 and 5583. Since $8000 > 7308$ and $8000 > 5583$, it IS a supercell. ✓
• 5583 (pos 8): neighbours 8000 and 52. Since $5583 < 8000$, it is NOT a supercell.
• 52 (pos 9): only left neighbour is 5583. Since $52 < 5583$, it is NOT a supercell.

Final Answer: The supercells are 6828, 9435, and 8000.

Given: The coloured (supercell) positions are positions 1 (5346), 4 (1258), and 8 (9635). We must fill the blanks with 4-digit numbers so that ONLY these three cells are supercells.

Concept: A cell is a supercell only if it is greater than both its neighbours. The non-coloured cells must NOT be supercells.

Strategy:

• Position 1 (5346) must be greater than position 2. So position 2 < 5346.
• Position 4 (1258) must be greater than positions 3 and 5. So positions 3 and 5 must be less than 1258, i.e., between 1000 and 1257.
• Position 8 (9635) must be greater than positions 7 and 9. So positions 7 and 9 < 9635.
• Non-coloured cells (2, 3, 5, 6, 7, 9) must not be supercells.

Sample filling:

• Position 2: 4000 (less than 5346 ✓; must not be supercell, so position 3 ≥ 4000)
• Position 3: 1100 (less than 1258 ✓; less than 4000, so position 2 = 4000 > 1100, position 2 is not a supercell since 4000 < 5346 ✓)
• Position 5: 1200 (less than 1258 ✓)
• Position 6: 3000 (greater than 1200, so position 5 is not a supercell ✓; position 6 must not be supercell, so position 7 ≥ 3000)
• Position 7: 8000 (greater than 3000, so position 6 is not a supercell ✓; position 7 must not be supercell, so 8000 < 9635 ✓ but position 6 = 3000 < 8000, so position 7 would be a supercell — adjust)
• Adjust position 7: 9000 (less than 9635 ✓). Position 6 = 3000 < 9000, so position 7 is not a supercell only if position 6 > 9000 — contradiction.
• Better: Position 6: 9100, Position 7: 9200 — but then 9200 > 9635 is false, and 9200 > 9100 makes position 7 a potential supercell.

Simpler valid filling:

Verification:

• 5346: right neighbour 2000 < 5346 ✓ (supercell)
• 2000: neighbours 5346 > 2000, not supercell ✓
• 1100: neighbours 2000 > 1100, not supercell ✓
• 1258: neighbours 1100 < 1258 and 1050 < 1258 ✓ (supercell)
• 1050: neighbours 1258 > 1050, not supercell ✓
• 2500: neighbours 1050 < 2500 and 9100 > 2500, not supercell ✓
• 9100: neighbours 2500 < 9100 and 9635 > 9100, not supercell ✓
• 9635: neighbours 9100 < 9635 and 1000 < 9635 ✓ (supercell)
• 1000: neighbour 9635 > 1000, not supercell ✓

Answer: One valid filling is: | 5346 | 2000 | 1100 | 1258 | 1050 | 2500 | 9100 | 9635 | 1000 |

Given: In the idli-vada game, 'idli' is said for multiples of 3, 'vada' for multiples of 5, and 'idli-vada' for common multiples (multiples of both 3 and 5, i.e., multiples of 15).

Concept: 'Idli-vada' is said at every common multiple of 3 and 5. The LCM of 3 and 5 is 15, so 'idli-vada' is said at every multiple of 15.

The multiples of 15 are: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, …

Counting: 1st → 15, 2nd → 30, 3rd → 45, …, 10th → 150.

$$10 \times 15 = 150$$

'Idli-vada' is said for the 10th time at the number 150.

Given: Perimeter = 14 cm, Breadth (b) = 2 cm

Formula: Perimeter of a rectangle = $2 \times (l + b)$

Working:
$$14 = 2 \times (l + 2)$$
$$\frac{14}{2} = l + 2$$
$$7 = l + 2$$
$$l = 7 - 2 = 5 \text{ cm}$$

Answer: Length = 5 cm

Given: 3 guavas together weigh 1 kg and all are of the same size.

Concept: When 1 whole unit is divided equally among 3 parts, each part = $\frac{1}{3}$ of the whole.

$$\text{Weight of each guava} = 1 \div 3 = \frac{1}{3} \text{ kg}$$

Answer: Each guava will roughly weigh $\dfrac{1}{3}$ kg.

Given: The central line AB = 8 cm, and the wave is made of half circles placed along AB.

Concept: A half circle (semicircle) is drawn with its diameter lying on the central line. If the total length AB = 8 cm is divided equally into two half circles, each half circle has a diameter of 4 cm.

Working:

• Each half circle sits on a portion of AB as its diameter.
• Diameter of each half circle = $\frac{8}{2} = 4$ cm
• Radius = $\frac{\text{Diameter}}{2} = \frac{4}{2} = 2$ cm

Answer:

• The radius to be taken in the compass = 2 cm
• The length of AX (the diameter of the first half circle) = 4 cm

Given: Figures at the start of the chapter (rangoli/kolam-type patterns) and a picture of a cloud.

Concept: A line of symmetry divides a figure into two mirror-image halves that exactly overlap when folded.

Answer:

• The decorative figures (like rangoli patterns) at the start of the chapter generally have lines of symmetry — they can have multiple lines of symmetry (vertical, horizontal, and diagonal) depending on the specific figure.
• A cloud does not have a line of symmetry because its boundary is irregular and uneven; no fold line will make both halves overlap exactly.

Conclusion: The rangoli/decorative figures have lines of symmetry; the cloud does not.

Given: Starting Floor = +2, Movement = -3.

Using the formula: Starting Floor + Movement = Target Floor

$$( +2) + (-3) = -1$$

You will reach Floor -1.