CBSE Class 8 Mathematics — NCERT Solutions

Concept: A perfect square can only end in the digits 0, 1, 4, 5, 6, or 9. A number ending in 2, 3, 7, or 8 is never a perfect square.

(i) 2032 — ends in 2 → Not a perfect square.

(ii) 2048 — ends in 8 → Not a perfect square.

(iii) 1027 — ends in 7 → Not a perfect square.

(iv) 1089 — ends in 9 (possible perfect square). Check: $33^2 = 1089$. ✓ → Is a perfect square.

Answer: (i) 2032, (ii) 2048, and (iii) 1027 are not perfect squares.

Concept: The last digit of a square depends only on the last digit of the base number.

• $64^2$: last digit of base = 4 → last digit of square = $4 \times 4 = 16$ → last digit 6
• $108^2$: last digit of base = 8 → last digit of square = $8 \times 8 = 64$ → last digit 4 ✓
• $292^2$: last digit of base = 2 → last digit of square = $2 \times 2 = 4$ → last digit 4 ✓
• $36^2$: last digit of base = 6 → last digit of square = $6 \times 6 = 36$ → last digit 6

Both $108^2$ and $292^2$ end in 4. Among the options, $108^2$ and $292^2$ have last digit 4.

Answer: $108^2$ and $292^2$ have last digit 4.

Given: A sheet of paper is folded 10 times. Initial thickness = $\nu$.

Concept: Each fold doubles the thickness. So after 1 fold, thickness = $2\nu$; after 2 folds = $2^2\nu$; after $n$ folds = $2^n\nu$.

Working: After 10 folds, thickness = $2^{10} \times \nu = 2^{10}\nu$.

Answer: Option (v) $2^{10}\nu$ is correct.

The other options are incorrect because:

• (i) $10\nu$ represents additive (not multiplicative) growth.
• (ii) $10 + \nu$ is just addition.
• (iii) $2 \times 10 \times \nu = 20\nu$ is linear, not exponential.
• (iv) $2^{10}$ does not include the initial thickness $\nu$.
• (vi) $10^2\nu = 100\nu$ is not the correct base or exponent.

Given: Number = 32400.

Method: Prime factorisation by successive division.

$$32400 \div 2 = 16200$$
$$16200 \div 2 = 8100$$
$$8100 \div 2 = 4050$$
$$4050 \div 2 = 2025$$
$$2025 \div 5 = 405$$
$$405 \div 5 = 81$$
$$81 \div 3 = 27$$
$$27 \div 3 = 9$$
$$9 \div 3 = 3$$
$$3 \div 3 = 1$$

So, $32400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 3 \times 3 \times 3 \times 3$.

In exponential form:
$$\boxed{32400 = 2^4 \times 5^2 \times 3^4}$$

Given: Numbers are represented by collections of sticks (one stick = one unit).

Addition: Place both collections of sticks together into one pile. The resulting collection represents the sum.

Example: ||| + |||| = ||||||| (3 + 4 = 7 sticks)

Subtraction: To subtract a smaller collection B from a larger collection A, pair each stick in B with one stick in A and remove those pairs. The remaining sticks in A represent the difference.

Example: ||||| − ||| → remove 3 pairs → || (5 − 3 = 2 sticks)

Multiplication: To multiply collection A by collection B, make as many copies of collection A as there are sticks in collection B, then combine all copies into one pile.

Example: ||| × || → make 2 copies of ||| → ||| ||| → combine → |||||| (3 × 2 = 6 sticks)

Division: To divide collection A by collection B, repeatedly remove a sub-collection of size equal to B from A and count how many times this can be done. The count (itself represented as a collection of sticks) is the quotient. Any remaining sticks form the remainder.

Example: |||||||| ÷ || → remove || four times → quotient = |||| (8 ÷ 2 = 4)

Thus all four arithmetic operations can be performed purely with physical stick collections, without using Hindu numerals or number names.

Given: Method 2 uses single letters (e.g., a, b, c, …) for numbers. We extend it using strings of letters.

One possible extension (positional/place-value style):

Assign each letter a value: a = 1, b = 2, c = 3, …, z = 26.

For numbers beyond 26, use two-letter strings where the first letter represents the 'tens-like' position and the second the 'units-like' position. For example, treat it like a base-26 system:

• 'aa' = 1×26 + 1 = 27
• 'ab' = 1×26 + 2 = 28
• 'az' = 1×26 + 26 = 52
• 'ba' = 2×26 + 1 = 53
• 'zz' = 26×26 + 26 = 702

For numbers beyond 702, use three-letter strings, and so on. In general, an $n$-letter string can represent numbers up to $26^n + 26^{n-1} + \cdots + 26$.

Another possible extension (additive style):

Keep single letters for 1–26, then use repeated letters: 'aa' = 27, 'aaa' = 28, etc. (i.e., each extra 'a' adds 1 beyond 26). This is simpler but less efficient.

Conclusion: There are many valid ways. The key idea is to use combinations of symbols systematically so that every number has a unique representation, which is the foundation of any number system.

Concept: In a rectangle, all four corner angles are 90°. When a diagonal is drawn, it divides each 90° corner into two parts. The angles formed by the diagonal are related by the properties of parallel lines (alternate interior angles) and the angle-sum property of triangles.

(i) Given: One of the angles formed by the diagonal inside the rectangle is 35°.

Since the corner angle of a rectangle = 90°, the other part of that corner = 90° − 35° = 55°.

By alternate interior angles (diagonal acts as a transversal between parallel sides):

• The angle alternate to 35° = 35°
• The angle alternate to 55° = 55°

So the four angles inside the rectangle (formed by the diagonal) are: 35°, 55°, 35°, 55°.

(ii) Given: One of the angles formed by the diagonal inside the rectangle is 55°.

Other part of the corner = 90° − 55° = 35°.

By alternate interior angles:

• Angle alternate to 55° = 55°
• Angle alternate to 35° = 35°

So the four angles inside the rectangle are: 55°, 35°, 55°, 35°.

A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

(i) 123
Sum of digits $= 1 + 2 + 3 = 6$
6 is not divisible by 9.
∴ 123 is not divisible by 9.

(ii) 405
Sum of digits $= 4 + 0 + 5 = 9$
9 is divisible by 9.
∴ 405 is divisible by 9.

(iii) 8888
Sum of digits $= 8 + 8 + 8 + 8 = 32$
32 is not divisible by 9.
∴ 8888 is not divisible by 9.

(iv) 93547
Sum of digits $= 9 + 3 + 5 + 4 + 7 = 28$
28 is not divisible by 9.
∴ 93547 is not divisible by 9.

(v) 358095
Sum of digits $= 3 + 5 + 8 + 0 + 9 + 5 = 30$
30 is not divisible by 9.
∴ 358095 is not divisible by 9.

Given: The middle number of a 3×3 frame in the multiplication grid is pq, meaning the row header is p and the column header is q.

In the multiplication grid, each entry is the product of its row number and column number. If the centre cell corresponds to row p and column q, then:

• The row above has row number (p−1), and the row below has row number (p+1).
• The column to the left has column number (q−1), and the column to the right has column number (q+1).

Therefore, the nine entries of the 3×3 frame are:

$$\begin{array}{|c|c|c|}\hline (p-1)(q-1) & (p-1)q & (p-1)(q+1) \\ \hline p(q-1) & pq & p(q+1) \\ \hline (p+1)(q-1) & (p+1)q & (p+1)(q+1) \\ \hline\end{array}$$

Expanding each:

• Top-left: $pq - p - q + 1$
• Top-centre: $pq - q$
• Top-right: $pq - p + q - 1$ (wait, let us keep them in factored form as that is cleaner)

The expressions in factored form are:

For example, using the 3×3 frame with centre 4×6 = 24 (p=4, q=6):

• The frame entries are: 3×5=15, 3×6=18, 3×7=21, 4×5=20, 4×6=24, 4×7=28, 5×5=25, 5×6=30, 5×7=35, which matches the grid.