Class 8 CBSE Maths: Cubes, Coordinate Geometry, Factorisation & Identities: 18 MCQs with answers

3

A perfect cube is $n \times n \times n$, so every prime appears three times over — its exponent must be a multiple of 3.

9261

$9261 = 3^3 \times 7^3 = 21^3$, so all exponents are multiples of 3. The others have primes with exponents not divisible by 3.

$-10$

The abscissa is the $x$-coordinate, which is $-10$ in the ordered pair $(-10, -7)$.

Pull out the HCF of coefficients and lowest powers of common variables

Factorisation always begins by removing the highest common factor — this is the step that makes the remaining expression manageable.

$a^2 + 2ab + b^2$

The square of a sum expands to $a^2 + 2ab + b^2$ — the middle term is positive.

True

A number is a perfect cube if and only if every prime in its prime factorisation has an exponent that is a multiple of 3 — this single test decides every cube question in the chapter.

The HCF $12x$ was not fully removed

The HCF of $24$ and $36$ is $12$, not $6$, so $6x$ is not the full HCF. The correct factorisation is $12x(2x^2 - 3y)$.

$67$

Use $x^2 + y^2 = (x-y)^2 + 2xy = 49 + 18 = 67$.

$6400$

Using $a^2 - b^2 = (a+b)(a-b)$: $(82+18)(82-18) = 100 \times 64 = 6400$.

$(a-3)(x-y)$

Since $3 - a = -(a-3)$, the expression becomes $x(a-3) - y(a-3) = (a-3)(x-y)$.

$(3, 0)$

Diagonals of a rhombus bisect each other. Midpoint of $AC$ is $(3, 2)$, so $D = (3, 0)$.

$21$ m

$9261 = 3^3 \times 7^3 = 21^3$, so the side is $21$ m.

8

Volume scales as the cube of the side, so doubling the side multiplies volume by $2^3 = 8$.

$38809$

Write $197 = 200 - 3$: $(200-3)^2 = 40000 - 1200 + 9 = 38809$.

$81x^4 - 16y^4$

First pair gives $9x^2 - 4y^2$. Multiplying by $9x^2 + 4y^2$ gives $81x^4 - 16y^4$.

$10000$

Recognise the perfect square: $64x^2 + 144xy + 81y^2 = (8x + 9y)^2$. Substituting gives $(88 + 12)^2 = 100^2 = 10000$.

The expression evaluates to $10000$

The expression can be recognized as a perfect square: $(8x + 9y)^2$. Substituting $x = 11$ and $y = \frac{4}{3}$ gives $8(11) + 9(\frac{4}{3}) = 100$, so the expression evaluates to $100^2 = 10000$.

Only points A, B, and C are collinear

To check for collinearity, calculate the slopes: $AB = 2$, $BC = 2$, and $CD = 0$. Points A, B, and C share the same slope, indicating they are collinear, while point D has a different slope, confirming it is not collinear with the others.