3 Marks Question

Question 13 Marks

Simplify $3x (4x – 5) + 3$ and find its values for $x = \frac{1}{2}$

Answer

We have $3x (4x - 5) + 3$
$= 3x (4x) - 3x(5) + 3 = 12x^2- 15x + 3$
Now putting $x = \frac{1}{2}$ in above equation, we get
$= 12x^2- 15x + 3$
$= 12\left(\frac{1}{2}\right)^{2}-15\left(\frac{1}{2}\right)+3$
$= 12 \times \frac{1}{4}-\frac{15}{2} + 3$
$= 3 - \frac{15}{2} + 3$
$= 6 - \frac{15}{2}$
$=\frac{12-15}{2}$
$=\frac{-3}{2}$

View full question & answer→

Question 23 Marks

Complete the table of products.

$\frac{\text { First monomial } \rightarrow}{\text { Second monomial } \downarrow}$	$2x$	$-5y$	$3x^2$	$-4xy$	$7x^2y$	$-9x^2y^2$
$2x$	$4x^2$	-	-	-	-	-
$-5y$	-	-	$-15x^2y$	-	-	-
$3x^2$	-	-	-	-	-	-
$-4xy$	-	-	-	-	-	-
$7x^2y$	-	-	-	-	-	-
$-9x^2y^2$	-	-	-	-	-	-

Answer

After doing the product, we get

$\frac{\text { First monomial } \rightarrow}{\text { Second monomial } \downarrow}$	$2x$	$-5y$	$3x^2$	$-4xy$	$7x^2y$	$-9x^2y^2$
$2x$	$4x^2$	$-10xy$	$6x^3$	$-8x^2y$	$14x^3y$	$-18x^3y^2$
$-5y$	$-10xy$	$25y^2$	$-15x^2y$	$20xy^2$	$-35x^2y^2$	$45x^2y^3$
$3x^2$	$6x^3$	$-15x^2y$	$9x^4$	$-12x^3y$	$21x^4y$	$-27x^4y^2$
$-4xy$	$-8x^2y$	$20xy^2$	$-12x^3y$	$16x^2y^2$	$-28x^3y^2$	$36x^3y^3$
$7x^2y$	$14x^3y$	$-35x^2y^2$	$21x^4y$	$-28x^3y^2$	$49x^4y^2$	$-63x^4y^3$
$-9x^2y^2$	$-18x^3y^2$	$45x^2y^3$	$-27x^4y^2$	$36x^3y^3$	$-63x^4y^3$	$81x^4y^4$

View full question & answer→

Question 33 Marks

Subtract: $3pq(p – q)$ from $2pq(p + q).$

Answer

We have $3pq(p – q) = 3p^2q – 3pq^2....(i)$
$2pq(p + q) = 2p^2q + 2pq^2......(ii)$
Subtracting eq$(i)$ from eq$(ii)$,

Thus, the difference is $-p^2q + 5pq^2$

View full question & answer→

Question 43 Marks

Find the volume of each rectangular box with given length, breadth and height.

	Length	Breadth	Height
$1$	$2ax$	$3by$	$5cz$
$2$	$m^2n$	$n^2p$	$p^2m$
$3$	$2q$	$4q^2$	$8q^3$

Answer

As we know that Volume of a cuboid = length \times breadth \times height
Hence, volume of first rectangular $box = (2ax) \times (3by) \times (5cz)$
$= 2 \times 3 \times 5 \times (ax) \times (by) \times (cz) = 30abcxyz$
Volume of $2^{nd}$ rectangular box = $m^{2} n \times n^{2} p \times p^{2} m$
$=\left(m^{2} \times m\right) \times\left(n \times n^{2}\right) \times\left(p \times p^{2}\right)=m^{3} n^{3} p^{3}$
Volume of $3^{rd}$ rectangular box= $2 q \times 4 q^{2} \times 8 q^{3}$
$=2 \times 4 \times 8 \times q \times q^{2} \times q^{3}=64 q^{6}$

View full question & answer→

MATHS 3 Marks Question

Test yourself on this topic