Exercise 3.1 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.1$
Question $1 .$
Complete the table.

Answer:

Question $2 .$
Find the product of the terms.
(i) $-2 m n,(2 m)^{2},-3 m n$
(ii) $3 x^{2} y,-3 x y^{3}, x^{2} y^{2}$
Answer:
(i) $(-2 m n) \times(2 m)^{2} \times(-3 m n)=(-2 m n) \times 2^{2} m^{2} \times(-3 m n)=(-2 m n) \times 4 m^{2} \times(-3 m n)$
$\begin{aligned}
&=(-)(+)(-)(2 \times 4 \times 3)\left(\mathrm{m} \times \mathrm{m}^{2} \times \mathrm{m}\right)(\mathrm{n} \times \mathrm{n}) \\
&=+24 \mathrm{~m}^{4} 4 \mathrm{n}^{2}
\end{aligned}$
(ii) $\left(3 x^{2} y\right) \times\left(-3 x y^{3}\right) \times\left(x^{2} y^{2}\right)=(+) \times(-) \times(+) \times(3 \times 3 \times 1)\left(x^{2} \times x \times x^{2}\right) \times\left(y \times y^{3} \times y^{2}\right)$ $=-9 x^{5} y^{6}$

Question $3 .$
If $1=4 p q^{2}, b=-3 p^{2} q h=2 p^{3} q^{3}$ then, find the value of $1 \times b \times h$.
Answer:
Given $1=4 p q^{2}$
$\begin{aligned}
&b=-3 p^{2} q \\
&h=2 p^{3} q^{3} \\
&1 \times b \times h=\left(4 p q^{2}\right) \times-3 p^{2} q \times 2 p^{3} q^{3} \\
&=(+)(-)(+)(4 \times 3 \times 2)\left(p \times p^{2} \times p^{3}\right)\left(q \times q^{2} \times q^{3}\right) \\
&=-24 p^{6} q^{6}
\end{aligned}$

Question $1 .$
Expand
(i) $5 x(2 y-3)$
(ii) $-2 p\left(5 p^{2}-3 p+7\right)$
(iii) $3 \mathrm{mn}\left(\mathrm{m}^{3} \mathrm{n}^{3} \quad 5 \mathrm{~m}^{2} \mathrm{n}+7 \mathrm{mn}^{2}\right)$
(iv) $x^{2}(x+y+z)+y^{2}(x+y+z)+z^{2}(x-y-z)$
Answer:
(i) $5 x(2 y-3)$
$\begin{aligned}
&5 x(2 y-3)=(5 x)(2 y)-(5 x)(3) \\
&=(5 \times 2)(x \times y)-(5 \times 3) x \\
&-10 x y-15 x
\end{aligned}$
(ii) $-2 p\left(5 p^{2}-3 p+7\right)$
$\begin{aligned}
&-2 \mathrm{p}\left(5 \mathrm{p}^{2}-3 \mathrm{p}+7\right)=(-2 \mathrm{p})\left(5 \mathrm{p}^{2}\right)+(-2 \mathrm{p})(-3 \mathrm{p})+(-2 \mathrm{p})(7) \\
&=\left[(-)(+)(2 \times 5)\left(\mathrm{p} \times \mathrm{p}^{2}\right)\right]+[(-)(-)(2 \times 3)(\mathrm{p} \times \mathrm{p})]+(-)(+)(2 \times 7) \mathrm{p} \\
&=-10 \mathrm{p}^{3}+6 \mathrm{p}^{2}-14 \mathrm{p}
\end{aligned}$
(iii) $3 m n\left(m^{3} n^{3}-5 m^{2} n+7 m^{2}\right)$
$\begin{aligned}
&3 m n\left(m^{3} n^{3}-5 m^{2} n+7 m n^{2}\right)=(3 m n)\left(m^{3} n^{3}\right)+(3 m n)\left(-5 m^{3} n\right)+(3 m n)\left(7 m^{3}\right) \\
&=(3)\left(m \times m^{3}\right)\left(n \times n^{3}\right)+(+)(-)(3 \times 5)\left(m \times m^{2}\right)(n \times n)+(3 \times 7)(m \times m)\left(n \times n^{2}\right) \\
&=3 m^{4} n^{4} \quad 15 m^{3} n^{2}+21 m^{2} n^{3}
\end{aligned}$
(iv) $x^{2}(x+y+z)+y^{2}(x+y+z)+z^{2}(x-y-z)$
$\begin{aligned}
&x^{2}(x+y+z)+y^{2}(x+y+z)+z^{2}(x-y-z)=\left(x^{2} \times x\right)+\left(x^{2} \times y\right)+\left(x^{2} \times z\right)+\left(y^{2} \times x\right)+\left(y^{2} \times y\right)+\left(y^{2} \times\right. \\
&z)+\left(z^{2} \times x\right)+z^{2}(-y)+z^{2}(-z) \\
&=x^{3}+x^{2} y+x^{2} z+x y^{2}+y^{3}+y^{2} z+x z^{2}-y z^{2}-z^{3} \\
&=x^{3}+y^{3}-z^{3}+x^{2} y+x^{2} z+x y^{2}+z y^{2}+x z^{2}-y z^{2}
\end{aligned}$

Question $5 .$
Find the product of
(i) $(2 \mathrm{x}+3)(2 \mathrm{x}-4)$
(ii) $\left(y^{2}-4\right)\left(2 y^{2}+3 y\right)$
(iii) $\left(m^{2}-n\right)\left(5 m^{2} n^{2}-n^{2}\right)$
(iv) $3(x-5) \times 2(x-1)$
Answer:
(i) $(2 x+3)(2 x-4)$
$(2 x+3)(2 x-4)=(2 x)(2 x-4)+3(2 x-4)$
$=(2 x \times 2 x)-4(2 x)-3(2 x)-3(4)$
$=4 x^{2}-8 x+6 x-12$
$=4 x^{2}+(-8+6) x-12$
$=4 x^{2}-2 x-12$
(ii) $\left(y^{2}-4\right)\left(2 y^{2}+3 y\right)$
$\left(y^{2}-4\right)\left(2 y^{2}+3 y\right)=y^{2}\left(2 y^{2}+3 y\right)-4\left(2 y^{2}\right)-4(3 y)$
$=y^{2}\left(2 y^{2}\right)+y^{2}(3 y)-4\left(2 y^{2}\right)-4(3 y)$
$=2 y^{4}+3 y^{3}-8 y^{2}-12 y$
(iii) $\left(m^{2}-n\right)\left(5 m^{2} n^{2}-n^{2}\right)$
$\left(m^{2}-n\right)\left(5 m^{2} n^{2}-n^{2}\right)=m^{2}\left(5 m^{2} n^{2}-n^{2}\right)-n\left(5 m^{2} n^{2}-n^{2}\right)$
$=m^{2}\left(5 m^{2} n^{2}\right)+m^{2}\left(-n^{2}\right)-n\left(5 m^{2} n^{2}\right)+(-)(-) n\left(n^{2}\right)$
$=5 m^{4} n^{2}-m^{2} n^{2}-5 m^{2} n^{3}+n^{3}$
(iv) $3(x-5) \times 2(x-1)$
$3(x-5) \times 2(x-1)=(3 \times 2)(x-5)(x-1)$
$=6 \times[x(x-1)-5(x-1)]$
$6[x . x-x .1-5 x+(-1)(-) 51]$
$=6\left[x^{2}-x-5 x+5\right]=6\left[x^{2}+(-1-5) x+5\right]$
$=6\left[x^{2}-6 x+5\right]=6 x^{2}-36 x+30$
 

Question $6 .$
Find the missing term.
(i) $6 x y x$ $=-12 x^{3} y$
Answer:
$6 x y \times\left(-2 x^{2}\right)=-12 x^{3} y$

(ii)
$x\left(-15 m^{2} n^{3} p\right)=45 m^{3} n^{3} p^{2}$
Answer:
$-3 m p \times\left(-15 m^{2} n^{3} p\right)=45 m^{3} n^{3} p^{2}$
(iii) $2 y\left(5 x^{2} y-\ldots+3 \quad\right)=10 x^{2} y^{2}-2 x y+6 y^{3}$
Answer:
$2 y\left(5 x^{2} y-x+3 y^{2}\right)=10 x^{2} y^{2}-2 x y+6 y^{3}$

Question $7 .$
Match the following

(A) iv, v, ii, i, iii
(B) $\mathrm{v}, \mathrm{iv}, \mathrm{iii}, \mathrm{ii}, \mathrm{i}$
(C) iv, v, ii, iii, i
(D) iv, v, iii, ii, i
Añawer:
(C) iv, v, ii, iii, i
a) - iv
b) $-v$
c) $-\mathrm{ii}$
d) - iii
e) $\mathrm{i}$
 

Question $8 .$
A car moves at a uniform speed of $(x+30) \mathrm{km} / \mathrm{hr}$. Find the distance covered by the car in $(y+2)$ hours.
(Hint: distance $=$ speed $\times$ time).
Answer:
Speed of the car $=(x+30) \mathrm{km} / \mathrm{hr}$.
Time $=(y+2)$ hours
Distance $=$ Speed $\times$ time
$=(x+30)(y+2)=x(y+2)+30(y+2)$
$=(x)(y)+(x)(2)+(30)(y)+(30)(2)$
$=x y+2 x+30 y+60$
Distance covered $=(x y+2 x+30 y+60) \mathrm{km}$
 

Objective Type Questions
Question 9.
The product of $7 p^{3}$ and $\left(2 p^{2}\right)^{2}$ is
(A) $14 p^{12}$
(B) $28 \mathrm{p}^{7}$
(C) $9 p^{7}$

(D) $11 \mathrm{p}^{12}$
Answer:
(B) $28 \mathrm{p}^{7}$
 

Question $10 .$
The missing terms in the product $-3 \mathrm{~m}^{3} \mathrm{n} \times 9()=__\mathrm{m}^{4} \mathrm{n}^{3}$ are
(A) $\mathrm{mn}^{2}, 27$
(B) $m^{2} n, 27$
(C) $m^{2} n^{2},-27$
(D) $m n^{2},-27$
Answer:
(A) $\mathrm{mn}^{2}, 27$
 

Question $11 .$
If the area of a square is $36 x^{4} y^{2}$ then, its side is
(A) $6 x^{4} y^{2}$
(B) $8 x^{2} y^{2}$
(C) $6 x^{2} y$
(D) $-6 x^{2} y$
Answer:
(C) $6 x^{2} y$

Question $12 .$
If the area of a rectangle is $48 \mathrm{~m}^{2} \mathrm{n}^{3}$ and whose length is $8 \mathrm{mn}^{2}$ then, its breadth is
(A) $6 \mathrm{mn}$
(B) $8 m^{2} n$
(C) $7 \mathrm{~m}^{2} \mathrm{n}^{2}$
(D) $6 \mathrm{~m}^{2} \mathrm{n}^{2}$
Answer:
(A) $6 \mathrm{mn}$

Question $13 .$
If the area of a rectangular land is $\left(a^{2}-b^{2}\right)$ sq-units whose breadth is $(a-b)$ then, its length is
(A) $a-b$
(B) $a+b$
(C) $a^{2}-b$
(D) $(a+b)^{2}$
Answer:
(B) $a+b$