Exercise 3.2 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
Fill in the blanks

Question $2 .$
Say True or False
(i) $8 x^{3} y \div 4 x^{2}=2 x y$
Answer:
True
(ii) $7 \mathrm{ab}^{3} \div 14 \mathrm{ab}=2 \mathrm{~b}^{2}$
Answer:
False

Question $3 .$
Divide
(i) $27 y^{3}$ by $3 y$
(ii) $x^{3} y^{2}$ by $x^{2} y$
(iii) $45 x^{3} y^{2} z^{4}$ by $(-15 x y z)$
(iv) $(3 x y)^{2}$ by $9 x y$
Answer:
(i) $27 \mathrm{y}^{3}$ by $3 \mathrm{y}$
$\frac{27 y^{3}}{3 y}=\frac{27}{3} y^{3-1}=9 \mathrm{y}^{2}$
(ii) $x^{3} y^{2}$ by $x^{2} y$
$\frac{x^{3} y^{2}}{x^{2} y}=x^{3-2} y^{2-1}=x^{1} y^{1}=x y$
(iii) $45 x^{3} y^{2} z^{4}$ by $(-15 x y z)$
$\frac{45 x^{3} y^{2} z^{4}}{-15 x y z}=\frac{45}{-15} x^{3-2} y^{2-1} y^{4-1} z^{4-1}=-3 x^{2} y z^{3}$
(iv) $(3 x y)^{2}$ by $9 x y$
$\frac{(3 x y)^{2}}{3 \times(3 x y)}=\frac{(3 x y)^{2}}{3 \times(3 x y)}=\frac{1}{3}(3 x y)^{2-1}=\frac{1}{3} \quad 3 x y=x y$

Question $4 .$
Simplify
(i) $\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}$
(ii) $\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}$

Answer:
(i) $\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}$
$\begin{aligned}
&\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}=3 m^{2-1}+2 m^{4-3} \\
&=3 m+2 m \\
&=(3+2) m \\
&=5 m
\end{aligned}$
(ii) $\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}$ $\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}=\frac{14}{2} \mathrm{p}^{5-2} \mathrm{q}^{3-1}-\frac{12}{3} \mathrm{p}^{3} \mathrm{q}^{4-3}$ $=7 \mathrm{p}^{3} \mathrm{q}^{2}-4 \mathrm{p}^{3} \mathrm{q}$
$\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}=\frac{14}{2} \mathrm{p}^{5-2} \mathrm{q}^{3-1}-\frac{12}{3} \mathrm{p}^{3} \mathrm{q}^{4-3}$
$=7 \mathrm{p}^{3} \mathrm{q}^{2}-4 \mathrm{p}^{3} \mathrm{q}$

Question $5 .$
Divide:
(i) $32 y^{2}-8 y z$ by $2 y$
(ii) $\left(4 m^{2} n^{3}+16 m^{4} n^{2}-m n\right)$ by $2 m n$
(iii) $5 x y^{2}-18 x^{2} y^{3}+6 x y$ by $6 x y$
(iv) $81\left(\mathrm{p}^{4} \mathrm{q}^{2} \mathrm{r}^{3}+2 \mathrm{p}^{3} \mathrm{q}^{3} \mathrm{r}^{2}-5 \mathrm{p}^{2} \mathrm{q}^{2} \mathrm{r}^{2}\right)$ by $(3 \mathrm{pqr})^{2}$
Answer:
(i) $32 y^{2}-8 y z$ by $2 y$
$\frac{32 y^{2}-8 y z}{2 y}=\frac{32 y^{2}}{2 y}-\frac{8 y z}{2 y}=\frac{32}{2} y^{2-1}-\frac{8}{2} y^{1-1} z=16 y-4 z$

(ii) $\left(4 m^{2} n^{3}+16 m^{4} n^{2}-m n\right)$ by $2 m n$
$\begin{aligned}
\frac{4 m^{2} n^{3}+16 m^{4} n^{2}-m n}{2 m n} &=\frac{4 m^{2} n^{3}}{2 m n}+\frac{16 m^{4} n^{2}}{2 m n}-\frac{m n}{2 m n} \\
&=\frac{4}{2} m^{2-1} n^{3-1}+\frac{16}{2} m^{4-1} n^{2-1}-\frac{1}{2} m^{1-1} n^{1-1} \\
&=2 m^{1} n^{2}+8 m^{3} n^{1}-\frac{1}{2} m^{0} n^{0} \\
&=2 m n^{2}+8 m^{3} n-\frac{1}{2}
\end{aligned}$
(iii) $5 x y^{2}-18 x^{2} y^{3}+6 x y$ by $6 x y$
$\begin{aligned}
\frac{5 x y^{2}-18 x^{2} y^{3}+6 x y}{6 x y} &=\frac{x y\left(5 y-18 x y^{2}+6\right)}{6 x y} \\
&=\frac{5 y-18 x y^{2}+6}{6}
\end{aligned}$
$\begin{aligned}
&\text { (iv) } 81\left(p^{4} q^{2} r^{3}+2 p^{3} q^{3} r^{2}-5 p^{2} q^{2} r^{2}\right) \text { by }(3 p q r)^{2} \\
&\frac{81\left(p^{4} q^{2} r^{3}+2 p^{3} q^{3} r^{2}-5 p^{2} q^{2} r^{2}\right)}{(3 p q r)^{2}}
\end{aligned}$

$\begin{aligned}
&=\frac{81\left(p^{2} q^{2} r^{2}\right)\left(p^{2} r+2 p q-5\right)}{9\left(p^{2} q^{2} r^{2}\right)} \\
&=\frac{81}{9}\left(p^{2} q^{2} r^{2}\right)^{1-1}\left(p^{2} r+2 p q-5\right) \\
&=9\left(p^{2} r+2 p q-5\right)=9 p^{2} r+18 p q-45
\end{aligned}$
 

Question $6 .$
Identify the errors and correct them.
(i) $7 y^{2}-y^{2}+3 y^{2}=10 y^{2}$
Answer: $7 y^{2}-y^{2}+3 y^{2}=10 y^{2}=(7-1+3) y^{2}$ $=(6+3) y^{2}$ $=9 y^{2}$
(ii) $6 x y+3 x y=9 x^{2} y^{2}$
Answer:
$6 x y+3 x y=(6+3) x y$
$=9 x y$

(iii) $\mathrm{m}(4 \mathrm{~m}-3)=4 \mathrm{~m}^{2}-3$
Answer:
$m(4 m-3)=m(4 m)+m(-3)$
$-4 m^{2} \quad 3 m$
(iv) $(4 n)^{2}-2 n+3=4 n^{2}-2 n+3$
Answer:
$(4 n)^{2}-2 n+3=16 n^{2}-2 n+3$
(v) $(x-2)(x+3)=x^{2}-6$
Answer:
$\begin{aligned}
&(x-2)(x+3)=x(x+3)-2(x+3) \\
&=x(x)+(x) \times 3+(-2)(x)+(-2)(3) \\
&=x^{2}+3 x-2 x-6 \\
&=x^{2}+x-6
\end{aligned}$
(vi) $-3 \mathrm{p}^{2}+4 \mathrm{p}-7=-\left(3 \mathrm{p}^{2}+4 \mathrm{p}-7\right)$
Answer:
$-3 p^{2}+4 p-7=-\left(3 p^{2}-4 p+7\right)$

Question $7 .$
Statement A: If $24 \mathrm{p}^{2} \mathrm{q}$ is divided by $3 \mathrm{pq}$, then the quotient is $8 \mathrm{p}$.
Statement B: Simplification of $\frac{(5 x+5)}{5}$ is $5 x$.
(i) Both $\mathrm{A}$ and $\mathrm{B}$ are true

(ii) $\mathrm{A}$ is true but $\mathrm{B}$ is false
(iii) $\mathrm{A}$ is false but $\mathrm{B}$ is true
(iv) Both $A$ and $B$ are false

Answer:
(ii) $\mathrm{A}$ is true but $\mathrm{B}$ is false

Question 8

Statement A: $4 \mathrm{x}^{2}+3 \mathrm{x}-2=2\left(2 \mathrm{x}^{2}+\frac{3 x}{2}-1\right)$
Statement B: $(2 \mathrm{~m}-5)-(5-2 \mathrm{~m})=(2 \mathrm{~m}-5)+(2 \mathrm{~m}-5)$
(i) Both $\mathrm{A}$ and $\mathrm{B}$ are true
(ii) $\mathrm{A}$ is true but $\mathrm{B}$ is false
(iii) $\mathrm{A}$ is false but $\mathrm{B}$ is true
(iv) Both $A$ and $B$ are false
Answer:
(i) Both $\mathrm{A}$ and $\mathrm{B}$ are true