Exercise 3.4 - Chapter 3 Algebra Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.4$
Miscellaneous Practice Problems
Question $1 .$
$6^{2} \times 6^{\mathrm{m}}=6^{5}$, find the value of ' $\mathrm{m}$ '
Solution:
$6^{2} \times 6^{m}=6^{5}$
$6^{2+\mathrm{m}}=6^{5} \text { [Since } \mathrm{a}^{\mathrm{m}} \times \mathrm{a}^{\mathrm{n}}=\mathrm{a}^{\mathrm{m}+\mathrm{n}} \text { ] }$
$6^{2} \times 6^{m}=6^{5}$ $6^{2+m}=6^{5}$ [Since $a^{m} \times a^{n}=a^{m}$ Equating the powers, we get
$\begin{aligned}
&2+\mathrm{m}=5 \\
&\mathrm{~m}=5-2=3
\end{aligned}$

Question 2 .
Find the unit digit of $124^{128} \times 126^{124}$
Solution:
In $124^{128}$, the unit digit of base 124 is 4 and the power is 128 (even power).
Therefore, unit digit of $124^{128}$ is 4 .
Also in $126^{124}$, the unit digit of base 126 is 6 and the. power is 124 (even power).
Therefore, unit digit of $126^{124}$ is 6 .
Product of the unit digits $=6 \times 6=36$
$\therefore$ Unit digit of the $124^{128} \times 126^{124}$ is 6 .

Question $3 .$
Find the unit digit of the numeric expression: $16^{23}+71^{48}+59^{61}$
Solution:
In $16^{23}$, the unit digit of base 16 is 6 and the power is 23 (odd power).
Therefore, unit digit of $16^{23}$ is 6 .
In $71^{48}$, the unit digit of base 71 is 1 and the power is 48 (even power).
Therefore, unit digit of $71^{48}$ is 1 .
Also in $59^{61}$, the unit digit of base 59 is 9 and the power is 61 (odd power).

Therefore, unit digit of $59^{61}$ is 9 .
Sum of the unit digits $=6+1+9=16$
$\therefore$ Unit digit of the given expression is 6 .

Question $4 .$
Find the value of
$\frac{(-1)^{6} \times(-1)^{7} \times(-1)^{8}}{(-1)^{3} \times(-1)^{5}}$
Solution:
$\begin{aligned} \frac{(-1)^{6} \times(-1)^{7} \times(-1)^{8}}{(-1)^{3} \times(-1)^{5}} &=\frac{(-1)^{6+7+8}}{(-1)^{3+5}}=\frac{(-1)^{21}}{(-1)^{8}}=(-1)^{21-8} \text { [By Quotient rule] } \\ &=(-1)^{13}=-1[\text { Since the power } 13 \text { is odd positive number }] \end{aligned}$
$\therefore \frac{(-1)^{6} \times(-1)^{7} \times(-1)^{8}}{(-1)^{3} \times(-1)^{5}}=1$

Question $5 .$
Identify the degree of the expression, $2 \mathrm{a}^{3}$ be $+3 \mathrm{a}^{3} \mathrm{~b}+3 \mathrm{a}^{3} \mathrm{c}-2 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}$
Solution:
The terms of the given expression are $2 a^{3} b c, 3 a^{3} b+3 a^{3} c-2 a^{2} b^{2} c^{2}$
Degree of each of the terms: $5,4,4,6$.
Terms with the highest degree: $-2 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}$
Therefore degree of the expression is 6 .
 

Question 6 .
If $\mathrm{p}=-2, \mathrm{q}=1$ and $\mathrm{r}=3$, find the value of $3 \mathrm{p}^{2} \mathrm{q}^{2} \mathrm{r}$.
Solution:
Given $\mathrm{p}=-2 ; \mathrm{q}=1 ; \mathrm{r}=3$
$\therefore 3 \mathrm{p}^{2} \mathrm{q}^{2} \mathrm{r}=3 \times(-2)^{2} \times(1)^{2} \times(3)$
$=3 \times(-2 \times 1)^{2} \times(3)\left[\right.$ Since $\left.a^{m} \times b^{m}=(a \times b)^{m}\right]$
$=3 \times(-2)^{2} \times(3)$
$=3 \times(-1)^{2} \times 2^{2} \times 3$
$=3^{1+1} \times 1 \times 4\left[\right.$ Since $\left.a^{m} \times a^{n}=a^{m+n}\right]$
$=3^{2} \times 4=9 \times 4$
$\therefore 3 \mathrm{p}^{2} \mathrm{q}^{2} \mathrm{r}=36$

Challenge Problems
Question $7 .$
LEADERS is a WhatsApp group with 256 members. Every one of its member is an admin for their own WhatsApp group with 256 distinct members. When a message is posted in LEADERS and everybody forwards the same to their own group, then how many members in total will receive that message?
Solution:
Members of the groups LEADERS $=256$
Members is individual groups of the members of LEADERS $=256$
Total members who receive the message

$\begin{aligned}
&=256 \times 256=2^{8} \times 2^{8} \\
&2^{8+8}=2^{16} \\
&=65536
\end{aligned}$
Totally 65536 members receive the message.
 

Question $8 .$
Find $x$ such that $3^{x+2}=3^{x}+216$.
Solution:
Given $3^{x+2}=3^{x}+216 ; 3^{x+2}=3^{x}+216$
Dividing throught by $3^{\mathrm{x}}$, we get

Equating the powers of same base
$\begin{aligned}
\frac{3^{x+2}}{3^{x}} &=\frac{3^{x}}{3^{x}}+\frac{2^{3} \times 3^{3}}{3^{x}} \\
3^{x+2-x} &=3^{x-x}+\left(2^{3} \times 3^{3-x}\right) \\
3^{2} &=3^{0}+2^{3} \times 3^{3-x} \\
3^{2}-3^{0} &=2^{3} \times 3^{3-x} \\
9-1 &=2^{3} \times 3^{3-x} \\
8 &=2^{3} \times 3^{3-x} \\
\frac{2^{3}}{2^{3}} &=3^{3-x} \\
2^{3-3} &=3^{3-x} \\
2^{0} &=3^{3-x} \\
1 &=3^{3-x} \\
3^{0} &=3^{3-x}
\end{aligned}$
$\begin{aligned}
&0=3-x \\
&x=3
\end{aligned}$

Question $9 .$
If $X=5 x^{2}+7 x+8$ and $Y=4 x^{2}-7 x+3$, then find the degree of $X+Y$.
Solution:
Given $x=5 x^{2}+7 x+8$
$\mathrm{X}+\mathrm{Y}=5 \mathrm{x}^{2}+7 \mathrm{x}+8+\left(4 \mathrm{x}^{2}-7 \mathrm{x}+3\right)$
$=\left(5 x^{2}+4 x^{2}\right)+(7 x-7 x)+(8+3)$
$=x^{2}(5+4)+x(7-7)+(8+3)=9 x^{2}+11$
Degree of the expression is 2 .

Question $10 .$
Find the degree of $\left(2 \mathrm{a}^{2}+3 \mathrm{ab}-\mathrm{b}^{2}\right)-\left(3 \mathrm{a}^{2}-\mathrm{ab}-3 \mathrm{~b}^{2}\right)$
Solution:
$\begin{aligned}
&\left(2 a^{2}+3 a b-b^{2}\right)-\left(3 a^{2}-a b-3 b^{2}\right) \\
&=\left(2 a^{2}+3 a b-b^{2}\right)+\left(-3 a^{2}+a b+3 b^{2}\right) \\
&=2 a^{2}+3 a b-b^{2}-3 a^{2}+a b+3 b^{2} \\
&=2 a^{2}-3 a^{2}+3 a b+a b+3 b^{2}-b^{2} \\
&=2 a^{2}-3 a^{2}+a b(3+1)+b^{2}(3-1) \\
&=-a^{2}+4 a b+2 b^{2}
\end{aligned}$
Hence degree of the expression is 2 .

Question 11 .
Find the value of $w$, given that $x=4, y=4, z=-2$ and $w=x^{2}-y^{2}+z^{2}-x y z$
Solution:
Given $x=3 ; y=4$ and $z=-2$.
$\begin{aligned}
&\mathrm{w}=\mathrm{x}^{2}-\mathrm{y}^{2}+\mathrm{z}^{2}-\mathrm{xyz} \\
&\mathrm{w}=3^{2}-4^{2}+(-2)^{2}-(3)(3)(-2) \\
&\mathrm{w}=9-16+4+24 \\
&\mathrm{w}=37-16 \\
&\mathrm{w}=21
\end{aligned}$

Question 12 .
Simplify and find the degree of $6 x^{2}+1-\left[8 x-\left\{3 x^{2}-7-\left(4 x^{2}-2 x+5 x+9\right)\right\}\right]$
Solution:
$\begin{aligned}
&6 x^{2}+1-\left[8 x-\left(3 x^{2}-7-\left(4 x^{2}-2 x+5 x+9\right)\right\}\right] \\
&=6 x^{2}+1-\left[8 x-\left\{3 x^{2}-7-4 x^{2}-2 x+5 x+9\right\}\right] \\
&\left.=6 x^{2}+1-\left[8 x-3 x^{2}+7+4 x^{2}-2 x+5 x+9\right\}\right] \\
&=6 x^{2}-1-\left[8 x+3 x^{2}-7-4 x^{2}+2 x-5 x-9\right] \\
&\left.=6 x^{2}+3 x^{2}-4 x^{2}-8 x+2 x-5 x-1-7-9\right] \\
&=x^{2}(6+3-4)+x(8+2-5)-15 \\
&=5 x^{2}-11 x-15
\end{aligned}$
Degree of the expression is 2 .
 

Question $13 .$
The two adjacent sides of a rectangle are $2 x^{2}-5 x y+3 z^{2}$ and $4 x y-x^{2}-z^{2}$. Find the perimeter and the degree of the expression.
Solution:
Let the two adjacent sides of the rectangle as
$1=2 x^{2}-5 x y+3 z^{2}$ and $b=4 x y-x^{2} y+3 z^{2}$

Perimeter of the rectangle
$\begin{aligned}
&=2(1+b)=2\left(2 x^{2}-5 x y+3 z^{2}+4 x y-x^{2}-z^{2}\right) \\
&=4 x^{2}-10 x y+6 z^{2}+8 x y-2 x^{2}-2 z^{2}
\end{aligned}$
$\begin{aligned}
&=4 x^{2}-2 x^{2}-10 x y+8 x y+6 z^{2}-2 z^{2} \\
&=x^{2}(4-2)+x y(-10+8)+z^{2}\left(6-2 z^{2}\right)
\end{aligned}$
Perimeter $=2 x^{2}-2 x y+4 z^{2}$
Degree of the expression is 2 .