Exercise 3.3 - Chapter 3 Algebra Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.3$
Miscellaneous Practice problems
Question 1.
Using identity, find the value of
(i) $(4.9)^{2}$
(ii) $(100.1)^{2}$
(iii) $(1.9) \times(2.1)$
Solution:
(i) $(4.9)^{2}$
$(4.9)^{2}=(5-0.1)^{2}$
Substituting $a=5$ and $b=0.1$ in
$(a-b)^{2}=a^{2}-2 a b+b^{2}$, we have
$(5-0.1)^{2}=5^{2}-2(5)(0.1)+(0.1)^{2}$
$(4.9)^{2}=25-1+0.01=24+0.01$
$(4.9)^{2}=24.01$
(ii) $(100.1)^{2}$
$(100.1)^{2}=(100+0.1)^{2}$
Substituting $a=100$ and $b=0.1$ in
$(a+b)^{2}=a^{2}+2 a b+b^{2}$, we have
$(100+0.1)^{2}=(100)^{2}+2(100)(0.1)+(0.1)^{2}$
$(100.1)^{2}=10000+20+0.01$
$(100.1)^{2}=10020.01$

(iii) $(1.9) \times(2.1)$
$(1.9) \times(2.1)=(2-0.1) \times(2+0.1)$
Substituting $a=100$ and $b=0.1$ in
$(a-b)(a+b)=a^{2}-b^{2}$ we have
$(2-0.1)(2+0.1)=2^{2}-(0.1)^{2}$
$(1.9) \times(2.1)=4-0.01$
(9.9) $(2.1)=3.99$
 

Question $2 .$
Factorise: $4 \mathrm{x}^{2}-9 \mathrm{y}^{2}$
Solution:
$4 x^{2}-9 y^{2}=22 x^{2}-3^{2} y^{2}=(2 x)^{2}-(3 y)^{2}$
Substituting $a=2 x$ and $b=3 y$ in
$\left(a^{2}-b^{2}\right)=(a+b)(a-b)$, we have
$(2 x)^{2}-(3 y)^{2}=(2 x+3 y)(2 x-3 y)$
$\therefore$ Factors of $4 x^{2}-9 y^{2}$ are $(2 x+3 y)$ and $(2 x-3 y)$
 

Question 3.
Simplify using identities
(i) $(3 p+q)(3 p+r)$
(ii) $(3 \mathrm{p}+\mathrm{q})(3 \mathrm{p}-\mathrm{q})$
Solution:
(i) $(3 \mathrm{p}+\mathrm{q})(3 \mathrm{p}+\mathrm{r})$
Substitute $x=3 p, a=q$ and $b=r$ in
$(x+a)(x+b)=x^{2}+x(a+b)+a b$
$(3 p+q)(3 p+r)=(3 p)^{2}+3 p(q+r)+(q \times r)$
$=3^{2} p^{2}+3 p(q+r)+q r$
$(3 p+q)(3 p+r)=9 p^{2}+3 p(q+r)+q r$
(ii) $(3 p+q)(3 p-q)$
Substitute $a=3 p$ and $b=q$ in
$(a+b)(a-b)=a^{2}-b^{2}$, we have $(3 p+q)(3 p-q)=(3 p)^{2}-q^{2}=32 p^{2}-q^{2}$ $(3 P+q)(3 p-q)=9 p^{2}-q^{2}$
$\begin{aligned}
&(3 p+q)(3 p-q)=(3 p)^{2}-q^{2}=32 p^{2}-q^{2} \\
&(3 p+q)(3 p-q)=9 p^{2}-q^{2}
\end{aligned}$

Question $4 .$
Show that $(x+2 y)^{2}-(x-2 y)^{2}=8 x y$.
Sólutionn:
$\begin{aligned}
&\text { LHS }=(x+2 y)^{2}-(x-2 y)^{2} \\
&=x^{2}+(2 \times x \times 2 y)+(2 y)^{2}-\left[x^{2}-(2 \times x \times 2 y)+(2 y)^{2}\right] \\
&=x^{2}+4 x y+4 y^{2}-\left[x^{2}-4 x y+2^{2} y^{2}\right] \\
&=x^{2}+4 x y+4 y^{2}-x^{2}+4 x y-4 y^{2} \\
&=x^{2}-x^{2}+4 x y+4 x y+4 y^{2}-4 y^{2} \\
&=x^{2}(1-1)+x y(4+4)+y^{2}(4-4) \\
&=0 x^{2}+8 x y+0 y^{2}=8 x y=\text { RHS } \\
&\therefore(x+2 y)^{2}-(x-2 y)^{2}=8 x y \\
&{\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}(a-b)^{2}=a^{2}-2 a b+b^{2}\right]}
\end{aligned}$
 

Question 5
The pathway of a square paddy field has $5 \mathrm{~m}$ width and length of its side is $40 \mathrm{~m}$. Find the total area of its pathway. (Note: Use suitable identity)
Solution:
Given side of the square $=40 \mathrm{~m}$
Also width of the pathway $=5 \mathrm{~m}$
$\therefore$ Side of the larger square $=40 \mathrm{~m}+2(5) \mathrm{m}=40 \mathrm{~m}+10 \mathrm{~m}=50 \mathrm{~m}$
Area of the path way = area of large square - area of smaller square
$=50^{2}-40^{2}$

Challenge Problems
Question 1.
If $X=a^{2}-1$ and $Y=1-b^{2}$, then find $X+Y$ and factorize the same.
Solution:
Given $X=a^{2}-1$
$\begin{aligned}
&Y=I-b^{2} \\
&X+Y=\left(a^{2}-1\right)+\left(1-b^{2}\right) \\
&=a^{2}-1+1-b^{2}
\end{aligned}$
We know the identity that $a^{2}-b^{2}=(a+b)(a-b)$
$\therefore \mathrm{X}+\mathrm{Y}=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$
 

Question $2 .$
Find the value of $(x-y)(x+y)\left(x^{2}+y^{2}\right)$.
Solution:
We know that $(a-b)(a+b)=a^{2}-b^{2}$
Put $a=x$ and $b=y$ in the identity (l) then
$(x-y)(x+y)=x^{2}-y^{2}$
Now $(x-y)(x+y)\left(x^{2}+y^{2}\right)=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)$
Again put $a=x^{2}$ and $b=y^{2}$ in (1)
We have $\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)=\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}=x^{4}-y^{4}$
So $(x-y)(x+y)\left(x^{2}+y^{2}\right)=x^{4}-y^{4}$

Question $3 .$
Simplify $(5 x-3 y)^{2}-(5 x+3 y)^{2}$
Solution:
We have the identities $(a+b)^{2}=a^{2}+2 a b+b^{2}$ $(a-b)^{2}=a^{2}-2 a b+b^{2}$ So $(5 x-3 y)^{2}-(5 x+3 y)^{2}=(5 x)^{2}-(2 \times 5 x$ $=5^{2} x^{2}-30 x y+3^{2} y^{2}-\left[5^{2} x^{2}-30 x y+3^{2} y\right.$ $=25 x^{2}-30 x y+9 y^{2}-\left[25 x^{2}+30 x y+9 y^{2}\right]$ $=25 x^{2}-30 x y+9 y^{2}-25 x^{2}-30 x y-9 y^{2}$ $=x^{2}(25-25)-x y(30+30)+y^{2}(9-9)$ $=0 x^{2}-60 x y+0 y^{2}=-60 x y$ $\therefore(5 x-3 y)^{2}-(5 x+3 y)^{2}=-60 x y$
$\begin{aligned}
&(a+b)^{2}=a^{2}+2 a b+b^{2} \\
&(a-b)^{2}=a^{2}-2 a b+b^{2} \\
&\text { So }(5 x-3 y)^{2}-(5 x+3 y)^{2}=(5 x)^{2}-(2 \times 5 x \times 3 y)+(3 y)^{2} \\
&=5^{2} x^{2}-30 x y+3^{2} y^{2}-\left[5^{2} x^{2}-30 x y+3^{2} y^{2}\right] \\
&=25 x^{2}-30 x y+9 y^{2}-\left[25 x^{2}+30 x y+9 y^{2}\right] \\
&=25 x^{2}-30 x y+9 y^{2}-25 x^{2}-30 x y-9 y^{2} \\
&=x^{2}(25-25)-x y(30+30)+y^{2}(9-9) \\
&=0 x^{2}-60 x y+0 y y^{2}=-60 x y \\
&\therefore(5 x-3 y)^{2}-(5 x+3 y)^{2}=-60 x y
\end{aligned}$

Question $4 .$
Simplify : (i) $(a+b)^{2}-(a-b)^{2}$
(ii) $(a+b)^{2}+(a-b)^{2}$
Solution:
Applying the identities
$\begin{aligned}
&(a+b)^{2}=a^{2}+2 a b+b^{2} \\
&(a-b)^{2}=a^{2}-2 a b+b^{2}
\end{aligned}$

$\begin{aligned}
&\text { (i) } \begin{array}{l}
(a+b)^{2}-(a-b)^{2}=a^{2}+2 a b+b^{2}-\left[a^{2}-2 a b+b^{2}\right] \\
=a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2} \\
=a^{2}(1-1)+a b(2+2)+b^{2}(1-1) \\
=0 a^{2}+4 a b+0 b^{2}=4 a b \\
(a+b)^{2}-(a-b)^{2}=4 a b
\end{array}
\end{aligned}$
$\text { (ii) } \begin{aligned}
&(a+b)^{2}+(a-b)^{2}=a^{2}+2 a b+b^{2}+\left(a^{2}-2 a b\right. \\
&=a^{2}+2 a b+b^{2}+a^{2}-2 a b+b^{2} \\
&=a^{2}(1+1)+a b(2-2)+b^{2}(1+1) \\
&=2 a^{2}+0 a b+2 b^{2}=2 a^{2}+2 b^{2}=2\left(a^{2}+b^{2}\right) \\
&\therefore(a+b)^{2}-(a-b)^{2}=2\left(a^{2}+b^{2}\right)
\end{aligned}$
 

Question $5 .$
A square lawn has a $2 \mathrm{~m}$ wide path surrounding it. If the area of the path is $136 \mathrm{~m}^{2}$, find the area of lawn.
Solution:
Let the side of the lawn $=a \mathrm{~m}$
then side Of big square $=(a+2(2)) \mathrm{m}$
$=(\mathrm{a}+4) \mathrm{m}$

Area of the path - Area Of large square - Area of smaller square
$\begin{aligned}
&136=(a+4) 2-a 2 \\
&136=a^{2}+(2 \times a \times 4)+4^{2}-a^{2} \\
&136=a^{2}+8 a+16-a^{2} \\
&136=8 a+16 \\
&136=8(a+2)
\end{aligned}$
Dividing by 8
$17=\mathrm{a}+2$
Subtracting 2 on both sides
$17-3=a+2-2$

$\begin{aligned}
&15=\mathrm{a} \\
&\therefore \text { side of small square }=15 \mathrm{~m} \\
&\text { Area of square }=(\text { side } \times \text { side }) \text { Sq. units } \\
&\therefore \text { Area of the lawn }=(15 \times 15) \mathrm{m}^{2}=225 \mathrm{~m}^{2} \\
&\therefore \text { Area of the lawn }=225 \mathrm{~m}^{2}
\end{aligned}$

Question $6 .$
Solve the following inequalities.
(i) $4 \mathrm{n}+7 \geq 3 \mathrm{n}+10$, $\mathrm{n}$ is an integer
(ii) $6(x+6) \geq 5(x-3)$, $x$ is a whole number.
(iii) $-13 \leq 5 x+2 \leq 32$, $x$ is an integer.
Solution:
(i) $4 n+7 \geq 3 n+10, n$ is an integer.
$\begin{aligned}
&4 n+7-3 n \geq 3 n+10-3 n \\
&n(4-3)+7 \geq 3 n+10-3 n \\
&n(4-3)+7 \geq n(3-3)+10 \\
&n+7 \geq 10
\end{aligned}$
Subtracting 7 on both sides
$\mathrm{n}+7-7 \geq 10-7$
$\mathrm{n} \geq 3$
Since the solution is an integer and is greater than or equal to 3 , the solution will be 3 , $4,5,6,7, \ldots .$
$\mathrm{n}=3,4,5,6,7, \ldots$
(ii) $6(x+6) \geq 5(x-3)$, $x$ is a whole number.
$6 x+36 \geq 5 x-15$
Subtracting $5 x$ on both sides
$\begin{aligned}
&6 x+36-5 x \geq 5 x-15-5 x \\
&x(6-5)+36 \geq x(5-5)-15 \\
&x+36 \geq-15
\end{aligned}$
Subtracting 36 on both sides
$x+36-36 \geq-15-36$
$x \geq-51$
The solution is a whole number and which is greater than or equal to $-51$
$\therefore$ The solution is $0,1,2,3,4, \ldots$
$x=0,1,2,3,4, \ldots$

(iii) $-13 \leq 5 x+2 \leq 32, x$ is an integer.
Subtracting throughout by 2
$-13-2 \leq 5 x+2-2 \leq 32-2$
$-15 \leq 5 x \leq 30$
Dividing throughout by 5
$\frac{-15}{5} \leq \frac{5 x}{5} \leq \frac{30}{5}$
$-3 \leq x \leq 6$
$\therefore$ Since the solution is an integer between $-3$ and 6 both inclusive, we have the solution as $-3,-2,-1,0,1,2,3,4,5,6$.
i.e. $x=-3,-2,0,1,2,3,4,5$ and 6 .