Exercise 3.4 - Chapter 3 Algebra Term 1 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 $1 .$
Subtract $-3 a b-8$ from $3 a b-8$. Also subtract $3 a b+8$ from $-3 a b-8$
Solution:
Subtracting $-3 \mathrm{ab}-8$ from $3 \mathrm{ab}+8$
$=3 a b+8-(-3 a b-8)=3 a b+8+(3 a b+8)$
$=3 \mathrm{ab}+8+3 \mathrm{ab}+8=(3+3) \mathrm{ab}+(8+8)$
$=6 \mathrm{a} \mathrm{b}+16$
Also subtracting $3 a b+8$ from $-3 a b-8$
$=-3 \mathrm{ab}-8-(3 \mathrm{ab}+8)=-3 \mathrm{ab}-8+(-3 \mathrm{ab}-8)=-3 \mathrm{ab}-8-3 \mathrm{ab}-8$
$=[(-3)+(-3)] a b+[(-8)+(-8)]=-6 a b+(-16)$
$=-6 \mathrm{ab}-16$

Question 2 .
Find the perimeter of a triangle whose sides are $x+3 y, 2 x+y, x-y$
Solution:
Perimeter of a triangle $=$ Sum of three sides
$=(x+3 y)+(2 x+y)+(x-y)$
$=x+3 y+2 x+y+x-y$
$=(1+2+1) x+(3+1+(-1)) y=4 x+3 y$
$\therefore$ Perimeter of the triangle $=4 x+3 y$

Question $3 .$
Thrice a number when increased by 5 gives 44 . Find the number.
Solution:
Let the required number be $x$.
Thrice the number $=3 \mathrm{x}$.
Thrice the number increased by $4=3 x+5$

Given $3 x+5=44$
$\begin{aligned}
&3 x+5-5=44-5 \\
&3 x=39 \\
&\frac{3 x}{3}=\frac{39}{3} \\
&x=13
\end{aligned}$
$\therefore$ The required number $=13$
 

Question $4 .$
How much smaller is $2 a b+4 b-c$ than $5 a b-3 b+2 c$.
Solution:
To find the answer we have to find the difference.
Here greater number $5 \mathrm{ab}-3 \mathrm{ab}+2 \mathrm{c}$.
$\therefore$ Difference $=5 \mathrm{ab}-3 \mathrm{~b}+2 \mathrm{c}-(2 \mathrm{ab}+4 \mathrm{~b}-\mathrm{c})=5 \mathrm{ab}-3 \mathrm{~b}+2 \mathrm{c}+(-2 \mathrm{ab}-4 \mathrm{~b}+\mathrm{c})$
$=5 \mathrm{a} \mathrm{b}-3 \mathrm{~b}+2 \mathrm{c}-2 \mathrm{ab}-4 \mathrm{~b}+\mathrm{c}$
$=(5-2) a b+(-3-4) b+(2+1) c=3 a b+(-7) b+3 c$
$=3 \mathrm{ab}-7 \mathrm{~b}+3 \mathrm{c}$
It is $3 \mathrm{ab}-7 \mathrm{~b}+3 \mathrm{c}$ smaller.

Question $5 .$
Six times a number subtracted from 40 gives $-8$. Find the number.
Solution:
Let the required number be $x$. Six times the number $=6 x$.
Given $40-6 x=-8$
$\begin{aligned}
&-6 x+40-40=-8-40 \\
&-6 x=-48 \\
&\frac{-6 x}{-6}=\frac{-48}{-6} \\
&x=8
\end{aligned}$
$\therefore$ The required number is 8 .

Challenge Problems
Question $6 .$
From the sum of $5 x+7 y-12$ and $3 x-5 y+2$, subtract the sum of $2 x-7 y-1$ and $-6 x+3 y+9$
Solution:
Sum of $5 x+7 y-12$ and $3 x-5 y+2$.
$=5 x+7 y-12+3 x-5 y+2=(5+3) x+(7-5) y+((-12)+2)$
$=8 x+2 y-10$.
Again Sum of $2 x-7 y-1$ and $-6 x+3 y+9$
$=2 x-7 y-1+(-6 x+3 y+9)=2 x-7 y-1-6 x+3 y+9$
$=(2-6) x+(-7+3) y+(-1+9)$
$=-4 x-4 y+8$
Now $8 x+2 y-10-(-4 x-4 y+8)$
$=8 x+2 y-10+(4 x+4 y-8)$
$=8 x+2 y-10+4 x+4 y-8$
$=(8+4) x+(2+4) y+((-10)+(-8))$
$=12 \mathrm{x}+6 \mathrm{y}-18$

Question $7 .$
Find the expression to be added with $5 \mathrm{a}-3 \mathrm{~b}-2 \mathrm{c}$ to get $\mathrm{a}-4 \mathrm{~b}-2 \mathrm{c}$ ?
Solution:
To get the required expression we must subtract $5 \mathrm{a}-3 \mathrm{~b}+2 \mathrm{c}$ from $\mathrm{a}-4 \mathrm{~b}-2 \mathrm{c}$.
$\begin{aligned}
&\therefore a-4 b-2 c-(5 a-3 b+2 c)=a-4 b-2 c+(-5 a+3 b-2 c) \\
&=a-4 b-2 c-5 a+3 b-2 c \\
&=(1-5) a+(-4+3) b+(-2-2) c \\
&=-4 a-b-4 c
\end{aligned}$

Question $8 .$
What should be subtracted from 2m + 8n + 10 to get – 3m + 7n + 16?
Solution:
To get the expression we have to subtract $-3 \mathrm{~m}+7 \mathrm{n}+16$ from $2 \mathrm{~m}+8 \mathrm{n}+10$.
$\begin{aligned}
&(2 m+8 n+10)-(-3 m+7 n+16)=2 m+8 n+10+3 m-7 n-16 \\
&=(2+3) m+(8-7) n+(10-16) \\
&=5 m+n-6
\end{aligned}$

Question $9 .$
Give an algebraic equation for the following statement:
"The difference between the area and perimeter of a rectangle is 20 ".
Solution:
Let the length of a rectangle $=1$ and breadth $=b$ then Area $=l b$; Perimeter $=2(1+b)$
Area $-$ Perimeter $=20$
$\therefore \mathrm{lb}-2(\mathrm{l}+\mathrm{b})$

Question 10 .
Add : $2 a+b+3 c$ and $a+\frac{1}{3} b+\frac{2}{5} c$

Solution:
$\begin{aligned}
2 a+b+3 c+\left(a+\frac{1}{3} b+\frac{2}{5} c\right) &=2 a+b+3 c+a+\frac{1}{3} b+\frac{2}{5} c \\
&=(2+1) a+\left(1+\frac{1}{3}\right) b+\left(3+\frac{2}{5}\right) c \\
&=3 a+\left(\frac{3}{3}+\frac{1}{3}\right) b+\left(\frac{15}{5}+\frac{2}{5}\right) c \\
&=3 a+\frac{4}{3} b+\frac{17}{5} c
\end{aligned}$