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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions and Answers
Exercise 3.1
Question $1 .$
Simplify the following and write the answer in Exponential form.
(i) $3^{2} \times 3^{4} \times 3^{8}$
(ii) $6^{15} \div 6^{10}$
(iii) $a^{3} \times a^{2}$
(iv) $7^{x} \times 7^{2}$
(v) $\left(5^{2}\right)^{3} \div 5^{3}$
Solution:
(i) $3^{2} \times 3^{4} \times 3^{2}=3^{2+4+8}=3^{14}\left[\because a^{m} \times a^{n}=a^{m+n}\right]$
So $3^{2} \times 3^{4} \times 3^{2}=3^{14}$
(ii) $6^{15} \div 6^{10}=6^{15-10}=6^{5}\left[\because a^{m} \times a^{n}=a^{m-n}\right]$
(iii) $a^{3} \times a^{2}=a^{3+2}=a^{5}$
(iv) $7^{x} \times 7^{2}=7^{x+2}$
(v) $\left(5^{2}\right)^{3} \div 5^{3}=5^{2 \times 3} \div 5^{3}=5^{6} \div 5^{3}\left[\because\left(\mathrm{a}^{\mathrm{m}}\right)^{\mathrm{n}}=\mathrm{a}^{\mathrm{m} \times \mathrm{n}}\right]$
 

Question $2 .$
Express the following as a product of factors only in exponential form
(i) $108 \times 192$
(ii) 270
(iii) $729 \times 64$

(iv) 768
Solution:
(i) $108 \times 192$
$\begin{aligned}
&108=2 \times 2 \times 3 \times 3 \times 3 \\
&192=2 \times 2 \times 2 \times 2 \times 2 \times 2
\end{aligned}$

$\begin{aligned}
&\therefore 108 \times 192=(2 \times 2 \times 3 \times 3 \times 3) \times \\
&(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3) \\
&=2^{8} \times 3^{4}
\end{aligned}$
(ii) 2700
(ii) 270
We have $270=2 \times 3 \times 3 \times 3 \times 5$

$\begin{aligned}
&=2^{1} \times 3^{3} \times 5^{1} \\
&=2 \times 3^{3} \times 5 \\
&\text { So } 270=2 \times 3^{3} \times 5
\end{aligned}$
(iii) $729 \times 64$
$\begin{aligned}
&729=3 \times 3 \times 3 \times 3 \times 3 \times 3 \\
&64=2 \times 2 \times 2 \times 2 \times 2 \times 2
\end{aligned}$

$\begin{aligned}
&729 \times 64=(3 \times 3 \times 3 \times 3 \times 3 \times 3) \\
&\times(2 \times 2 \times 2 \times 2 \times 2 \times 2) \\
&=3^{6} \times 2^{6} \\
&\therefore 729 \times 64=3^{6} \times 2^{6}
\end{aligned}$
(iv) 768
We have $768=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

Question $3 .$
Identify the greater number, wherever possible in each of the following.
(i) $5^{3}$ or $3^{5}$
(ii) $2^{8}$ or $8^{2}$
(iii) $100^{2}$ or $2^{1000}$
(iv) $2^{10}$ or $10^{2}$
Solution:
(i) $5^{3}$ or $3^{5} 5^{3}=5 \times 5 \times 5=125$
$3^{5}=3 \times 3 \times 3 \times 3 \times 3=243$
$243>125 \therefore 3^{5}>5^{3}$
(ii) $2^{8}$ or $8^{2} 2^{8}=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=256$
$\begin{aligned}
&\text { (ii) } 2^{8} \text { or } 8^{2} 2^{8}=2 \times 2 \times \\
&8^{2}=8 \times 8=64 \\
&256>64 \therefore 2^{8}>8^{2}
\end{aligned}$

(iii) $100^{2}$ or $2^{1000}$
We have $100^{2}=100 \times 100=10000$
$\begin{aligned}
&2^{100}=\left(2^{10}\right)^{10}=(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2)^{10} \\
&=(1024)^{10}=\left[(1024)^{2}\right]^{5} \\
&=(1024 \times 1024)^{5}=(1048576)^{5} \\
&\text { Since } 1048576>10000 \\
&(1048576)^{5}>10000 \\
&\text { i.e., }(1048576)>100^{2} \\
&\left(2^{10}\right)^{10}>100^{2} \\
&2^{100}>100^{2}
\end{aligned}$
(iv) $2^{10}$ or $10^{2}$
We have $2^{10}=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=1024$
$10^{2}=10 \times 10=100$
Since $11024>100$
$2^{10}>10^{2}$
 

Exercise $3.2$
Question $1 .$
Find the unit digit of the following exponential numbers.
(i) $255^{223}$
(ii) $8111^{1000}$
(iii) $4866^{431}$
Solution:
(i) $255^{223}$

Unit digit of base 255 is 5 and power is $223 .$
Thus the unit digit of $255^{223}$ is $5 .$
(ii) $8111^{1000}$
Unit digit of base 8111 is 1 and power is 1000 .
Thus the unit digit of $8111^{1000}$ is $1 .$
(iii) $4866^{431}$
Unit digit of base 4866 is 6 and power is 431 .
Thus the unit digit of $4866^{431}$ is $6 .$
 

Question 2 .
Find the unit digit of the numbers
(i) $1844^{671}$
(ii) $1564^{100}$
Solution:
(i) $1844^{671}$
Unit digit of base 1844 is 4 and the power is 671 (odd power)
Therefore unit digit of $1844^{671}$ is 4
(ii) $1564^{100}$
Unit digit of base 1564 is 4 and the power is 100 (even power)
Therefore unit digit of $1564^{100}$ is 6 .

(iii) $4866^{431}$
Unit digit of base 4866 is 6 and power is 431 .
Thus the unit digit of $4866^{431}$ is 6 .

Question $2 .$
Find the unit digit of the numbers
(i) $1844^{671}$
(ii) $1564^{100}$
Solution:
(i) $1844^{671}$
Unit digit of base 1844 is 4 and the power is 671 (odd power)
Therefore unit digit of $1844^{671}$ is 4
(ii) $1564^{100}$
Unit digit of base 1564 is 4 and the power is 100 (even power)
Therefore unit digit of $1564^{100}$ is 6 .

Question $3 .$
Find the unit digit of the numbers
(i) $999^{222}$
(ii) $1549^{777}$
Solution:

(i) $999^{222}$
Unit digit of base 999 is 9 and the power is 222 (even power).
Therefore, unit digit of $999^{222}$ is $1 .$
(ii) $1549^{777}$
Unit digit of base 1549 is 9 and the power is 777 (odd power).
Therefore, unit digit of $1549^{777}$ is 9 .
 

Question $4 .$
Find the unit digit of $1549^{101}+6541^{20}$
Solution:
$1549^{101}+6541^{20}$
In $1549^{101}$, the unit digit of base 1549 is 9 and power is 101 (odd power).
Therefore, unit digit of the $1549^{101}$ is 9 .
In $6541^{20}$, the unit digit of base 6541 is 1 and power is 20 (even power).
Therefore, unit digit of the $6541^{20}$ is 1 .
$\therefore$ Unit digit of $1549^{101}+6541^{20}$ is $9+1=10$
$\therefore$ Unit digit of $1549^{101}+6541^{20}$ is 0 .

Exercise $3.3$
Question $1 .$
Find the degree of the following polynomials.
(i) $x^{5}-x^{4}+3$
(ii) $2-y^{5}-y^{3}+2 y^{8}$
(iii) 2
(iv) $5 x^{3}+4 x^{2}+7 x$
(v) $4 x y+7 x^{2} y+3 x y^{3}$
Solution:
(i) $x^{5}-x^{4}+3$
The terms of the given expression $\operatorname{are}^{5},-x^{4}, 3$.
Degree of each of the terms: $5,4,0$
Term with highest degree: $\mathrm{x}^{5}$
Therefore degree of the expression in 5 .
(ii) $2-\mathrm{y}^{5}-\mathrm{y}^{3}+2 \mathrm{y}^{8}$
The terms of the given expression are $2,-y^{5},-y^{3}, 2 y^{8}$.
Degree of each of the terms: $0,2,3,8$.
Term with highest degree: $2 \mathrm{y}^{8}$
Therefore degree of the expression in 8 .
(iii) 2
Degree of the constant term is 0 .
$\therefore$ Degree of 2 is 0 .
(iv) $5 x^{3}+4 x^{2}+7 x$
The terms of the given expression are $5 x^{3}, 4 x^{2}, 7 x$

Degree of each of the terms: $3,2,1$
Term with highest degree: $5 x^{3}$
Therefore degree of the expression in $3 .$
(v) $4 x y+7 x^{2} y+3 x y^{3}$
The terms of the given expression are $4 x y, 7 x^{2} y, 3 x y^{3}$
Degree of each of the terms: $2,3,4$
Term with highest degree: $3 x y^{3}$
Therefore degree of the expression in 4 .
 

Question $2 .$
State whether a given pair of terms in like or unlike terms.
(i) 1,100
(ii) $-7 x, \frac{5}{2} x$
(iii) $4 \mathrm{~m}^{2} \mathrm{p}, 4 \mathrm{mp}^{2}$
(iv) $12 \mathrm{xz}, 12 \mathrm{x}^{2} \mathrm{z}^{2}$
Solution:
(i) 1,100 is a pair of like terms. $\left[\because 1=x^{0}\right.$ and $\left.100=100 x^{0}\right]$
(ii) $-7 \mathrm{x}, \frac{5}{2} \mathrm{x}$ is a pair of like terms.
(iii) $4 \mathrm{~m}^{2} \mathrm{p}$, Amp is a pair of unlike terms.
(iv) $12 \mathrm{xz}, 12 \mathrm{x}^{2} \mathrm{z}^{2}$ is a pair of unlike terms.

Question $3 .$
Subtract $5 \mathrm{a}^{2}-7 \mathrm{ab}+5 \mathrm{~b}^{2}$ from $3 \mathrm{ab}-2 \mathrm{a}^{2}-2 \mathrm{~b}^{2}$ and find the degree of the expression Solution:
(i) We have $\left(3 a b-2 a^{2}-2 b^{2}\right)-\left(5 a^{2}-7 a b+5 b^{2}\right)$
$\begin{aligned}
&=3 \mathrm{ab}-2 \mathrm{a}^{2}-2 \mathrm{~b}^{2}-5 \mathrm{a}^{2}+7 \mathrm{ab}-5 \mathrm{~b}^{2} \\
&=(3 \mathrm{ab}+7 \mathrm{ab})+(-2-5) \mathrm{a}^{2}+(-2-5) \mathrm{b}^{2} \\
&=10 \mathrm{ab}+(-7) \mathrm{a}^{2}+(-7) \mathrm{b}^{2} \\
&=10 \mathrm{ab}-7 \mathrm{a}^{2}-7 \mathrm{~b}^{2}
\end{aligned}$
Degree of the expression is 2 .
 

Question $4 .$
Add $x^{2}-y^{2}-1, y^{2}-1-x^{2}, 1-x^{2}-y^{2}$ and find the degree of the expression.
Solution:
We have $\left(x^{2}-y^{2}-1\right)+\left(y^{2}-1-x^{2}\right)+\left(1-x^{2}-y^{2}\right)$
$=x^{2}-y^{2}-1+y^{2}-1-x^{2}+1-x^{2}-y^{2}$
$\begin{aligned}
&=\left(x^{2}-x^{2}-x^{2}\right)+\left(-y^{2}+y^{2}-y^{2}\right)+(-1-1+1) \\
&=(1-1-1) x^{2}+(-1+1-1) y^{2}+(-2+1) \\
&=(-1) x^{2}+(-1) y^{2}+(-1)=-x^{2}-y^{2}-1
\end{aligned}$
Degree of the expression is 2 .

Question 5 .
Find the degree of the terms
(i) $x^{2}$
(ii) $4 x y z$
(iii) $\frac{7 x^{2} y^{4}}{x y}$
(iv) $\frac{x^{2} \times y^{2}}{x \times y^{2}}$
Solution:
We have
(i) $x^{2}$
The exponent in $\mathrm{x}^{2}$ is $2 . \therefore$ Degree of the term is 2 .
(ii) $4 \mathrm{xyz}$
In $4 x y z$ the sum of the powers of $x, y$ and $z$ as $3 .$
(iii) $\frac{7 x^{2} y^{4}}{x y}$
We have $\frac{7 x^{2} y^{4}}{x y}=7 \mathrm{x}^{2-1} \mathrm{y}^{4-1}=7 \mathrm{x}^{1} \mathrm{y}^{3}$ [Since $\frac{a^{m}}{a^{n}}=a m-\mathrm{n}$ ]
In $7 \mathrm{x}^{1} \mathrm{y}^{3}$ the sum of the poweres of $\mathrm{x}$ and $\mathrm{y}$ is $4(1+3=4)$ Thus degree of the expression is 4 .
(iv) $\frac{x^{2} \times y^{2}}{x \times y^{2}}$
We have $\frac{x^{2} \times y^{2}}{x \times y^{2}}=x^{2-1} y^{2-2}=x^{1} y^{0}=x^{1}\left[\because y^{0}=1\right]$ The exponent of the expression is $1 .$