In Text Questions Try These (Text Book Page No.44,48,54,55) - Chapter 3 Algebra Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Text Questions : Chapter 3 Algebra Term 2 Class 7th std Maths Guide Samacheer Kalvi Solutions
Exercise $3.1$
Try These (Text book Page No. 44)
Question $1 .$
Observe and complete the following table. First one is done for you.

Solution:

Try These (Text book Page No. 46)
Question $1 .$
Simplify and write the following in exponential form.
1. $2^{3} \times 2^{5}$
2. $\mathrm{p}^{2} \times \mathrm{P}^{4}$
3. $x^{6} \times x^{4}$
4. $3^{1} \times 3^{5} \times 3^{4}$
5. $(-1)^{2} \times(-1)^{3} \times(-1)^{5}$
Solution:
1. $2^{3} \times 2^{5}=2^{3+5}=2^{8}$ [since $a^{m} \times a^{n}=a^{m+n}$ ]
2. $\mathrm{p}^{2} \times \mathrm{p}^{4}=\mathrm{p}^{2+4}=\mathrm{p}^{6}\left[\right.$ since $\left.\mathrm{a}^{\mathrm{m}} \times \mathrm{a}^{\mathrm{n}}=\mathrm{a}^{\mathrm{m}+\mathrm{n}}\right]$
3. $x^{6} \times x^{4}=x^{6}+4=x^{10}\left[\right.$ since $\left.a^{m} \times a^{n}=a^{m+n}\right]$
4. $3^{1} \times 3^{5} \times 3^{4}=3^{1+5} \times 3^{4}\left[\right.$ since $\left.a^{m} \times a^{n}=a^{m+n}\right]$
$=3^{6} \times 3^{4}\left[\right.$ since $\left.a^{m} \times a^{n}=a^{m+n}\right]$
$=3^{10}$
5. $(-1)^{2} \times(-1)^{3} \times(-1)^{5}$
$=(-1)^{2+3} \times(-1)^{5}\left[\right.$ Since $\left.a^{m} \times a^{n}=a^{m+n}\right]$
$=(-1)^{5} \times(-1)^{5}$
$=(-1)^{5+5}\left[\right.$ Since $\left.a^{m} \times a^{n}=a^{m+n}\right]$
$=(-1)^{10}$

Try These (Text book Page No. 48)
Question $1 .$
Simply the following.
1. $23^{5} \div 23^{2}$
2. $11^{6} \div 11^{3}$
3. $(-5)^{3} \div(-5)^{2}$
4. $7^{3} \div 7^{3}$
5. $15^{4} \div 15$

Solution:
1. $2^{3} \div 2^{5} \quad=2^{3-5}=23^{3} \quad$ [since $\frac{a^{m}}{a^{n}}=a^{m-n}$ ]
2. $11^{6} \div 11^{3} \quad=\quad 11^{6-3}=11^{3} \quad$ [since $\frac{a^{m}}{a^{n}}=a^{m-n}$ ]
3. $(-5)^{3} \div(-5)^{2}=(-5)^{3-2}=(-5)^{1} \quad$ [since $\left.\frac{a^{m}}{a^{n}}=a^{m-n}\right]$
4. $7^{3} \div 7^{3} \quad=\quad 7^{3-3}=7^{0}=1 \quad$ [since $\frac{a^{m}}{a^{n}}=a^{m-n} ; a^{0}=1$ ]
5. $15^{4} \div 15 \quad=15^{4} \div 15^{1}=15^{4-1}=15^{3} \quad$ [since $\frac{a^{m}}{a^{n}}=a^{m-n}$ ]
 

Try These (Text book Page No. 48)
Question $1 .$
Simplify and write the following in exponent form.
1. $\left(3^{2}\right)^{3}$
2. $\left[(-5)^{3}\right]^{2}$
3. $\left(20^{6}\right)^{2}$
4. $\left(10^{3}\right)^{5}$
Solution:
1. $\left(3^{2}\right)^{3}=3^{2 \times 3}=3^{6}\left[\right.$ since $\left.\left(a^{m}\right)^{n}=a^{m \times n}\right]$
2. $[(-5)]^{2}=(-5)^{3 \times 2}=(-5)^{6}\left[\right.$ since $\left.\left(a^{m}\right)^{n}=a^{m \times n}\right]$
3. $\left(20^{6}\right)^{2}=20^{6 \times 2}=20^{12}$ [since $\left(\mathrm{a}^{\mathrm{m}}\right)^{\mathrm{n}}=\mathrm{a}^{\mathrm{m} \times \mathrm{n}}$ ]
4. $\left(10^{3}\right)^{5}=10^{3 \times 5}=10^{15}\left[\right.$ since $\left.\left(a^{m}\right)^{n}=a^{m \times n}\right]$

Question $2 .$
Express the following exponent numbers using $a^{m} \times b^{m}=(a \times b)^{m}$.
(i) $5^{2} \times 3^{2}$
(ii) $x^{3} \times y^{3}$
(iii) $7^{4} \times 8^{4}$
Solution:
(i) $5^{2} \times 3^{2}=(5 \times 3)^{2}=15^{2}\left[\right.$ since $\left.\mathrm{a}^{\mathrm{m}} \times \mathrm{b}^{\mathrm{m}}=(\mathrm{a} \times \mathrm{b})^{\mathrm{m}}\right]$
(ii) $x^{3} \times y^{3}=(x \times y)^{3}=(x y)^{3}$
(iii) $7^{4} \times 8^{4}=(7 \times 8)^{4}=56^{4}$
 

Question $3 .$
Simplify the following exponent numbers by using $\left(\frac{a}{b}\right)^{\mathrm{m}}=\frac{a^{m}}{b^{w}}$
(i) $5^{3} \div 2^{3}$
(ii) $(-2)^{4} \div 3^{4}$
(iii) $8^{6} \div 5^{6}$
(iv) $6^{3} \div(-7)^{3}$
Solution:
(i) $5^{3} \div 2^{3}=\left(\frac{5}{2}\right)^{3}-\left[\right.$ Since $\left.\frac{a^{\mathrm{m}}}{b^{m}}=\left(\frac{a}{b}\right)^{\mathrm{m}}\right]$
(ii) $(-2)^{4} \div 3^{4}=\left(\frac{-2}{3}\right)^{4}$
(iii) $8^{6} \div 5^{6}=\left(\frac{8}{6}\right)^{6}$
(iv) $6^{3} \div(-7)^{3}=\left(\frac{6}{-7}\right)^{3}$

Exercise $3.2$
Try These (Text book Page No. 54)
Question $1 .$
Find the unit digit of the following exponential numbers:
(i) $106^{21}$
(ii) $25^{8}$
(iii) $31^{18}$
(iv) $20^{10}$
Solution:
(i) $106^{21}$ Unit digit of base 106 is 6 and the power is 21 and is positive. Thus the unit digit of $106^{21}$ is 6 .
(ii) $25^{8}$ Unit digit of base 25 is 5 and the power is 8 and is positive. Thus the unit digit of $25^{8}$ is 5 .
(iii) $31^{18}$ Unit digit of base 31 is 1 and the power 18 and is positive. Thus the unit digit of $31^{18}$ is $1 .$
(iv) $20^{10}$ Unit digit of base 20 is 0 and the power 10 and is positive. Thus the unit digit of $20^{10}$ is 0 .
 

Try These (Text book Page No. 55)
Question $1 .$
Find the unit digit of the following exponential numbers:
(i) $64^{11}$
(ii) $29^{18}$
(iii) $79^{19}$

(iv) $104^{32}$
Solution:
(i) $64^{11}$ Unit digit of base 64 is 4 and the power is 11 (odd power).
$\therefore$ Unit digit of $64^{11}$ is 4 .
(ii) $29^{18}$ Unit digit of base 29 is 9 and the power is 18 (even power). Therefore, unit digit of $29^{18}$ is $1 .$
(iii) $79^{19}$ Unit digit of base 79 is 9 and the power is 19 (odd power). Therefore, unit digit of 7919 is $9 .$
(iv) $104^{32}$ Unit digit of base 104 is 4 and the power is 32 (even power). Therefore, unit digit of $104^{32}$ is $6 .$
 

Exercise $3.3$
Try These (Text book Page No. 35)

Question $1 .$
Complete the following table:

Solution:

Question $2 .$
Identify the like terms from the following:
(i) $2 \mathrm{x}^{2} \mathrm{y}, 2 \mathrm{xy}^{2}, 3 \mathrm{xy}^{2}, 14 \mathrm{x}^{2} \mathrm{y}, 7 \mathrm{yx}$
(ii) $3 x^{3} y^{2}, y^{3} x, y^{3} x^{2},-y^{3} x, 3 y^{3} x$
(iii) $11 \mathrm{pq}$, -pq, $11 \mathrm{pqr},-11 \mathrm{pq}, \mathrm{pq}$
Solution:
(i) $2 x^{2} y, 2 x y^{2}, 3 x y^{2}, 14 x^{2} y, 7 y x$
(a) $2 x^{2} y$ and $14 x^{2} y$ are like terms.
(b) $2 x y^{2}$ and $3 x y^{2}$ are like terms.
(ii) $3 x^{3} y^{2}, y^{3} x, y^{3} x^{2},-y^{3} x, 3 y^{3} x$
(a) $y^{3} x,-y^{3} x$ and $3 y^{3} x$ are like terms.
(iii) $11 \mathrm{pq}$, -pq, $11 \mathrm{pqr},-11 \mathrm{pq}, \mathrm{pq}$
(a) $11 \mathrm{pq},-\mathrm{pq},-\mathrm{pq}$ and $\mathrm{pq}$ are like terms.