S.chand publication New Learning Composite mathematics solution of class 8 Chapter 5 Algebraic Expressions Exercise 5F

 Exercise 5F

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Q1 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 1

Divide:

(a) 7a+7b by 7

Sol :
$=\frac{7a+7b}{7}=\frac{7(a+b)}{7}$
=a+b

(b) ax+ay by a

Sol :

$=\frac{ax+ay}{a}=\frac{a(x+y)}{a}$
=x+y

(c) 6y–2 by 2

Sol :

$=\frac{6y-2}{2}=\frac{2(3y-1)}{2}$
=3y-1

(d) b–a by b

Sol :
$=\frac{b-a}{b}=\dfrac{b\left(1-\frac{a}{b}\right)}{b}$
$=1-\frac{a}{b}$

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Q2 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 2

(a) 8y–8 by -8

Sol :
$=\frac{8y-8}{-8}=\frac{8(y-1)}{-8}$
=-y+1

(b) 6x⁴+3x³+3x² by 3x²

Sol :
$=\frac{6x^4+3x^3+3x^2}{3x^2}$

$=\frac{3x^2(2x^2+x+1)}{3x^2}$

=2x²+x+1

(c) 25x⁸–20x⁵ by 5x³y³

Sol :
$=\frac{25x^8-20x^5}{5x^3y^3}=\frac{5x^5(5x^3-4)}{5x^3y^3}$

$=\frac{x^2}{y^3}(5x^3-4)$

$=\frac{5x^4}{y^3}-\frac{4x^2}{y^3}$

(d) x²y²z²–xyz+1 by xyz

Sol :

$=\frac{x^2y^2z^2-xyz+1}{xyz}=\frac{x^2y^2z^2}{xyz}-\frac{xyz}{xyz}+\frac{1}{xyz}$

$xyz-1+\frac{1}{xyz}$

$=xyz+\frac{1}{xyz}-1$

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Q3 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 3

Express each quotient as a sum.

(a) $\frac{35a^5+28a^4b^2-14a^3b^3}{-7a^2}$

Sol :

$=\frac{35a^5+28a^4b^2-14a^3b^3}{-7a^2}$

$=-5a^3-4a^2b^2+2ab^3$

(b) $\frac{16x^2y-48xy^2+8x^2y^2}{8x^2y^2}$

Sol :

$=\frac{16x^2y-48xy^2+8x^2y^2}{8x^2y^2}$

$=\frac{2}{y}-\frac{6}{x}+1$

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Q4 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 4

Evaluate

$\frac{x}{y}$ when $=\frac{x+y}{y}=21$

Sol :

$=\frac{x+y}{y}=21$

x+y=21y

x=21y-y=20y

$\frac{x}{y}=20$