Selina solutions

# Question 1

Q1) Divide:

(i) -16ab^2c\ by\ 6abc

(ii) 25x^2y\ by\ -5y^2

(iii) 8x\ +\ 24\ by\ 4

(iv) 4a^2-a\ by\ a

(v) 8m-16\ by\ -8

(vi) -50+40p\ by\ 10p

(vii) 4x^3-2x^2\ by\ -x

(viii) 10a^3-15a^2b\ by\ -5a^2

(ix) 12x^3y-8x^2y^2+4x^2y^3\ \div4xy

(x) 9a^4b-15a^3b^2+12a^2b^3\ \div-3a^2b

Solutiion 1:

(i) -16ab^2c\ by\ 6abc

= \frac{-16ab^2c}{6abc}

= \frac{-16}{6}a^{_{\left(1-1\right)}}b^{\left(2-1\right)}c^{\left(1-1\right)}

=\frac{-8}{3}b

(ii) 25x^2y\ by\ -5y^2

= \frac{25x^2y}{-5y^2}

= \frac{25}{-5}x^2y^{\left(1-2\right)}

= -\frac{5}{y}x^2

(iii) 8x\ +\ 24\ by\ 4

= \frac{8x+24}{4}=\frac{4\left(2x+6\right)}{4}

= 2x+6

(iv) 4a^2-a\ by\ a

= \frac{4a^2-a}{a}=\frac{a\left(4a-1\right)}{a}=4a-1

(v) 8m-16\ by\ -8

= \frac{8m-16}{-8}=\frac{8\left(m-2\right)}{-8}=2-m

(vi) -50+40p\ by\ 10p

=

\frac{-50+40p}{10p}=\frac{10\left(-5+4p\right)}{10p}=\frac{-5}{p}+4

(vii) 4x^3-2x^2\ by\ -x

= 4x^3-2x^2\div-x

= -\left\{4x^{\left(3-1\right)}-2x^{\left(2-1\right)}\right\}

= -4x^2+2x

(viii) 10a^3-15a^2b\ by\ -5a^2

= 10a^3-15a^2b\div-5a^2

= 5\left(2a^3-3a^2b\right)\div-5a^2

= -\left(2a^{\left(3-2\right)}-3a^{\left(2-2\right)}b\right)

= -2a+3b

(ix) 12x^3y-8x^2y^2+4x^2y^3\ \div4xy

=

\frac{12}{4}x^{\left(3-1\right)}y^{\left(1-1\right)}-\frac{8}{4}x^{\left(2-1\right)}y^{\left(2-1\right)}+\frac{4}{4}x^{\left(2-1\right)}y^{\left(3-1\right)}

= 3x^2-2xy+xy^2

(x) 9a^4b-15a^3b^2+12a^2b^3\ \div-3a^2b

= \frac{9}{-3}a^{\left(4-2\right)}b^{\left(1-1\right)}-\frac{15}{-3}a^{\left(3-2\right)}b^{\left(2-1\right)}+\frac{12}{-3}a^{\left(2-2\right)}b^{\left(3-1\right)}

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

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