Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour

3.Work out the following divisions.
(i) \((10x – 25) ÷ 5\)

(ii) \((10x – 25) ÷ (2x – 5)\)

(iii) \(10y (6y + 21) ÷ 5 (2y + 7)\)

(iv) \(9x^2y^2 (3z – 24) ÷ 27xy (z – 8)\)

(v) \(96abc (3a – 12) (56 – 30) ÷ 144 (a – 4) (b – 6)\)


Answer :

(i)Given:\((10x – 25) ÷ 5\)

\(=\frac{10x-25}5=\frac{5\times (2x-5)}5\)

\(=2x-5\)

(ii)Given:\((10x – 25) ÷ (2x – 5)\)

\(=\frac{10x-25}{2x – 5}=\frac{5\times (2x-5)}{2x – 5}\)

\(=5\)

(iii)Given:\(10y (6y + 21) ÷ 5 (2y + 7)\)

\(=\frac{10y (6y + 21)}{ 5 (2y + 7)}=\frac{5\times 2\times y\times 3\times (2y + 7)}{ 5\times (2y + 7)}\)

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

(iv)Given: \(9x^2y^2 (3z – 24) ÷ 27xy (z – 8)\)

\(=\frac{9x^2y^2 (3z – 24)}{ 27xy (z – 8)}=\frac{9\times x^2y^2 \times 3\times (z-8)}{ 9\times 3\times xy \times (z -8)}\)

\(=\frac{xy}1=xy\)

(v)Given: \(96abc (3a – 12) (56 – 30) ÷ 144 (a – 4) (b – 6)\)
\(=\frac{96abc (3a – 12) (56 – 30)}{ 144 (a – 4) (b – 6)}=\frac{48\times 2 \times abc \times 3\times (a-4)\times 5 \times (b-6)}{48\times 3\times (a-4) \times (b-6)}\)

\(=\frac{2 \times abc \times 5}1=10abc\)

NCERT solutions of related questions for Factorization

NCERT solutions of related chapters class 8 maths

NCERT solutions of related chapters class 8 science