Question
Find the common factors of the given term: $12x, 36$
✓
Answer
$12x, 36$
$12x = 2 \times 2 \times 3 \times x$
$36 = 2 \times 2 \times 3 \times 3$
Common prime factors are $2$ (occurs twice) and $3$.
$\therefore H.C.F. = 2 \times 2 \times 3 = 12$
$12x = 2 \times 2 \times 3 \times x$
$36 = 2 \times 2 \times 3 \times 3$
Common prime factors are $2$ (occurs twice) and $3$.
$\therefore H.C.F. = 2 \times 2 \times 3 = 12$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Factorise the expressions: $6xy – 4y + 6 – 9x$
→Solve:
$ x^5-16 x^3 $
→$ x^5-16 x^3 $
Find the smallest number by which $243$ must be multiplied to obtain a perfect cube.
→Making use of the cube root table, find the cube root $0.27$
→Use the distributivity of multiplication of rational numbers over their addition to simplify:$\frac{3}{4}\times\Big(\frac{8}{9}-40\Big)$
→Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
What can you say about the angle sum of a convex polygon with $7$ number of sides?
→| Figure |  |
 |
 |
 |
| Side | $3$ | $4$ | $5$ | $6$ |
| Angle sum | $180^\circ$ | $2 \times 180^{\circ}$ $=(4-2) \times 180^{\circ}$ |
$3 \times 180^{\circ}$ $=(5-2) \times 180^{\circ}$ |
$4 \times 180^{\circ} $ $=(6-2) \times 180^{\circ}$ |
Diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus? Give a figure to justify your answer.
Find the cube root of each of the following natural numbers: $74088000$
→Solve: $5x + 7 = 2x - 8$
→Find the gain or loss percent when. $CP = Rs. 620$ and $SP = Rs. 713$
→