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Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If $\displaystyle f\left ( x \right )=\left ( x+2 \right )\left ( x^{2}+8x+15 \right )$ and $\displaystyle g\left ( x \right )=\left ( x+3 \right )\left ( x^{2}+9x+20 \right )$ then find the HCF of $f(x)$ and $g(x)$.

  1. $x + 3$
  2. $\displaystyle x^{2}+8x+15$
  3. $x + 4$
  4. $\displaystyle x^{2}+9x+20$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Prime factorisation of $ (x+2)({x}^{2}+8x+15) = (x+2)  \times [(x+3) \times (x+5)] $
Prime factorisation of $ (x+3)({x}^{2}+9x+20) = (x+3) \times [(x+4) \times (x+5)] $
So,HCF $  =  (x+3) \times (x+5) = ({x}^{2}+8x+15) $

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If the HCF of the polynomials $f(x)$ and $g(x)$ is $4x - 6$, then $f(x)$ and $g(x)$ could be :

  1. $2, 2x - 3$
  2. $8x - 12, 2$
  3. $\displaystyle 2\left ( 2x-3 \right )^{2},4\left ( 2x-3 \right )$
  4. $\displaystyle 2\left ( 2x+3 \right ),4\left ( 2x+3 \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, HCF $ = 4x-6 = 2(2x-3) $

Since HCF needs to be a factor of both the polynomials, clearly only option C with polynomials $ 2({2x-3)}^{2} , 4(2x-3) $  have both factors $ 2 $ and $ (2x-3) $

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

HCF of $120, 144$ and $216$ is:

  1. $38$
  2. $24$
  3. $120$
  4. $144$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The HCF of $120,144,216$ is

$120= 2 \times 2 \times 2 \times 3 \times 5 $
$144= 2 \times 2 \times 2 \times 2 \times 3 \times 3 $
$216= 2 \times 2 \times 2\times 3 \times 3 \times 3 $
Common factor is $2\times 2\times 2\times 3=24$ 
Hence, the HCF of $120,144$ and $216$ is $24$.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

Find the greatest number which divides 120, 165 and 210 exactly leaving remainders 5, 4 and 3 respectively

  1. 7

  2. 5

  3. 23

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The required number will be the H.C.F of (120 - 5), (165 - 4) and (210 - 3) i.e. H.C.F. of 115
161 and 207
$\displaystyle \therefore $ Required number = H.C.F. of 115, 161 and 207 = 23

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

HCF of $24, 36$ and $92$ is:

  1. $24$
  2. $36$
  3. $12$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The HCF of $24,36,92$ can be found by factorising all three numbers:

$24= 2 \times 2 \times 2 \times 3 $
$36= 2 \times 2 \times 3 \times 3 $
$92 =2 \times 2 \times 23 $
Now, common factors are $2$ and $2$
So, HCF is $ 2 \times 2=4$
Hence, the answer is $4$.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

What is the least number by which $825$ must be multiplied in order to produce a multiple of $715 ?$

  1. $13$
  2. $15$
  3. $11$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$825=3\times5\times5\times 11$

$715=5\times11\times13$
In the factor of both numbers, $13$ is not common. 
Hence, the least number by which $825$ must be multiplied in order to produce a multiple of $715=13.$