Tag: hcf

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

When teams of same size are formed from three groups of $512, 430$ and $489$ students separately $8, 10$ and $9$ students respectively are left out What could be the largest size of the team?

  1. $6$
  2. $12$
  3. $18$
  4. $20$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

It is given that $8,10$ and $9$ students are respectively left out from the three separate groups $512,430$ and $489$ when the teams of same size are formed.


Number of students taken from first group are $512-8=504$
Number of students taken from second group are $430-10=420$
Number of students taken from third group are $489-9=480$

Now, we factorize $504,420$ and $480$ as follows:

$504=2\times 2\times 2\times 3\times 3\times 7\ 430=2\times 2\times 3\times 5\times 7\ 480=2\times 2\times 2\times 2\times 2\times 3\times 5$

Therefore, the HCF of $504,420$ and $480$ is:

HCF$\left( 504,430,480 \right) =2\times 2\times 3=12$

Hence, the largest size of the team is of $12$ students.

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

If HCF of $m$ and $n$ is $1,$ then what are the HCF of $m + n, m$ and HCF of $m - n, n$ respectively? 

$\displaystyle \left ( m> n \right )$

  1. $1$ and $2$
  2. $2$ and $1$
  3. $1$ and $1$

  4. cannot be determined
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let us consider an example.
Let $m =16$ and $n =9$ be relatively prime numbers.
So, $m+n=25$. The HCF of $25$ and $16$ is $1$. 

$m-n=7$. The HCF of $7$ and $9$ is $1$.
Similarly, if we take other values for $m$ and $n,$ we get the same answer. 
Therefore, option $C$ is correct.
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

The GCD of $\displaystyle \frac{3}{16}$,$\displaystyle \frac{5}{12}$,$\displaystyle \frac{7}{18}$ is 

  1. $\displaystyle \frac{105}{48}$
  2. $\displaystyle \frac{1}{4}$
  3. $\displaystyle \frac{1}{48}$

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The greatest common divisor is same as the highest common factor that is GCD is same as HCF and,

HCF of two or more fractions is given by HCF of Numerators divided by LCM of Denominators 

HCF of the numerators $(3,5,7)=1$
LCM of the denominators $(16,12,18)=2\times 2\times 2\times 2\times 3\times 3=144$

Therefore, 

HCF$\left( \dfrac { 3 }{ 16 } ,\dfrac { 5 }{ 12 } ,\dfrac { 7 }{ 18 }  \right) =\dfrac { 1 }{ 144 }$

Hence, GCD of $\dfrac { 3 }{ 16 } ,\dfrac { 5 }{ 12 } ,\dfrac { 7 }{ 18 }$ is $\dfrac { 1 }{ 144 }$
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If (x + 6) is the HCF of $\displaystyle p\left ( x \right )=x^{2}-a$ and $\displaystyle q\left ( x \right )=x^{2}-bx+6$ then $\displaystyle \frac{p\left ( x \right )}{q\left ( x \right )}$ in its lowest terms is______

  1. $\displaystyle \frac{x-6}{x-2}$
  2. $\displaystyle \frac{x+6}{x+1}$
  3. $\displaystyle \frac{x-6}{x-1}$
  4. $\displaystyle \frac{x-6}{x+1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since (x + 6) is the HCF, (x + 6) must be a factor of both polynomials. For p(x) = x^2 - a, x = -6 makes p(-6) = 36 - a = 0, so a = 36. Thus p(x) = (x - 6)(x + 6). For q(x) = x^2 - bx + 6, x = -6 makes 36 + 6b + 6 = 0, so 6b = -42, b = -7. Thus q(x) = x^2 + 7x + 6 = (x + 6)(x + 1). The ratio p(x)/q(x) simplifies to (x - 6)/(x + 1).

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

Two positive numbers have their HCF as $12$ and their sum is $84$. Find the number of pairs possible.

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

As the HCF is $ 12 $, the numbers can be written as $ 12x $ and $ 12y $, where x and y are co-prime to each other.
So, $ 12x + 12y = 84 => x + y = 7 $

The pair of numbers that are co-prime to each other and sum up to $7$ are $(2, 5), (1,6), (3,4)$.
Hence, only $ 3 $  pairs of such numbers are possible.
 The numbers are $ (24, 60), (12,72) $ and $ (36,48) $

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) $