Tag: highest common factor (h.c.f.)

Questions Related to highest common factor (h.c.f.)

Find $HCF$ by finding factors:
$75, 79$ and $89$.

  1. $2$

  2. $3$

  3. $1$

  4. $4$


Correct Option: C
Explanation:
Factorization of the following 
$75 = 5 \times 5 \times 3\times 1$
$79=1 \times 79$
$89 = 1 \times 89$
Since, the common factor is $1$ this implies that
$HCF = 1$

Hence, the correct option is $C$

Manukaka distribute $96$ marbles among the children of a class in such a way that each child got equal number of marble. In the same class, sameway he also distributed $72$ chocolates. No chocolate and marble is left. How many maximum students are there in this class so that it is possible?

  1. $22$

  2. $24$

  3. $26$

  4. $28$


Correct Option: B
Explanation:
Factorization of the following.
Maximum students are $ =24$
$72 = (1,2,3,4,6,8,9,12,72)$
$24 = (1,2,3,4,6,8,12,24)$
Since, The common factor is $1,2,3,4$
$H.C.F = 24$
Hence, The correct option is $B$

Find HCF by using prime factor method:
$66$ and $88$.

  1. $21$

  2. $23$

  3. $24$

  4. $22$


Correct Option: D
Explanation:

Factorization of the following

$66 = 1 \times 3 \times 2 \times 11$
$88 = 1 \times 2 \times 2 \times 2 \times 11$
Since, the common factor is $1,2,11$ this implies that 
$HCF=22$
Hince, the correct option $D$

Find HCF by using prime factor method:
$25$ and $55$.

  1. $5$

  2. $3$

  3. $2$

  4. $4$


Correct Option: A
Explanation:

Factorization of the following

$25 = 1 \times 5 \times 5$
$55 = 1 \times 5 \times 11$
Since. the common factor is $1,5$ this implies that 
$HCF=5$
Hence, the correct option $A$

What is the HCF of $13$ and $22$?

  1. $13$

  2. $22$

  3. $1$

  4. $286$


Correct Option: C
Explanation:
Factorization of the following.
HCF=Highest common factor
 so  HCF between $\left( {13,22} \right)$
$ \Rightarrow 13 = 1,13\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - eq.\left( 1 \right)$
$ \Rightarrow 22 = 1,2,11,22\,\,\,\,\,\,\,\,\,\,\, - eq.\left( 2 \right)$  
$(1)$ &$(2)$ equation common is $1$.
Since, The common factor is $1$. This implies that
 then HCF  is $1$.
Hence, the correct option is $C$.

G.C.D of $4$ and $19$ is _________.

  1. $1$

  2. $4$

  3. $19$

  4. $76$


Correct Option: A
Explanation:

There is no common factor between $4$ and $19$. Hence, G.C.D. of $4$ and $19$ is $1$.

The two numbers nearest to 10000 which are exactly divisible by each of 2, 3, 4, 5, 6 and 7, are _____.

  1. 9660, 10080

  2. 9320, 10080

  3. 9660, 10060

  4. 10340, 10080


Correct Option: A
Explanation:

The numbers which are exactly divisible by 2, 3, 4, 5, 6 and 7 are the multiples of the LCM of the given numbers.
$\therefore$  LCM = 2 x 2 x 3 x 5 x 7 = 420
Now, dividing 10000 by 420, we get remainder = 340
$\therefore$  Number just less than 10000 and exactly divisible by the given numbers = 10000 - 340 = 9660
Number just greater than 10000 and exactly divisible by the given numbers = 10000 + (420 - 340) = 10080

If the HCF of 85 and 153 is expressible in the form $85n -153$, then the value of n is:

  1. 3

  2. 2

  3. 4

  4. 1


Correct Option: B
Explanation:

$85 = 5 \times  17$

$153 = 3 \times  3 \times 17$
So HCF will be 17
$17 = 85n-153$ from here we get n=2
So the correct answer is option B

If the HCF of 85 and 153 is expressible in the form 85n $-$ 153, then value of n is :

  1. 3

  2. 2

  3. 4

  4. 1


Correct Option: B
Explanation:

HCF of $85\  and\  153 = 17$

Now given HCf can be expressed in the gorm of $85n-153$
So $17=85n-173$
On solving the above equation we get $n=2$
So correct answer will be option B

Choose the correct answer form the alternatives given.
What is the HCF of $(x^4 \, - \, x^2 \, - \, 6) \, and \, (x^4 \, - \, 4x^2 \, + \, 3)$? 

  1. $x^2$ - $3$

  2. $x + 2$

  3. $x + 3$

  4. $x^2$ + $3$


Correct Option: A
Explanation:

$\displaystyle (x^4 \, - \, x^2 \, - \, 6) \, = \, (x^2 \, - \, 3) (x^2 \, + \, 2)$
$\displaystyle (x^4 \, - \, 4x^2 \, + \, 3) \, = \, (x^2 \, - \, 3) (x^2 \, - \, 1)$
HCF is = $x^2$ - $3$