Tag: common factors and hcf

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

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

  1. $5$
  2. $3$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A$

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

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

  1. $13$
  2. $22$
  3. $1$
  4. $286$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$.
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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. $x^2$ + $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 = $x^2$ - $3$

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

The greatest common divisor of $878787878787$ and $787878787878$ equals.

  1. $3$
  2. $9$
  3. $27$
  4. $101010101010$
  5. $303030303030$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

$787878787878)878787878787(1\ \quad \quad \quad \quad \quad  -\underline { 787878787878 } \ \quad \quad \quad \quad \quad \quad \quad 90909090909)787878787878(8\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \underline { -727272727272 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 60606060606)90909090909(1\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -60606060606 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 30303030303)60606060606(2\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -60606060606 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0$

$\therefore$ G.D.C = 30303030303

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

Three bells, toll at intervals of $36$ sec, $40$ sec and $48$ sec respectively. They start ringing together at particular time. They next toll together after

  1. $6$ minutes
  2. $12$ minutes
  3. $18$ minutes
  4. $24$ minutes
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

G.C.D of $36,40,48=720\Rightarrow 720sec=12min$
$\therefore$ Next time when three balls toll together is after $12$ mins

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

The G.C.D. of two whole numbers is $5$ and their L.C.M. is $60$. If one of the numbers is $20$, then the other number would be

  1. $23$
  2. $13$
  3. $16$
  4. $15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If we are given two numbers $N _1$ and $N _2$ and their $G.C.D$ and $L.C.M.$.

then by property of numbers $N _1$$\times$$N _2=G.C.D$ $\times$ $L.C.M.$

Here Given:
$N _1=20$
$G.C.D.=5$
and $L.C.M=60$
Let, $N _2=x$

then from  above relation
$20$$\times$$x=5$$\times$$60$

$=>x=\dfrac{300}{20}$

$=>x=15$