Tag: hcf

Questions Related to hcf

What is the H.C.F . of $348$ and $319$

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

  1. $348$

  2. $319$

  3. $56$

  4. $29$

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Correct Option: D
Explanation:

$348=$$2\times$$2\times$$3\times$$29$

$319=11$$\times29$
thus $HCF=29$

HCF of $144$ and $198$ is

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

  1. $9$

  2. $18$

  3. $6$

  4. $12$

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Correct Option: A
Explanation:

HCF of 144 and 198 is:

$ 198 = 2 \times 3^2 \times 11$
$ 144 = 2^4 \times 3^2$

Highest Common factor is 9.
Option A is correct

HCF of $24 $ and $36$ is ..............

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

  1. $6$

  2. $4$

  3. $9$

  4. $12$

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Correct Option: D
Explanation:
$2\underline {|24} {\;} 2\underline {|36}$
$2\underline {|12} {\;} 2\underline {|18}$
$2\underline {|6} {\;} 3\underline {|9}$
$3\underline {|3} {\;} 3\underline {|3}$
    1    1
$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore H.C.F=2\times 2 \times 3=12$

HCF of the numbers $36$ and $144$ is 

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

  1. $36$

  2. $144$

  3. $4$

  4. $2$

Show answer
Correct Option: A
Explanation:

Factors of the given numbers are,

$36= 2\times 2 \times 3 \times 3$
$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 $
$\therefore $ HCF of $36$ and $144$ $ = 2 \times 2 \times 3 \times 3 = 36$


The number of ordered pairs $(a, b)$ of positive integers, such that $a + b = 90$ and their greatest common division is $6$, equals

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

  1. $5$

  2. $4$

  3. $8$

  4. $10$

Show answer
Correct Option: C
Explanation:

Let's look at products of $6$
$6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90$
The pairs whose sums equal to $90$ are:
$6 + 84=90$
$12 + 78=90$
$18 + 72=90$
$66 + 24=90$
$60 + 30=90$
$54 + 36=90$
$48 + 42=90$
Total number of pairs are $7$.
But there can be  $7$ more pairs when the numbers are reversed.
$\therefore$ total number of ordered pair are $14$

If $x =2^3 \times 3 \times 5^2,Y=2^2 \times 3^3$, then HCF(x,y)is:

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

  1. 12

  2. 108

  3. 6

  4. 36

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Correct Option: A
Explanation:

$x= 2^3 \times 3 \times 5^2$
$y = 2^2 \times 3^3$
HCF (x,) $=2^2 \times 3$= 12

The product of two numbers is $2240$ and their HCF is $14$. Which of the following is not the possible pair.

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

  1. $(14,160)$

  2. $(28,80)$

  3. $(42, 80)$

  4. $(56,40)$

Show answer
Correct Option: C
Explanation:

Pair in Option C is not possible.
because in option C pair is $(42,80)$
Since, Product of numbers $=42\times 80 =3360$
Option C is correct.

G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ is

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

  1. $(a+b-c)^6$

  2. $(a+b-c)^{10}$

  3. $(a+b-c)^2$

  4. $(a+b-c)^4$

Show answer
Correct Option: D
Explanation:

Since, $(a + b -c)^6 = (a + b -c)^4 \times (a + b -c)^2 $
$\therefore$  G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ = $(a + b -c)^4$
$\because (a + b -c)^4$ is greatest common in $(a + b -c)^4$ and $(a + b -c)^6$.
Option D is correct.

Find G.C.D of $20x^2-9x+1$ and $5x^2-6x+1$

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

  1. (x-1)

  2. (5x-1)

  3. (5x+1)

  4. None of these

Show answer
Correct Option: B
Explanation:

Let, $p(x) = 20x^2-9x+1$ and $q(x) = 5x^2-6x+1$
$p(x) = 20x^2-9x+1$
         $=20x^2-5x-4x+1$
         $=5x(4x-1)-1(4x-1)$
         $=(5x-1)(4x-1)$
and
$q(x) = 5x^2-6x+1$
         $=5x^2-5x-x+1$
         $=5x(x-1)-1(x-1)$
         $=(5x-1)(x-1)$
$\therefore$ G.C.D of $p(x)$  and  $q(x)=(5x-1)$.
Option B is correct.

The G.C.D of $x^3+x^2-x-1$ and $x^2-1$ is

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

  1. $x^2-1$

  2. $x+1$

  3. $x^3-1$

  4. $x-1$

Show answer
Correct Option: A
Explanation:

Let $p(x) = x^3+x^2-x-1$ and $ q(x) = x^2-1$
$p(x) = x^3+x^2-x-1$
         $ = x^2(x+1)-1(x+1) $
         $ = (x^2-1) (x+1) $
         $ = (x-1)(x+1)(x+1) $
and
$ q(x) = x^2-1$
         $= (x+1)(x-1) $
$\therefore $ G.C.D of $p(x)$ and $q(x)$ =$ (x+1)(x-1) = x^2 - 1 $
Option A is correct.