Quantitative Aptitude · Mathematics

HCF and LCM

233 Questions

Improve your quantitative aptitude speed by solving these HCF and LCM questions. The problems involve finding the highest common factor and least common multiple for sets of numbers. These calculations form a core part of the mathematics section in SSC, banking, and railway recruitment exams.

Finding HCFFinding LCMGreatest common factorLeast common multipleNumber properties

HCF and LCM Questions

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

If two positive integers $a$ and $b$ are written as $a=x^3y^2$ and $b=xy^3$; $x, y$ are prime numbers, then HCF of $a$ and $b$ is

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

Given,

$a={  x}^{3  }{ y }^{2  } = x\times x\times x\times y\times y$

$b={  x}{ y }^{3  }         =x\times y\times y\times y$

H.C.F of $a,b$  = ${  x}{ y }^{2  } $

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.)

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.)

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.)

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.)

Find $HCF$ by finding factors:
$6$ and $8$.

  1. $4$
  2. $6$
  3. $2$
  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Factorization of the following.

$6 = 3 \times 2\times 1$
$8 = 2 \times 2 \times 8\times 1$

Since, The common factor is $2$.This implies that
$H.C.F = 2$

Hence,the correct option is $C$

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

Choose the correct answer from the alternatives given.
LCM of $\dfrac 14 $and $\dfrac 18$ is

  1. $\dfrac12$
  2. $1$
  3. $\dfrac 14$
  4. $\dfrac 18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know that,
LCM of dfrac tions = $\dfrac { LCM\ of\ numerators}{HCF\ of\ denominators}$
= $\dfrac { LCM (1,1)}{ HCF(4,8)} = \dfrac {1}{4}$