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 real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

If HCF of numbers $408$ and $1032$ can be expressed in the form of $1032x -408 \times 5$, then find the value of $x$.

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$408=2\times2\times2\times3\times17$

$1032=2\times2\times2\times3\times43$

Hence, $HCF=2\times2\times2\times3=24$

Now, $1032x-408\times5=24\Rightarrow 1032x=2064\Rightarrow x=2$

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

Find the LCM and HCF of the following integers by the prime factorization mass

  1. 12, 15 and 21

  2. 17, 23, and 29

  3. 8, 9 and 25

  4. 72 and 108

  5. 72 and 108

  6. 306 and 657
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The question asks for the LCM and HCF of integers but provides a list of sets. Option A (12, 15, 21) is a valid set of integers for which LCM and HCF can be calculated.

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

When the HCF of $468$ and $222$ is written in the form of  $ 468 x + 222y$ then the value of $ x$ and $y$ is 

  1. $x =-9 \ and \ y =19$
  2. $x =9 \ and \ y = -19$
  3. $x =9\ and \ y = 19$
  4. $x =-9 \ and \ y =- 19$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

HCF of $468$ and $222$
$468 = \left(222 \times 2\right) + 24$
$222 = \left(24\times\ 9\right) + 6$
$24 = \left(6\times\ 4\right) + 0$
$\therefore HCF = 6$

$6 = 222 - \left(24\times\ 9\right)$
$ = 222 - \left[\left(468 -222 \times 2\right) \times\ 9\right]  $ [where $468 = 222 \times 2 + 24$]
$ = 222 - \left[468 \times 9 -222 \times 2 \times 9\right]$
$= 222 - \left(468 \times9\right) - \left(222\times 18\right)$
$ = 222 + \left(222 \times 18\right) - \left(468 \times9\right)$
$= 222\left[1 + 18\right]-  468 \times 9$
$= 222 \times19-  468 \times 9$
$  = 468 \times -9 + 222\times 19$
$\therefore x=-9$ and $y=19$.

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

The HCF of $136 ,170 \ and \ 255$ is 

  1. $13$
  2. $15$
  3. $17$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

136)170(1

  -    136
-------------------
          34)136(4
                136
----------------------------
                 0</div>

34)255(7
   -  238
-------------------
        17)34(2
             34
------------------------
              0

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

The H.C.F. of two expressions is x and their L.C.M is $ \displaystyle x^{3}-9x  $  IF one of the expression is $ \displaystyle x^{2}+3x  $  then,the other expression is 

  1. $ \displaystyle x^{2}-3x $
  2. $ \displaystyle x^{3}-3x $
  3. $ \displaystyle x^{2}+9x $
  4. $ \displaystyle x^{2}-9x $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let two expressions $p(x)$ and $q(x)$ then

$p(x)\times q(x)=L.C.M.\ \times\ H.C.F.$

Since $p(x)=x^2+3x$
$(x^2+3x)\times q(x)=(x^3-9x) \times\ x$
$(x^2+3x)\times q(x)=(x^2-3x) \times\ (x^2+3x)$

$q(x)=(x^2-3x)$
Hence, this is the required solution.

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

The H.C.F. of the numbers $16.5, 0.90$ and $15$ is

  1. $16.5$
  2. $0.90$
  3. $15$
  4. $0.3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$1650 = 2\times 3\times 5^{2}\times 11$
$90 = 2\times 3^{2} \times 5^{1}$
$1500 = 2^{2} \times 3^{1} \times 5^{3}$
H.C.F. of $1650, 90$ and $1500$ is $2\times 3\times 5 = 30$

Therefore, H.C.F. of $16.5, 0.90$ and $15$ is $0.30$.
So, option D is correct.