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 introduction to multiples multiples lcm

LCM of numbers 1, 2, 3 is equal to their

  1. product

  2. division

  3. sum

  4. difference
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$2, 3$ are primes.
$\therefore$ Each number has no factor other than $1$ and itself.
$\therefore$ Their LCM is the product of the numbers.
$\therefore$ LCM of $1,2,3=2\times 3=6$.
Also here $1+2+3=6$.
Answer- Option A and Option C.

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

The HCF of $3^5, 3^9$, and $3^{14}$ is

  1. $3^5$
  2. $3^9$
  3. $3^{14}$
  4. $3^{21}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To  find  the  Highest  Common  Factor  (HCF)  of  two  or  more  numbers, 
find  prime  factors  of  the  numbers , and  then  identify  the  common  prime  factors.
Then  the  HCF  is  the  product  of  the  common  prime  factors.
$3^{5}= 1 \times 3^{5}$
$3^{9} = 1 \times 3^{5} \times 3^{4}$
$3^{14} = 1 \times 3^{5} \times 3^{9}$
Hence the HCF is $ 3^{5}.$
Another  method  to  find  the  answer  is  to find  the  largest  divisor  of all  three  numbers,
 which  is $ 3^{5} $

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Choose the most appropriate option.
The traffic lights at three different signal points change after every $45$ seconds, $75$ seconds and $90$ seconds respectively. If all change simultaneously at $7:20:15$ hours, then they will change again simultaneoulsy at.

  1. $7:27:30$ hours
  2. $7:28:00$ hours
  3. $7:27:50$ hours
  4. $7:27:45$ hours
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The $3$ signals (at $3$ points) change every $45 s,\, 75 s,\, 90 s$

So, they will change simultaneously for a common time, which is the common multiple or $L.C.M$ of the three

$\Rightarrow$  $45 = 5\times 9 = 3^2 \times 5$ 

$\Rightarrow$  $75 = 3\times 25 = 3\times 5^2$

$\Rightarrow$  $90 = 9\times 10 = 2\times 3^2\times 5$

$L.C.M= 2 \times 3^2 \times 5^2 = 2\times 9\times 25 = 450s$

So, they will change simultaneously every $450s$ or $7\,mins\,30 \,sec$

$\Rightarrow$  So, next they will change together at $7:27:45$ hours.
Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Select the correct option.
The HCF and the LCM of $12, 21, 15$ respectively are

  1. $3, 140$
  2. $12, 420$
  3. $3, 420$
  4. $420, 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Numbers $= 12, 15, 21$


$12 =  2 \times 2 \times 3$


$15 = 3 \times 5$

$21 = 3 \times 7$

HCF = Product of smallest power of each common prime factor $= 3' = 3$
LCM = Product of greatest power of each prime factor 

$2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$

$(C) \,\, 3, 420$

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

Determine the HCF of $a^2 - 25, a^2 -2a -35$ and $a^2+12a+35$

  1. (a-5)(a+7)

  2. (a+5)(a-7)

  3. (a-7)

  4. (a+5)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, $a^2 - 25 = (a-5)(a+5) $
$ a^2 -2a -35 = a^2 -7a +5a -35 $
                         $= a(a-7)+5(a-7) $
                         $= (a+5)(a-7) $
and
$a^2+ 12a + 35 =a^2 +7a +5a +35 $
                          $=(a+7)(a+5) $
Clearly HCF of $a^2 - 25, a^2 -2a -35$ and $a^2+ 12a + 35$ i.e $ (a-5)(a+5), (a+5)(a-7)$ and $(a+7)(a+5)$ is $a+5$
Option D is correct.

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

The LCM of 54 90 and a third number is 1890 and their HCF is 18 The third number is

  1. 36

  2. 180

  3. 126

  4. 108
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given the LCM two numbers 54, 90 and third number is 1890 and HCF is 18

Let the number is 18x because one factor is also 18 the common factor HCF
Then factor 54,90 ,18 =$18\times 3,18\times 5,18\times 18\times x$

$\therefore 18\times 3\times 5\times x=1890\Rightarrow 270x=1890\Rightarrow x=7$
Then third number is $18\times 7=126$ 

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

HCF of the two numbers =

  1. Product of numbers + their LCM

  2. Product of numbers - their LCM
  3. Product of numbers $\times$ their LCM
  4. Product of numbers $\div$ their LCM

  5. Answer required
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The product of highest common factor $(H.C.F.)$ and lowest common multiple $(L.C.M.)$ of two numbers is equal to the product of two numbers.

If two numbers are $a$ and $b$ then
$HCF\ \times LCM=a\times b$ 

Hence,
$HCF=\dfrac{ab}{LCM}$

For example: 
Let $a=10\Rightarrow 2\times 5$ and $b=15\Rightarrow 3\times 5$
So, the $LCM$ of both numbers $=2\times 3\times 5\Rightarrow 30$
Then 
$HCF=\dfrac{10\times 15}{30}\Rightarrow 5$

Hence,
$HCF=\dfrac{Product\ of\ numbers}{Their\ LCM}.$

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

What is the HCF of $4x^{3} + 3x^{2}y - 9xy^{2} + 2y^{3}$ and $x^{2} + xy - 2y^{2}$?

  1. $x - 2y$
  2. $x - y$
  3. $(x + 2y)(x - y)$
  4. $(x - 2y)(x - y)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $f(x, y) = 4x^{2} + 3x^{2}y - 9xy^{2} + 2y^{3}$
and $g(x, y) = x^{2} + xy - 2y^{2}$
Clearly $(x - y) = 0$ or $x = y$ satisfy both equation, since for $x = y$ makes both equations zero. 

Hence, $(x - y)$ is the H.C.F. 
Similarly $(x + 2y) = 0$ i.e., $x = -2y$ also makes both equations zero. 
Thus, $(x + 2y)(x - y)$ is the H.C.F.

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

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ Given\quad that\quad product\quad of\quad the\quad number\quad is\quad 5400=30\times 3\times 2\times 30.\ \therefore \quad Possible\quad pairs\quad as\quad per\quad the\quad requirment\quad are-\ (1)\quad 30\times (3\times 2\times 30)=30\times 180\ (2)\quad (30\times 3)\times (2\times 30)=90\times 60\ \therefore \quad Total\quad number\quad of\quad pairs=2\quad \quad (Ans) $