0
7365

# HCF and LCM Questions for SSC CGL PDF:

Download SSC CGL HCF and LCM questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important HCF and LCM objective questions for SSC exams.

Question 1: The H.C.F. and L.C.M. of, two numbers are 8 and 48 respectively. If one of the numbers is 24, then the other number is

a) 48

b) 36

c) 24

d) 16

Question 2: Two numbers are in the ratio 3:4. Their L.C.M. is 84. The greater number is

a) 21

b) 24

c) 28

d) 84

Question 3: The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is

a) 2/35

b) 3/25

c) 4/35

d) 2/25

Question 4: L.C.M. of two numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers?

a) 140

b) 80

c) 60

d) 70

Question 5: Product of two coprime numbers is 117. Then their LCM is

a) 9

b) 13

c) 39

d) 117

Question 6: HCF and LCM of two numbers are 11 and 825 respectively. If one number is 275 find the other number.

a) 53

b) 45

c) 33

d) 43

Question 7: What is the LCM (least common multiple) of 57 and 93?

a) 1767

b) 1567

c) 1576

d) 1919

Question 8: What is the HCF (highest common factor) of 57 and 513?

a) 10

b) 57

c) 3

d) 27

Question 9: The two numbers are 63 and 77, HCF is 7, Find the LCM.

a) 668

b) 693

c) 674

d) 680

Question 10: What is the HCF (highest common factor) of 77 and 275?

a) 12

b) 11

c) 7

d) 25

Question 11: The two numbers are 55 and 99, HCF is 11, What is their LCM?

a) 486

b) 479

c) 476

d) 495

Question 12: What is the HCF (highest common factor) of 133 and 112?

a) 15

b) 7

c) 19

d) 16

Question 13: Find the HCF of $\frac{2}{3}, \frac{1}{2}, \frac{3}{5}$

a) 6

b) $\frac{1}{30}$

c) $\frac{1}{15}$

d) $\frac{1}{10}$

Question 14: What is the LCM of 64 and 56?

a) 448

b) 488

c) 484

d) 408

Question 15: What is the HCF of 7/9, 2/3, 5/8 and 7/12?

a) 1/18

b) 1/36

c) 1/144

d) 1/72

Question 16: The LCM of two numbers is 4 times their HCF. The sum of LCM and HCF is 125. If one of the numbers is 100, then the other number is

a) 5

b) 25

c) 100

d) 125

Question 17: Find the HCF of 2/3 , 6/8, 7/12 and 2/5

a) 1/120

b) 1/60

c) 1/30

d) 1/240

Question 18: Find the HCF of 8/3, 12/7 and 13/12.

a) 1/504

b) 1/126

c) 1/84

d) 1/1008

Question 19: Find the LCM of 3/4, 5/8 and 9/27.

a) 15

b) 45

c) 1/24

d) 1/48

Question 20: Find the HCF of 0.8, 0.125, 0.625 and 0.5.

a) 0.1

b) 1/40

c) 1/20

d) 1/80

Question 21: Find the HCF of 3/8, 17/19 and 21/23.

a) 1/2392

b) 1/3192

c) 1/3128

d) 1/3496

Question 22: The sum of two numbers is 7 and the sum their squares is 23, their product is equal to:

a) 10

b) 11

c) 12

d) 13

Question 23: The difference between two numbers is 1146. When we divide the larger number by smaller we get 4 as quotient and 6 as remainder. Find the larger number.

a) 1526

b) 1431

c) 1485

d) 1234

Question 24: The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is

a) 4302

b) 4032

c) 4023

d) 4203

Question 25: A number between 1000 and 2000 which when divided by 30, 36 & 80 gives a remainder 11 in each case is

a) 1451

b) 1641

c) 1712

d) 1523

Given:-
Numbers- First = 24
Second = x (suppose)
H.C.F. of numbers = 8
L.C.M. of numbers = 48
As we know:
H.C.F.* L.C.M. = Product of numbers
Hence
48*8 = 24*x
x = 16

Let the numbers be 3x, 4x
LCM of 3x and 4x is = 12x
So the number 84 is divisible by 12
$\frac{84}{12}$ = 7
The numbers are 7×3 = 21 , 7x 4 = 28
The greatest number is 28

let’s say numbers are $x$ and $y$
hence sum of the reciprocals will be  $\frac{1}{x} + \frac{1}{y}$
or $\frac{x +y}{xy}$
as $x+y$ = 36 (given)
and $xy$ = $HCF \times LCM$
= $3 \times 105 = 315$
after putting the values we will get summation of reciprocals equals to $\frac{4}{35}$

We assume that numbers are $hr_1$ and $hr_2$ (where h= H.C.F. of numbers and $r_1$ and $r_2$ are prime factors)
So L.C.M. will be = $hr_1r_2$ = 120
or $r_1r_2$ = 12
So $r_1$ =4 and $r_2$ = 3 ; numbers will be 40 and 30, sum is 70
or $r_1$ = 12 and $r_2$ = 1 ; numbers will be 120 and 10, sum is 130
Hence only option D justifies.

Let the two numbers be a,b.
Hence a * b = L.C.M(a,b) * G.C.D(a,b)
It is given that a,b are co-primes, implies G.C.D(a,b) = 1
Hence from the above equation we get L.C.M(a,b) = a*b = 117

Let the number = $x$

HCF = 11 and LCM = 825

Product of HCF and LCM = Product of the two numbers

=> $x \times 275 = 11 \times 825$

=> $x = \frac{11 \times 825}{275}$

=> $x = \frac{825}{25} = 33$

=> Ans – (C)

Prime factorization of 57 = 3 $\times$ 19

Prime factorization of 93 = 3 $\times$ 31

=> L.C.M. of 57 and 93 = 3 $\times$ 19 $\times$ 31

= 57 $\times$ 31 = 1767

=> Ans – (A)

Factors of 57 = 1 , 3 , 19 , 57

Factors of 513 = 1 , 3 , 9 , 19 , 27 , 57 , 171 , 513

The common factors are = 1 , 3 , 19 , 57

=> Highest common factor = 57

=> Ans – (B)

H.C.F. (a,b) $\times$ L.C.M. (a,b) = $a \times b$

The numbers a = 63 and b = 77 and HCF = 7

=> L.C.M. = $\frac{a \times b}{HCF}$

= $\frac{63 \times 77}{7} = 63 \times 11$

= 693

Factors of : 77 = 1 , 7 , 11 , 77

275 = 1 , 5 , 11 , 25 , 55 , 275

The common factors are 1 and 11

and HCF = 11

=> Ans – (B)

Let the LCM = $x$

Numbers are = 55 , 99

Also, product of numbers = HCF $\times$ LCM

=> $55 \times 99 = 11 \times x$

=> $x = \frac{55 \times 99}{11} = 5 \times 99$

=> $x = 495$

=> Ans – (D)

Prime factorization of

133 = 7 $\times$ 19

112 = $2^4$ $\times$ 7

There is only 1 common factor, and thus the HCF (highest common factor) = 7

=> Ans – (B)

HCF = HCF of numerators/ LCM of denominators

HCF of 1, 2, 3 = 1

LCM of 3, 2, 5 = 30

Hence, $\frac{1}{30}$

(diagram)

so LCM of 64 & 56 is = 8*8*7 = 448

So the answer is option A.

HCF of fractions = (HCF of the numerators)/ (LCM of the denominators).

HCF of 7,2,5 and 7 is 1.

LCM of 9, 3, 8 and 12 is 72.
Therefore, the HCF of the given fractions is 1/72.

Option D is the right answer.

Let one of the numbers = $x$ and other number = 100

Let L.C.M = $L$ and H.C.F = $H$

According to ques, => $L=4H$ ————–(i)

and $L+H=125$

Substituting value from equation (i), we get : $4H+H=5H=125$

=> $H=\frac{125}{5}=25$

=> $L=4\times25=100$

Thus, product of numbers = $L\times H$

=> $100\times x=100\times25$

=> $x=25$

=> Ans – (B)

HCF of fractions = HCF of the numerators / LCM of the denominators.
All the fractions must be in their empirical form to apply this formula.

6/8 can be written as 3/4.
The given fractions are 2/3, 3/4, 7/12 and 2/5.
HCF of the numerators = 1.
LCM of (3,4,12 and 5) is 60.
Therefore, the HCF of the given fractions is 1/60.
Hence, option B is the right answer.

HCF of fractions = HCF of the numerators / LCM of the denominators.
All the fractions must be in their empirical form to apply this formula.

HCF of the numerators (8,12,13) is 1.
LCM of the denominators (3,7,12) is 84.

Therefore, the HCF of the given fractions is 1/84. Hence, option C is the right answer.

LCM of fractions = LCM of the numerators / HCF of the denominators.
All the fractions must be in their empirical form to apply this formula.
9/27 can be reduced to 1/3.
Therefore, the given fractions are 3/4, 5/8 and 1/3.
LCM of the numerators (3,5,1) is 15.
HCF of the denominators (4,8,3) is 1.
Therefore, the LCM is 15/1 = 15.
Hence, option A is the right answer.

HCF of fractions = HCF of the numerators / LCM of the denominators.
All the fractions must be in their empirical form to apply this formula.
0.8 = 4/5
0.125 = 1/8
0.625 = 5/8
0.5 = 1/2

The given numbers can be written as 4/5, 1/8, 5/8, 1/2.
HCF of the numerators = 1.
LCM of the denominators (5,8,2) = 40.
Therefore, the HCF of the given numbers will be 1/40 or 0.025.
Therefore, option B is the right answer.

HCF of fractions = HCF of the numerators / LCM of the denominators.
All the fractions must be in their empirical form to apply this formula.

The HCF of (3,17,21) is 1.
The LCM of (8,19,23) is 3496.
Therefore, the HCF of the given fractions is 1/3496.
Hence, option D is the right answer.

Let the numbers be $x$ and $y$

It is given that $x^2+y^2=23$ ———–(i)

Also, $x+y=7$

Squaring both sides, we get :

=> $x^2+y^2+2xy=49$

=> $23+2xy=49$

=> $2xy=49-23=26$

=> $xy=\frac{26}{2}=13$

$\therefore$ Product of the numbers = 13

=> Ans – (D)

Let the smaller number be $x$ and the larger number = $(x+1146)$

According to ques, on dividing the larger term by smaller one,

=> $(x+1146)=4x+6$

=> $4x-x=1146-6$

=> $3x=1140$

=> $x=\frac{1140}{3}=380$

$\therefore$ Larger number = $380+1146=1526$

=> Ans – (A)

LCM of 12,18,21,32 is 252

Multiples of 252 between 4000 and 5000 are 4032, 4284, 4536, 4788.

4032 is present in the options.

Hence, option B is the correct answer.

$\therefore$ Number which gives a remainder 11 in each case (1440 + 11) = 1451