Question

# Set Fn gives all factors of n. Set Mn gives all multiples of n less than 1000. Which of the following statements is/are true i. F108 ∩ F84 = F12 ii. M12 ∪ M18 = M36 iii. M12 ∩ M18 = M36 iv. M12 ⊂ (M6 ∩ M4)

CAT 2021

A
B
C
D
##### All statements are true
Solution

Correct option is

(B)

(i) $F_{108}cap&space;F_{84}$

this is the set of all numbers that are factors of 108 and 84

hence we have to find common factors which , HIGHEST COMMON FACTOR OF 84 and 108

HCF (84,108)=12

hence this is true

(ii)$M_{12}$

will have numbers {12 , 24 , 36,....} but these number will not come in $M_{36}$ ,

hence this statement is false

(iii) $M_{12}$$M_{18}$- This is the set of all common multiples of 12 and 18 ,

so , LCM(12 , 18 ) = 36 , hence this statement is also true

(iv) LCM(6,4)= 12 , and every set is a subset of itself , Hence this is also true

so , correct option is B (i ,iii,iv)