MCQ: HCF & LCM - 2 - SSC CGL MCQ

Here, HCF = 13
Let the numbers be 13x and 13y where x and y are Prime to each other.
Now, 13x × 13y = 2028

The possible pairs are : (1, 12), (3, 4), (2, 6)
But the 2 and 6 are not co-prime.
∴ The required no. of pairs = 2

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × second number = HCF × LCM
⇒ 52 × second number = 4 × 520

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × second number = HCF × LCM
⇒ 84 × second number = 12 × 336

HCF of two-prime numbers = 1
∴ Product of numbers = their LCM = 117
117 = 13 × 9 where 13 & 9 are co-prime. L.C.M (13,9) = 117.

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM
⇒ 75 × Second number = 15 × 225

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
H.C.F. of the two 2-digit numbers = 16
Hence, the numbers can be expressed as 16x and 16y, where x and y are prime to each other.
Now,
First number × second number = H.C.F. × L.C.M.
⇒ 16x × 16y = 16 × 480

The possible pairs of x and y, satisfying the condition xy = 30 are : (3, 10), (5, 6), (1, 30), (2, 15)
Since the numbers are of 2-digits each.
Hence, admissible pair is (5, 6)
∴ Numbers are : 16 × 5 = 80 and 16 × 6 = 96

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,

Obviously, numbers are 111 and 37 which satisfy the given condition.
Hence, the greater number = 111

HCF = 12
Numbers = 12x and 12y
where x and y are prime to each other.
∴ 12x × 12y = 2160

= 15 = 3 × 5, 1 × 15
Possible pairs = (36, 60) and (12, 180)

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
We know that,
First number × Second number = LCM × HCF

Let the number be 15x and 15y, where x and y are co –prime.
∴ 15x × 15y = 6300

So, two pairs are (7, 4) and (14, 2)

HCF of two numbers is 8.
This means 8 is a factor common to both the numbers. LCM is common multiple for the two numbers, it is divisible by the two numbers. So, the required answer = 60

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,

LCM = 2 × 2 × 2 × 3 × 5
Hence, HCF = 4, 8, 12 or 24
According to question
35 cannot be H.C.F. of 120.

Let the numbers be 6x and 6y where x and y are prime to each other.
∴ 6x × 6y = 216

∴ LCM = 6xy = 6 × 6 = 36