By Hit and Trial
From option (d), we can say that the HCF of 36 and 24 is 12 and it is also satisfies the given condition a > b > 12

Correct Option: D

By Hit and Trial
From option (d), we can say that the HCF of 36 and 24 is 12 and it is also satisfies the given condition
a > b > 12
∴ a = 36 and b = 24

Let LCM = m, HCF = n,

According to the question,
m = 12n, .......(i)
and m + n = 403.......(ii)

Correct Option: C

Let LCM = m, HCF = n,

According to the question,
m = 12n, .......(i)
and m + n = 403.......(ii)
⇒ 12n + n = 403 [from Eq.(i)]
⇒ 13n = 403
∴ n = 403/13 = 31
∴ m = 12 x 31 = 372

Let the another number = k
∴ 93 x k = 372 x 31
[as product of two numbers HCF x LCM]
⇒ k = (372 x 31)/93 = 124

Let HCF = N
According to the question, LCM = 20N
Given that, HCF + LCM = 2520
⇒ N + 20N = 2520
⇒ N = 2520 / 21 = 120
Now, LCM = 20N = 20 x 120 = 2400

We know that,
1st number x 2nd number = HCF x LCM

Correct Option: D

Let HCF = N
According to the question, LCM = 20N
Given that, HCF + LCM = 2520
⇒ N + 20N = 2520
⇒ N = 2520 / 21 = 120
Now, LCM = 20N = 20 x 120 = 2400

We know that,
1st number x 2nd number = HCF x LCM
⇒ 2nd number = (LCM x HCF) / (1st number)
= (120 x 2400) / 480 = 600
∴ Required answer = 600 x 3 = 1800

Given that,
x = 130, y = 305, z = 245
a = 6, b = 9, c = 17

According to the formula,
Required number = HCF of [(130 - 6), (305 - 9), (245 - 17)]

Correct Option: A

Given that,
x = 130, y = 305, z = 245
a = 6, b = 9, c = 17

According to the formula,
Required number = HCF of [(130 - 6), (305 - 9), (245 - 17)]
= HCF of 124, 296, 228 = 4.

LCM of 16, 18 and 20 = 720
∴ Required number = 720k + 4
Where, k is a natural number to be divisible by 7, (720 k + 4) will be a multiple of 7.

Correct Option: A

LCM of 16, 18 and 20 = 720
∴ Required number = 720k + 4
Where, k is a natural number to be divisible by 7, (720 k + 4) will be a multiple of 7.
Smallest value of k = 4
∴ Required number = 720 x 4 + 4 = 2884