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 numbers be 10N, 12N, 15N and 18N.
Then, LCM = 180N and HCF = N
Hence, required LCM = 180 x 3 = 540
Let the numbers are 3N, 4N and 5N.
Where, N = HCF
Then, LCM = 60N
According to the question,
60N = 1200
? N = 20
Given,
LCM = 1989, HCF = 13
1st number = 117 and
2nd number = ?
According to the formula,
Product of LCM and HCF = Product of two numbers
? 1989 x 13 = 117 x ?
? = (1989 x 13)/117
? = 221
Given, LCM of two number = 2376
HCF of two number = 33
One of the number = 297
? (HCF of two numbers) x (LCM of two numbers) = (First number) x (Second number)
? Second number = (33 x 2376)/297
= 264
LCM of two coprimes is equal to their product.
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
? 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)]
= 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.
Smallest value of k = 4
? Required number = 720 x 4 + 4 = 2884
Time at which they meet at starting point
= LCM of 2, 4 and 5.5
= 44 h
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