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HCF (A, B) = 12
LCM (A, B) = 180
We know that,
HCF (A, B) × LCM (A, B) = A × B
Also, A – B = 30
Or, B = A – 30 ……… (I)
ATQ:
A × (A – 30) = 12 × 180
Or, A² – 30A = 2160
Or, A² – 30A – 2160 = 0
Or, A² – 60A + 36A – 2160 = 0
Or, A(A – 60) + 36(A – 60) = 0
Or, (A – 60)(A + 36) = 0
So, A = 60 or A = -36
Since, A cannot be negative, we discard A = -36.
So, A = 60
And, B = 60 – 30 = 30
Therefore, required sum = (60 + 30) = 90
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