Source.. dont know... found it on a old file in my computer
Q.) is a+b<c+d , the LCM of a & b is equal to the HCF (GCD ) of c & d
A.) a,b,c,d are multiples of 20
B.) a+c < b+d
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DS problem
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amitamit2020
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Can B be the answer? Since we are calculating HCF and LCM I assume a, b, c, d are integers. (Can negative numbers have HCF and LCM, I don't know. Can anyone tell?)
So, LCM of a and b will be >= to (greater of a and b).
HCF of c and d will be <= to (smaller of c and d).
Since, LCM = HCF:
a + b < = 2 LCM = 2 HCF < = c + d
Hence, a + b <= c + d
Equality may hold true ONLY if a, b, c, d are all equal.
Statement B says a + c < b + d.
Hence, a, b, c, d are not equal.
Therefore, a + b < c + d.
'A' can anyway be eliminated immediately by intuition. It is a useless piece of information.
Hence, my answer is B. What's the OA?
So, LCM of a and b will be >= to (greater of a and b).
HCF of c and d will be <= to (smaller of c and d).
Since, LCM = HCF:
a + b < = 2 LCM = 2 HCF < = c + d
Hence, a + b <= c + d
Equality may hold true ONLY if a, b, c, d are all equal.
Statement B says a + c < b + d.
Hence, a, b, c, d are not equal.
Therefore, a + b < c + d.
'A' can anyway be eliminated immediately by intuition. It is a useless piece of information.
Hence, my answer is B. What's the OA?
