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Bunuel
If the least common multiplier of positive integers A and B is 120 and the ratio of A and B is 3:4, what is the largest common divisor of A and B?

A. 8
B. 9
C. 10
D. 12
E. 15

The least common multiplier of A and B is 120, the ratio of A and B is 3:4, what is the largest common divisor?
HCF=?

We know HCF * LCM = The number A *B (product)
LCM = 120 and \(\frac{A}{B} =\frac{3x}{4x}\)

=> HCF*120=3x*4x

HCF=\(\frac{12x^2}{120}\) = \(\frac{x^2}{10}\)

Trying out options to solve the equation could help. Since HCF ≠ fraction, the option which equals to integer is the answer.

A. 8 =\(\frac{8^2}{10}\) = integer? NO

B. 9 =\(\frac{9^2}{10}\) = integer? NO

C. 10 =\(\frac{10^2}{10}\) = integer? YES

D. 12 =\(\frac{12^2}{10}\) = integer? NO

E. 15 =\(\frac{15^2}{10}\) = integer? NO

Therefore, IMO answer is option C
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We know that A*B=120x where 'x' is an integer.
We also know that A=3x and B=4x since A:B=3:4. Thus, A*B=12x^2
Therefore, 120x=12x^2....> x=10
Thus, A=30 & B=40

HCF of A and B is the product of their common divisors.
Divisors of A are 3, 2 & 5
Divisors of B are 2,2,2 & 5.
Divisors common to A and B are 2 & 5.
HCF of A and B is 2*5=10.

ANS: C
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avigutman

Would you mind sharing your approach to this question? I'm stumped by the ratio part.
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