sanju09 wrote:If x, y, and z are positive integers, and x is a multiple of 3, for which of the following must z and 18 have a common factor greater than 1?
I. (x/9) + (y/6) = (z/18)
II. (x/6) + (y/9) = (z/18)
III. (x/6) + (y/5) = (z/30)
A. I only
B. II only
C. I and II only
D. I an III only
C. I, II, and III
To see the situation more clearly, we can plug in values.
Be sure to satisfy the constraint that x is multiple of 3.
Since statement I is included in four of the five answer choices, it is almost certain that statement I must be included in the correct answer choice.
To save time, start with statement II.
II: (x/6) + (y/9) = (z/18)
To clear the fractions, multiply both sides of the equation by the GREATEST of the 3 denominators (18).
18 (x/6 + y/9) = 18 (z/18)
3x + 2y = z.
Case 1: If x=3 and y=1, then z = 3(3) + 2(1) = 11.
z=11 and 18 do not have a common factor greater than 1.
Here, it does not have to be true that z and 18 have a common factor greater than 1.
Eliminate any answer choice that includes statement II.
Eliminate B, C and E.
III: (x/6) + (y/5) = (z/30)
To clear the fractions, multiply both sides of the equation by the GREATEST of the 3 denominators (30).
30 (x/6 + y/5) = 30 (z/30)
5x + 6y = z.
Case 1: If x=3 and y=1, then z = 5(3) + 6(1) = 21.
z=21 and 18 are both divisible by 3.
Case 2: If x=6 and y=2, then z = 5(6) + 6(2) = 42.
z=42 and 18 are both divisible by 3 and 6.
In each case, z and 18 have a common factor greater than 1.
One more to be safe:
Case 3: If x=9 and y=5, then z = 5(9) + 6(5) = 75.
z=75 and 18 both divisible by 3.
In every case, z and 18 have a common factor greater than 1.
Thus, the correct answer choice must include statement III.
Eliminate A.
The correct answer is
D.
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