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Bunuel
What is the ratio of x:y:z given xyz not equal to 0?

(1) x + y = 2z
(2) 2x + 3y = z

Given: xyz ≠ 0

Question: x:y:z = ?

Statement 1: x + y = 2z

Case 1: If x = 1 and y = 1 then z = 1 hence x:y:z = 1:1:1
Case 2: If x = 1 and y = 3 then z = 2 hence x:y:z = 1:3:2
Inconsistent answers hence
NOT SUFFICIENT (if you wish to learn the strategy to approach DS questions then you may watch 9 videos on my youtube channel in DS playlist)

Statement 2: 2x + 3y = z
Case 1: If x = 1 and y = 1 then z = 5 hence x:y:z = 1:1:5
Case 2: If x = 1 and y = 2 then z = 8 hence x:y:z = 1:2:8
Inconsistent answers hence
NOT SUFFICIENT

COmbining the statements
x + y = 2z
2x + 3y = z
i.e. x+y = 2(2x+3y)
i.e .3x+5y = 0
i.e. x:y = -5:3

If x = -5, y = 3 then z = -1 hence x:y:z = -5:3:-1
If x = 5, y = -3 then z = 1 hence x:y:z = 5:-3:1

SUFFICIENT

Answer: Option C

The two ratios -5:3:-1 and 5:-3:1 wouldn't it imply that this two are two different sets of ratio meaning we still don't have an unique value to answer the question ?

Check for x:y, y:z, and x:z in both cases.

In x:y:z= -5:3:-1

x:y= -5:3
y:z= -3:1
x:z= 5:1

In x:y:z= 5:-3:1
x:y= -5:3
y:z= -3:1
x:z= 5:1

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Here

(x+y)/2 = 2x+3y
3x+5y=0
so x and y has to be opposite signs.

So -ve in the values wont get cancelled out when we take ratios.

Are -ve numbers allowed in ratio, always thought ratios had to be +ve?
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