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Algebra - Ratios

Expert replies
by tushnir540 » Mon Mar 26, 2012 7:09 pm
Robots X, Y, and Z each assemble components at their respective constant rates. If rx is the ratio of robot X's constant rate to robot Z's constant rate and ry is the ratio of robot Y's constant rate to robot Z's constant rate, is robot Z's constant rate the greatest of the three?

(1) rx < ry
(2) ry < 1

This is problem 44 in Data Sufficiency from the 12th Edition. For (2), even though we aren't given that the rates are positive, are we always supposed to assume that the rate is positive for rate questions? My question is revolves around (2): ry=Y/Z. Y/Z < 1 => Y < Z.

Since we don't know the sign of the rates (i.e. whether one of the robot's is actually disassembling, and thus the rate is <0 ), how are we able to multiply a variable over an inequality?

I guess we are going on an assumption that the rates are positive, but are never actually told that they are all positive. I guess my question is whether we are correct to assume that a rate is positive even if it isn't clearly stated? Thanks!
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Source: — Data Sufficiency |

by Anurag@Gurome » Mon Mar 26, 2012 7:37 pm
Let us assume that the rates of robots X, Y, and Z be x, y, and z respectively.
Given: r(x) = x : z and r(y) = y : z

We have find if Z's constant rate is the greatest of the three.

(1) r(x) < r(y) implies x : z < y : z or x < y; NOT sufficient.

(2) r(y) < 1 implies y : z < 1 or y < z; NOT sufficient.

Combining (1) and (2), we know that x < y and y < z, which implies x < y < z; SUFFICIENT.

The correct answer is C.
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by spartacus1412 » Tue Mar 27, 2012 12:05 am
C indeed.
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by tushnir540 » Tue Mar 27, 2012 5:30 am
I believe the point of my post was missed. I wasn't curious on how to solve the problem. I was curious as to whenever we are given a rate problem, are we always to assume that the rate, and thus the inputs into the rate, are positive?
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by icemanKK » Sun Apr 15, 2012 6:22 am
I think it should be safe to assume rate as +ve. Cause if it were actually negative it would probably mean that whoever/whatever did stuff with a -ve rate is "not doing productive work" but is in all probability "doing destructive work" ..... Meaning if all the entities were working together, the person/machine with -ve rate is actually hampering productivity ....

What I have given above, could well be a result of unnecessary over-thinking !!! We do need expert advice.... Surely there should be a more logical reason for not assuming rate as -ve or 0.
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by bittu.0807 » Sun Apr 15, 2012 7:27 am
I believe we are over thinking here, rates can not be negative.
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