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805+ Level|   Algebra|      
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Emdad
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If yx/z^2 = 27 and y^2/(zx) = 1/(3√3), what is the ratio of z to y? Updated on: 07 Oct 2021, 12:03
Originally posted by Emdad on 19 May 2021, 07:48.
Last edited by Emdad on 07 Oct 2021, 12:03, edited 4 times in total.
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95% (hard)

Question Stats:

40% (02:28) correct 60% (02:38) wrong based on 185 sessions

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If \(\frac{yx}{z^2} = 27\) and \(\frac{y^2}{zx} =\frac{ 1}{3√3}\), what is the ratio of z to y?

A. √3
B. √3/3
C. 1/3
D. 3√3
E. None of these
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B
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Bunuel
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If yx/z^2 = 27 and y^2/(zx) = 1/(3√3), what is the ratio of z to y? 25 May 2021, 01:21
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Emdad
If yx/z² = 27 and y²/zx = 1/(3√3), what is the ratio of z to y?
A. √3
B. √3/3
C. 1/3
D. 3√3
E. None of these
Show more

First of all, it should be y^2/(zx) not y^2/zx because y^2/zx means y^2/z*x. Next, the OA should be B, not A.

So, we have: \(\frac{yx}{z^2 }= 27\) and \(\frac{y^2}{zx} = \frac{1}{3√3}\)

Multiply the expression: \(\frac{yx}{z^2 }*\frac{y^2}{zx} = \frac{27}{3√3}\);

\(\frac{y^3}{z^3 } = \frac{9}{√3}\);

\(\frac{y^3}{z^3 } = \frac{(√3)^4}{√3}\);

\(\frac{y^3}{z^3 } = (√3)^3\);

\(\frac{y}{z } = √3\);

So, \(\frac{z}{y } = \frac{1}{√3}=\frac{√3}{3}\).

Answer: B.
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ZulfiquarA
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If yx/z^2 = 27 and y^2/(zx) = 1/(3√3), what is the ratio of z to y? 25 May 2021, 00:58
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This route have more calculation but I did it because not in any world I would think of multiplying those two equations like the other person did above (he/she is on another level or I am just too dumb to do it)
yx/z^2=27, y^2/zx=1/3(root3)
Get z from first equation= yx/z=27z = z=yx/27z
Get y from second equation= y/zx=1/3(root3)y = y=zx/3(root3)y
Put them in ratio of z to y
z/y=yx/27z/zx/3(root3)y
z/y=27z^2/3(root3)y^2, Multiple Z with Z^2 & Y with Y^2
z^3/y^3=27/3(root3) = z/y=1/(root3)

This one is long and would love to use the multiplying shortcut but I can't find any trigger that would make me multiply them. Could someone share what made them multiply it.
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If yx/z^2 = 27 and y^2/(zx) = 1/(3√3), what is the ratio of z to y? 25 May 2021, 02:22
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ZulfiquarA
This route have more calculation but I did it because not in any world I would think of multiplying those two equations like the other person did above (he/she is on another level or I am just too dumb to do it)
yx/z^2=27, y^2/zx=1/3(root3)
Get z from first equation= yx/z=27z = z=yx/27z
Get y from second equation= y/zx=1/3(root3)y = y=zx/3(root3)y
Put them in ratio of z to y
z/y=yx/27z/zx/3(root3)y
z/y=27z^2/3(root3)y^2, Multiple Z with Z^2 & Y with Y^2
z^3/y^3=27/3(root3) = z/y=1/(root3)

This one is long and would love to use the multiplying shortcut but I can't find any trigger that would make me multiply them. Could someone share what made them multiply it.
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We have two independent equations but three variables.

We need to know already that substitution is not the only way to eliminate the extra variable.

We also have options to add, subtract, divide and multiply both equations or even multiples of both equations to eliminate an extra variable.

It is easy to "see" if substitution or addition, or subtraction could work for simple linear equations.

I suppose the main "trigger" here is realising that these non-linear equations are in the form of ratios, and either multiplication or division could eliminate one variable.

A secondary "trigger" is realising that checking if multiplication or division can indeed eliminate the third variable is quicker than substitution.

I would say practice would help give some intuition in avoiding long solution methods. I say this because I solve some medium-hard questions in less than 90 seconds because I have experience solving that class of problems, and some easy-medium questions I solve in over 3 minutes because I used the first minute to think about my life choices. :)
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If yx/z^2 = 27 and y^2/(zx) = 1/(3√3), what is the ratio of z to y? 18 Nov 2024, 08:08
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Ilanchezhiyan

Bunuel
Emdad
If yx/z2 = 27 and y2/zx = 1/(3√3), what is the ratio of z to y?
A. √3
B. √3/3
C. 1/3
D. 3√3
E. None of these

First of all, it should be y^2/(zx) not y^2/zx because y^2/zx means y^2/z*x. Next, the OA should be B, not A.

So, we have: \(\frac{yx}{z^2 }= 27\) and \(\frac{y^2}{zx} = \frac{1}{3√3}\)

Multiply the expression: \(\frac{yx}{z^2 }*\frac{y^2}{zx} = \frac{27}{3√3}\);

\(\frac{y^3}{z^3 } = \frac{9}{√3}\);

\(\frac{y^3}{z^3 } = \frac{(√3)^4}{√3}\);

\(\frac{y^3}{z^3 } = (√3)^3\);

\(\frac{y}{z } = √3\);

So, \(\frac{z}{y } = \frac{1}{√3}=\frac{√3}{3}\).

Answer: B.
Why dowe have to multiply the two equations. It never occurred to me. I was trying to find the value of y and z. And I got lost in a series of never ending calculations.😅 Why multiply the two equations?
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Usually, with two equations and three unknowns, it's not possible to solve for each unknown (though there are exceptions, but this isn't one of them).

In this case, we are tasked with finding the value of z/y, so the goal is to eliminate x from the equations. Multiplication and division are operations that can cancel out unknowns, which should have served as a hint to multiply the two equations.
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If yx/z^2 = 27 and y^2/(zx) = 1/(3√3), what is the ratio of z to y? 19 Nov 2024, 02:23
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Understood sir. Thanks for the clarification.
Bunuel
Ilanchezhiyan


Why dowe have to multiply the two equations. It never occurred to me. I was trying to find the value of y and z. And I got lost in a series of never ending calculations.😅 Why multiply the two equations?

Usually, with two equations and three unknowns, it's not possible to solve for each unknown (though there are exceptions, but this isn't one of them).

In this case, we are tasked with finding the value of z/y, so the goal is to eliminate x from the equations. Multiplication and division are operations that can cancel out unknowns, which should have served as a hint to multiply the two equations.
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