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Bunuel
If x = 3y = 4z, which of the following must equal 6x ?

I. 18y
II. 3y + 20z
III. (4y + 10z)/3

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Checking the options.
I. 18y
Given ; x = 3y
\(\frac{18y}{3y}\) * x = 6x (Must be True)

II. 3y + 20z
Given 3y = x; 4z = x
\(\frac{20z}{4z}\) * x = 5x
3y + 20z = x + 5x = 6x (Must be True)

III. \(\frac{(4y + 10z )}{3}\)
\((\frac{4}{3}x + \frac{10}{4}x )/3\) = \((\frac{4}{3}x + \frac{5}{2}x)/3\) = \((\frac{8x + 15x}{6})/3\) = \(\frac{23x}{18}\)(Not True)

Answer (D) I and II only ...
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Bunuel
If x = 3y = 4z, which of the following must equal 6x ?

I. 18y
II. 3y + 20z
III. (4y + 10z)/3

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Test values. Because x is the greatest number, use LCM of 3, 4, and 6 (because x gets multiplied by 6) for x.

Let x = 12, y = 4, z = 3

6x = 72

I. 18 y?

Yes. 18*4 = 72. Eliminate (B) and (C)

II. 3y + 20z?

3y = 12, and 20z = 60. (12 + 60) = 72. Correct. And you're done -- eliminate answers A and E because they do not include II. I am not that brave. Checking III.

III. \(\frac{(4y + 10z)}{3}\)?

4y = 16, and 10z = 30. \(\frac{(16 + 30)}{3} =\frac{46}{3}\) -- and is nowhere near 72. Reject.

Answer D
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Given x=3y=4z or y = x/3 and z = x/4

Since we have a clear relation of x with both y and z, its easy to test each statement by converting in terms of x.

I. 18y = 18*x/3 = 6x, hence TRUE

II. 3y + 20z = 3*x/3 + 20*x/4 = x+5x = 6x, hence TRUE

III. (4y + 10z)/3 = (4*x/3 + 10*x/4)/3 = 4x/9 + 5x/6 = 23x/18, so NOT TRUE

Only I and II are true, hence D answer
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Bunuel
If x = 3y = 4z, which of the following must equal 6x ?

I. 18y
II. 3y + 20z
III. (4y + 10z)/3

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

x = 3y = 4z = 12

Or, x = 12 ; y = 4 & z = 3

So, 6x = 72


Now, check the options -

I. 18y = 18*4 = 72
II. 3y + 20z = 3*4 + 20*3 = 72
III. (4y + 10z)/3 = (4*4 + 10*3)/3 = 46/3

Hence, the correct answer must be (D) I and II only
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Bunuel
If x = 3y = 4z, which of the following must equal 6x ?

I. 18y
II. 3y + 20z
III. (4y + 10z)/3

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only


We are given that x = 3y = 4z and need to determine what must equal 6x. From the given information, we see that 6x = 18y = 24z. Let’s analyze each Roman numeral:

I. 18y

Since 6x = 18y, Roman numeral I is true.

II. 3y + 20z

Since 3y = x and 20z = 5(4z) = 5x, 3y + 20z = x + 5x = 6x; Roman numeral II is also true.

III. (4y + 10z)/3

Let’s express 4y and 10z in terms of x.

4y = (4/3)3y = (4/3)x

10z = (10/4)4z = (5/2)x

Now, we have:

(4y + 10z)/3 = ((4/3)x + (5/2)x)/3 = [(23/6)x]/3 = (23/18)x.

Clearly, this expression is not equal to 6x.

Thus, only Roman numerals I and II are true.

Answer: D
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If x = 3y = 4z, which of the following must equal 6x ?

I. 18y
II. 3y + 20z
III. (4y + 10z)/3

Option 1: 18Y
Since we know that X=3Y
18Y=6X

This is valid.

Option 2:3Y+20Z
X=3y
X=4Z , 20Z=5X
3Y+20Z = X+5X = 6X

This is valid

Option 3:(4Y+10z)/3
You get X less than 6. Not valid

Only 1 and 2 are correct

Ans: D
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Bunuel
If x = 3y = 4z, which of the following must equal 6x ?

I. 18y
II. 3y + 20z
III. (4y + 10z)/3

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only


it's given,

x=3y=4z

we know that the value of z , 3y and 4z must be equal. Now to make the value equal, we can find out the LCM of x, 3y and 4z. LCM is 12
So,

x=12
y=4
z=3

The value of 6x = 6*12=72

Now scan the answer choice that fits.

Option A: 18y=18*4=72

option B: 3y+20z = 3*4 + 20*4 = 72

option c: (4y+10z)/3 = (4*3 + 10*4)/3 = (12+40)/3 = Fractional value and less than the value of 6x. Eliminate it.

Thus Option D is the best answer.
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x = 3y = 4z
y = x/3 and z = x/4

1) 18y = 18*x/3 = 6x
2) 3y +20z = 3*x/3 + 20*x/4 = 6x
3) (4y + 10z)/3 =4*x/9 + 10*x/12 = 46x/36

Ans D
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Bunuel
If x = 3y = 4z, which of the following must equal 6x ?

I. 18y
II. 3y + 20z
III. (4y + 10z)/3

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

X=3Y=4Z

X=3y
X/3=Y ,

X=4Z
X/4=z

Try to insert above values in the option , you will find only I & II works !
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lets assume x = 3y = 4z = 12
we choose 12 as it is the LCM of 3 and 4

then we can say that x= 12 , y = 4 , z = 3.
question here is which of the following must equal 6x?
6 x = 6 * 12 = 72

So after substituting values of y and z in the options ,we should get 72
I. 18y = 18* 4= 72 Hence True
II. 3y + 20z = 3*4 + 20*3= 72 So its true
III. (4y + 10z)/3 = (4*4 + 10* 3)/3 = 46 /3 So its NOT TRUE

Option D ( I and II only) is the correct answer.

Thanks,
Clifin J Francis,
GMAT SME
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