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

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If $$x=3y=4z$$ Which of the following must equal 6x?

$$I.18y$$
$$II.3y+20z$$
$$III.\frac{(4y+10z)}{3}$$

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

The OA is D.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

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by DavidG@VeritasPrep » Fri Nov 24, 2017 12:12 pm
LUANDATO wrote:If $$x=3y=4z$$ Which of the following must equal 6x?

$$I.18y$$
$$II.3y+20z$$
$$III.\frac{(4y+10z)}{3}$$

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

The OA is D.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
If x = 3y = 4z, then 6x = 18y = 24z

I. Well, we can see above that 6x = 18y, so this is in.
II. 3y + 20z can be rephrased as 3y + 5* (4z), if 3y = 4z, then we can substitute, and get 3y + 5 *(3y) = 3y + 15y = 18y. So this one is also in.

No need to test III, as we don't have the option of selecting I, II, and III. The answer must be D
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by EconomistGMATTutor » Sat Nov 25, 2017 7:37 am
If $$x=3y=4z$$ Which of the following must equal 6x?

$$I.18y$$
$$II.3y+20z$$
$$III.\frac{(4y+10z)}{3}$$

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

The OA is D.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
Hi LUANDATO,
Let's take a look at your question.

$$x=3y=4z$$
We need to check which of the given expressions are equivalent to 6x. Let's evaluate them one by one.

$$I. 18y$$
Let's rewrite y in terms of x.
Since,
$$x\ =\ 3y$$ $$y=\frac{x}{3}$$
Replace value of y in 18y, we get,
$$=18\left(\frac{x}{3}\right)$$
$$=6x$$

$$II.3y+20z$$
From x=3y=4z, we get,
$$y=\frac{x}{3},\ z=\frac{x}{4}$$
Plugin the values in II, we get:
$$=3\left(\frac{x}{3}\right)+20\left(\frac{x}{4}\right)$$
$$=x+5x=6x$$

$$III.\frac{(4y+10z)}{3}$$
Replace values of y and z, we get
$$=\frac{4\left(\frac{x}{3}\right)+10\left(\frac{x}{4}\right)}{3}$$
$$=\frac{\frac{4x}{3}+\frac{5x}{2}}{3}$$
$$=\frac{\frac{8x+15x}{6}}{3}$$
$$=\frac{23x}{6\times3}=\frac{23x}{18}$$

Therefore, only I and II are equal to 6x.
Hence, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Scott@TargetTestPrep » Fri Dec 15, 2017 6:04 am
LUANDATO wrote:If $$x=3y=4z$$ Which of the following must equal 6x?

$$I.18y$$
$$II.3y+20z$$
$$III.\frac{(4y+10z)}{3}$$

(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

Scott Woodbury-Stewart
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[email protected]

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