Let's plug in some NICE VALUES that satisfy the given information...

Last year the manufacturer sold twice as many units of Q as P.
So, let's say the manufacturer sold 1 P and 2 Q's

What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?
TOTAL revenue = 1($20.00) + 2($17.00)
= $20.00 + $34.00
= $54

There were 3 units sold.
So, average price = $54/3 = $18

Answer: E

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Revenue last year from the sale of P is 20*1 = 20
Revenue last year from the sale of Q is 17*2 = 34

Total Revenue from the sale of P and Q last year is 54

Average Revenue from the sale of P and Q last year is \(\frac{54}{2+1} = 18\)

Hence, answer will be (E) 18
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Ratio of quantity of P and Q sold, P:Q = 1:2.

Thus, average revenue per unit = (20*P + 17*Q)/(P+Q) = (20*1 + 17*2)/(2 + 1) = 54/3 = $ 18

Hence E answer

We can compute a weighted average to solve. Let’s assume that 2 units of Q and 1 unit of P were produced last year. So the total revenue is 2 x 17 + 1 x 20 = $54, and thus the average revenue per unit is thus 54/3 = $18.

Answer: E
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If the weights were the same we would get an average of (17 + 20) / 2 = 18.5.

But we have more of the product that costs 17$, so the average would be less than 18.5, only 18 works - E

17+17+20/3=18 was per unit revenue
Where,
Q=17, twice
P=20

Answer is E

Posted from my mobile device

And I got stuck with the TRAP answer B. 27.
After reading properly the second time I got it has to be divided by 3.
Why is this happening with me?
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Let's plug in some NICE VALUES that satisfy the given information...

Last year the manufacturer sold twice as many units of Q as P.
So, let's say the manufacturer sold 1 P and 2 Q's


There were 3 units sold.
So, average price = $54/3 = $18

Answer: E


GMATPrepNow

how do we deduce this statement? I get confused with this statement and it often appears on word problems.

can you help me understand this and its various iterations on the exam?

Which statement are you referring to?

Cheers,
Brent
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Solving through logic: A/B are out as average cannot be more than 20$
Since no of Q > no of P, avg will be more near 17. If No of Q = No of P avg would be 18.5 (option D) which we eliminate since it's not the case.

C is out as it is more near 20.

Logic leaves us with E.

Let p = units of product P ($20/unit) sold and q = units of product Q ($17/unit) sold
Also, q = 2p
Therefore, Average = \(\frac{20p + 17p}{p + q}\) ==> \(\frac{20p + 17(2p)}{p + 2p}\) ==> \(\frac{20p + 34p}{3p}\) ==> \(\frac{54p}{3p}\) ==> $18 (answer is E)

GMATPrepNow

I often get the below logic wrong. Could you please help?

Last year the manufacturer sold twice as many units of Q as P.
So, let's say the manufacturer sold 1 P and 2 Q's

This kind of wording can be tricky at times.

Last year the manufacturer sold TWICE as many units of Q as P = the number of Q's sold is TWICE the number of P's sold = = the number of Q's sold = 2(the number of P's sold)

Likewise, Joe has TWICE as many dogs as cats = the number of dogs is TWICE the number of cats = = the number of dogs = 2(the number of cats)

Does that help?
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Thank you very much. That helps!

Hello Experts,
EMPOWERgmatRichC, VeritasKarishma, IanStewart, Bunuel, chetan2u, ArvindCrackVerbal, GMATGuruNY, AaronPond, GMATinsight, ccooley
^^ this is the actual question. What if the question is modified a bit? Could you help me by removing other choices with the EASIEST way in the new version that i modified?
A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is $20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of P as Q. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?

A. $28.50
B. $27.00
C. $19.00
D. $18.50
E. $18.00
Thanks__

Asad

The average revenue cant go beyond 17 and 20 so two options A and B eliminated

twice as many units of P as Q

i.e Average should be twice as close to price of P than to the price of Q i.e. Options D and E eliminated because the price is either midpoint 18.5 or close to Q

P = 20, Q = 17

i.e. Answer will be Option C
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Hi Asad,

The Answer choices to this question are really helpful - and they're "spread out" in a way that creates a great 'concept shortcut' so that we really don't have to do much math to answer this question.

To start, in simple terms we're taking the average of a certain number of 17s and a certain number of 20s.... so the average MUST be between 17 and 20. Next, IF we had the SAME number of 17s and 20s, then the average would be (17+20)/2 = 37/2 = 18.5..... However, we do NOT have the same number of each. In your example, we have MORE 20s, which will INCREASE the average to something GREATER than 18.5 (but not greater than 20, since that's not mathematically possible here). There's only one answer that makes sense....



GMAT assassins aren't born, they're made,
Rich
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It doesn't matter what the question is. Think this way:
P - Selling price is $20
Q - Selling price is $17

If equal quantities of the two are sold, the average selling price would be right in the middle, i.e. 18.5.

Now, if you sell P more, avg selling price will move toward 20.
But if you sell Q more, avg selling price will move toward 17.

So if Q are more than P in number, avg will lie between 17 and 18.5 (exclusive if both are sold)
If P are more in number than Q, avg will lie between 18.5 and 20 (exclusive if both are sold)
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