Let the regular prices be a, b, and c, so that a > b > c.

Basically the questions: is \(0.2a+0.1b+0.1c>0.15(a+b+c)\)? --> is \(a>b+c\)?

(1) The regular price of the most expensive item was $50 and the regular price of the next most expensive item was $20 --> \(a=50\), \(b=20\), \(c\leq{20}\) (as the second most expensive item was $20 then the least expansive item, the third one, must be less than or equal to 20). So the question becomes: is \(50>20+c\) --> is \(c<30\)? As we got that \(c\leq{20}\), hence the above is always true. Sufficient.

(2) The regular price of the least expensive item was $15. Clearly insufficient.

Answer: A.
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THe Q asks for the which of discounts will be greater 20%, 10%, 10% on A, B, C resp or 15% on (A+B+C). How can we say its same as 50> 20+C ? Now I m totally lost.

The questions asks: is \(0.2a+0.1b+0.1c>0.15(a+b+c)\)? After simplifying we'll get that the question becomes: is \(a>b+c\)?

(1) says \(a=50\) and \(b=20\). Also we concluded that: \(c\leq{20}\). Now, if we substitute the values of a and b we'll get that the question boils down to: is \(50>20+c\)? or is \(c<30\)?

Hope it's clear.
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Thanks. Its clear now.

Just how do you know to keep A on one side & B & C on other? This simplied the Ans a lot.

It really doesn't matter how you write it: \(a>b+c\), \(a-b-c>0\), \(a-b>c\), ... You still will get the same result. For me \(a>b+c\) was the most "attractive" form, so I chose this one.
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Bunuel: if the first item is 50, the next 2 are 20. If I take 20% of 50 I get 10 and 10% of 20 I get 2. Total discount is 14 and total price is 90. 14/90>.15. What's am I doing wrong?

Total discount = $14.
The sum of the regular prices of the 3 items = $90.
15% of $90 = $13.5.

$14 > $13.5.
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I believe this question is a perfect example of an "Easy C" trap. Just by realizing that A&B would be sufficient and make the problem way too easy, we can eliminate C,D and E. This is a simple rule that could improve one's chances on 700+ problems

Fortunately or Unfortunately, the OA is C for this one. I'm staring at the answer on GMATPREP software :/ Attaching screenshot

The OA for the question discussed above is A, NOT C.

The question you attached is not the one discussed above, it's another one discussed here: https://gmatclub.com/forum/henry-purcha ... 94280.html
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Hi Bunuel could you please elaborate this simplification:
0.2a+0.1b+0.1c>0.15(a+b+c)0.2a+0.1b+0.1c>0.15(a+b+c) ? --> is a>b+ca>b+c?

\(0.2a+0.1b+0.1c>0.15(a+b+c)\);

\(0.2a+0.1b+0.1c>0.15a+0.15b+0.15c\);

\(0.05a>0.05b+0.05c\);

\(a>b+c\).
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Bunuel : I somehow got lost in the wording of this question. The Q only says Henry purchased 3 items . No where does it mention that the other 2 items he purchased are the second most expensive item or the least expensive item.
How can we assume from statement one that Henry has indeed purchased 2nd most expensive item> He could have purchased any 2 of the other items right?

I think you misunderstood the question, if I got your doubt right.

So, Henry purchased 3 items in total, not more. Say regular prices of these 3 items were $a, $b, and $c, so that a > b > c. He received a 20 percent discount off the regular price of the most expensive item, so $a item, he bought for $0.8a. For the remaining 2 items, so for the items which cost $b and $c, he got 10 percent discount, so Henry purchased these 2 items for $0.9b and $0.9c.

Does this make sense?
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We can treat this as a MIXTURE PROBLEM.
How can we combine a 20% solution (the higher discount) with a 10% solution (the lower discount) to yield a mixture that is more than 15% (the total discount)?
If we use EQUAL amounts of 20% solution and 10% solution, the resulting solution will be exactly 15%.
Implication:
To yield a solution that is MORE THAN 15%, we must use MORE OF the 20% solution and LESS OF the 10% solution.
In other words:
The price of the most expensive item (the 20% solution) must be greater than the total cost of the two cheaper items (the 10% solution).:

Question stem, rephrased:
Is the price of the most expensive item greater than the total cost of the two cheaper items?

Statement 1: The regular price of the most expensive item was $50, and the regular price of the next most expensive item was $20.
Maximum total cost of the two cheaper items = 20+20 = 40.
Since the price of the most expensive item ($50) is greater than the maximum total cost of the 2 cheaper items ($40), the answer to the rephrased question stem is YES.
SUFFICIENT.

Statement 2: The regular price of the least expensive item was $15.
No way to determine whether the price of the most expensive item is greater than the total cost of the 2 cheaper items.
INSUFFICIENT.


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Hey ! This is the question from official GMAT practice test paper #3 and the answer is C although I ticked on A. Can someone please clear this?

The OA for the question discussed above is A, NOT C.

The question you are talking about is discussed here: https://gmatclub.com/forum/henry-purcha ... 94280.html
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