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.
_________________

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.
_________________

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.
_________________

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.
_________________

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
_________________

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?

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

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

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

\(a>b+c\).
_________________

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?
_________________

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.


_________________

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
_________________

Basically the question is whether the sum of the prices of the two less expensive items is less than the price of the most expensive item.

Weighted averages is very good here:

2*10%----------15%----------20%

The total percentage will be to the right of 15% if the most expensive item costs more than the other two items together.