Henry purchased 3 items during a sale. He received a 20 percent discount of the regular price of the most expensive item and a 10 percent discount off the regular price of the other 2 items. Was the total amount of the 3 discounts greater than 15 percent of the sum of regular prices of the 3 items?

1) The regular price of the most expensive item was $50, and the regular price of the next most expensive item was $20.
2) The regular price of the least expensive item was $15

All these are GMAT prep questions.[/quote]

Hi, Vikram, what is your actual question? I'm happy to explain the whole problem from start to finish if you need me to, but I'd like you to try to do a little of the hard thinking, if possible. :)

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

It is easy to prove that statement A is alone sufficient. Lets assume that the items cost x, y and z. In statement A, we are given that x = 50 and y = 20. So, the total discount = (10 + 2 + 0.1z) = 12 + 0.1z. where the original total cost = (50 + 20 + z) = 70 + z

Hence avg. discount = (12 + 0.1z) * 100 / (70 + z) = 100 * { [(7 + 0.1z)/(70+z)] + [5 / (70+z)] }
= 100 * [0.1 + 5/(70+z)]
= 10 + 500/(70+z)

But since z is the least expensive item, z must be less than 20 and greater than 0.
Hence for z = 0 , avg discount = 17.14% and for z = 20, avg. disc = 15.55% Hence in all situations avg. disc is greater than 15%.
----------------------------------------------------------------------------------------------------------------------------------------------------------------

Now lets test for statement B along. Here by the same logic given above,
Avg. discount = 100 * (0.2x + 0.1y + 1.5) / (x + y + 15)
Since z = 15, both x and y must be greater than 15
Lets consider three different cases -
Case 1:
If both x = y = 15, avg. disc = 600/45 = 13.33%

Case 2:
If x = y = 1000,000 avg. disc = 30000150/2000015 = 14.999%
Note: When both x and y are sufficiently large and almost equal, the avg. discount tends to the ratio of 3/20 = 15%

Case 3:
If x = 1000,000 and y = 1000, avg disc = (200,000 + 100 + 1.5)*100/1001001.5 = 20010150/1001001.5 = 19.99%
Note: If x is a very high value and y is comparitively low, the ratio would tend to become 20%

Hence the value varies from 13.33% to nearly 20%. Hence statement B alone is not sufficient.
----------------------------------------------------------------------------------------------------------------------------------------------------------------

Hence the correct answer is A.

That's confusing! And since it says sums...can we find the solution using sums rather than using averages. I used sum to find A sufficient, and for B...my reasoning is that we do not know the limits for how expensive the goods are. They may balance each other out (and be equal), or one may be greater or less than the other.......... thus making it insufficient. I am not quite sure though. Help please!

A = Most Expensive
B = Less expensive than A
C = Less expensive than A (could be same price as B, info does not specify)

From stem: Is .20A + .10B + .10C > .15(A + B + C) ?

(1) .20(50) + .10(20) + .10C > .15(50+20+C) ?

We know that C has to be anywhere from ($0 - $20)

So test out some numbers within that range. For example if C = $20, we get .20(50) + .10(20) + .10(20) > .15(50+20+20)
= $14 > $13.50 and if continue to test numbers less than $20 this will still apply.

So (1) is suff

(2) .20(A) + .10(B) + .10(15) > .15(A+B+15)

Clearly we need more info to derive an answer so Insuff.

Answer = A

Hope this helps!

From the question, Let I1 > I2, I3 where I1 is the most expensive item and I1 and I2 be the least expensive ones.

After, 20% off on I1 and 10% off on I2, I3, their prices reduce to .8I1, .9I2 and .9I3 respectively

Now converting the wording in the question to mathematical expression we need to find - >

.2I1 + .10I2 + .10I3 > .15(I1 + I2 +i3)

solving, we get I1 > I2 + I3 ( This is actually the question......we need to determine)

1) I1 =50 and I2 or I3 = 20, therefore if I2 or I3 =20 (the 2nd expensive one), the least expensive cannot be more than 20
so A answers the question...as YES

2) is insufficient as it only tells about one number.

From the question, Let I1 > I2, I3 where I1 is the most expensive item and I1 and I2 be the least expensive ones.

After, 20% off on I1 and 10% off on I2, I3, their prices reduce to .8I1, .9I2 and .9I3 respectively

Now converting the wording in the question to mathematical expression we need to find - >

.2I1 + .10I2 + .10I3 > .15(I1 + I2 +i3)

solving, we get I1 > I2 + I3 ( This is actually the question......we need to determine)

1) I1 =50 and I2 or I3 = 20, therefore if I2 or I3 =20 (the 2nd expensive one), the least expensive cannot be more than 20
so A answers the question...as YES

2) is insufficient as it only tells about one number.

perfect.

this problem is also a classic C TRAP.

a c trap is a problem that's clearly written to be difficult, but on which both statements taken together are VERY CLEARLY sufficient.
and by VERY CLEARLY i don't mean "after i solve this equation, and move that over there, then ... oh yeah, my memorized rule of thumb tells me they're sufficient" - i mean it's OBVIOUS. (examples follow, for those of you who have your og's handy.)
on these problems, you can rest assured that (c) is not the answer. also, because the two statements together are sufficient, you can also strike answer (e).
this leaves only (a), (b), (d). and in the dream situation, in which one of the answers by itself is clearly insufficient, then you can guess that the other one must be sufficient - and you'll be right a startlingly high percentage of the time.

you should NOT use the "c trap" approach as a PRIMARY method - i'm sure that, one fine day, a difficult "c trap" problem will come along to which the answer actually is 'c' (although i've yet to see one) - but, rather, as an AID TO GUESSING.
still, if you get into a guessing situation, the c trap is one of the strongest weapons in your arsenal.

this problem:
* identify the problem as a c trap: if you take the two statements together, then you have the prices of ALL the items in the problem. if that's the case, then the answer to the prompt question is clearly either "yes" or "no"; hence, sufficient.
kill (c) and (e) and narrow the choices to (a), (b), and (d).

* statement (2) is insufficient. this isn't ridiculously obvious, but the presence of two remaining unknowns should convince you (remember that you're in guessing mode here).
kill (b) and (d).

answer = a.

".2I1 + .10I2 + .10I3 > .15(I1 + I2 +i3)

solving, we get I1 > I2 + I3 ( This is actually the question......we need to determine)"


Can you go over how you solved this. thank you

[editor: this is routine algebra; maybe you're intimidated by the decimals. the process is the same as for an inequality like
4x + 2y + 2z > 3(x + y + z)
i.e., just distribute the right-hand expression and then collect like terms.]

Instr,
After simpl.all the math taking the stem along with stat.1, the question becomes is "the least expensive item less than 30", and since we know that it must be true according to the data presented, one can conclude stat.1 is sufficient.
Thanks for reply in advance.

to what, exactly, are you looking for a “reply”? you haven't asked any actual question -- or anything else to which one might conceivably “reply” -- in your post.

Hello Ron,
My question is. Is it correct to conclude that the actual qyestion is ( based on the stem and statement 1) " is the least expensive item cost less than 30", I arrived to this conclussion after doing all the math as mentioned in other above posted replies.
Thanks.

whether the least expensive item is less than $30 is not a question that should be asked at all once you bring in statement 1, which tells you the least expensive item is no more than $20..

_________________
Tim Sanders
Manhattan GMAT Instructor