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Henry purchase 3 items during a sale

by Xbond » Fri Aug 14, 2009 12:17 am
Hi there,

I would like to know the simplest process to resolve this DS. Explanation is requested.

Henry purchase 3 items during a sale. He received a 20 percent discount off the regular price of the most
expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total
amount of the 3 discounts greater than 15 percent of the sum of the 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 lease expensive item was $15.
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Source: — Data Sufficiency |
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by ashis979 » Fri Aug 14, 2009 4:08 pm
This is your standard "C-trap" question.

X=most expensive item
Y= second most expensive item
Z= least expensive item, however, Z can be equal to Y, since the question does not specify

So, question is asking: Is 0.2X+0.1Y+0.1Z>0.15(X+Y+Z)?

Stmt 1:
X=$50, so discount=$10
Y=$20, so discount=$2

Back to the stem: Is 10+2+0.1Z>0.15($50+$20+Z)?

The most Z can be is $20, assuming it cost the same as Y.
So, when Z=$20, total discount=$14 and total cost=$90
$14/$90>15%?.....YES ($13.5/$90 would be exactly 15%)
If Z=$0, then total discount=$12 and total cost=$70
$12/$70>15%...YES ($10.5/$70 would be exactly 15%)

If you continue to test for any value of Z less than 20, you will see that the sum of the discounts will remain greater than 15%. But since you know that the sum of discounts, when Z=20 (the maximum possible value for Z) is greater than 15%, you need not do any further calculations. So, the answer is A.....Stmt 1 alone is SUFFICIENT.

Stmt 2: Doesn't tell you anything, except that the smallest discount was $1.50...NOT SUFFICIENT.
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by anand0408 » Sat Aug 15, 2009 2:35 am
for this, maybe a more mathematical approach might help:

we basicaly have to find if the discount amount sum is greater than 15% of the total original price:

Let the Most expensive item cost: x
Let the other 2 cost y and z:

Therefore: Total original price = (x+y+z)

Also, discounts:

On X = 0.2x
On y = 0.1y
On z = 0.1z

therefore: sum of discounts = 0.2x+0.1y+0.1z

so we have to find:

is: (0.2x+0.1y+0.1z) > 0.15(x+y+z)
=> 0.2x+0.1y+0.1z > 0.15x+0.15y+0.15z
=> 0.05x>0.05y+0.05z
=>x>y+z

So basically we need to find out if the most expensive item is costlier than the other 2 put together. If yes, we are sorted.

Stmt 1 says most expensive is $50 while next highest is $ 20. So if we assume that both the others are $ 20 each also, their combined sum ($40) is less than $50. Hence it is sufficient.

Hence A:
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by Xbond » Mon Aug 17, 2009 5:17 am
many thks guys

OA is A
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by jayavignesh » Wed Mar 09, 2011 3:11 am
opt A
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by GMATGuruNY » Wed Mar 09, 2011 4:31 am
Xbond wrote:Hi there,

I would like to know the simplest process to resolve this DS. Explanation is requested.

Henry purchase 3 items during a sale. He received a 20 percent discount off the regular price of the most
expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total
amount of the 3 discounts greater than 15 percent of the sum of the 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.
This is a weighted average question.
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 the 20% solution and the 10% solution, the resulting solution will be exactly 15%.
Thus, 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 combined prices of the two cheaper items (the 10% solution).

Rephrased, the question is asking:

Is the price of the most expensive item greater than the combined prices 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.
The combined prices of the 2 cheaper items cannot be greater than 20+20 = 40.
Thus, the $50 price of the most expensive item must be greater than the combined prices of the 2 cheaper items.
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 combined prices of the 2 cheaper items.
Insufficient.

The correct answer is A.
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by bblast » Sat Jul 16, 2011 10:29 am
WOW Mitch !! this kind of thinking and solution one can only expect from you.
Cheers !!

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by sushantgupta » Sat Jul 16, 2011 9:01 pm
We know the price of most expensive = 50 next most expensive = 20


lets say 3 most expensive is 1

so total discount = 10 + 2 + .1 = $12.1
15% of 71 = 10.65

so if even at the lowest price the % is more than 15 hence statement 1 is sufficient.
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by tailoc » Tue Sep 27, 2011 12:51 am
Say a,b,c : price of each items a>b>c
from stem: 20%a+10%(b+c)>15%(a+b+c) ??? or a>b+c ???
statement 1: a=50 b=20, with c which is smaller than or equal b => always a>b+c sufficient
statement 2: clearly insufficient

OA: A
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by ronnie1985 » Thu Mar 29, 2012 7:06 am
Given: Most expensive = x3, discount = 20%. Others x1 and x2, discount = 10%.
Reqd: Total discount > 15%

S1. x3 = 50, x2 = 20. => Min discount = 10+2 = 12.
If cost of other item = 19, discount = 1.9 and hence the total price paid = 40+18+17.1 = 75.1
Total discount = ((89-75.1) / 89)*100 > 16%. For any value of third item < 19, the ratio (58+0.9x1)/(58+x1) > 15. Hence, sufficient.

S2. The cost of third item does not give any details about the total deal, hence not sufficient.

(A) is answer.
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by moussaobeid » Sun May 27, 2012 5:09 am
GMATGuruNY wrote:
Xbond wrote:Hi there,

I would like to know the simplest process to resolve this DS. Explanation is requested.

Henry purchase 3 items during a sale. He received a 20 percent discount off the regular price of the most
expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total
amount of the 3 discounts greater than 15 percent of the sum of the 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.
This is a weighted average question.
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 the 20% solution and the 10% solution, the resulting solution will be exactly 15%.
Thus, 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 combined prices of the two cheaper items (the 10% solution).

Rephrased, the question is asking:

Is the price of the most expensive item greater than the combined prices 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.
The combined prices of the 2 cheaper items cannot be greater than 20+20 = 40.
Thus, the $50 price of the most expensive item must be greater than the combined prices of the 2 cheaper items.
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 combined prices of the 2 cheaper items.
Insufficient.

The correct answer is A.
Thanks Mitch awesome way of thinking
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by moussaobeid » Sun May 27, 2012 5:11 am
GMATGuruNY wrote:
Xbond wrote:Hi there,

I would like to know the simplest process to resolve this DS. Explanation is requested.

Henry purchase 3 items during a sale. He received a 20 percent discount off the regular price of the most
expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total
amount of the 3 discounts greater than 15 percent of the sum of the 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.
This is a weighted average question.
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 the 20% solution and the 10% solution, the resulting solution will be exactly 15%.
Thus, 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 combined prices of the two cheaper items (the 10% solution).

Rephrased, the question is asking:

Is the price of the most expensive item greater than the combined prices 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.
The combined prices of the 2 cheaper items cannot be greater than 20+20 = 40.
Thus, the $50 price of the most expensive item must be greater than the combined prices of the 2 cheaper items.
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 combined prices of the 2 cheaper items.
Insufficient.

The correct answer is A.
Thanks Mitch awesome way of thinking
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by rajeshsinghgmat » Mon Apr 15, 2013 1:13 am
C the aNSWER.

x=0.2*A + 0.1*(B+C) - 0.15*(A+B+C)

Is x > 0?

x=0.05(A - B - C)

From I

A = 50, B = 20

x=0.05*(30-C)

FROM II

x=0.05*(A-B-15)
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by jaspreetsra » Sun Dec 28, 2014 2:43 am
IMO:A
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by thang » Wed Apr 29, 2015 2:00 am
GMATGuruNY wrote:
Xbond wrote:Hi there,

I would like to know the simplest process to resolve this DS. Explanation is requested.

Henry purchase 3 items during a sale. He received a 20 percent discount off the regular price of the most
expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total
amount of the 3 discounts greater than 15 percent of the sum of the 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.
This is a weighted average question.
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 the 20% solution and the 10% solution, the resulting solution will be exactly 15%.
Thus, 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 combined prices of the two cheaper items (the 10% solution).

Rephrased, the question is asking:

Is the price of the most expensive item greater than the combined prices 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.
The combined prices of the 2 cheaper items cannot be greater than 20+20 = 40.
Thus, the $50 price of the most expensive item must be greater than the combined prices of the 2 cheaper items.
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 combined prices of the 2 cheaper items.
Insufficient.

The correct answer is A.

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