Henry purchase 3 items during a sale

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Xbond: Master | Next Rank: 500 Posts; Posts: 275; Joined: Wed Jul 02, 2008 4:19 am; Thanked: 4 times

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: Beat The GMAT — Data Sufficiency | Fri Aug 14, 2009 12:17 am

ashis979: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Tue Jul 28, 2009 8:12 pm; Thanked: 4 times

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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anand0408: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sun Jun 21, 2009 10:30 am; Thanked: 2 times

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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Xbond: Master | Next Rank: 500 Posts; Posts: 275; Joined: Wed Jul 02, 2008 4:19 am; Thanked: 4 times

by Xbond » Mon Aug 17, 2009 5:17 am

many thks guys

OA is A

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jayavignesh: Senior | Next Rank: 100 Posts; Posts: 45; Joined: Tue Mar 30, 2010 7:00 pm

by jayavignesh » Wed Mar 09, 2011 3:11 am

opt A

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Sat Jul 16, 2011 10:29 am

WOW Mitch !! this kind of thinking and solution one can only expect from you.

Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

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sushantgupta: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Sun Mar 27, 2011 9:54 am; Thanked: 2 times

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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tailoc: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Wed Jun 15, 2011 11:38 am; Thanked: 1 times

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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ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members; Location: Ahmedabad; Thanked: 31 times

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.

Follow your passion, Success as perceived by others shall follow you

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moussaobeid: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Sat Feb 25, 2012 4:38 am

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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moussaobeid: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Sat Feb 25, 2012 4:38 am

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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rajeshsinghgmat: Master | Next Rank: 500 Posts; Posts: 171; Joined: Tue Jan 08, 2013 7:24 am; Thanked: 1 times

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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jaspreetsra: Master | Next Rank: 500 Posts; Posts: 164; Joined: Sat Sep 20, 2014 10:26 pm; Thanked: 1 times

by jaspreetsra » Sun Dec 28, 2014 2:43 am

IMO:A

Hard work brings success!

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thang: Master | Next Rank: 500 Posts; Posts: 111; Joined: Wed Oct 31, 2007 2:50 am; Thanked: 11 times

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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