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.
Henry purchase 3 items during a sale
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.
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.
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:
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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This is a weighted average question.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.
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
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WOW Mitch !! this kind of thinking and solution one can only expect from you.
Cheers !!
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Quant 47-Striving for 50
Verbal 34-Striving for 40
My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
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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.
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.
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
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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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.
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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Thanks Mitch awesome way of thinkingGMATGuruNY wrote:This is a weighted average question.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.
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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Thanks Mitch awesome way of thinkingGMATGuruNY wrote:This is a weighted average question.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.
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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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)
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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GMATGuruNY wrote:This is a weighted average question.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.
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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