Please help me with this one.
OA is C
I lost this with the time constraint and can't find the shorter way to solve it.
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N1 = Model P
N2 = Model Q
N1 + N2 = 40
N1/N2 = (q-141)/(141-p)
N1/40-N1 = (q-141)/(141-p)
To find: N1
Statement 1:
p = q - 30
if we put "p" in equation.. we will be stuck with variables "q"
INSUFFICIENT
Statement 2:
If we use p = 120 then we will be stuck with "q" and vice versa
INSUFFICIENT
Combining.
We can get both p & Q
As, p = q - 30 So, if p = 120 .. q = 150
N1/(40-N1) = 9/21
N1(7) = 3(40-N1)
N1 = 12
and, if q=120, p=90
N1/(40-N1) = -21/51
17N1 = -7(40-N1)
We will get -ve value.. So, not valid.
Answer [spoiler]{C}[/spoiler]
N2 = Model Q
N1 + N2 = 40
N1/N2 = (q-141)/(141-p)
N1/40-N1 = (q-141)/(141-p)
To find: N1
Statement 1:
p = q - 30
if we put "p" in equation.. we will be stuck with variables "q"
INSUFFICIENT
Statement 2:
If we use p = 120 then we will be stuck with "q" and vice versa
INSUFFICIENT
Combining.
We can get both p & Q
As, p = q - 30 So, if p = 120 .. q = 150
N1/(40-N1) = 9/21
N1(7) = 3(40-N1)
N1 = 12
and, if q=120, p=90
N1/(40-N1) = -21/51
17N1 = -7(40-N1)
We will get -ve value.. So, not valid.
Answer [spoiler]{C}[/spoiler]
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Total revenue for all 40 televisions = 40*141 = 5640.Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P TV or A Model Q TV. Each Model P sold for $p and each model Q sold for $q. The average selling price of the 40 televisions was $141. How many of the 40 televisions were Model P Televisions?
1 - Model P sold for $30 less than the Model Q Televisions
2 - Either p = 120 or q = 120.
Since the average price = 141, one price must be LESS than 141, while the other price must be GREATER than 141.
(Unless p=q=141, which is highly unlikely.)
Statement 1: Model P sold for $30 less than Model Q.
Thus, p< 141, while q>141.
Check the ONLY case that also satisfies statement 2:
Case 1: p=120 and q=150.
To evaluate this case, use ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.
Step 1: Plot the 3 prices on a number line, with the prices for the two models on the ends and the average price in the middle.
P 120-------------141------------150 Q
Step 2: Calculate the distances between the prices.
P 120-----21------141-----9------150 Q
Step 3: Determine the ratio in the mixture.
The required ratio of Model P televisions to Model Q televisions is equal to the RECIPROCAL of the distances in red.
P:Q = 9:21 = 3:7 = 12:28.
Thus, if 12 Model P televisions are sold for $120 each, and 28 Model Q televisions are sold for $150 each -- for a total of 40 televisions -- the total revenue will be $5640:
(12*120) + (28*150) = 5640.
Case 2: Reverse the distances from Case 1 and plot the new prices for P and Q on the ends of the number line
P 132-----9------141-----21------162 Q
In this case, P:Q = 21:9 = 7:3 = 28:12.
Thus, if 28 Model P televisions are sold for $132 each, and 12 Model Q televisions are sold for $162 each -- for a total of 40 televisions, with a price difference of $30 between the 2 models -- the total revenue will still be $5640:
(28*132) + (12*162) = 5640.
Since both cases are possible, INSUFFICIENT.
Statement 2: Either p = 120 or q = 120.
Case 1 also satisfies statement 2.
Case 3: If p and q swap positions on the number line in Case 1 -- so that q=120 and p=150 -- the result will be that 12 Model Q televisions are sold for $120 each, while 28 Model P televisions are sold for $150 each.
Since both cases are possible, INSUFFICIENT.
Statements combined:
Only Case 1 satisfies both statements, implying that 12 Model P televisions are sold for $120 each.
SUFFICIENT.
The correct answer is C.
For two similar problems, check here:
https://www.beatthegmat.com/ratios-fract ... 15365.html
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I have seen many problems by you are done by alligation method and I must say you are champ in it . Even I am expert in alligation method. But what I lack is the skill to judge which question you can solve by alligation. If its a mixture problem clearly I would go for it. But question such as this one, I did it other way as Rahul has done above.GMATGuruNY wrote:Total revenue for all 40 televisions = 40*141 = 5640.Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P TV or A Model Q TV. Each Model P sold for $p and each model Q sold for $q. The average selling price of the 40 televisions was $141. How many of the 40 televisions were Model P Televisions?
1 - Model P sold for $30 less than the Model Q Televisions
2 - Either p = 120 or q = 120.
Since the average price = 141, one price must be LESS than 141, while the other price must be GREATER than 141.
(Unless p=q=141, which is highly unlikely.)
Statement 1: Model P sold for $30 less than Model Q.
Thus, p< 141, while q>141.
Check the ONLY case that also satisfies statement 2:
Case 1: p=120 and q=150.
To evaluate this case, use ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.
Step 1: Plot the 3 prices on a number line, with the prices for the two models on the ends and the average price in the middle.
P 120-------------141------------150 Q
Step 2: Calculate the distances between the prices.
P 120-----21------141-----9------150 Q
Step 3: Determine the ratio in the mixture.
The required ratio of Model P televisions to Model Q televisions is equal to the RECIPROCAL of the distances in red.
P:Q = 9:21 = 3:7 = 12:28.
Thus, if 12 Model P televisions are sold for $120 each, and 28 Model Q televisions are sold for $150 each -- for a total of 40 televisions -- the total revenue will be $5640:
(12*120) + (28*150) = 5640.
Case 2: Reverse the distances from Case 1 and plot the new prices for P and Q on the ends of the number line
P 132-----9------141-----21------162 Q
In this case, P:Q = 21:9 = 7:3 = 28:12.
Thus, if 28 Model P televisions are sold for $132 each, and 12 Model Q televisions are sold for $162 each -- for a total of 40 televisions, with a price difference of $30 between the 2 models -- the total revenue will still be $5640:
(28*132) + (12*162) = 5640.
Since both cases are possible, INSUFFICIENT.
Statement 2: Either p = 120 or q = 120.
Case 1 also satisfies statement 2.
Case 3: If p and q swap positions on the number line in Case 1 -- so that q=120 and p=150 -- the result will be that 12 Model Q televisions are sold for $120 each, while 28 Model P televisions are sold for $150 each.
Since both cases are possible, INSUFFICIENT.
Statements combined:
Only Case 1 satisfies both statements, implying that 12 Model P televisions are sold for $120 each.
SUFFICIENT.
The correct answer is C.
For two similar problems, check here:
https://www.beatthegmat.com/ratios-fract ... 15365.html