Data Sufficiency

Hello,

Can you please assist with this:

Quick Sell Outlet sold a total for 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.

OA: C


I tried to solve as follows:

Let the number of model P TV's sold = x
Let the number of model Q TV's sold = y

Given, 40 = x + y - Eq. 1

Average Selling price = 141
=> 141 = x($p) + y($q) / 40
=> xp + yq = 5640 - Eq. 2

To determine: x = ?

1) p = q - 30 - Insuff.

2) p = 120 or q = 120 - Insuff.


1 and 2:

p = 120 = > 120 = q - 30 => q = 150
q = 120 => p = 120 - 30 => p = 90


At this point I got stuck though. Can you please help with this?

Thanks,
Sri

gmattesttaker2 wrote:Hello,

Can you please assist with this:

Quick Sell Outlet sold a total for 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.

OA: C


I tried to solve as follows:

Let the number of model P TV's sold = x
Let the number of model Q TV's sold = y

Given, 40 = x + y - Eq. 1

Average Selling price = 141
=> 141 = x($p) + y($q) / 40
=> xp + yq = 5640 - Eq. 2

To determine: x = ?

1) p = q - 30 - Insuff.

2) p = 120 or q = 120 - Insuff.


1 and 2:

p = 120 = > 120 = q - 30 => q = 150
q = 120 => p = 120 - 30 => p = 90


At this point I got stuck though. Can you please help with this?

Thanks,
Sri

Hi Sri,
You came almost close to the solution,
we have two equations,
P+Q = 40 --> 1 and P*p +Q*q = 141*40 --> 2

and you found
p = 120 = > 120 = q - 30 => q = 150
q = 120 => p = 120 - 30 => p = 90

substitute above two in 2nd equation,

you will get,
P(120) + Q(150) = 3*47*40 => 4P+5Q = 47*4
P(90) + Q(120) = 3*47*40 => 3P+4Q = 47*4

so we now have two different 2nd equations,

try to solve 1st with above two eqn,
where you can see the negative value for red equation which is not possible.

Green goes correct and P = 12

Regards,
Uva

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.

Total revenue for all 40 televisions = 40*141 = 5640.
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