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A company makes and sells two products, P and Q. The costs per unit of

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A company makes and sells two products, P and Q. The costs per unit of making and selling P and Q are $8.00 and $9.50, respectively, and the selling prices per unit of P and Q are $10.00 and $13.00, respectively. In one month the company sold a total of 834 units of these products. Was the total profit on these items more than $2,000?

(1) During the month, more units of P than units of Q were sold.
(2) During the month, at least 100 units of Q were sold.

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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 13 Jun 2016, 23:16
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Bunuel wrote:
A company makes and sells two products, P and Q. The costs per unit of making and selling P and Q are $8.00 and $9.50, respectively, and the selling prices per unit of P and Q are $10.00 and $13.00, respectively. In one month the company sold a total of 834 units of these products. Was the total profit on these items more than $2,000?

(1) During the month, more units of P than units of Q were sold.
(2) During the month, at least 100 units of Q were sold.



Here is now I will approach it:

Profit on per unit of P = $2
Profit on each unit of Q = $3.50
The company sold a total of 834 units.
If all units were P, revenue = 834 * 2 = 1668
If all units were Q, revenue = 834 *3.5 = more than 2400

Had they been split evenly between P and Q, it would have sold 417 units of each.
This would give a revenue of 417*2 + 417*3.5 = 834 + (something above 1200) = more than 2000

(1) During the month, more units of P than units of Q were sold.
There were more units of P. If say there were 418 units of P and 416 of Q, the revenue would remain more than 2000. If there were mostly P sold, revenue would go below 2000. Not sufficient.

(2) During the month, at least 100 units of Q were sold.
100 units of Q give an additional 100*1.5 = 150 dollars only. So a split such as 734-100 would give a revenue of 1668+ 150 (less than 2000).
Again P and Q split could be anything such as 417-417 in which case revenue would be more than 2000.
Not sufficient

Using both statements, the split of P-Q can be 418-416 (more than 200 revenue) or 734-100 (revenue less than 2000)

Answer (E)
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 25 Oct 2015, 11:20
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Profit per unit of P = 10-8 = 2
Profit per unit of Q = 13-9.5 = 3.5

Let number of units of P sold = p
number of units of Q sold = q
Total number of units sold ,p+q= 834

is 2p + 3.5 q > 2000 ?

1.
p> q
If p = 834
then 2p = 1668 < 2000

If p = 418 , q = 416
2* 418 + 416 * 3.5 , which will be greater than 2000

Insufficient

2.
q=100
100 * 3.5 + 734* 2 <2000


If q= 834
834*3.5 > 2000

Combining
Still insufficient
Answer E

We can also solve this using weighted average
If total profit in selling 834 items is 2000
Then profit per item = 2000/ 834 = 2.4

2 --2.4-----3.5

1. If the p=number of units of P sold is equal to number of units of Q=q, then profit per item will be the average of 2 and 3.5 that is 2.75 .
If p is marginally greater than q , then profit per item will be marginally less than 2.75.
As p increases , then profit per item will move towards 2 .

Not sufficient

2.Similarly , when q is equal to 100 or marginally greater than 100 , then profit per item will be near 2 . The distance will be in the inverse proportion of number of items sold.
When q increases , profit per unit will per unit will increase and reach 3.5 (when q will reach 834)

Combining, we get
Still insufficient
Answer E
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 25 Oct 2015, 10:11
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Bunuel wrote:
A company makes and sells two products, P and Q. The costs per unit of making and selling P and Q are $8.00 and $9.50, respectively, and the selling prices per unit of P and Q are $10.00 and $13.00, respectively. In one month the company sold a total of 834 units of these products. Was the total profit on these items more than $2,000?

(1) During the month, more units of P than units of Q were sold.
(2) During the month, at least 100 units of Q were sold.

Kudos for a correct solution.


p+q = 834
we need to know whether 2p + 3.5q > 2,000
(1) During the month, more units of P than units of Q were sold.
=> p>q
=> 2p + 3.5q >5.5q
But we do not know the value of q => insufficient
(2) During the month, at least 100 units of Q were sold.
=> q>=100, p <=734
=> 2p + 3.5q >= 2p + 350
Insufficient
(1) + (2): p>834/2 = 417 => 2p + 3.5q >= 2p + 350 >834+350

Do not know the valute of 2p + 3.5q

Insufficient

Ans: E
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 01 Nov 2015, 15:41
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I think the answer is 'A,' I am unsure why 'E' is the answer. The stem basically asks is "2P + 3.5Q > 2,000" and the stem also states 834 unites were sold. Stmt A states that "p > q" so the lowest values to make that statement true is 418 (p) and 416 (q) which gives us 836 + 1,456 which is greater than 2,000.

Where is the hole in my thinking?
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 01 Nov 2015, 20:47
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AS7x wrote:
I think the answer is 'A,' I am unsure why 'E' is the answer. The stem basically asks is "2P + 3.5Q > 2,000" and the stem also states 834 unites were sold. Stmt A states that "p > q" so the lowest values to make that statement true is 418 (p) and 416 (q) which gives us 836 + 1,456 which is greater than 2,000.

Where is the hole in my thinking?



AS7x ,

Statement1 : p> q
p can take any value from 418 to 834 .Let's consider the below 2 cases-


Case 1.If p = 834
then 2p = 1668 < 2000

Case 2.If p = 418 , q = 416
2* 418 + 416 * 3.5 , which will be greater than 2000

Hope it helps!! :)
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 15 Dec 2015, 00:07
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Bunuel, Skywalker18 :
My biggest problem is understanding the implied inequality. It takes me sometime to grab the direction of inequalities and then form equations. It's specific to these types of problems involving "word problems+DS+inequality" combo. I think I need more practice. Can you please suggest more similar problems? Thanks again
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 15 Dec 2015, 22:25
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A company makes and sells two products, P and Q. The costs per unit of making and selling P and Q are $8.00 and $9.50, respectively, and the selling prices per unit of P and Q are $10.00 and $13.00, respectively. In one month the company sold a total of 834 units of these products. Was the total profit on these items more than $2,000?

(1) During the month, more units of P than units of Q were sold.
(2) During the month, at least 100 units of Q were sold.

In the original condition, it is the question frequently given in the Gmat Math test, which is "2 by 2" like the table below.
첨부
In the above a+b=834, there are 2 variables(a,b) and 1 equation(a+b=834), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer
In 1), despite a>b, a,b)=(450,384) -> yes, (a,b)=(750, 134) -> no, which is not sufficient.
In 2), despite b>=100, (a,b)=(450,384) -> yes, (a,b)=(750, 134) -> no, which is not sufficient.
In 1) & 2), (a,b)=(450,384) -> yes, (a,b)=(750, 134) -> no, which is not sufficient. Therefore, the answer is E.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 13 Jun 2016, 09:19
Skywalker18 wrote:
Profit per unit of P = 10-8 = 2
Profit per unit of Q = 13-9.5 = 3.5

Let number of units of P sold = p
number of units of Q sold = q
Total number of units sold ,p+q= 834

is 2p + 3.5 q > 2000 ?

1.
p> q
If p = 834
then 2p = 1668 < 2000

If p = 418 , q = 416
2* 418 + 416 * 3.5 , which will be greater than 2000

Insufficient

2.
q=100
100 * 3.5 + 734* 2 <2000


If q= 834
834*3.5 > 2000

Combining
Still insufficient
Answer E

We can also solve this using weighted average
If total profit in selling 834 items is 2000
Then profit per item = 2000/ 834 = 2.4

2 --2.4-----3.5

1. If the p=number of units of P sold is equal to number of units of Q=q, then profit per item will be the average of 2 and 3.5 that is 2.75 .
If p is marginally greater than q , then profit per item will be marginally less than 2.75.
As p increases , then profit per item will move towards 2 .

Not sufficient

2.Similarly , when q is equal to 100 or marginally greater than 100 , then profit per item will be near 2 . The distance will be in the inverse proportion of number of items sold.
When q increases , profit per unit will per unit will increase and reach 3.5 (when q will reach 834)

Combining, we get
Still insufficient
Answer E

Using 2p+3.5q>2000 and knowing that p+q=834 we can substitute q for p and get one equation with one variable: p=834-q => 1668-2q+3.5q>2000 Hence the question basically asks if 1.5q>332 => q>221. Using both statement together it is clear that q can be less that 221 or more than 221.
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 27 Oct 2016, 09:51
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Bunuel wrote:
A company makes and sells two products, P and Q. The costs per unit of making and selling P and Q are $8.00 and $9.50, respectively, and the selling prices per unit of P and Q are $10.00 and $13.00, respectively. In one month the company sold a total of 834 units of these products. Was the total profit on these items more than $2,000?

(1) During the month, more units of P than units of Q were sold.
(2) During the month, at least 100 units of Q were sold.


From the stem we get:

Profit of P = 10 – 8 = 2
Profit of Q = 13 – 9.5 = 3.5

P + Q = 834 => 2P = 2(834 – Q) => 2P = 1668 – 2Q

Question : 2P + 3.5Q > 2000 ?
Substitute 2P in the question: (1668 – 2Q) + 3.5Q > 2000?

The question becomes: is Q > 221.3?

1) P > Q = > 417 > Q > 0, INSUFFICIENT
2) Q > 100 => INSUFFICIENT
1 and 2 => 417 > Q> 100 => INSUFFICIENT. If Q is 300 the answer is YES; if Q is 200 the answer is NO.

Answer is E
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 11 Nov 2017, 05:32
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Hello together,

I understand the question and I understand the way to get the result as well.
But my problem is, that I need too much time on such tasks. Because I have to build up the equations.
Does anybody know a strategy to recognize the result due to hidden clues in a minimum of the time? Like "if in both satements are something with "at least" or "more than" it is not solvable"?
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 24 Jan 2018, 04:41
Bumping Bunuel chetan2u niks18 for an alternate solution.
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 24 Jan 2018, 11:43
adkikani wrote:
Bumping Bunuel chetan2u niks18 for an alternate solution.


Hi adkikani

As this DS question has lot of information in the question stem, you need to use every information possible before moving on to the statements.

As it turns out my approach to this question was the same as posted by cledgard above.

the question boils down to whether q>221.3 and then reading both the statements become simple.

There might be other simpler approaches which other experts might be able to explain. :-) :-)
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A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 18 Apr 2018, 15:40
Given:
P + Q = 834
Profit P = $2
Profit Q = $3.5

i) P > Q. The official guide says no information about quantity of P or Q is provided; hence, this is incorrect. I disagree. You do not need either variable to solve for this. The max P can achieve is when Q is 0 (i.e. P = 834 units sold). Hence, total profit will be 834 * $2 = $1668. This by itself is sufficient to answer that the profit is not more than $2k.
However, nothing says that Q = 0; thus, let's try the other extreme (i.e. min value of P which occurs when P and Q are as close as possible or 834 / 2 + 1 = P and 834 / 2 - 1 = Q (P > Q). Here, profit = p_min (418) * $2 + q_max (416) * $3.50 = $2292 which is > $2k.

Since we receive 2 different answers that make the stem T (case 1) and F (case 2), this statement is insufficient.
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Re: A company makes and sells two products, P and Q. The costs per unit of  [#permalink]

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New post 09 Oct 2018, 15:26
Let's talk strategy here. Many explanations of Quantitative questions focus blindly on the math, but remember: the GMAT is a critical-thinking test. For those of you studying for the GMAT, you will want to internalize strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time. The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. Ready? Here is the full "GMAT Jujitsu" for this question:

The target of this question is "total profit", and we know that profit is equal to the selling price minus the cost. Thus, the profit per unit for product P would be \($_P = 10-8 = 2\), and the profit per unit for product Q would be \($_Q = 13-9.5 = 3.5\). The total profit would therefore be (where \(P\) and \(Q\) represent the number of units of each product):

\(T=P($_P )+Q($_Q)\)
\(T=P(2)+Q(3.5)\)

Now that we have set up the basic equation, we need to pause in our discussion here. Many people spend too much time on Data Sufficiency questions because they think they need to get to the bitter end. The question asks “Was the total profit on these items more than $2,000.00?” This is a “Yes/No” question – a very common structure for Data Sufficiency problems. The fundamental trap for problems like these is to bait you into thinking that you actually need to solve for every value. You don’t. As soon as you have enough information to conclude whether you can come up with only one answer to the question, you can move on. For “Yes/No” questions, if you can think of two situations (or two variable inputs) that are consistent with all of the problem’s constraints but come up with different answers to the question, you know a statement is insufficient. In my classes, I call this strategy “Play Both Sides.”

Let’s analyze each statement, and you will see what I mean. Statement #1 tells us that there are more units of \(P\) than there are units of \(Q\). Now, look at the shape of the equations. From the question stem, we know that \(P+Q=834\). “834” is a bit of a messy number, so approximating this would make our analysis a lot easier. First, let’s assume that \(P\) is roughly equal to \(Q\), making both values around \(400\). Thus,

\(T=P(2)+Q(3.5)≈400(2)+400(3.5)≈800+1400≈2200\)

So, it is possible to exceed \($2,000\), and we know we can answer “Yes” to the question. The next question we ask ourselves is: Can we play both sides? Is it possible to have a solution where the total is under \($2,000\)? If \(P\) dominated the relationship (in other words, if \(P\) were around \(800\), then we would have less than \($2,000\). After all, \(800*2\) is only equal to \($1,600\). We are not even close to the boundary of \($2,000\), so approximating is sufficient here. Since one scenario would clearly give us a combined profit of less than \($2,000\) while the other would give a total profit of more than \($2,000\), Statement #1 is insufficient by itself.

Statement #2 tells us that \(Q\) is “at least 100 units.” We have already shown by our analysis of Statement #1 that if \(Q\) is around \(400\), then the total will exceed \($2,000\). There is no reason to redo this. As we “Play Both Sides”, we already know that there is a solution where \(Q\) is “at least 100 units” that allows us to answer [color=#00a651]“Yes”[/color] to the original question. So, we need to know if we can show that the total may be less than \($2,000\). The phrase “at least” is massive leverage. This is a minimum amount, so let’s see what a minimum amount would do – in other words, if \(Q=100\). Again, approximating here is more enough. If \(Q=100\), then \(P≈700\). Thus:

\(T≈700(2)+100(3.5)≈1400+350≈1750\)

Since we can also show that the total can be less than \($2,000\), we have “played both sides” and Statement #2 is insufficient.

We now need to look at combining them to see if they are sufficient. But we have already done all the analysis we need. The hypothetical situations we created for Statements #1 and #2 show two possibilities where (1) P can be greater than Q and (2) Q is at least 100. Since we can “play both sides” of this Yes/No question even when the statements are simultaneously true, the answer is “E”.

Now, let’s look back at this problem from the perspective of strategy. This problem can teach us several patterns seen throughout the GMAT. First, minimize your math. You only need to do enough math to prove the answer to the question. With Yes/No questions, this often means “playing both sides” – coming up with two possible solutions that follow the rules of the problem, but give different answers to the question. Second, notice how approximation can often cut through otherwise messy math. This saves time and energy, allowing you to focus on other, potentially harder, questions. And that is how you think like the GMAT.
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