A merchant paid $300 for a shipment of x identical calculators. The

Question

A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?

A. 24
B. 25
C. 26
D. 28
E. 30

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Is this question correct? Shipping cost given but cost of each calculator missing.....

A. 24
B. 25
C. 26
D. 28
E. 30

PS45502.01

Is this question correct? Shipping cost given but cost of each calculator missing.....

Bunuel · Accepted Answer

"A merchant paid $300 for a shipment of x identical calculators" means that there was a shipment of x calculators which cost merchant $300.

The average cost of each calculator is  \frac{300}{x} ;
The average price of each calculator is  \frac{300}{x}+5 ;
Number of calculators sold is  x-2 ;
The total revenue is  300+120=420 ;

So,  (x-2)*(\frac{300}{x}+5)=420 . At this point it's better to substitute values. Answer choice E fits:  (30-2)*(\frac{300}{30}+5)=28*15=420 .

Answer: E.

Hope it's clear.

achatterjee14 · Answer

Try plugging in the answer:-
1st Try selecting options which are divisible by 300
If you select 30 as x.Therefore avg cost of each calculator 300/30 i.e 10$.
Set aside 2 .So we have 28 calc.
Each calc sold at 5$ more i.e 10+5=15$ & 15x28=$420
Difference is $420-$300=120
E) is the answer hope it helps

BrentGMATPrepNow · Answer

APPROACH #1 : Testing the answer choices 
  STRATEGY : As with all GMAT Problem Solving questions, we should immediately ask ourselves,   Can I use the answer choices to my advantage?   
In this case, it's possible to test the answer choices, AND I can see how we might eliminate some answer choices first (more on this later). 
Now we should give ourselves about 20 seconds to identify a faster approach. 
Well, we can also solve the question algebraically, but that approach could take a while for this tricky question. 
Since I feel like I can eliminate 2 or 3 answer choices first, I'm going to test the given values.

Given: The merchant sells EACH of the remaining calculators for $5 more than the average cost of the x calculators  
So, for example, if the correct answer is C (x = 26) , then the average cost of each of the 26 calculators will be a complete mess, since 26 doesn't divide nicely into $300
We get: average cost per calculator = $300/26 = $11.538461...
This means the merchant would have to sell each of the remaining 24 calculators at a price of $16.538461..., which makes no sense.

Since answer choices C and D don't divide nicely into $300, I'm not going to test these values. 
 Aside: If the merchant bought 24 calculators (answer choice A), the average price per calculator would be $12.50, which isn't a bad to work with. So if answer choices B and E don't work out, I'll know the correct answer must be A.

Let's test answer choice B ( x = 25 )
If it costs $300 to purchase 25 calculators, then the average cost per calculator = $300/25 = $12
This means the merchant will sell each of the remaining 23 calculators at a price of $17
So, the total revenue = (23)($17) 
  I already know this answer is incorrect, because the question tells us that the total revenue was $120 more than the cost of the shipment. In other words, the total revenue must be $420, and since (23)($17) definitely does not equal $420, we can eliminate answer choice B.

Now let's test answer choice E ( x = 30 )
If it costs $300 to purchase 30 calculators, then the average cost per calculator = $300/30 = $10
This means the merchant will sell each of the remaining 28 calculators at a price of $15
So, the total revenue = (28)($15) = $420
Voila!!
The correct answer is E.

APPROACH #2 : Algebra 
The merchant originally buys x calculators for $300
So, the average purchase cost =  300/x  dollars per calculator

Later, the calculators are sold for  $5 more  than the average purchase cost of  300/x  dollars
So, the resell price is  (300/x) + 5  dollars per calculator

How many were sold? 
Well, the merchant began with x calculators, but used 2 as demonstrators, which means the merchant sold  x - 2  calculators.

Finally, the merchant's profit was $120 (after the $300 investment). So, the total revenue was  $420

We can now write an equation:   (x - 2) = 420

IMPORTANT : This is an awful equation to solve. At this point, it may be faster to try plugging in the answer choices.

Or we can solve the equation as follows: 
Start with:   (x - 2) = 420  
Expand:  300 - (600/x) + 5x - 10 = 420 
Multiply both sides by x to get:  300x - 600 + 5x² - 10x = 420x 
Simplify:  5x² - 130x - 600 = 0 
Divide both sides by 5 to get:  x² - 26x - 120 = 0 
Factor:  (x - 30)(x + 4) = 0 
So,  x = 30  or  x = -4

Since x can't be negative, we know that  x = 30

Answer: E

jlgdr · Answer

I'd go with backsolving on this one though I must admit that originally I tried to solve algebraically and was stuck with variables and quadratics We will have (300/x + 5)(x-2) = 420 Then begin replacing, I started with 25 cause 26 would not yield an integer when divided by 300 but it didn't work, so I realized I needed a bigger number and the only factor of 30 that could work was E So E Hope it helps Cheers! J

PareshGmat · Answer

Answer = E

Total Calculators = x

Cost per calculator =  \frac{300}{x}

Calculators sold = x-2

Selling price per calculator =  \frac{300}{x} +5

Total amount received =  (\frac{300}{x} + 5) * (x-2) = 420

Solving,

x^2 - 26x - 120 = 0

x = 30

EMPOWERgmatRichC · Answer

Hi All,

The information in the prompt include a variety of nice 'round' numbers, so it's likely that the number of calculators in the shipment is also a round number AND divides evenly into 300. Only two of the answers 'fit' that idea - 25 and 30, so let's TEST THE ANSWERS and see if either of those values fits all of the information given in the prompt.

IF.... X = 30 and there are 30 calculators in the shipment....

The merchant paid $300/30 = $10 per calculator 
2 calculators were used for demonstrations, so 30 - 2 = 28 calculators were sold 
Each of the 'sold' calculators brought in $10+$5 = $15 
The total revenue was ($15)(28) = $420 
The total profit was $420 - $300 = $120

This is an exact match for what we were told, so this MUST be the answer.

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

chrtpmdr · Answer

How to you avoid taking too long on factoring the quadratic equation? 
I somehow always slightly get stressed when I don't have a first sight quadratic equation and I tend to overcomplicate, even though this one was quite solvable.

EMPOWERgmatRichC · Answer

Hi chrtpmdr,

Most GMAT questions are written so that they can be solved in more than one way; in many cases, the 'math approach' actually takes longer than a more Tactical approach would take. You bring up a really important point about how factoring a quadratic equation could take more time than you want it to. In this question, you don't actually "need" to work in that way to get the correct answer - so you have to be careful about assuming that 'your way' to approach a question is the only way. You might inadvertently be making everything harder for yourself (and costing you time and points as a result).

GMAT assassins aren't born, they're made,
Rich

ScottTargetTestPrep · Answer

The average price of the cost of the x calculators is 300/x dollars.

(x - 2) calculators were sold for (300/x) + 5 dollars each, for a total revenue of:

(x - 2)  = 300 + 5x - (600/x) - 10 = (300x - 600)/x + 5x - 10

Since the total revenue from the sale of the calculators was $120 more than the cost of the shipment:

(300x - 600)/x + 5x - 10 = 120 + 300

Multiplying by x, we have:

300x - 600 + 5x^2 - 10x = 120x + 300x

5x^2 - 130x - 600 = 0

x^2 - 26x - 120 = 0

(x - 30)(x + 4) = 0

x = 30 or x = -4

Since x can’t be negative, x = 30.

Answer: E

CrackverbalGMAT · Answer

Hello chrtpmdr,

In my opinion, you should  never try to SOLVE  an equation, whatever type it may be, on a GMAT Quant question.

In questions on equations, very often we solve to find the unknown/s. As such,  the answer options given with the question will also represent the value of the unknowns we are trying to obtain . Therefore, solving a linear equation/quadratic in the step-by-step approach, by factorizing/taking things common/solving them as simultaneous equations smacks of a very ‘text-book’ kind of approach.

That’s not what GMAT is testing you on. Sure, GMAT expects you to be able to develop variables, construct equations and solve them. But it does not expect you to solve them in a certain set way. In fact, it probably wants you to try out innovative approaches by plugging in values – based on your judgement or from the options. 
If you look at Cubic equations, there’s absolutely no way of knowing the first root until you plug in values like 0, 1 and -1. So,  you need to understand that Math alone will not help you ace topics in GMAT Quant. You need to put on your logical reasoning hat as well .

But, unfortunately, that’s the bane of many test takers. They end up depending too much on Math to bail them out. They end up assuming that, solving an equation a particular way it was taught in school, is the best way of solving it. Really, there is no pride in solving equations the conventional way IMHO. I’d rather leverage the answer options and logic to figure out which options may fit in. I’d rather use the time to do this than breaking my head about the factors.

To answer your question about how to quickly factorise, you need to observe the constant and think of all the ways in which you can break it down into a product of 2 of its own factors.

For example, 120 = 20*6 / 30*4/12*10/15*8/24*5/60*2/120*1/40*3.

When you do this, it will become easy for you to pick the pair that will give you the middle term when added. I’m not saying that you will have to write out all the pairs all the time. If you inculcate this habit when you are starting off with your prep on Quadratics, by the time you factorise 10 to 15 quadratics, this would probably have become muscle memory for you. And that’s  the idea of prepping for GMAT, building good muscle memory.

If you did not want to deal with this quadratic, you could have dealt with the more preliminary equation which has been equated to 420. Observe that, out of the options,  only 28 and 30 are factors of 420 . Remember that we are trying to find the number of calculators (which cannot be a fraction), so  I’d naturally chose 30 since (30-2) should also be a factor of 420 . Note that this happened more sub-consciously with me because of my muscle memory.

So, it’s really not about being strong with Math alone. If you think of it that way, I’d not be surprised if you are stressed in such situations. From now on, try to incorporate a few of these things in your Quant prep and you will see yourself saving on time on effort.

Hope that helps!

wishmasterdj · Answer

While the question is straight forward, I was thinking of an alternate approach for this. Please advise on what I am doing wrong (since I dont get 30 as the answer):

delta between SP and CP = 5
Number of units sold = x-2
Profit = 120

5 * (x-2) = 120

This gives us 26, which is wrong. But I was wondering what my mistake is in this approach..

EMPOWERgmatRichC · Answer

Hi wishmasterdj,

Your approach ignores the fact that the total revenue has to be $120 more than the cost of the SHIPMENT (re: $300) - and since 2 of the calculators were NOT sold, that impacts what the SP of each unit has to be. This means that it's not enough to solve for total profit (and your equation is focused on the $120 profit); you have to solve for a total of $420 of revenue.

GMAT assassins aren't born, they're made,
Rich

crampipan · Answer

(t + 300/x) * (x-2) - 300 = 120
5x + 300x/x - 600/x = 430
5x - 600/x = 130
5x^2 + 600 = 130x
5x^2 - 130x = 600
x^2 - 26x = 120
x^2 - 26x + 169 = 289 (add 169 to both sides)
(x-13)^2 = 289
(x - 13) = sqrt(289)
(x - 13) = 17
x = 30

Crytiocanalyst · Answer

cost of each calculator is 300/x
The average price of each calculator is 300x+5;
Total calculators sold is x−2;
The total =>300+120=420;

(x−2)∗(300/x+5)=420. 
by simple brute force method we get to the answer 
Hence IMO E

MBAHOUSE · Answer

A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?

A. 24
B. 25
C. 26
D. 28
E. 30

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Is this question correct? Shipping cost given but cost of each calculator missing.....

kokus · Answer

The cost of the calculators is $300 and they are sold for $420 (since the profit is $120 more than the cost). Therefore, the number of sold units must be divisible by the revenue ($420) and the cost ($300). Only E fits.

Gemmie · Answer

p: cost price of a calculator

x * p = 300
(x-2) * (p+5) = 300 + 120 = 420

From the answer choices, x is among 24, 25, 26, 28, 30

300 is divisible by 25 and 30 only
300 = 25 * 12 = 30 * 10

27 * 17 ends with digit 9
28 * 15 = 420

=> x = 30

obedear · Answer

How I arrived at the answer:

Through the given information we know that total revenue was $120 more than the cost which is $300, so the total revenue must be $420.

Revenue = price * quantity, so we have

420 = (300/x + 5) * x-2. 
At this point, if we break down 420 into its prime factors, we can see:

2^2*3*5*7=(300/x + 5) * x-2.

Now, x will be the answer choices, so quickly scanning, we can see what will not be a factor of the right hand side.

A - If x-2 = 22, then x-2 has an 11 in it, which 420 does not have. Cannot be this choice.
B - if x-2 = 23, then x-2 has a 23 in it, which 420 does not have. Cannot be this choice.
C - If x-2 = 24, then x-2 has a 2^3 in it, which 420 does not have (it only has 2)
D - If x-2 = 26, then x-2 has a 13 in it, which 420 does not have.
E - If x-2 = 28, then x-2 has a 2^2*7 in it, which 420 does have. By process of elimination, E must be the answer.

Only problem with this approach is that this assumes both are integers, and the price does not necessarily have to be an integer.

A merchant paid $300 for a shipment of x identical calculators. The

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