A dealer originally bought 100 identical batteries at a tota

Question

A dealer originally bought 100 identical batteries at a total cost of q dollars. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many dollars was each battery sold?

(A) 3q/200
(B) 3q/2
(C) 150q
(D) q/100
(E) 150/q

(A) 3q/200
(B) 3q/2
(C) 150q 
(D) q/100
(E) 150/q

Bunuel · Accepted Answer

A dealer originally bought 100 identical batteries at a total cost of q dollars. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many dollars was each battery sold?

(A) 3q/200
(B) 3q/2
(C) 150q
(D) q/100
(E) 150/q

(A) 3q/200
(B) 3q/2
(C) 150q 
(D) q/100
(E) 150/q

ALGEBRAIC APPROACH: 
The cost of 100 batteries is q dollars, thus the cost of 1 battery is q/100 dollars. Since the selling price is 50% greater than the cost price than the selling price is q/100*1.5=q/100*3/2=3q/200.

Answer: A.

NUMBER PLUGGING APPROACH: 
Say q=$200, then the cost of 1 battery is q/100=$2. 
The selling price is 2*1.5=$3.

Now, plug q=200 in the answers to see which yields $3. Only answer choice A works.

Answer: A.

noTh1ng · Answer

I don't understand how q can also be used as the price.

I set up an equation like 100 * x = q; with x being the price per battery. Therefore price per battery would be 100x/q...

I just don't get how a equation would be set up without another variable for the price per battery... If anybody could help?

EgmatQuantExpert · Answer

Dear  noTh1ng

The correct equation to write from the highlighted part above is: x = q/100.

That is,  price per battery  = q/100

The mistake that you did was in the red part.

Please let me know if something is still not clear.

Best Regards

Japinder

EMPOWERgmatRichC · Answer

Hi All,

This question can also be solved by TESTing VALUES.

We're told that 100 identical batteries cost a total of Q dollars.

Let's TEST....
Q = 200
So we bought 100 batteries for $200; that is $2/battery.

Next, we're told that each battery was sold at 50 percent above the original cost per battery...

Buy price = $2/battery
Sell price = (1.5)(2) = $3/battery

We're asked for how many dollars each battery was sold.....so we're looking for an answer that equals 3 when Q = 200...

Answer A: 3Q/200 = 600/200 = 3 This IS a match
Answer B: 3Q/2 = 600/2 = 300 NOT a match
Answer C: 150Q = 30,000 NOT a match
Answer D: Q/100 = 200/100 = 2 NOT a match
Answer E: 150/Q = 150/200 = 3/4 NOT a match

Final Answer:  A

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

apoorv601 · Answer

I think its easier to calculate like 1.5*q/100 = 3q/200

EMPOWERgmatRichC · Answer

Hi apoorv601,

One of the great things about most GMAT questions is that they can be approached in a variety of different ways. Ultimately, when dealing with any individual question, you want to know MORE than one way to answer the question AND you want to use whatever method is fastest and easiest for you. If you think that an Algebra-based approach is easiest for this prompt, then that's fine, BUT an Algebra-based approach will NOT be the easiest approach for every prompt. To deal with that contingency on Test Day, you have to be practicing other approaches now, so that you'll be flexible enough to not get 'stuck' during the actual Exam.

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

ScottTargetTestPrep · Answer

We are given that 100 batteries cost a TOTAL of q dollars. We are also given that EACH battery was sold at 50% above the original cost. The first thing we must do is create an equation for q. Remember that q is the TOTAL COST. So if we make b = the original cost per battery we can say:

100 x b = q

b = q/100

We now have the original cost per battery in terms of q. Next, we determine the selling price when we increase the cost by 50%. To calculate this increase we simply multiply q/100 by 1.5. We have:

(q/100) x 1.5

(q/100) x 3/2 = 3q/200

Answer: A

If you don't like working with variables, you could instead substitute a convenient number for q. Normally I would not suggest the plugging-in method in a problem such as this; however, since we have only one variable, the method will be sufficient.

Let's say we make the total cost of the 100 batteries q = 200. This is a convenient number that will work well with the numbers presented in the problem. We can now set up a similar equation to what we did above, where b = the original price per battery.

100 x b = 200

b = 200/100

b = 2

The cost per battery is $2. Now we need to show the selling price per battery by increasing $2 by 50%.

2 x 1.5 = 3

The answer 3 is the selling price of the battery.

The last step is to now plug our value of q = 200 into each answer choice to see which one provides us with a value of 3. This will yield the correct answer.

(A) 3q/200

(3 x 200)/200 = 3

This IS equal to 3.

(B) 3q/2

(3 x 200)/2 = 600/2 = 300

This IS NOT equal to 3.

(C) 150q

150 x 200 = 30,000

This IS NOT equal to 3.

(D) q/100

200/100 = 2

This IS NOT equal to 3.

(E) 150/q

150/200 = 15/20 = ¾

This IS NOT equal to 3.

Answer choice A is the only one that is equal to 3.

BaddyB · Answer

ATQ- 100 batteries = q dollars
therefore, cost of each battery= q/100

Now, each battery was sold 50% above the OG cost i.e. q/100* 50/100 = q/200 (profit per battery)

Question is asking selling price of each battery. So, SP= Actual cost per battery + Profit per battery
 = q/100 + q/200
 = 2q+q/ 200 (making the denominator same)
 = 3q/200
Hence, Option (A)- Answer

ArnauG · Answer

The dealer originally bought 100 identical batteries at a total cost of q dollars. To find the cost per battery when each battery is sold at 50 percent above the original cost, we need to divide the total cost by the number of batteries.

Let's denote the cost per battery as x dollars. The original cost of 100 batteries is q dollars, so we have the equation: 100x = q

Now, when each battery is sold at 50 percent above the original cost, the selling price per battery will be 1.5 times the original cost per battery:

Selling price per battery = 1.5x

We're asked to express the selling price per battery in terms of q. To find the selling price per battery in terms of q, we can substitute the value of x from the first equation into the second equation:

Selling price per battery = 1.5 * (q/100) = (1.5q)/100 = 3q/200

Therefore, the selling price of each battery, in terms of q, is (A) 3q/200.

harvinderkaurprep · Answer

While taking any values, why are we taking q as $200 only? If we were to take q as $100, we get original cost per battery as $1. Which implies the selling price is now $1.5 (with a 50% increase).  Making the option E as the correct choice since 150/100 = 1.5.   Can someone please help me understand the flaw in my thought process?

Bunuel · Answer

For q = 100, both A and E yiled 1.5. For the plug-in method, it might happen that for some particular number(s), more than one option may give a "correct" answer. In this case, just pick some other numbers and check again these "correct" options only.Hope it's clear.

totaltestprepNick · Answer

A https://www.youtube.com/watch?v=6tR82XKyJPk Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience tutoring@totaltestprep.net

A dealer originally bought 100 identical batteries at a tota

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