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# A dealer originally bought 100 identical batteries at a tota

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Retired Moderator
Joined: 25 Feb 2013
Posts: 1220
Location: India
GPA: 3.82
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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23 Mar 2018, 10:08
1
dave13 wrote:
Bunuel wrote:
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

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.

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.

hello niks18,

q/100*1.5 this part means that selling price of each item is increaed by 50% - shouldnt 1.5q/100 be the answer? ;?

and this one q/100*3/2 what does it mean ? where from do we get 2/3 and what is it

Hi dave13

the highlighted part is actually $$\frac{q}{100}*1.5$$. This should resolve your first query.

For the second query, highlighted in red above, I would suggest you try to find out the answer yourself because this is a basic mathematical operations. Try and find out how to convert decimals into fractions. This topic might not be covered in GMAT quant books but you will get detailed explanations regarding mathematical operations in any primary level maths book.
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Joined: 09 Mar 2016
Posts: 1287
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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24 Mar 2018, 01:06
1
Hi niks18 many thanks for a professionally expressed math hint

i got it now how it solved, i actually learnt it, but i for some reason forget some concepts i learnt before ... Anyway here is the solutuon

$$\frac{q}{100}*1.5$$

$$\frac{q}{100}*\frac{150}{100}$$

$$\frac{q}{100}*\frac{3}{2}$$

$$\frac{3q}{200}$$

have a great weekend
Intern
Joined: 20 Aug 2018
Posts: 4
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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21 Sep 2018, 09:22
ScottTargetTestPrep wrote:
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

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

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.

Pls explain how 50%=1.50 it should be 0.5

Posted from my mobile device
Retired Moderator
Joined: 25 Feb 2013
Posts: 1220
Location: India
GPA: 3.82
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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21 Sep 2018, 09:29
1
SuryaChandra wrote:
ScottTargetTestPrep wrote:
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

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

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.

Pls explain how 50%=1.50 it should be 0.5

Posted from my mobile device

Hi SuryaChandra

Here 2 is increased by 50% i.e. 2+2*0.5=2*(1+0.5)=2*1.5
Re: A dealer originally bought 100 identical batteries at a tota &nbs [#permalink] 21 Sep 2018, 09:29

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