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What am I supposed to be seeing?

Expert replies
by jzw » Mon Mar 05, 2012 8:09 am
Number 84 in official guide problem solving. Obviously they don't expect anyone to plug in here and do the math, it would take forever.

What's the esoterica I'm supposed to know?

"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 + 50
(e) 150/q
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Source: — Problem Solving |

by GMATGuruNY » Mon Mar 05, 2012 9:57 am
jzw wrote:Number 84 in official guide problem solving. Obviously they don't expect anyone to plug in here and do the math, it would take forever.

What's the esoterica I'm supposed to know?

"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 + 50
(e) 150/q
Let q=200.
Since q is the cost for 100 batteries, the cost per battery = 200/100 = 2.
Since each battery is sold at 50% above cost, the selling price of each battery = 3. This is our target.

Now we plug q=200 into the answers to see which yields our target of 3.

A quick scan of the answers shows that only A works:
3q/200 = 3(200)/200 = 3.

The correct answer is A.
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by Neo Anderson » Mon Mar 05, 2012 10:15 am
clearly A. 3q/200

original price of 100 batteries is q, thus price per battery is q/100
sold at 50% higher value than cost price =>(q/100) X (1+ 50/100)

=> (q/100) X (3/2)= 3q/200
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by krusta80 » Mon Mar 05, 2012 10:22 am
jzw wrote:Number 84 in official guide problem solving. Obviously they don't expect anyone to plug in here and do the math, it would take forever.

What's the esoterica I'm supposed to know?

"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 + 50
(e) 150/q
Original purchase price = q
cost per battery = q/100

selling price per battery = 1.5 * cost per battery = 1.5 * q / 100 = 3q/200

A it is
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by jzw » Mon Mar 05, 2012 3:23 pm
maybe this little nugget should go under a different topic header - i find that after i bombard myself with 4 hours of math, i start misreading questions. my stupid misread here was, that i thought they were saying that each successive battery was purchased @ 50% more of the preceding battery's cost?!?!?!

i guess i figure this test is that sadistic that it's not beneath them to ask that. so there i am thinking, "well, if we start @ a buck, the next one is $1.50, then $2.25, etc... and i'm thinking - there's no way they expect us to do this math.

thanks to krusta80 and GMATGuruNY for explaining it in idiot proof fashion :)
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