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Factoring and Simplifying algebraic expressions

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by pullagurla » Thu Sep 02, 2010 10:31 pm
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 : please give solution by picking numbers method !
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Source: — Problem Solving |

by Rahul@gurome » Fri Sep 03, 2010 12:05 am
pullagurla 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+ 50
(E)
150/q : please give solution by picking numbers method !
Let 100 batteries' total cost = $100 implies 1 battery costs $1.
Each battery was sold at 50 percent above the original cost per battery implies new cost of 1 battery = $(50/100)1 + $1 = 150/100 = 3/2
Generalizing, Cost of 1 battery = $q/100 + (50/100)(q/100) = q/100 + q/200 = 3q/200

The correct answer is [spoiler](A)[/spoiler].
Rahul Lakhani
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