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It costs g cents a mile for gasoline and m cents a mile for all other

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It costs g cents a mile for gasoline and m cents a mile for all other  [#permalink]

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New post 22 Oct 2017, 05:33
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A
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C
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Difficulty:

  45% (medium)

Question Stats:

48% (00:43) correct 52% (00:57) wrong based on 44 sessions

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It costs g cents a mile for gasoline and m cents a mile for all other costs to run a car. How many dollars will it cost to run the car for 100 miles?

(A) (g + m)/100
(B) 100g + 100m
(C) g + m
(D) g + .1m
(E) g

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It costs g cents a mile for gasoline and m cents a mile for all other  [#permalink]

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New post 22 Oct 2017, 13:57
Bunuel wrote:
It costs g cents a mile for gasoline and m cents a mile for all other costs to run a car. How many dollars will it cost to run the car for 100 miles?

(A) (g + m)/100
(B) 100g + 100m
(C) g + m
(D) g + .1m
(E) g

Algebra

\(\frac{gCents}{1mi} +\frac{mCents}{1mi} = \frac{(g+m)Cents}{1mi}\)

\(\frac{(g+m)Cents}{1mi}*100 miles * \frac{1dollar}{100cents}=\)

\((g + m)\) dollars

Answer C

Choose values

Let g = 2 cents
Let m = 2 cents
It costs 2 + 2 = 4 cents per mile to run the car

\(\frac{(4)Cents}{1mi}*100 miles = 400 cents\)

How many dollars?

\(400 cents *\frac{1dollar}{100cents}=\) 4 dollars

Using g = 2, m = 2, I need the choice that yields 4 as answer, because the question prompt includes the dollars:

(A) (g + m)/100 = 4/100. NO

(B) 100g + 100m = 400 + 400. NO

(C) g + m = 2 + 2 = 4. MATCH

(D) g + .1m = 2 + .2 = 2.2 NO

(E) g = 2. NO

Answer C
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Re: It costs g cents a mile for gasoline and m cents a mile for all other  [#permalink]

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New post 22 Oct 2017, 17:54
1 mile= g cents (For gasoline)
100 mile= 100g cents------(1)

1 mile= m cents (For gasoline)
100 mile= 100m cents------(2)

Add (1) and (2)

Total cost= 100g+ 100m cents= 100(g+m)cents

Now 1 cents= \(\frac{1}{100}\) dollar

100(g+m) cents= 100(g+m)cents*\(\frac{1}{100}\)

= (g+m) cents

Answer: C.
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Re: It costs g cents a mile for gasoline and m cents a mile for all other  [#permalink]

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New post 22 Oct 2017, 21:07
Costs for 1 mile = g + m (cents)
=> costs for 100 miles = 100 * (g +m) cents
1 dollar = 100 cents
Hence, total cost IN DOLLAR = 100(g+m)/ 100 = g+m => answer C.

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Re: It costs g cents a mile for gasoline and m cents a mile for all other  [#permalink]

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New post 22 Oct 2017, 21:48
varun4s wrote:
1 mile= g cents (For gasoline)
100 mile= 100g cents------(1)

1 mile= m cents (For gasoline)
100 mile= 100m cents------(2)

Add (1) and (2)

Total cost= 100g+ 100m cents= 100(g+m)cents

Now 1 cents= \(\frac{1}{100}\) dollar

100(g+m) cents= 100(g+m)cents*\(\frac{1}{100}\)

= (g+m) cents
Answer: C.

varun4s , I think you mean (g + m) dollars?

The question asks for dollars.

100(g+m)cents, then

\(100(g+m) cents * \frac{1 dollar}{100 cents}\)

The units "cents" cancel. Dollars are left.
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It costs g cents a mile for gasoline and m cents a mile for all other  [#permalink]

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New post 22 Oct 2017, 21:59
IMO Option C.

Total costs for 1 mile= g+m/100
total costs for 100 miles= g+m
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It costs g cents a mile for gasoline and m cents a mile for all other &nbs [#permalink] 22 Oct 2017, 21:59
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