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22 Oct 2017, 06:33
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C
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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
(A) (g + m)/100
(B) 100g + 100m
(C) g + m
(D) g + .1m
(E) g
22 Oct 2017, 14:57
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Bunuel
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(A) (g + m)/100
(B) 100g + 100m
(C) g + m
(D) g + .1m
(E) g
\(\frac{gCents}{1mi} +\frac{mCents}{1mi} = \frac{(g+m)Cents}{1mi}\)
\(\frac{(g+m)Cents}{1mi}*100 miles * \frac{1dollar}{100cents}=\)
\((g + m)\) 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
22 Oct 2017, 18:54
Kudos
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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.
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.
