Find the average to reduce the expenditure

algebra-precalculus

Alan drives an average 100 miles each week .His car can travel an average of 25 miles per gallon of gasolines .Alan would to reduce his expenditure on gasoline by $\$5$
Which equation of the following can Alan use to determine how many fewer average miles ,m , he should drive each week ? Assuming that the gallon of gasoline costs $\$4$
$$a) \frac{25}{4}m = 95$$
$$ b) \frac{25}{4}m = 5$$
$$c)\frac{4}{25}m = 95$$
$$d) \frac{4}{25} m = 5$$
My turn : The cost of gasoline per week is $\$16$
So he wants to reduce the cost to be $\$11 $ per week
Then $$\frac{100}{16}= \frac{m}{11}= 68.75$$
But this answer does not math with any one of the choices

Best Answer

In both (a) and (c) the only way to get "95" from the given data is to subtract 5 from 100. But "5" was the desired savings in dollars and 100 was the distance driven in miles. Subtracting dollars from miles is meaningless.

In (b) 25/4 is "miles per gallon" divided by "dollars per gallon" so "miles per dollar". The "5" on the right side is 5 dollars hoped to be saved. multiplying "miles per dollar" by m "miles" will not give "dollars".

In (d) 4/25 is "dollars per gallon" divided by "miles per gallon" so "dollars per mile". Here multiplying "dollars per mile" by m "miles" does give dollars! So this one is worth looking at. If he drives m miles and gets 25 miles per gallon, then he uses m/25 gallons. At 4 dollars per gallon that cost (4/25)m dollars. But that should be equal to the 16- 5= 11 dollars he wants to spend, not the 5 dollars he wants to save.

You are right. None of these is correct.m

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