Greg's Arithmetic Algebra 6 class - The taxi Question- (Solved !)

GRE Quant Problem Solving
1

If a taxicab charges x cents for the first 1/9 mile and x/5 cents for each additional 1/9 mile or fraction thereof, what is the charge, in cents, for a ride of y miles, where y is a whole number?

(A) x + (xy-x)/45
(B) x - (xy-x)/45
(C) 2x+9y/5
(D) x + (9x-y)/5
(E) x + (9xy-x)/5

I’m not able to reach the answer through algebra way. Can anyone please help me with that??
Ans - E

2

cost for first 1/9 mile + cost for remaining miles after first 1/9 mile
x + cost for remaining miles after first 1/9 mile
x + cost per mile * # of remaining miles after first 1/9 mile
x + (x/5 * 9) * (y-1/9)
x + x/5 * (9y-1)
x +(9xy-x)/5

3

Hey, thanks for the reply buddy.

Let me know if my thought process is right?
I’m assuming the taxi guy takes x cents for the 1st 1/9th part of the mile regardless of the total “y” miles. That’s the reason the equations comes out to be (9y-1) * x/5. or else it would be the conventional one 8y/9 *x/5.
I was actually stuck with this for a solid hr.
Cause if you read the question you would assume that it meant 1/9th part of the “y” miles. Instead, what it meant is 1/9th of just the first mile.

4

Right, I think we’re on the same page now. The question specifies the price for the first 1/9 mile of the entire trip, then every additional 1/9 mile after that. In order for the question to be instead referring to the first 1/9 mile of every mile, I think the question would need to have the phrase “for every/each mile”.

Consider the case where the question instead asks “what is the charge, in cents, for a ride of z 1/9th miles, where z is a whole number”. I think then you would NOT make the same assumption, even though that section portion of the sentence does not make any statements about how the taxicab splits its charges.

5

I agree.
The question needs to be presented in an unequivocal manner.

6

After spending over an hour to understand how to solve this algebraically , I found the solution
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For 1st 1/9th mile charge is x cents

Now in remaining miles i.e. y-1/9
every 1/9th is charged as x/5 (reading problem every addition 1/9th mile or fraction thereof. )

So we need to calculate how many 1/9th are there in remaining miles
divide (y-1/9)/1/9 =9y-1
and then multiply by x/5.
Hence Total is x+ (9y-1) *x/5
Therefore ans is E