If a taxi driver charges x cents for the first quarter-mile of a trip

Since 1 mile has 4 quarter-miles, a trip of y miles has 4y quarter-miles, and after the first quarter-mile that charges x cents, there are 4y - 1 additional quarter-miles that charge x/5 cents each. Thus, the total cost is:

x + (x/5)(4y - 1)

5x/5 + (4xy - x)/5

(4x + 4xy)/5

Answer: B
_________________

let Y be 1 mile
x= 20 cents

as per given info
in 1 mile we have 4 quarter mile
so for the first mile x=20 cents and x/5 cents =20/5=4 cents for every additional quarter mile
so total =20+4+4+4=32 cents charge for complete ride.

we just substitute our values in the given options and we get our ans B.

Answer = B. (4x + 4xy)/5

Let y = 1, which constitutes total 4 quarters

1st quarter cost = x

Remaining 3 quarters cost \(= \frac{3x}{5}\)

Total cost \(= \frac{3x}{5} + x = \frac{8x}{5}\)

Placing y = 1 in the OA, only option B fits in

ans ..B..
total quarters miles in y miles=4y...
cost @x/5 cents= 4xy/5 cents...
in this total, first quarter is x cents... so add x and subtract x/5..
ans 4xy/5+4x/5
_________________

I tried to solve in the following manner, but my values are satisfying option D, i want to know where am i going wrong.

x= 5 cents
y= 4 mile
For the first mile, charge should be x(5 cents) and for the remaining 3 miles charge should be x/5 or 1 cent for each mile.
Total charge= 5 cents + 1 cent *(3 miles) = 8 cents

Answer choice D, (4x + xy)/5 gives 8 cents and hence i marked D, however answer B is correct.

Experts please help!!

Hi ash99,
your method is correct but you have not understood the Q properly..
the prices are for quarter mile and not for entire mile..
so taking the values as you have decided..
total charge=5+(4*3*1(4*3 is the number of quarter in last 3 milesand 1 cent is the price)+3*1(3 is the number of quarter in first mile as the first Quarter of it has the price 5 , which has already been considered)....=20..
B should fit in
hope it helped
_________________

y=1 mile, x=5 cents and x/5=1 cent. so y trip will cost 8 cents. plug in x and y and solve in 30 secs. B

In y miles,how many 1/4 miles are there?

y/(1/4)=4y.

Now for 1 part of y,the charge is x cents.

So remaining parts or 4y-1=(4y-1)*x/5=4xy-x/5

Total charge=x+4xy-x/5= 4xy+4x/5

2 Methods:

(1st Method) Test a Few Y= Whole Number Mile cases to see how many (X/5) Charges need to be applied

case 1: Y = 4

1st --- there are 16 Total (1/4th) Increments in Y = 4 Miles

2nd ---- we need to take away just ONE of these 16 increments ----- 15 Increments will be charged on top of the Flat Fee

Total No. of (1/4th) Miles Charged at (X/5) cents -----> 4 * (Y = 4 miles) - 1 = 15


case 2: Y = 6

1st ----- there are 24 Total (1/4th) Increments in Y = 6 miles

2nd----- we need to take ONE of these 24 Increments away because a Flat Fee is charged for the FIRST --- 23 Increments will be charged in excess of the Flat Fee

Total No. of (1/4th) Mile Increments charged at (X/5) Cents = 4 * (Y = 6) - 1 = 23



Thus, the Total Charge is:

X cent for the FIRST (1/4th) Mile

+

(X/5) cents for Each 1/4th Miles Increment beyond that which, since Y = a Whole Number, is measured by: (4Y - 1)


Total Charge = (X) + (X/5) * (4Y - 1) =

(5X)/(5) + (4XY - X)/(5) =

(4X + 4YX)/(5)

-B-


Method 2:

Set up the Algebraic Equation at the outset:

X cents is charged for the FIRST (1/4th) Mile

How many (1/4th) Mile Increments are remaining if the Total Miles = Y = Whole Number?

[(Y - (1/4) ]
____________
(1/4)

=

[ (4Y - 1) / (4) ] / [ (1) / (4) ]

=

4Y - 1 = No of (1/4th) Mile Increments charged at X/5 Cents per


Total Charge = X + (X/5) * (4Y - 1) ------> Solve as above

As soon as we see variables in the question that are repeated in the answer choices, I'm thinking about Plugging In.
x=5
y=1
Cost is 8 cents.

Check the answer choices:
A. (5+5)/125 Nope.
B. (20 + 20)/5 = 40/5 = 8 Keep it.
C. (20+5)/500 Nope.
D. (20+5)/5 = 25/5 = 5 Nope.
E. 5/25 Nope.

Answer choice B.
Less than 30 seconds and no risk of making a silly mistake with algebra. I truly don't see why students don't value this approach MORE than the "real" math.


ThatDudeKnowsPluggingIn

Hi,

My approach to this question is very different.

Total miles = y
Charges for 1st mile, which is "y/4" = x cents.
Additional miles left = "3y/4". So, charges for them = 3y/4*x/5

Total charges = x + 3xy/20

Where am I going wrong on this? Kindly help me on this.

Hi experts,
I employed the same method. Can someone elaborate on this please as to where we went wrong?
Thanks

Hi think you are being too mechanical in the process.
1 mile has 4 quarters -- then y miles will have 4y quarters

see the solution provided by jefftargettestprep sir...you will know where you are wrong

Charges for the first mile (not 1/4th of the total distance) is x cents.
for 1st mile -> x cents
remaining distance is (y-1/4) miles
dividing the remaining distance into quarter miles will give you 4(y-1/4) = 4y-1
so (4y-1)x X/5 cents + X cents will give you the answer.
Hope this helps!

Anupama_1090

I know now where I was going wrong. Let me share with you.

So, let's of think of 1 year and how many quarters it has. There are 4 quarters in 1 year. So, "Quarter 1 = 1/4". But the only reason I am able to take 4 in the denominator is because I know the value "1 year (It has 4 quarters)". But can I take the value of "Quarter 1 = 1/4" for 2 years or 3 years? I cannot.

In regard to the question, I do not know the value of "y". But I can infer the same thing from the above analysis I did with 1 year. Just to understand replace "y miles" with "y years".

Pointers:

1. y years have 4y quarters (1 year has 4 quarters)
2. Taking "y/4" as I did earlier is assuming that there are only 4 quarters in y years.

From here on, the solution provided by JeffTargetTestPrep becomes much clear. You can follow it if there is any doubt further.

Total Charge= Charge for first quarter-mile + Charge for remaining distance
= x + (4y-1)(x/5)
= (4x + 4xy)/5
Hence B

Can someone please elaborate on this!! I am facing the same issue