# Question: Taxi Ride

## Comment on Taxi Ride

### Sir can you explain why did

Sir can you explain why did you take the cost as

Cost = x + y(t-2)

what i didn't understood is (t-2) term,

### Let's look at some examples,

Let's look at some examples, and try to identify a pattern.

If the TOTAL trip is 5 miles, then t = 5.
Here, the fare is x dollars for the first 2 miles, and y dollars per mile for the 3 remaining miles.
Where did we get 3 from?
We got the 3 by taking the total trip (t = 5 miles) and SUBTRACTING the first 2 miles.
In other words, t - 2

If the TOTAL trip is 6 miles, then t = 6.
Here, the fare is x dollars for the first 2 miles, and y dollars per mile for the 4 remaining miles.
Where did we get 4 from?
We got the 4 by taking the total trip (t = 6 miles) and SUBTRACTING the first 2 miles.
In other words, t - 2

If the TOTAL trip is 12 miles, then t = 12.
Here, the fare is x dollars for the first 2 miles, and y dollars per mile for the 10 remaining miles.
Where did we get 10 from?
We got the 10 by taking the total trip (t = 12 miles) and SUBTRACTING the first 2 miles.
In other words, t - 2

And so on.

### I plugged in the values x=10

I plugged in the values x=10 y=5 and t=20 and got an out put of 100 dollars but none of the options evaluated to be 100.is there anything wrong with the values i chose

### If x = 10, y = 5, and t = 20,

If x = 10, y = 5, and t = 20, then the output is, indeed, \$100

Plug those values into answer choice E to get:

E) x + y(t - 2) = 10 + 5(20 - 2)
= 10 + 5(18)
= 10 + 90
= 100

Cheers,
Brent

### How do you go about choosing

How do you go about choosing "nice" numbers to plug in? I used x=10, y=6, t=8

### Those numbers will also work.

Those numbers will also work. You'll just get a different output.

Cheers,
Brent

### I find algebraic methods

I find algebraic methods easier though.

First two miles cost is x.
Remaining distance is : t-x
cost of remaining distance = remaining distance*cost/mile = (t-x)*y
so, total distance = x+(t-2)*y

### I agree; in most cases, an

I agree; in most cases, an algebraic approach is usually faster.