Problem Solving

A telephone company charges x cents for the first minute of a call and charges for any additional time at the rate of y cents per minute. If a certain call costs $5.55 and lasts more than 1 minute, which of the following expressions represents the length of that call, in minutes?

(A) $$\frac{555-x}{y}$$

(B) $$\frac{555+x-y}{y}$$

(C) $$\frac{555-x+y}{y}$$

(D) $$\frac{555-x-y}{y}$$

(E) $$\frac{555}{x+y}$$

OA is C
How do I tackle this question? I need a simple approach method from an Expert, please. Thanks in anticipation

Roland2rule wrote:A telephone company charges x cents for the first minute of a call and charges for any additional time at the rate of y cents per minute. If a certain call costs $5.55 and lasts more than 1 minute, which of the following expressions represents the length of that call, in minutes?

(A) $$\frac{555-x}{y}$$

(B) $$\frac{555+x-y}{y}$$

(C) $$\frac{555-x+y}{y}$$

(D) $$\frac{555-x-y}{y}$$

(E) $$\frac{555}{x+y}$$

OA is C
How do I tackle this question? I need a simple approach method from an Expert, please. Thanks in anticipation

Hello Roland.

Let's take a look.

Let's suppose that the call lasts t minutes. Now, let's write $5.55 as 555 cents, then

for the first minute = x cents.

for the rest of the call = y(t-1).

Hence, $555 cents = x+y(t-1). This implies $$x+y\left(t-1\right)=555\ \Leftrightarrow\ \ \ yt-y=555-x\ \Leftrightarrow\ \ yt=555-x+y\ \Leftrightarrow\ \ t\ =\ \frac{555-x+y}{y}.$$ This is why the correct answer is the option C.

I hope it helps.

I'm available if you'd like a follow-up.