Question

Charu’s phone company charges $35\;$ cents each minute of use during peak hours and $15\;$ cents each minute of use during non-peak hours. If Charu’s phone company charged her $7.90\;$ dollars for a half an hour phone call, what is the greatest number of minutes charged at peak hour rates?
A. $7$
B. $13\;$
C. $15\;$
D. $17\;$
E. $21\;$

Answer 1

Hint: Assume the number of minutes in peak hours as $x$ and that of non-peak hours as $y$. When we add both of these variables we need to get $30\;$ minutes as it is given in the question that the phone company charged Charu for half an hour. Now write an equation with the prices and evaluate to get the number of minutes of peak and non-peak hours.

Complete step-by-step solution:
Now let’s assume the below notations.
Let the number of minutes in peak hours in a phone company be $x$ .
Let the number of minutes in non-peak hours in a phone company be $y$ .
Now it is given in the question that the phone company charged Charu for half an hour. It means $30\;$ minutes.
Now we can write the equation as,
$ \Rightarrow x + y = 30$
It is also given that the phone company charges $35\;$ cents for a minute during peak hours which is equal to $35x\;$ . It also charged $15\;$ cents for a minute during non-peak hours which is also equal to $15\;x$ .
The phone company has charged Charu $7.90\;$ dollars, since the rates are in units of cents, we now convert the dollars into cents.
$ \Rightarrow 1\$ = 100$ cents
$ \Rightarrow 7.90\$ = 790$ cents.
Now we can put these quantities into an equation. We can ignore the units while solving the equations.
$ \Rightarrow 35x + 15y = 790$
Now we have two equations which are to be solved. They are,
$ \Rightarrow x + y = 30;35x + 15y = 790$
Consider the first equation, We can write $x + y = 30$ also as $x = 30 - y$
Now substitute this in the second equation.
$ \Rightarrow 35(30 - y) + 15y = 790$
Now expand the brackets by multiplying the constant outside them.
$ \Rightarrow 1050 - 35y + 15y = 790$
Now perform operations on the same degree terms.
$ \Rightarrow 1050 - 20y = 790$
Again, evaluate the constant in the equation.
$ \Rightarrow 1050 - 790 = 20y$
$ \Rightarrow 260 = 20y$
Now divide the whole equation with $20$
$ \Rightarrow \dfrac{{260}}{{20}} = y$
$ \Rightarrow y = 13$
Now put this value back in the first equation to get the value of $x$
$ \Rightarrow x + 13 = 30$
$ \Rightarrow x = 17$

$\therefore $ The number of minutes in peak hours is $17\;$ minutes and the number of minutes in non-peak hours is $13\;$ minutes which is option $D$ .

Note: Make sure the units are the same for the quantities while putting them in a polynomial equation. If they aren’t of the same units, convert them first and then write the equation. Never forget to write the units beside the constant after the entire calculation or evaluation is done.