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The charge for a telephone call between City R and City S

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by VJesus12 » Mon Mar 19, 2018 4:01 am
The charge for a telephone call between City R and City S is $0.42 for each of the first 3 minutes and $0.18 for each additional minute. A certain call between these two cities lasted for x minutes, where x is an integer. How many minutes long was the call?

(1) The charge for the first 3 minutes of the call was $0.36 less than the charge for the remainder of the call.

(2) The total charge for the call was $2.88.

The OA is D.

I liked this question but I couldn't solve it. May someone shows me how to solve it? Thanks in advanced.
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Source: — Data Sufficiency |
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by Vincen » Mon Mar 19, 2018 6:30 am
VJesus12 wrote:The charge for a telephone call between City R and City S is $0.42 for each of the first 3 minutes and $0.18 for each additional minute. A certain call between these two cities lasted for x minutes, where x is an integer. How many minutes long was the call?

(1) The charge for the first 3 minutes of the call was $0.36 less than the charge for the remainder of the call.

(2) The total charge for the call was $2.88.

The OA is D.

I liked this question but I couldn't solve it. May someone shows me how to solve it? Thanks in advanced.
Let's write the things we know:
- $0.42 for each of the first 3 minutes.
- $0.18 for each additional minute.
- A Certain call lasted for x minutes, where x is an integer.

Now let's use each statement.

(1) The charge for the first 3 minutes of the call was $0.36 less than the charge for the remainder of the call.

We have that $$\left(First\ 3\ \text{minutes}\right)\cdot0.42\ =\left(Rest\ of\ the\ call\right)\cdot0.18-0.36$$ $$3\cdot0.42\ =\left(x-3\right)\cdot0.18-0.36\ \Rightarrow\ \ 1.26+0.36=\left(x-3\right)\cdot0.18\ \Rightarrow\ \ \frac{1.62}{0.18}=x-3$$ $$\Rightarrow\ x=9+3\ =12.$$ Since we got the value of x, this statement is SUFFICIENT.

(2) The total charge for the call was $2.88.

$$\left(First\ 3\ \text{minutes}\right)\cdot0.42\ +\left(Rest\ of\ the\ call\right)\cdot0.18=2.88$$ $$3\cdot0.42\ +\left(x-3\right)\cdot0.18=2.88\ \Rightarrow\ \ \left(x-3\right)\cdot0.18=2.88-1.26\ \Rightarrow\ x-3=\frac{1.62}{0.18}$$ $$\Rightarrow\ x=9+3\ =12.$$ Here we get the value of x, hence this statement is SUFFICIENT.

Hence, the answer is D.

I hope it helps you.
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by ErikaPrepScholar » Sat Mar 31, 2018 3:32 pm
The charge for a telephone call between City R and City S is $0.42 for each of the first 3 minutes and $0.18 for each additional minute. A certain call between these two cities lasted for x minutes, where x is an integer. How many minutes long was the call?

(1) The charge for the first 3 minutes of the call was $0.36 less than the charge for the remainder of the call.

(2) The total charge for the call was $2.88.

Quick note on this problem: we don't need to solve for exactly how long the call was. We only need to determine that we can solve it. We know that the first 3 minutes of the call cost $0.42 each, or $1.26 in total. We know that the remaining minutes of the call cost $0.18 apiece. This means that if we know the cost of the minutes in the call after 3 minutes, we can divide by $0.18 to get the number of minutes after 3 minutes. Then we can add 3 to get the total minutes in the call.

So if we can find the cost of the minutes in the call after 3 minutes, the statement will be sufficient.

Statement 1

This tells us that the charge for the minutes in the call after 3 minutes is $1.26 + $0.36 = $1.62. We could divide by $0.18 to get minutes after 3 and then add 3, but we don't need to - we know that we could. Sufficient.

Statement 2

This tells us that the charge for the minutes in the call after 3 minutes is $2.88 - $1.26 = $1.62. Again, we know we can solve. Sufficient.
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