1. Two word problems.

First:

A company purchases a billboard on the side of the highway for $10,000. For each person that calls the number on the billboard and fills out an application, the company receives$100. 15% of people who call the number actually fill out an application. What is the minimum number of calls that need to be made so that the company can make a profit off the billboard?

Second:

A company sends out an email to 1,000,000 people to register for an event. The email has a 8% open rate, 5% click-through rate, 200 registrations at $10/registration. What is the cost per click? I had a hard time with these two at a recent interview. Thanks in advance. 2. Re: Help me solve two word problems ... For the first one, Let x be the amount of people filling out an application Let y be the total amount of people calling Solve the system: 100x = 10000 0.15y = x What do you get? 3. Re: Help me solve two word problems ... x = 100 y = 667 people Correct? 4. Re: Help me solve two word problems ... I'm pretty sure. But why don't you check it yourself? I.e. is 100 equal to 15% of 667? 5. Re: Two word problems. Hello, llckll! A company purchases a billboard on the side of the highway for$10,000.
For each person that calls the number on the billboard and fills out an application,
15% of people who call the number actually fill out an application.

What is the minimum number of calls that need to be made
so that the company can make a profit off the billboard?

$\text{Let }x\text{ = number of calls made.}$

$\text{15\% of them will fill out an application.}$
. . $0.15x\text{ applications are filled.}$

$\text{Each application earns \100.}$
. . $\text{The company gets: }\:100(0.15x) \,=\,15x\text{ dollars}$

$\text{This amount must be at least \10,000.}$
. . $15x\:\ge \:10,\!000 \quad\Rightarrow\quad x \:\ge\:666\tfrac{2}{3}$

$\text{Therefore, there must be at least 667 calls.}$

6. Re: Two word problems.

WHY are about 99% of problems called "word problems"?!

7. Re: Two word problems.

Do you have a confidence interval for that proportion Wilmer?

8. Re: Two word problems.

Originally Posted by pickslides
Do you have a confidence interval for that proportion Wilmer?
That sounds like a "word problem"!