12.5k views
4 votes
Zoé is working two summer making $7 per hour babysitting and making $15 per hour clearing tables. In given week, She can make a maximum of 14 total hours in must earn at least $130. If Zoe worked 4 hours babysitting, determine all possible values for the number of who I was clearing the tables that she must work to meet her requirements. Your answer should be a comma separate list of values. there are no possible solutions, submit an empty answer

User Ingconti
by
5.9k points

1 Answer

0 votes

Answer:

Zoe can clear tables for 7, 8 , 9 and 10 hours to meet her requirements.

Explanation:

Here according to the question:

Earning per hour of Zoe from babysitting = $7 per hour

Earning per hour of Zoe from clearing tables = $15 per hour

Let us assume the number of hours Zoe babysits = m hours

Also, let us assume the number of hours Zoe clears table = n hours

Now,as given : She can make a maximum of 14 total hours

SO, the number of hours worked at ( Babysitting + Clearing Tables) ≤ 14

or, m + n ≤ 14 ....... (1)

Now, the total earning after working m hours babysitting

= m x ( Cost of babysitting per hour) = m ($7) = 7 m

And, the total earning after working n hours clearing tables

= n x ( Cost of clearing tables per hour) = n ($15) = 15 n

Now, total earning after worming both jobs = 7 m + 15 n

But, She can must earn at least $130.

7 m + 15 n ≥ 130 ......... (2)

Now, given : Number of hours Zoe babysits = 4 hours

⇒ m = 4

So, putting the value of m = 4 in (1) , we get: n = 14 - 4 = 10

Also, putting m = 4, n = 10 in (2) , we get:

7 m + 15 n = 7 (4) + 15(10) = 28 + 150 = 178 > 130

Hence, (4,10) is the SOLUTION OF THE GIVEN SYSTEM.

now, m = 4, so, n can be 1, 2, 3, 4, 5, .... 9 (any of the values)

for, m = 4, n = 1: 7 m + 15 n = 28 + 15 = 43 ≯ 130

⇔ Equation (2) is not satisfied for (4,1)

Checking for all values of (m,n), we get:

The possible solutions are: (4 ,7) , (4,8) , (4,9) and (4, 10).

Hence, she can clear tables for 7, 8 , 9 and 10 hours to meet her requirements.

User Don Baechtel
by
6.8k points