Question

5. You are making your weekly mean plans and are working with the following constraints: It costs $8 to go out to dinner. It costs $5 to go out to lunch. You want to go out to dinner at least as many times as you go out to lunch. You can spend at most $42.

What is the greatest number of meals you can eat out?
3
4
5
6

Answer 1

The correct answer for this question would be option D.6. Given that per dinner, you can spend $8 and per lunch is $5. Given that you can spend at most $42, you can only eat out 6 times. 3 times for lunch and 3 times for dinner.
Since 8 times 4 would be $24 in total, and 5 x 3 would be $15 in total, so the overall would be $39. Hope this is the answer that you are looking for.

Answer 2

Let

x-------> the number of dinner

y-------> the number of lunch

we know that

`8x+5y \leq 42` -------> equation A

`x=y` ------> equation B

Substitute equation B in equation A

`8[y]+5y \leq 42`

`13y \leq 42`

`y \leq 42/13`

`y \leq 3.23`

so

the greatest number of lunch is
`y=3`

`x=y`

Hence

the greatest number of dinner is
`x=3`

therefore

the greatest number of meals is

`x+y=3+3=6`

the answer is

`6`