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

$6$