# Problem of the Week Problem B and Solution What's for Lunch?

## Problem

Sanji and his 34 classmates always bring sandwiches to school for lunch. One day, everyone brought either a jelly sandwich, or a ham and cheese sandwich, or a tuna salad sandwich.
If the number of students who brought jelly sandwiches ($$J$$) was twice the number who brought ham and cheese ($$H$$), and four times the number who brought tuna salad ($$T$$), how many sandwiches were there of each type?
Hint: What should the total number of sandwiches be?
You may find the table useful for the ‘guess and check’ method.

## Solution

Solution 1:

There are a total of $$34 + 1 = 35$$ students. Therefore, there must be 35 sandwiches. The completed table below reveals that the only possible combination is 5 tuna salad, 10 ham and cheese, and 20 jelly sandwiches, in order to give the correct total number of 35 sandwiches.

$$T$$ $$H$$ $$J$$ Total
1 2 4 7
2 4 8 14
3 6 12 21
4 8 16 28
5 10 20 35
6 12 24 42

Solution 2:

An alternate, algebraic solution is presented below. The algebra used in this solution may be beyond what students at this age have typically seen.
We are given that $$J = 4 \times T$$ and $$J = 2 \times H$$. This means that $$H = 2 \times T$$.
We now have $J + H + T = 4 \times T + 2 \times T + T= 7 \times T$ But $$J+H+T = 7\times T$$ is also equal to the total number of students.

Thus, $$35 = 7 \times T$$, which gives $$T = 5$$.
Since $$T=5$$, we get $$H =2\times 5 = 10$$, and $$J = 4 \times 5 =20$$.
Therefore, 5 students brought a tuna salad sandwich, 10 students brought a ham and cheese sandwich, and 20 students brought a jelly sandwich.