Every sack of sugar has the same weight. Every sack of flour has the same weight, but not necessarily the same as the weight of the sacks of sugar. Suppose that three sacks of sugar together with four sacks of flour weighs no more than 50 pounds, and that the weight of two sacks of flour is no more than 13 pounds more than the weight of three sacks of sugar. What is the largest possible weight (in pounds) of a sack of flour?
3s+4f <= 50
2f <= 13+3s
so,
4f <= 6s+26
3s+4f <= 3s+6s+26 <= 50
9s <= 24
s <= 8/3
So, since the smallest s gives the largest f in
3s+4f <= 50
8 + 4f <= 50
4f <= 42
f <= 21/2
Max f is 21/2
To find the largest possible weight of a sack of flour, let's break down the problem step by step:
Step 1: Assign variables
Let's use "s" to represent the weight of a sack of sugar and "f" to represent the weight of a sack of flour.
Step 2: Form equations based on the information given
From the given information, we can form the following equations:
3s + 4f ≤ 50 (equation 1) - Three sacks of sugar together with four sacks of flour weighs no more than 50 pounds.
2f ≤ 3s + 13 (equation 2) - The weight of two sacks of flour is no more than 13 pounds more than the weight of three sacks of sugar.
Step 3: Solve the equations
We have two inequalities here, so we can start by solving equation 2 for f:
2f ≤ 3s + 13
Divide both sides by 2:
f ≤ (3s + 13)/2
Now we substitute this value of f into equation 1:
3s + 4f ≤ 50
3s + 4[(3s + 13)/2] ≤ 50
Multiply both sides by 2:
6s + 4(3s + 13) ≤ 100
6s + 12s + 52 ≤ 100
18s + 52 ≤ 100
18s ≤ 48
s ≤ 48/18
s ≤ 8/3
Step 4: Find the largest possible weight of a sack of flour
Now we can substitute the value of s back into equation 2:
2f ≤ 3s + 13
2f ≤ 3(8/3) + 13
2f ≤ 8 + 13
2f ≤ 21
f ≤ 21/2
f ≤ 10.5
Therefore, the largest possible weight of a sack of flour is 10.5 pounds.