All sacks of sugar have the same weight. All sacks of flour also have the same weight, but not necessarily the same as the weight of the sacks of sugar. Suppose that two sacks of sugar together with five sacks of flour weigh no more than $60$ pounds, and that the weight of a sack of flour is no more than $8$ pounds more than the weight of two sacks of sugar. What is the largest possible weight (in pounds) of a sack of flour?

1 answer

Let \( s \) represent the weight of a sack of sugar and \( f \) represent the weight of a sack of flour.

We are given:

  1. The weight of two sacks of sugar together with five sacks of flour weighs no more than 60 pounds: \[ 2s + 5f \leq 60 \]

  2. The weight of a sack of flour is no more than 8 pounds more than the weight of two sacks of sugar: \[ f \leq 2s + 8 \]

We want to maximize \( f \). To do that, we will substitute for \( s \) in terms of \( f \) from the second inequality and use it in the first inequality.

From the second inequality: \[ f \leq 2s + 8 \] Rearranging gives: \[ 2s \geq f - 8 \quad \Rightarrow \quad s \geq \frac{f - 8}{2} \]

Substituting this expression for \( s \) into the first inequality: \[ 2\left(\frac{f - 8}{2}\right) + 5f \leq 60 \] This simplifies to: \[ f - 8 + 5f \leq 60 \] Combining like terms: \[ 6f - 8 \leq 60 \] Adding 8 to both sides: \[ 6f \leq 68 \] Dividing by 6: \[ f \leq \frac{68}{6} \approx 11.33 \]

Since \( f \) must be a whole number (as it represents a sack of flour), the largest integer value for \( f \) is 11. We will check if this satisfies both conditions.

Setting \( f = 11 \): From the second inequality: \[ 2s + 8 \geq 11 \quad \Rightarrow \quad 2s \geq 3 \quad \Rightarrow \quad s \geq 1.5 \]

Let’s find \( s \) when \( f = 11 \): Substituting \( f = 11 \) into the first inequality: \[ 2s + 5(11) \leq 60 \] \[ 2s + 55 \leq 60 \] \[ 2s \leq 5 \quad \Rightarrow \quad s \leq 2.5 \]

So we have: \[ 1.5 \leq s \leq 2.5 \]

The values of \( s \) within this range are permissible. We can select \( s = 2 \) for a simple integer choice.

Now, we need to verify if \( f = 11 \) satisfies both equations:

  1. Using \( s = 2 \):
    • For \( 2s + 5f \): \[ 2(2) + 5(11) = 4 + 55 = 59 \leq 60 \]
    • For \( f \): \[ 11 \leq 2(2) + 8 \quad \Rightarrow \quad 11 \leq 12 \]

Since both conditions are satisfied with \( s = 2 \) and \( f = 11 \), the largest possible weight of a sack of flour is: \[ \boxed{11} \]