Let \( s \) represent the weight of a sack of sugar and \( f \) represent the weight of a sack of flour.
We are given:
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The weight of two sacks of sugar together with five sacks of flour weighs no more than 60 pounds: \[ 2s + 5f \leq 60 \]
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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:
- 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: \[ \boxed{11} \]
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