Question

Two bags of flour have a total weight of 27lb. If one lb is removed from the lighter bag, the heavier bag will weigh twice as much as the lighter bag. What does the heavier bag weigh

Answer 1

Answer: $17\ lb$

Step-by-step explanation:

The complete exercise is: "Two bags of flour have a total weight of 27 pounds. If 1 pound is added to the heavier bag and 1 pound is removed from the lighter bag, the heavier bag will weigh twice as much as the lighter bag. What does the heavier bag weigh?"

Let be "l" the weight in pounds of the lighter bag and "h" the weigth in pounds of the havier bag.

These bags have a total weight of 27 pounds. Then:

$l+h=27$    [Equations 1]

 If 1 pound is added to the heavier bag and 1 pound is removed from the lighter bag, the heavier bag will weigh twice as much as the lighter bag. This is:

$h+1=2(l-1)$     [Equation 2]

Then, in order to find the weight of the heavier bag, the steps are:

-Solve for "l" from [Equations 1]:

$l+h=27\\\\l=27-h$

- Substitute the equation obtained into the [Equations 2] and solve for "h":

$h+1=2((27-h)-1)\\\\h+1=2(26-h)\\\\h=52-2h-1\\\\3h=51\\\\h=\frac{51}{3}\\\\h=17$

Answer 2

ANSWER:  

The weight of heavier bag is 17.33

SOLUTION:

Let the weight of heavier bag and lighter bag be a , b

Given that, two bags of flour have a total weight of 27lb

Hence we get a + b = 27  --- eqn (1)

And if 1lb is removed from the lighter bag, the heavier bag will weigh twice as much as the lighter bag.

Which means, a = 2(b-1) ---- eqn 2

Now, substitute a value in (1)

2(b-1) + b = 27

On multiplying the terms within the bracket we get

2b -2 + b = 27

3b = 27 + 2

3b = 29

b = $\frac{29}{3}$

substitute the value of “b” in eqn 2

$a=2\left(\frac{29}{3}-1\right)$

$a=2\left(\frac{29-3}{3}\right)$

$=2 \times \frac{26}{3}$

$=\frac{52}{3}=17.33$

Hence, the weight of heavier bag is 17.33