How can I find the weight of a sack of sugar in a two pan scale?

Chemist116

The problem is as follows:

A two pan scale is unbalanced. If a sack of sugar is put over the right pan, the sugar weighs $19\,kg$. If it is put on the left pan, it weighs $15\,kg$. What is the weight of the sack of sugar?

The alternatives given are as follows:

$\begin{array}{ll} 1.&17\,kg\\ 2.&16\,kg\\ 3.&18\,kg\\ 4.&15\,kg\\ \end{array}$

What I've attempted to do was to make this equation:

$\textrm{left pan + sugar = 15}$

$\textrm{right pan + sugar = 19}$

$\textrm{right pan - left pan = 4}$

But in this given situation I'm not given any additional equation to solve this riddle. How exactly can I find this unknown weight? Help please?

weirddave

The first equations should be:

$\textrm{left pan + sugar = 15 + right pan}$

$\textrm{right pan + sugar = 19 + left pan}$

mathman

Forum Staff
No matter how complicated you try to make, it still looks like a simple average = 17 kg.

Yooklid

Not a simple average which would be an arithmetic mean. This is a geometric mean, $\sqrt{15 \times 19}$ which is 16.88, still close enough to 17 that you would pick answer #1.

Weight of sugar: $m_S$

Weighing with sugar in right pan requires 19kg in left pan to balance:
$\dfrac{m_S}{b}=\dfrac{19}{a}$

Weighing with sugar in left pan requires 15kg in right pan to balance:
$\dfrac{m_S}{a}=\dfrac{15}{b}$

It appears initially that there are two equations in three unknowns, but the variables a and b can be reduced to a single variable c = a/b
which results in two equations in two unknowns.

Last edited:

skipjack

Forum Staff
Shouldn't the variables $a$ and $b$ be swapped in the balance equations?

Yooklid

Oops. Yes. $a$ and $b$ should be switched in either the diagram or the formula. Since the choice of $a$ and $b$ is a matter of convention, it doesn't affect the solution.
The weight is still 16.88 kg.

weirddave

If we use the weight of the pans as the cause of the unbalance:

l+s = 15+r
r+s = 19+l

p=r-l

s = 15 +r -l = 15 + p

r-l + s =19
p + s =19

p = 19 - s

s = 15 + 19 - s

2s = 34

s =17

Yooklid

Yes, that's right. So there's not really enough information given in the original problem, to calculate the weight exactly.