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

Jun 2017
399
6
Lima, Peru
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?
 
Apr 2014
1,013
348
UK
The first equations should be:

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

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

mathman

Forum Staff
May 2007
6,933
775
No matter how complicated you try to make, it still looks like a simple average = 17 kg.
 
Jul 2008
5,248
58
Western Canada
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.

Unbalanced Scale_2s.png

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
Dec 2006
21,482
2,472
Shouldn't the variables $a$ and $b$ be swapped in the balance equations?
 
Jul 2008
5,248
58
Western Canada
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.
 
Apr 2014
1,013
348
UK
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
 
Jul 2008
5,248
58
Western Canada
Yes, that's right. So there's not really enough information given in the original problem, to calculate the weight exactly.