Question 951209
THE FIFTH GRADER SAYS
Not counting that extra carbon atom (the "one more atom of carbon that oxygen" as the problem says,
there are {{{45-1=44}}} atoms.
Among those {{{44}}} atoms,
for each oxygen atom there are 2 hydrogen atoms and one carbon atom,
forming a group of 4 atoms.
so, among those {{{44}}} atoms, there are
{{{44/4=11}}} such groups of 4 atoms.
At {{{1}}} oxygen atom per group, there is a total of
{{{highlight(11)}}} oxygen atoms.
At {{{2}}} hydrogen atoms per group, there is a total of
{{{2*11=highlight(22)}}} hydrogen atoms.
At {{{1}}} carbon atom per group, there is a total of
{{{11}}} carbon atoms in the 4-atom groups,
plus that extra carbon atom, for a total of
{{{11+1=highlight(12)}}} carbon atoms.
 
THE CHEMIST says that may be sucrose (table sugar),
but it could also be lactose (milk sugar),
or any other disaccharide.
 
THE ALGEBRA TEACHER SAYS you need to apply what we have been learning about systems of linear equations.
Let's first define variables:
{{{c}}}= number of carbon atoms;
{{{h}}}= number of hydrogen atoms, and
{{{o}}}= number of oxygen atoms.
The phrase "twice as many atoms of hydrogen as oxygen" translates as
{{{h=2o}}} .
The phrase "one more atom of carbon that oxygen" translates as
{{{c=o+1}}} .
The phrase "there are 45 atoms" translates as
{{{h+c+o=45}}} .
Our system is
{{{system(h=2o,c=o+1,h+c+o=45)}}}
Substituting the expressions for {{{h}}} and {{{c}}} given by the first two equations into the third one, we get
{{{2o+(o+1)+o=45}}}<-->{{{2o+o+1+o=45}}}<-->{{{4o+1=45}}}<-->{{{4o=45-1}}}<-->{{{4o=44}}}<-->{{{o=44/4}}}-->{{{highlight(o=11)}}} oxygen atoms.
Substituting the value found for {{{o}}} into the first two equations, we get
{{{h=2*11}}}<--->{{{highlight(h=22)}}} hydrogen atoms, and
{{{c=11+1=highlight(12)}}} carbon atoms.