Question 935070
c for price of child ticket, g for price of grownup ticket.


{{{160c+90g=1600}}} accounting for sales revenue.


{{{16c+9g=160}}}, and expect that {{{c<g}}}.


{{{16c=-9g+160}}}
{{{c=-(9/16)g+10}}}
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substitute into the inequality:
{{{-(9/16)g+10<g}}}
{{{10<g+(9/16)g}}}
{{{160<16g+9g}}}
{{{160<25g}}}
{{{g>160/25}}}
{{{g>32/5}}} which is $6.40. This is a lower boundary for adult ticket price.


Now solve the equation for g instead; <i>MAYBE</i> it would help further(?).
{{{9g=160-16c}}}
{{{g=160/9-(16/9)c}}}
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substitute to the inequality,
{{{c<g}}}
{{{c<160/9-(16/9)c}}}
{{{9c<160-16c}}}
{{{9c+16c<160}}}
{{{25c<160}}}
{{{c<160/25}}}, which is the same boundary as for adult price, but just in the other direction.


As finely grained as you can find solutions, the ticket prices MUST be in whole penny quantities, including 0.  Children's tickets are priced as BELOW $6.40, and adult tickets are HIGHER than $6.40.  Several solutions to the two dimensional problem are possible as long as you take perfect hundredth of a dollar accuracy.