Jason and Kyle both choose a number from 1 to 10 at random. What is the probability that both numbers are odd?
2 answers:
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Number of child tickets bought is 20
<h3><u>Solution:</u></h3>Given that It cost 5 dollars for a child ticket and 8 dollars for a adult ticket
cost of each child ticket = 5 dollars
cost of each adult ticket = 8 dollars
Let "c" be the number of child tickets bought
Let "a" be the number of adult tickets bought
Total tickets sold were 110 bringing in 820 dollars
<em>Number of child tickets bought + number of adult tickets bought = 110</em>
c + a = 110 ----- eqn 1
<em><u>Also we can frame a equation as:</u></em>
Number of child tickets bought x cost of each child ticket + number of adult tickets bought x cost of each adult ticket = 820
5c + 8a = 820 -------- eqn 2
Let us solve eqn 1 and eqn 2 to find values of "c" and "a"
From eqn 1,
a = 110 - c ------ eqn 3
Substitute eqn 3 in eqn 2
5c + 8(110 - c) = 820
5c + 880 - 8c = 820
-3c = - 60
c = 20
Therefore from eqn 3,
a = 110 - 20 = 90
a = 90
Therefore number of child tickets bought is 20