In a cinema hall, 300 tickets were sold for a
total of 532,000. If the tickets were of two
denominations of 80 and 120, how many
of each were sold?
Answers
Step-by-step explanation:
\sf\large\underline{Given:}
Given:
\rm{Number\:of\: tickets=300}Numberoftickets=300
\rm{Total\:collection\:amount=Rs.32000}Totalcollectionamount=Rs.32000
\sf\large\underline{Solution:}
Solution:
\rm{\implies Let,\:The\: N\:of\:person\:who\: bought\:Rs.80\: tickets=x}⟹Let,TheNofpersonwhoboughtRs.80tickets=x
\rm{\implies The\: N\:of\: person\:who\: bought\:Rs.120\: tickets=y}⟹TheNofpersonwhoboughtRs.120tickets=y
\sf\underline{Given\:in\:1st\:case:}
Givenin1stcase:
\rm{\implies sum\:of\:2\: tickets=Total\:tickets}⟹sumof2tickets=Totaltickets
\rm{\implies x+y=300-------(ii)}⟹x+y=300−−−−−−−(ii)
\sf\underline{Given\:in\:2nd\:case:}
Givenin2ndcase:
\rm{\implies Sum\:of\:2\: denomination=32000}⟹Sumof2denomination=32000
\rm{\implies 80x+120y=32000-----(i)}⟹80x+120y=32000−−−−−(i)
Now, solving equation i and ii here]
\rm{\implies x+y=300}⟹x+y=300
\rm{\implies 80x+120y=32000}⟹80x+120y=32000
in eq ii multiplying by 80 ]
\rm{\implies 80x+120y=32000}⟹80x+120y=32000
\rm{\implies 80x+80y=24000}⟹80x+80y=24000
By solving we get here]
\rm{\implies 40y=8000}⟹40y=8000
\rm{\implies y=200}⟹y=200
Putting the value of y=200 in eq ii]
\rm{\implies x+y=300}⟹x+y=300
\rm{\implies x+200=300}⟹x+200=300
\rm{\implies x=300-200}⟹x=300−200
\rm{\implies x=100}⟹x=100
\sf\large{Hence,}Hence,
\rm{\implies The\: N\:of\:person\:who\: bought\:Rs.80\: tickets=100}⟹TheNofpersonwhoboughtRs.80tickets=100
\rm{\implies The\: N\:of\: person\:who\: bought\:Rs.120\: tickets=200}⟹TheNofpersonwhoboughtRs.120tickets=200