Archibald H.

asked • 06/07/14

How many children tickets were purchased?

At a State Fair, tickets for children cost $3.75 and tickets for adults cost $6.75.  How many tickets for children were purchased if 20 tickets were bought for $90.00.

3 Answers By Expert Tutors

By:

Alex S. answered • 06/07/14

Tutor
5 (12)

UConn Chemical Engineering Student Tutoring Math and Chemistry

See tutors like this
See tutors like this
This can be solved using a system of equations. The total amount of tickets is 20, which is the sum of children and adult tickets, modeled by the equation c + a = 20, where c represents the number of children tickets and a represents the number of adult tickets. The price per ticket is given as well as the total cost, modeled by the equation 3.75c + 6.75a = 90.00. There are two equations with two unknowns:
 
c + a = 20
3.75c + 6.75a = 90.00
 
Solve this system using the substitution method. 
Upvote • 0 Downvote
More
Report

Arthur D. answered • 06/07/14

Tutor
4.9 (283)

Forty Year Educator: Classroom, Summer School, Substitute, Tutor
About this tutor ›
About this tutor ›
($3.75)x+($6.75)(20-x)=$90.00
3.75x+135-6.75x=90
-3x+135=90
-3x=90-135
-3x=-45
x=(-45)/(-3)
x=15 tickets at $3.75
20-15=5 tickets at $6.75
check: 15*$3.75=$56.25
5*$6.75=$33.75
$56.25+$33.75=$90.00
Upvote • 0 Downvote
More
Report

Archibald H.

Thank you Arthur.  I find your answer the easiest, solving by substitution where I can eliminate one variable by relating it to the other where you substituted the number of tickets for x and x-20.
 
Thanks Arthur D.
Archie
Report

06/08/14

Arthur D.

tutor
You're very welcome, Archie.
Arthur D.
Report

06/09/14

Archibald H.

Arthur I have another question which I found in Math10.com which unfortunately is very confusing because of the phrasing of the person that submitted the problem and the lack of explanation of his different steps on solving.  Here is a quote:
Problem 11
A motorman should have taken a distance from town A to town B for exact time. Two hours after he left, he noticed that he covered 80 km and if he keeps that speed he will arrive in B with 15 min delay. So he increased the speed with 10km/h and arrived in town B 36 minutes earlier. Find:
a) The distance between the two towns;
b) The exact time that the motorman should have taken the distance from A to B
Solution:
We mark the distance from A to B with x km. Because the motorman took 80km for 2 hours his speed is V = 80/2 = 40 km/h. With that speed he would have taken the whole distance for x/40 h, delaying with 15min, i.e. the exact time is x/40 – 15/60 h. The rest of the distance (x - 80) km. he took with V = 40 + 10 = 50 km/h.
So the time he took the distance from A to B, is 2 +(x - 80)/50 h. and it is with 36 min. earlier than expected. Therefore the expected time is 2 + (x -80)/50 + 36/60 When we equalize the expressions for the expected time, we get the equation:
x/40 – 15/60 = 2 + (x -80)/50 + 36/60 <=> (x - 10)/40 = (100 + x - 80 + 30)/50 <=> (x - 10)/4 = (x +50)/5 <=> 5x - 50 = 4x + 200 <=> x = 250
So the searched distance is 250 km. The exact time we will find by substituting x with 250 in of the sides of the first equation, for example;
x/40 – 15/60 = 250/40 – 1/4 = 25/4 – 1/4 = 24/4 = 6 hours
Please help!
Archie
Report

06/09/14

Arthur D.

tutor
I solved the problem but it did not appear here.
Report

06/10/14

Karla R. answered • 06/07/14

Tutor
5.0 (137)

Let Me Help You With Your Math!(Elem Math thru Pre-Calc and SAT Math)

See tutors like this
See tutors like this
Let x=#child tickets at $3.75
      y=#adult tickets at $6.75
 
Then setting up 2 equations gives you:
 3.75x+6.75y=90
        x+y      =20
 
Solve by Elimination method(You can eliminate "y" by multiplying 2nd equation by -6.75):
3.75x+6.75y=90 ==> 3.75x+6.75y=90
-6.75(x + y =20)   ==>-6.75x-6.75y =-135
                                 -3.00x          =-45
                         
                                  -3.00x = -45  
                                  -3.00      -3.00
 
                                   x=15 child tickets
                                   y= 5 adult tickets
 
Upvote • 0 Downvote
More
Report

Gary W.

Very clever and easy way to solve this Karla with the elimination method.  You don't see that use often.
Report

06/07/14

Archibald H.

Karla thank you very much for your explanation.  If you can be so kind and again help me in this problem I found in Math10.com I would appreciate it.  As you can see the fellow that posted the problem has definite problems with the english language and for sure is not the best teacher since his steps in the solution are not well explained.
Problem 11
A motorman should have taken a distance from town A to town B for exact time. Two hours after he left, he noticed that he covered 80 km and if he keeps that speed he will arrive in B with 15 min delay. So he increased the speed with 10km/h and arrived in town B 36 minutes earlier. Find:
a) The distance between the two towns;
b) The exact time that the motorman should have taken the distance from A to B
Solution:
We mark the distance from A to B with x km. Because the motorman took 80km for 2 hours his speed is V = 80/2 = 40 km/h. With that speed he would have taken the whole distance for x/40 h, delaying with 15min, i.e. the exact time is x/40 – 15/60 h. The rest of the distance (x - 80) km. he took with V = 40 + 10 = 50 km/h.
So the time he took the distance from A to B, is 2 +(x - 80)/50 h. and it is with 36 min. earlier than expected. Therefore the expected time is 2 + (x -80)/50 + 36/60 When we equalize the expressions for the expected time, we get the equation:
x/40 – 15/60 = 2 + (x -80)/50 + 36/60 <=> (x - 10)/40 = (100 + x - 80 + 30)/50 <=> (x - 10)/4 = (x +50)/5 <=> 5x - 50 = 4x + 200 <=> x = 250
So the searched distance is 250 km. The exact time we will find by substituting x with 250 in of the sides of the first equation, for example;
x/40 – 15/60 = 250/40 – 1/4 = 25/4 – 1/4 = 24/4 = 6 hours
Please help,
Archie
Report

06/09/14

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.