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.

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Inactive Tutor answered • 06/07/14

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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. 
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Arthur D. answered • 06/07/14

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($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
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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
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06/08/14

Arthur D.

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You're very welcome, Archie.
Arthur D.
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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
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06/09/14

Arthur D.

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I solved the problem but it did not appear here.
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06/10/14

Inactive Tutor answered • 06/07/14

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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
 
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Inactive Tutor

Very clever and easy way to solve this Karla with the elimination method.  You don't see that use often.
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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
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06/09/14

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