Word problems i been having a problem with.

Question

At a college function, a total of 1,800 tickets were sold.  The tickets were priced at $5, $10 and $15, and the money collected that day was $1,900.  The sum of the number of $5 and $15 tickets sold were twice the amount of $10 tickets sold.  Find the number of each type of ticket sold for the college function.

Answer 1

I started participating in the WyzAnt Answers Forum to improve my information about how modern students learn, but I have learned more about the variety of tutors than I have learned about the learning styles of Students.

This problem, like a document yet to be spell-checked, has a typo.  Now, spell-check and grammar-check can autocorrect or suggest a correction (and AI is even better than that).  Using probabilities, we start by assuming the cost of tickets ($5, $10, $15) is likely correct, then the relationship of the sum of $5 and $10 tickets to the number (note: “amount” is ambiguous) of $10 tickets is correct.  So, that leaves possible errors in the number of tickets or else the total of the money.

Tutors (and computers) can easily calculate values both ways and conclude that “positive integer amounts” of tickets and cents are required (assume:  correct accounting but typo in reporting college function information).  So, $19,000 is better than 180 tickets [calculate it!] (or other typos).

Let:
        V = Number of $5 tickets
        X = Number of $10 tickets
        F = Number of $15 tickets          (I like single letters, so think some Roman Numerals)

Thus, the problem has “givens:” 
        V     +     X     +    F    =       1800           (eq1)
        5V   +   10X   +    15F    =  19000            (eq2)
        V     +                  F     =      2X                 (eq3)

Use elimination method (eq1 – eq3):
                     -X             =     2X – 1800
                        -3X        =   -1800            (subtract 2X from both sides)
                           X = 600                         (divide both sides by -3)

Now, we must use eq2, so calculate ( eq2   -  5*eq3):
                 5V   +   10X   +    15F    =  19000
                 5V   -    10X   +     5F     =    0       (modified eq3)
              ------------------------------------------
                           20X      +    10F    =  19000
                            12000   + 10F    = 19000       (using X=600)
                                       10F =  7000                 (subtract 12000 from both sides)
                                         F = 700

Now, using eq1:
                        V  +  600  +  700   = 1800
                        V  =  500                             (subtract 1300 from both sides)

Checking (very important):
     Is     500 + 600 + 700 = 1800   ?     yes

     Is    5(500) + 10(600) + 15(700) = 19000   ?
           2500     + 6000   +  10500  = 19000   ?     yes

     Is   500  +   700   =   2(600)    ?      yes

Answer 2

There seems to be an incorrect amount posted for the total amount of the tickets. It may be another value instead of $1900.  Then on average for 3 sets of tickets totalling 1800....600 of each set would be sold....$5×600=$3000, $15×600=$9000, $10×600=$6000....these amounts would total $18000...the number of tickets sold for $5and $15 total 1200....which is twice that of the $10 tickets sold of 600....the total amount of money for the tickets would also be twice...$12000 for the $5and $15 tickets...and $6000 for the $10 tickets...but if the total value is really $1900...I stand corrected...

Answer 3

I think you have a typo? Even if all of the tickets were the cheapest ones, that's $5 each, and you would make 1800(5), which would be $9000. Are you missing a 0 on the $1900? Something is wrong with this problem.

 

 

Answer 4

N5=Number of $5 tickets sold.

N10=Number of $10 tickets sold.

N15=Number of $15 tickets sold.

 

Equation 1:     (N5)+(N10)+(N15)=1800

 

Equation 2:     5(N5)+10(N10)+15(N15)=1900

 

Equation 3:     (N5)+(N15)=2(N10)

 

From 3:     (N5)=2(N10)-(N15)

 

Substitute in Equation 1 for (N5)

 

2(N10)-(N15))+(N10)+(N15)=1800

 

3(N10)=1800

 

(N10)=600

 

Substitute this in first two equations to get the following two equations in two unknowns.

 

(N5)+(N15)=1200

 

5(N5)+15(N15)=-4100

 

Divide second of these equations by 5

 

(N5)+3(N15)=-820

 

Subtract (N5)+(N15)=1200 from this equation

 

2(N15)=-820+1200=380

 

(N15)=190

 

Substitute for (N10) and (N15) in Equation 1

 

(N5)+600+190=1800

 

(N5)=1010