A high school sells tickets to a basketball game. Adult tickets cost

9a+5s=927, 9a=927-5s, a=103-5s/9=integer
5s/9=integer, s=multiple(9)
0<5s/9≤103, 0<5s≤9(103), 0<s≤9(103)/5=~185
multiples of 9 from 0 to 185: 180-0/9=20+1=21

Ans (D)

A high school sells tickets to a basketball game. Adult tickets cost $9 and student tickets cost $5. If n tickets are sold for a total revenue of $927, how many possible values are there for n?

A. 18
B. 19
C. 20
D. 21
E. 22

Let Number of adult tickets = $A
Number of student tickets = $S
n = A + S
Now, 9A + 5S = 927 where A and S are positive integers

Here, n would take various values, how many, shall be the answer we are looking for.
So, \(A(max) = \frac{927}{9} = 103\) (S = 0 and since 927 is a multiple of 9)
Thus n(min) = 103 (A + S = 103 + 0)

A would have minimum value when S has maximum value.
A(min) = \(\frac{27}{9} = 3\) where S(min) = \(\frac{900}{5} = 180\)
So, n(max) = 180 + 3 =183

Next lower value of n would be \(\frac{(900 - 45(LCM(5,9))}{5} +\frac{45}{9} + \frac{27}{9} = 171 + 8 = 179\)
Then next lower is \(= \frac{(900 - 2*45(LCM(5,9))}{5} + \frac{45*2}{9} + \frac{27}{9} = 162 + 13 = 175\)
So, values of n differ by 4 (183 -179 = 179 - 175 ...)

Hence, Total possible values of \(n = \frac{(183 - 103)}{4} + 1 = 20 + 1 = 21\)

Answer D.

927=9a+5s, where a=no of adults and s=no of students.

(a,s) = (103,0), (98,9),...,(3,180).
There are 21 sets of (a,s) and consequently, there are 21 unique n=a+s (e.g. 103, 107,...,183).

FINAL ANSWER IS (D)

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From the above information, we can form the following equation.
9y+5x=927, where y represents the number of adult tickets sold and x represents the number of student tickets sold.
We have a linear equation with a y-intercept of 103, and an x-intercept of 185.4.

Since the y-intercept is an integer, it means that 927 is a multiple of 9. This also implies that the possible values of x that satisfy the given equation are integral multiples of 9. In order words, we have the following possible values of x: 0, 9, 18, 27, ...,180. Note that the last term is 9a≤185, where a is an integer and this happens to be 9(20)=180.
The number of unique values of x that correspond to an integral value of y = 1+(180-0)/9 = 1+20 = 21.
There are therefore 21 ordered pairs of x and y, where x and y are positive integers, that satisfy the equation 9y+5x=927.

Hence the answer is D.

9x + 5y = 927
x+y =n
—> \(y = \frac{(927 —9x)}{5}\) — should be integer.

If x =3, then \(y= \frac{(927–27)}{5}= 180\)
If x= 8, then y = integer
If x= 13, then y = integer
....
If x= 103, then y= 0.

—> \(a_1= 3\), \(a_2= 8\) ... \(a_n= 103\)
—> 103 = 3 + (n—1)*5
100= 5*(n—1)
n—1 = 20
—> n= 21

The answer is D

Posted from my mobile device

A high school sells tickets to a basketball game. Adult tickets cost $9 and student tickets cost $5.

If n tickets are sold for a total revenue of $927, how many possible values are there for n?

Let the adult tickets sold be x.

9x + 5(n-x) = 927
4x + 5n = 927
n = (927 - 4x)/5 ; x is a positive integer; n>x

Possible values of (n, x) = {(183,3),(179,8),(175,13),(171,18),(167,23),(163,28),(159,33),(155,38),(151,43),(147,48),(143,53),(139,58),(135,63),(131,68),(127,73),(123,78),(119,83),(115,88),(111,93),(107,98),(103,103)} : 21 possible values of n

IMO D