Question

At the Wardlaw Hartridge School Christmas program, student tickets cost 3\ $and adult tickets cost twice as much. If a total of 200 tickets were sold and$900$$ was collected, how many student tickets were sold?
(a) 50
(b) 75
(c) 100
(d) 150

Answer 1

Let us assume that the number of student tickets is x in number then adult tickets are 200 – x in number because it is given that the total number of tickets are 200. The total money that is collected from selling 200 tickets is 900\ $so multiplying x number of tickets by the cost of 1 student ticket and adding the result of this with the multiplication of 200 – x by cost of 1 adult ticket equates this addition to$900$$. Solving this equation you will get the value of x which is the number of student tickets.

Complete step-by-step answer:
Let us assume that the number of student tickets is x. It is given that total 200 tickets were sold so the number of adult tickets is 200 – x.
The total money collected from the selling is 900\ $. It is given that the cost of a student ticket is$3$$and the cost of an adult ticket is twice that of student ticket which means the cost of the adult ticket is double of$3$ $or equal to$6$$. Multiplying the number of student tickets by cost of 1 student ticket we get,$3\left( x \right)$ $…….Eq. (1) Multiplying the number of adult tickets by cost of 1 adult ticket we get,$\begin{aligned}
& 6\left( 200-x \right) \\
& =\left( 1200-6x \right)$.........Eq.(2)\\\end{aligned}$Adding eq. (1) and eq. (2) and equate it to$900$$ we get,

$\begin{aligned} & \left( 3x+1200-6x \right)=900 \\\ & \Rightarrow 1200-3x=900 \\\ & \Rightarrow 300=3x \\\ \end{aligned}$
Dividing 3 on both the sides of the above equation we get,
$\begin{aligned} & \dfrac{300}{3}=x \\\ & \Rightarrow x=100 \\\ \end{aligned}$
We have assumed above the number of student tickets as x so from the above, we have got the value of x as 100. Hence, the number of student tickets is 100.
Hence, the correct option is (c).

Note: You can check whether the value of x that we have got above is correct or not by multiplying the value of x by 3\$$ and then multiplying 200 – x by 6$ $and then adding the result of both the multiplications and equating it to$900$$. We have got the value of x as 100.$\begin{aligned}
& 100\left(3\right)+100\left(6\right)=900$\&
\Rightarrow300$+600$=900$\&
\Rightarrow900$=900$\\
\end{aligned}$
As you can see that L.H.S is equal to R.H.S in the above equation so the value of x that we have solved in the above solution is correct.