At a college production of a play, 420 tickets were sold. The ticket prices were $8, $10, and $12, and the total income from ticket sales was $3900. How many tickets of each type were sold if the combined number of $8 and $10 tickets sold was 5 times the number of $12 tickets sold?

There were 220 tickets of $8, 130 tickets of $10 and 70 tickets of $12 were sold.

Step-by-step explanation:

Given,

Total number of tickets sold = 420

Total money collected = $3900

We need to find the number of each type of ticket.

Solution,

Let the number of $8 ticket be 'x'.

Also let the number of $10 ticket be 'y'.

And also let the number of $12 ticket be 'z'.

So the total number of tickets is equal to the sum of the number each type of ticket.

framing in equation form, we get;

Also given the combined number of $8 and $10 tickets sold was 5 times the number of $12 tickets sold.

So we can frame it as;

Now substituting the value of equation 2 in equation 1, we get;

On dividing both side by '6' using division property, we get;

Now we get from equation 1;

 ⇒ equation 3

Also we can say that;

Total money collected is equal to 8 multiplied by number of $8 ticket plus 10 multiplied by number of $10 ticket plus 12 multiplied by number of $12 ticket.

framing in equation form we get;

Now we will substitute value of z in above equation we get;

Now Dividing by 10 we get;

⇒ equation 4

subtracting equation 4 from equation 3 we get;

Dividing both side by 0.2 we get;

Now substituting the value of x in equation 3 we get.

Hence There were 220 tickets of $8, 130 tickets of $10 and 70 tickets of $12 were sold.

9.7 seconds to the nearest tenth.

Step-by-step explanation:

He falls for a distance of 1,500 ft.

We use the equation of motion  s = 1/2 * 32 * t^2   where s = distance and t = time.

1500 = 16 * t^2

t^2  = 93.75

t =  9.7 seconds answer.