At a local fair, the admission fees are $1.50 for children, $3 for college students, and $5 for adults, On Saturday, 1300 tickets were sold and brought in $3800,

The number of college student tickets sold were 100 less than the number of adults tickets sold,

How many children's tickets were sold?

Number of children = 280 , Number of adults = 460.

Number of students = 360

Step-by-step explanation:

Let x= Number of children, y = number of adults.

Number of students = y-100

Given: At a local fair, the admission fees are $1.50 for children, $3 for college students, and $5 for adults.

1.50 x + 3 (y-100) + 5 y =3800   (i)

x+y-100+y= 1300 (ii)

From (iii)

1.50x + 3y -300+5y = 3800

⇒ 1.50x + 8y = 4100              (iii)

From (ii)

x+2y= 1200       (*)

⇒ 4x + 8y = 4800    (iv)      [multiply 4 on both sides]

Eliminate (iii) from (iv) , we get

2.5x = 700

⇒ x= 280 [ Divide both sides by 2.5]

Put x= 280 in (*), we get

280+ 2 y= 1200

⇒ 2 y= 920

⇒ y = 460

Hence, Number of children = 280 , Number of adults = 460.

Number of students = 460-100 = 360

answer: 7270

step-by-step explanation:

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17