The Admission Fee of the Fair

Let $a$ be the number of adults and $c$ be the number of children, so $a+c=2200.$ Using the ticket cost information, we have $4a+1.5c=5050.$

Substituting $a=2200-c,$ we have $4(2200-c) + 1.5c = 5050,$ so $8800 - 2.5c = 5050.$ Thus, $c = \frac{8800-5050}{2.5} = 1500.$

Solve this problem by ; x + y = 2200 ----- (i) and 4x + 1.5y = 5050 ---- (ii)

put the value of "y = 2200 - x" in equation-ii

we get x = 1500.

Let $C$ be the number of children and $A$ be the number of adults. Then

$C+A=2200$ $\implies$ $A=2200-C$ $\color{#D61F06}(1)$

$1.5C+4A=5050$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ . We have

$1.5C+4(2200-C)=5050$

$1.5C+8800-4C=5050$

$8800-5050=2.5C$

$3750=2.5C$

$\boxed{1500=C}$

simply check all options...only "1500 = num of children" satisfies given conditions..

x + y = 2200 ----- (i) and 4x + 1.5y = 5050 ---- (ii) put the value of "y = 2200 - x" in equation-ii ,we get x = 1500.

1.5x + 4 (2200 -x ) =5050

I thought that ok 1.50 per child and 4.00 for adults well is you divide 1.50 into 5050 you get about 3366 then divide by two saying every adult has one child then you get about 1700 and I picked the closest answer

Solution!

let put x: number of children, y: number of adults

we have: 1.5x + 4y = 5050 (1) x + y= 2200 (2) we multiply the equation number 2 by 4 we get: 4x + 4y = 8800 then (2) - (1) = 2,5x = 5/2x=8800-5050=3750 x=1500 so the solution is: x=1500 children

it was realy simple just we have to check was to multiply the given numbers with 1.5 (cost of children's ticket) and then then subtract the product from 5050 (total money) the number which was remained was to be checked whether it was divisible with 4 (cost of adult's ticket) or not

solving two operating equations we can calculate the required numbers ..... the two operating equations are : ....... c+a=2200 .......& ...... 1.5c+4a=5050........so >> the results are : children = 1500 ..... and adults are : 700

Same equation I used as Abdul Basit Idrees, x+y=2200 and 1.5x+4y=5050 then using elimination method we can get x=1,500.

solving the equation like 4(2200-x)+1.5x=5050 here x is number of children

(1.5)x + (2200-x)4 = 5050 ....Solve for x..

c+a = 2200 (1) 1.5c+4a = 5050 (2) solving eq. 1, 2 c=1500 a=700

x + y = 2200; 1.5 x+ 4y = 5050; Solve the equation will get x= 1500 as Childrens .

by simultaneous equations... 1.5X +4Y = 5050 ----i X + Y = 2200 ----ii where X=number of children and Y=number of adults

X = 1500

=)

Hello,

let x = those children , y = those adults, by simultaneous equation,

x + y = 2200,

x = 2200 - y (1st),

1.5x + 4y = 5050 (2nd),

substitute (1st) into (2nd),

1.5(2200 - y) + 4y = 5050

2.5y = 5050 - 3300

y = 700

x = 2200 - 700 = 1500

therefore, 1500 children were going to the fair...

thanks...

Solve this system of equations x + y = 2200; 1.5x + 4y = 5050;

We got x = 700 (adults), y = 1500 (children)

Lets call adults x. Lets call children y

Given: 2200 people attend the fair. Using x and y write and equation to express this = x + y = 2200

Given: Child admission fee is $1.50

Adult admission fee is $4.00

The total amount collected is $5050.00

4x + 1.5y = 5050

Ok so now we have a system of equations that we can use to solve for x and y

x + y = 2200 AND 4x + 1.5y = 5050

To solve the equations I will demonstrate the substitution method.

Take the first equation and set it equal to x

x + y = 2200

subtract y from both sides

x + y - y = 2200 - y

x = 2200 - y

Now since we have shown that x equals 2200-y we can substitute that for x in the second equation and solve for y

4x + 1.5y = 5050

4(2200-y) + 1.5y = 5050

8800 - 4y + 1.5y = 5050

8800 - 2.5y = 5050

subtract 8800 from both sides

8800 - 8800 - 2.5y = 5050 - 8800

-2.5y = -3750

divide both sides by -2.5

-2.5y/-2.5 = -3750/-2.5

y = 1500

Answer: 1500 children attended the fair