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The cost of children’s ticket is $ 5
<h3><u>Solution:</u></h3>Let "c" be the cost of one children ticket
Let "a" be the cost of one adult ticket
Given that adult ticket to a museum costs 3$ more than a children’s ticket
<em>Cost of one adult ticket = 3 + cost of one children ticket</em>
a = 3 + c ------ eqn 1
<em><u>Given that 200 adult tickets and 100 children tickets are sold, the total revenue is $2100</u></em>
200 adult tickets x cost of one adult ticket + 100 children tickets x cost of one children ticket = 2100
200a + 100c = 2100 ------ eqn 2
<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "c"</u></em>
Substitute eqn 1 in eqn 2
200(3 + c) + 100c = 2100
600 + 200c + 100c = 2100
600 + 300c = 2100
300c = 1500
<h3>c = 5</h3>Thus the cost of children’s ticket is $ 5
The statement no solutions represents the simplified form of the given equation 6x + 14 = 3(2x + 5). Hence Option A is correct
<u>Solution:</u>
Given, equation is 6x + 14 = 3(2x + 5).
We have to find the correct options that represents the simplified form of the given equation and correctly describes the solution.
So, now let us simplify the given equation
⇒6x + 14 = 3(2x + 5) ⇒ 6x + 14 = 6x + 15 ⇒ 6x – 6x + 14 = 15 ⇒ 14 ≠ 15
As L.H.S not equals with R.H.S, no value of x can satisfy the equation and there will be no solution for given equation.
Hence, option A is correct.