Question

Tickets for the cinema cost £a for adults and £c for children. Mr and Mrs Kennedy go to the cinema with their four children. The total cost is £27. Mrs Jones takes her three children. The total cost is £17. Write two equations in a and c and solve them to find the cost of an adult ticket and of a child ticket.

Answer 1

Tickets for the cinema cost for adults= £a

Tickets for the cinema cost for adults= £c

ATQ

Mr and Mrs Kennedy go to the cinema with their four children.

So 2a + 4c = 27 .... (1)

Mrs Jones takes her three children.

So a + 3c = 17 ....(2)

solving equation 1 and 2 by eliminating we get

2a + 6c = 34 .... (2)

-2a + 4c = 27 .... (1)

2c = 7

c = 7/2 £

a = 13/2 £

Answer 2

$\bold{\underline{\underline{Answer:}}}$

Adult ticket = £ 13/2

Child ticket = £ 7/2

$\bold{\underline{\underline{Step\:-\:by\:-\:step\:explanation:}}}$

Given :

• Tickets for the cinema cost £a for adults and £c for children.
• Mr and Mrs Kennedy go to the cinema with their four children. The total cost = £27.
• Mrs Jones takes her three children. The total cost = £17.

To find :

• The cost of an adult ticket
• Cost of a child ticket.

Solution :

Let the cost of an adult ticket be a.

Let the cost of child ticket be c.

$\bold{\underline{\underline{As\:per\:the\:first\:condition:}}}$

• Mr and Mrs Kennedy go to the cinema with their four children. The total cost = £27.

Constitute the given condition in equation,

$\bold{2a\:+4c\:=27}$ ----> (1)

[Considering Mr and Mrs Kennedy as adults (2a) and their 4 children (4a)]

$\bold{\underline{\underline{As\:per\:the\:second\:condition:}}}$

• Mrs Jones takes her three children. The total cost = £17.

Constitute the given condition in equation,

$\bold{a\:+3c\:=17}$ ---> (2)

[Considering Mrs Jones (a) as adult and her three children (3c)]

Multiplying equation (2) by 2,

$\bold{2a\:+\:6c\:=\:34}$ ---->(3)

Solve equations 2 and 3, simultaneously by elimination method.

Subtract equation 3 from 2,

....+ 2a + 6c = 34

- ( + 2a + 4c = 27 )

-----------------------------

2c = 7

$\implies$ $\bold{c\:=\:{\frac{7}{2}}}$

Substitute $\bold{c\:=\:{\frac{7}{2}}}$ in equation 1,

$\implies$ $\bold{2a\:+\:4c\:=\:27}$

$\implies$ $\bold{2a\:+\:4\times\:{\frac{7}{2}}\:=\:27}$

$\implies$ $\bold{2a\:+\:{\frac{28}{2}}\:=\:27}$

$\implies$ $\bold{2a\:+\:14\:=\:27}$

$\implies$ $\bold{2a\:+\:=\:27\:-14}$

$\implies$ $\bold{2a\:+\:=\:13}$

$\implies$ $\bold{a\:=\{\frac{13}{2}}$