A group of friends decided to divide the $800 cost of a trip equally among themselves. When two of the friends decided not to go on the trip, those remaining still divided the $800, 800 cost equally, but each friend’s share of the cost increased by $20. How many friends were in the group originally?

Given : A group of friends decided to divide the $800 cost of a trip equally among themselves. When two of the friends decided not to go on the trip, those remaining still divided the $800, 800 cost equally, but each friend’s share of the cost increased by $20.

To find : How many friends were in the group originally?

Solution :

Let the number of friends be 'x'.

A group of friends decided to divide the $800 cost of a trip equally among themselves.

i.e. Cost of trip for each friend is [tex]\frac{800}{x}[/tex]

When two of the friends decided not to go on the trip, those remaining still divided the $800.

i.e. Cost of trip for each friend is [tex]\frac{800}{x-2}[/tex]

The increase in cost for each remaining friend is $20.

i.e. [tex]\frac{800}{x-2}-\frac{800}{x}=20[/tex]

Divide the equation by 20,

[tex]\frac{40}{x-2}-\frac{40}{x}=1[/tex]

Taking LCM,

[tex]\frac{40x-40(x-2)}{x(x-2)}=1[/tex]

[tex]\frac{40x-40x+80}{x(x-2)}=1[/tex]

[tex]\frac{80}{x^2-2x}=1[/tex]

Cross multiply,

[tex]80=x^2-2x[/tex]

[tex]x^2-2x-80=0[/tex]

Solving by middle term split,

[tex]x^2-10x+8x-80=0[/tex]

[tex]x(x-10)+8(x-10)=0[/tex]

[tex](x-10)(x+8)=0[/tex]

[tex]x=10,-8[/tex]

Rejecting x=-8.

Accepting x=10.

Therefore, The number of friends were in the group originally is 10.