Question

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?

Answer 1

Answer:

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

Step-by-step explanation:

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 $\frac{800}{x}$

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  $\frac{800}{x-2}$

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

i.e.  $\frac{800}{x-2}-\frac{800}{x}=20$

Divide the equation by 20,

$\frac{40}{x-2}-\frac{40}{x}=1$

Taking LCM,

$\frac{40x-40(x-2)}{x(x-2)}=1$

$\frac{40x-40x+80}{x(x-2)}=1$

$\frac{80}{x^2-2x}=1$

Cross multiply,

$80=x^2-2x$

$x^2-2x-80=0$

Solving by middle term split,

$x^2-10x+8x-80=0$

$x(x-10)+8(x-10)=0$

$(x-10)(x+8)=0$

$x=10,-8$

Rejecting x=-8.

Accepting x=10.

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