MATH SOLVE

4 months ago

Q:
# 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?

Accepted Solution

A:

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 [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.