Thread: quadratic problems

i need some help with these two quadratic problems:

- A family drives 240 km to their cottage. They can save fuel by reducing their average speed by 20 km/h and this will mean that the trip will take one hour longer. What is this lower average speed?

- JB bought some balls for $600. If the cost for each ball had been 25 cents less, she would have had 10 more balls for her $600. How much did she pay for each ball.

can you help me set up the equation?
thank you for your expertise

distance=240 km
let time= x hr
time speed= 240/x km/hr

new speed= (240/x)-20 kn/he
new time= x+1 hr
but distance is same = 240 km

so


240=((240/x)-20))*x+1
on solving.
x=3 or -4
-4 not-possible
time =3 hr.
initial speed = 80 km/hr
new lower speed =(80-20)km/hr=60 km/hr.

i guess thats the answer?

thanks so much
I set up a ratio but it kept coming out wrong

total money= $600
let no. of balls bought =x
cost of each ball = $600/x

new cost = $((600/x)-0.4)
total money=$600
no. of balls bught at reduced rate= 600/new cost.
new number of balls bought=x+10
so

x+10= 600/((600/x)-0.4)

u do the rest.

and thanks for the thanks.

i can see how this makes sense but when I do it, the x works out to 6, which makes them going pretty slow!

multiplied out:

240 = 240x/x - 20x + 240/x - 20x
the x cancels
240 = 240 - 40x + 240/x

multiply by x to get rid of denominator
240x = 240x - 40x^2 + 240x

combine
0 = -40x^2 + 240x


using quadratic formula
a= -40
b= 240
c=0 (can c equal 0?)

- 240 +- sq root 240^2 - 4*-40*0/ 2(-40)
- 240 +- 240 / -80
- 480/ -80
6

what did I do wrong?

x is the time taken. so now u will have to calculate speed = d/t

and c can be zero.