Question 424078
A trader bought some petrol for $500. He paid $x for each litre of petrol.
a) Find in term of x, an expression for the number of litres he bought,

number of liters = 500/x

b) Due to a leak, he lost 3 litres of petrol. He sold the remainder of the petrol for $1 per litre more than he paid for it. Write down an expression, in terms of x, for the sum of money he received,

((500/x)-3 )(x+1) = money received.

c) He made a profit of $20. 
((500/x)-3 )(x+1)-500 =20
(500-3x)/x*(x+1)-500=20
multiply by x
(500-3x)(x+1)-500x=20x
500x+500-3x^2-3x-500x=20x
500-3x^2-3x-20x=0
-3x^2-23x+500=0
/-1
3x^2+23x-500=0
i) Write down an equation in x to represent this information and show that it reduces to 3(x)^2 + 23 x -500 = 0. ii) Solve this equation, giving both your answers correct to one decimal place.
Find the roots of the equation by quadratic formula

a=  3 , b =23   , c = -500    .

b^2-4ac= 6529

{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}

{{{x1=(-23+sqrt(6529))/(6)}}}
x1= 9.63
Cost per litre =$ 9.6
 
...
{{{x2=(-23-sqrt(6529))/(6)}}}
This is negative. so ignore

d) Find, correct to the nearest whole number, how many litres of petrol he sold.

Each litre costs $9.6
he paid total $500
Number of litres = 500/9.6
Number of litres bought = 52
Number of litres sold 52-3 = 49