SOLUTION: 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, b) Due to a leak, he lost 3 lit

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 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, b) Due to a leak, he lost 3 lit      Log On


   



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,
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,
c) He made a profit of $20. 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.
d) Find, correct to the nearest whole number, how many litres of petrol he sold.
*I know it's a long question, but please try to understand I can't put the parts separately because then they wouldn't make sense. Please answer as soon as possible. Much appreciated :) =)

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
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 $20.
((500/x)-3 )(x+1)-500 =20
(500-3x)/x*(x+1)-500=20
multiply by x
(500-3x)(x+1)-500x=20x
500x+500-3x^2-3x-500x=20x
500-3x^2-3x-20x=0
-3x^2-23x+500=0
/-1
3x^2+23x-500=0
i) Write down an equation in x to represent this information and show that it reduces to 3(x)^2 + 23 x -500 = 0. ii) Solve this equation, giving both your answers correct to one decimal place.
Find the roots of the equation by quadratic formula
a= 3 , b =23 , c = -500 .
b^2-4ac= 6529
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-23%2Bsqrt%286529%29%29%2F%286%29
x1= 9.63
Cost per litre =$ 9.6

...
x2=%28-23-sqrt%286529%29%29%2F%286%29
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