I cannot solve the iii one. i tried but ended up getting 14000 rather than 42000
Answer (4)
80 because 10 8=80 that wouldn't work so 80 10=800 and 800/10=80.
800 is ten times as much as 80. We can see this by dividing 800 by 10 or by multiplying 80 by 10. Both methods give us the same result, confirming that the correct answer is 80.
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Jawaban:[tex] =\frac{6000}{x + 10}\\ = \frac{6000}{x} - \frac{6000}{x + 10} = \frac{3}{5} \\ reduces \: to \: = x^{2} + 10x - 42000 = 0 \\ x = 200 \\ \frac{3400}{210} = 16.19 \: litres[/tex]
(i)The price of petrol was x cents per litre, and we have $60 = 6 000 cents.So, that amount of money can buy V = 6 000/x litres of petrol.(ii)The price is now x + 10. Thus, the amount of petrol that can be bought in December 2013 is V = 6 000/(x + 10) litres.(iii)In November 2013, we can bought 1 3/7 = 10/7 more litres of petrol than in December 2013, or mathematically,[tex]\displaystyle \frac{6000}{x} = \frac{6000}{x + 10} + \frac{10}{7}[/tex]To eliminate denominators, we multiply both sides by 7x(x+10):[tex]\begin{aligned}7x\left(x+10\right)\times \frac{6\ 000}{x} &= 7x\left(x+10\right)\times \left( \frac{6\ 000}{x + 10} + \frac{10}{7} \right)\\7x\left(x+10\right)\times \frac{6\ 000}{x} &= 7x\left(x+10\right)\times \frac{6\ 000}{x + 10} + 7x\left(x+10\right)\times\frac{10}{7} \\7(x + 10) \times 6\ 000 &= 7x \times 6\ 000 + 10x(x + 10)\\42\ 000x + 420\ 000 &= 42\ 000x + 10x^2 + 100x\\0 &= 10x^2 + 100x - 420\ 000\\0 &= x^2 + 10x - 42\ 000\end{aligned}[/tex](iv)You can solve for x using factorization or by using the quadratic formula.(v)From the value of x obtained from part (iv), we can find the value of petrol in December 2013. Thus, we can find how many petrol can be bought with $34 then.I'm sorry in advance if there are any errors, hope this useful.
