Question

If the price of petrol is increased by Rs.5 per litre. One gets 2 litres less petrol spending Rs.1320.What is the increased price of the petrol ?

Answer 1

$let \: \: the \: \: earlier \: \: price \: \: of \: \: petrol \\ be \: \: Rs. \: x \: \: per \: \: litre \\ it \: \: is \: \: given \\ \frac{1320}{x + 5} = \frac{1320}{x } - 2 \\ \frac{1320}{x + 5} = \frac{1320 - 2x}{x } \\ 1320x = (x + 5)(1320 - 2x) \\ 1320x = 1320x - 2 {x}^{2} + 6600 - 10x \\ {x}^{2} + 5x - 3300 = 0 \\ (x + 60)(x - 55) = 0 \\ x = - 60 \: \: or \: \: 55 \\ since \: \: negative \: \: value \: \: for \: \: x \: \: is \: \: not \: \: admissible \\ x = 55 \\ therefore \: \: increased \: \: value \: \: of \: \: \\ petrol \: \: is \: \: 55 + 5 = Rs. \: 60$

Answer 2

$\huge\bf\green{\mid{\fbox{Answer:}}\mid}$

let the earlier price of petrol be Rs.x/ litre

it is given = $\frac{1320}{x+5} = \frac{1320}{x}-2$

$\frac{1320}{x+5} = \frac{1320-2x}{x}$

=> 1320x = (1320-2x)×(x+5)

=> 1320x = 1320x + 6600 - 2x² - 10x

=> 2x² + 10x - 6600 = 0

=> x² + 5x - 3300 = 0

=> x² + 60x - 55x -3300 = 0

=> x(x+60) -55(x+60) = 0

=> (x+60)(x-55) = 0

=> x+60 = 0 or x-55=0

=> x = -60 or x = 55

Since , cost cannot be negative ,

x = Rs. 55

If the price of petrol is increased by Rs.5 per litre.

x + 5

=> Rs. 55 + 5

=> Rs. 60 Ans.