Question
Vessels 'P' and 'Q' contain (4x - 10) litres and (x +
10) litres of petrol, respectively, and the price per litre of petrol in vessel 'P' and 'Q' is Rs. 24 and Rs. (2x - 8) respectively. The total cost of petrol in 'P' is equal to that in 'Q'. If 20% of the petrol is taken out from vessel 'P' and added to vessel Q, then determine the cost price of each litre of petrol in vessel 'Q' after the addition.Solution
ATQ, Total cost of petrol in vessel 'P' = Rs. (4x - 10) × 24 Total cost of petrol in vessel 'Q' = Rs. (x + 10) × (2x - 8) ATQ; (4x - 10) × 24 = (x + 10) × (2x - 8) 96x - 240 = 2x² + 20x - 8x - 80 96x - 240 = 2x² + 12x - 80 2x² - 84x + 160 = 0 x² - 42x + 80 = 0 x² - 40x - 2x + 80 = 0 x(x - 40) - 2(x - 40) = 0 (x - 40) (x - 2) = 0 'x' = 40 or 2 For 'x' = 2, (2x - 8) = -4, but since the price per litre cannot be negative, So, 'x' = 40 Quantity of petrol in vessel 'P' = 4x - 10 = 150 litres Quantity of petrol in vessel 'Q' = x + 10 = 50 litres Per litre price of petrol in vessel 'Q' = 2x - 8 = Rs. 72 New quantity of petrol in vessel 'Q' after addition = 50 + 0.2 × 150 = 80 litres New total cost of petrol in vessel 'Q' = 50 × 72 + 30 × 24 = Rs. 4320 Therefore, required cost price = (4320/80) = Rs.54
Simplify the following expressions and choose the correct option.
{[(13)² − (7)²] ÷ 12} × 4 = ?
(25)² × 4 ÷ 5 + (3)³ + 48=? + 425
?2 + 114 - 48 ÷ 2 × 5 = 163
182 + 10 × 12 - ? = 312
2/5 of 3/4 of 7/9 of 7200 = ?
If (3 × 144 – 252 ÷ 14) ÷ 18 = √1024 – x, then find the value of ‘x’.
12.50% of 1440 - 17 × 51 + 721 =?
[(15)³ × (8)²] ÷ (90 × 6) = ?²
?2 - (40% of 240) = 25 X 5
Simplify: 48 ÷ 4 × 3 + 5 × (6 − 2)