Question
A bus was travelling from Jaipur to Udaipur was delayed
by 15 minutes and made up for the delay on a section of 85 km travelling with a speed 12 km per hour higher than its normal speed. Find the approximate original speed of the bus?Solution
Let original speed = 'x' km/hr. Time to cover 85 km at normal speed = 85/x . Time to cover 85 km at increased speed (x+12) = 85/(x+12) Time difference = 15 minutes = 1/4 hours 
Simplify and solve: 
Solve quadratic equation: Using the quadratic formula, x = 58.15  x = 58.15 km/hr.
70.14% of 799.95 - 240.12 = ? + 40.17% of 299.95
(?)2 + 4.113 = 24.92 – 32.03Â
The greatest number that will divide 398,436, and 542 leaving 7, 11, and 15 as remainders, respectively, is:
58.03% of 1499.99 - ? % of 699.95 = 394.04
?% of (112.31 ÷ 13.97 × 90.011) = 359.98
What will be the approximate value of the following questions.
(√143.74 + 29.89% of 720.27) × (5/9 of 539.79) = ?
11.06 2 – 7.12 × 4.88 + 9.96 = 12.22 × ?Â
? = 49.97% of 38.09% of 1998.95
5.55% of 8120 – 66.66% of 540 = ? – 28% of 5500
(5/9 of 2699.81) + (49.88% of 144.18) - (2/7 of 489.77) = ?
