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Area of the track = length of track x breadth of track = 22 x breadth of track = 2200 So, breadth of track = 2200 ÷ 22 = 100 So, perimeter of the track = (22 + 100) x 2 = 244 metres So, speed of the car = (244/20) = 12.2 m/s Relative speed of the car with respect to the person = 12.2 + 4 = 16.2 m/s So, distance between the car and the person, 3 seconds before they cross each other = 16.2 x 3 = 48.6 metres
(15/8) x [6924 – 2124] + 910 = ? + 190
(1748 ÷ 8) + 76.8 × 35 =(? × 4) + (42 × 35.5)
2/9 of 5/8 of 3/25 of ? = 40
If x²- 5x + 1 = 0, what is the value of x² + 1/x2?
(70% of 480) ÷ 6 + ? = 45% of 3500 + 802+ 272
`(450 -: ?)/(2.5 xx 1.2)` = 250
116*2/3% of 18600 + 666*2/3% of 1290 = 457*1/7% of 1750 + 555*5/9% of 3150 + ?
[∛(91125/19683 )- ∛(3375/5832 ) ] × ∛(512/9261) = ? - √(484/3969)