A train covers a certain distance at a speed of 48 km/h and the next same distance with a speed of 60 km/h and the next same distance with a speed of 80 km/h. find the average speed?
Since the distance is same in all the conditions. Therefore, Average speed = 3xyz/(xy + yz + zx) Average speed = 3 x 48 x 60 x 80/(48 x 60 + 60 x 80 + 48 x 80) Average speed = 3 x 48 x 60 x 80/(2880 + 4800 + 3840) = 691200/11520 = 60 km/hr
78.89 × 81.03 – (16.83)² + 8.33% of 9602.87 = ? – 50.23
1220 ÷ 61 ÷ 5 + 450 of 20% - 70 = √ ?
4261 + 8234 + 2913 + 8217 + 6283 + 4172 =?
25% of 400 + 3 × 102 = ?2
104 × 21 ÷ 13 + ? % of 300 = 320 + 22
1549.8 ÷ 8.2 + 65.6 × 55 = (? × 4) + (42 × 30.5)
√( (664+ √(136+ √(59+ √(21+ √(7+ √81) ) ) ) ) ) = ?
[4(1/3) + 4(1/4)] × 24 – 62 = ?2
(5/7) of (7/11) of (3/5) of 52% of 4400 = ? - (44)2 + (50)2 - (62% of 1750) - (188 ÷ 9.4)
756 + 432 – 361 + ? = 990
