Ajay and Vishal left the place ‘X’ with different speeds which are in the ratio 9:5 respectively. Ajay left the place 36 seconds after Vishal. If after 72 seconds of the departure of Vishal, Ajay is 50 meters apart from Vishal, then find the speed of Vishal.
Let the speeds of Ajay and Vishal be 9x and 5x km/hr respectively. Distance covered by Vishal in 72 seconds = 5x x (5/18) x 72 = 100x m Distance covered by Ajay in 36 seconds = 9x x (5/18) x 36 = 90x m According to the question, => 100x – 90x = 50 => x = 5 Therefore the speed of Vishal = 5x = 25 km/hr
√784 × 3 + (713.99 ÷ 6.98) = ?% of 619.99
766/51 ÷ 387/42 × 121/13 = ?
648.13 ÷ 35.86 + 28.88 × 13.13 – 39.92% of 900.19 = ?
1359.98 ÷ 20.48 × 12.12 = ? × 4.16
? (30.03 - 17.98 × 15.99 ÷ 12.01) = 729.03
124% of 620.99 + 11.65% of 1279.23 = ?
480.11 ÷ 15.98 × 5.14 – 134.9 = √?
Solve the following expression and calculate the approximate value.
802 of
{44.78% of (11.67 × 54.78)} ÷ 2.65 = ? + 19.25
3.55% of 8120 – 66.66% of 540 = ? – 28% of 5500
