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ATQ,
Let the usual speed of the car be '4s' km/h and the distance be 'd' km. Let the usual time taken by the car be 't' hours. Using the relation:
Now, at three-fourths of the original speed, the speed becomes (3/4) × 4s = 3s The time taken at this speed:
Substitute 'd' from equation (I):
Multiply both sides by 3: 4t = 3t+1 t = 1 So, the usual time taken by the car is 1 hour or 60 mins
1000÷ 250 = ( 3√? × √1444) ÷ ( 3√512 × √361)
1540 ÷ 7 - 184 ÷ 8 = ?
12.232 + 29.98% of 539.99 = ? × 5.99
√256 * 3 – 15% of 300 + ? = 150% of 160
18 × 15 + 86 – 58 =? + 38
[5 X {(52 X 5) - 10} + 50 of 20] = ?
4(1/3) × 2(11/14) = 50% of ? + 86/11
(15 x 6 + 60% of 500 - 16 x 7) = ?
25% of 140 + 2 × 8 = ? + 9 × 5
If a nine-digit number 389x6378y is divisible by 72, then the value of √(6x + 7y) will be∶