Question
A is cycling at an average speed of 24 km/hr such that
he can reach a certain point at 12 noon. If he cycles at 48 km/hr, then he will reach the destination at 10 a.m. At what speed he should cycle to reach the destination at 11 a.m.Solution
Let the distance to be travelled be βxβ km Time taken when speed is 24 km/hr = (x/24) hours Time taken when speed is 48 km/hr = (x/48) hours According to the question, (x/24) β (x/48) = 2 Or, 2x β x = 2 Γ 48 Or, x = 96 km Time taken to travel 96 km with speed of 24 km/hr = 96/24 = 4 hours Therefore, he started at 12 β 4 = 8 a.m. Therefore, required speed to reach the destination at 11 a.m. or 3 hours = 96/3 = 32 km/hr
12 Γ 19 + 13 Γ 15 + 152 = ?% of 500
{(8Γ 8 + 3 Γ 39) - 620 Γ· 20} = ?
3(2/5) + 6(1/3) + 3(2/5) + 11(2/3) =?
What will come in the place of question mark (?) in the given expression?
?2 - 222 = 900 Γ· 4 - β1089
Find the value of βyβ if (2/7) Γ y = 0.79 β 1.77 Γ· 3.
Β β256 * 4 β 30% of 190 + ? = 110% of 220
65% of ? = 50 + 20 Γ 4
15(2/9) + 11(2/9) + 17(1/9) + 13(4/9) = ?