Question
On a certain day, 'P' leaves his
house and reaches his school 6 minutes early when traveling at a speed of 5 km/hr. The following day, he reduces his speed by 2 km/hr and ends up reaching his school 10 minutes late. What would be the speed of 'P' if he arrives at his school exactly on time?Solution
ATQ, Let time be t hours. And, distance between school and home = d So, 5 × (t – 6/60) = d --------(i) And, 3 × (t + 10/60) = d -------(ii) From (i) and (ii), we get 5 × (t – 6/60) = 3 × (t + 10/60) 5t - 1/2 = 3t + 1/2 5t – 3t = 1/2 + 1/2 2t = 1 hour, t = 1/2 hours Distance between school and home = 5 × (t – 1/10) = 5 × (4/10) = 2 km Therefore, required speed = 2/(1/2) = 4 km/h
562, 628, 698, ?, 850
40   41    37   46    ?      55    19
...10, 15, 25, 35, ?, 65
168, 178, ?, 273, 400, 618, 963
120, 130, 145, 165, ?, 220
16.12 × 15.94 + 654.92 – 344.83 = ?× 5.95
What will come in place of the question mark (?) in the following series?
121, 110, 97, ?, 61, 38
11   6   7   12   26    ?    205.5
- What will come in place of (?), in the given number series.
1, 4, 9, 16, ?, 36 98, 122, 182, 278, 410, ?
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