Question
If a person walks 20% more than of his usual speed,
reaches his distance 90 minutes before. If the destination is 459 km away, then the usual speed of a person is (in km/hr)?Solution
Let the usual speed be x km/hr New speed = 120% of x = 6x/5 km/hr Distance = 459 km According to the question 459/x – 459/(6x/5) = 90/60 459 x [(6 – 5)/6x] = 3/2 Therefore, x = 459 x (1/6) x (2/3) = 51 km/hr.
26% of 950 + 50/3% of 7962 = ? Â
- What will come in place of (?) in the given expression.
(84 + 36 ÷ 6) × 2 = ? 690 ÷ (75% of 460) = ? ÷ (50% of 160)

21% of 400 − 150 = ? − 77
Evaluate: 320 − {18 + 4 × (21 − 9)}
What will come in the place of question mark (?) in the given expression?
(50 × 6 ÷ 12) × 9 = ?
(24% of 500 - 20) ÷ 25 = ? ÷ 4
√ (12+√ (12+√ (12+ ⋯ ∞ ))Â
