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ATQ,
30 minutes = (30/60) = 0.5 hours
So, usual speed of Asha = (40.5 / 1.5) = 27 km/h
Increased speed of Asha = 27 + 3 = 30 km/h
Therefore, distance covered by her in 4 hours = 30 × 4 = 120 km
8(3/4) + 5(1/6) – 4(3/4) = ?
20 ×33 + 12 × 23 - 40 ÷ 15-1 + ? = 50
(750 / 15 × 15 + 152 + 20% of 125) = ?3
√ (12+√ (12+√ (12+ ⋯ ∞ ))
(60 × 8 ÷ 10) × 5 = ?
(√529 + 63 /8)% of 800 = ?% of 250
1 + 1 + 1/2+ 1/3 + 1/6 + 1/4 is equal to ____
68% of 450 – 1008 ÷ 14 + 516 ÷ 43 =?