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ATQ, Let the speed of train Q = ‘p’ m/s So, the speed of train P = p + 32.4 × (5/18) = (p + 9) m/s Also let the lengths of trains P and Q are ‘a’ m and ‘b’ m respectively. Since, both the trains cross each other in 10 seconds. So, (a+b)/(2p+9)=10 a + b = 20p + 90 ------------- (1) Since, train P crosses the 60 m long tunnel in 10 seconds. So, (a+60)/(p+9)=10 a + 60 = 10p + 90 ------------- (2) Since, train Q crosses the 60 m long tunnel in 18 seconds. So, (b+60)/p=18 p = (b+60)/18--------(3) From equations (1) and (3): 18a + 18b = 20b + 1200 + 1620 9a – b = 1410 -------------- (4) From equations (2) and (3): 18a + 1080 = 10b + 600 + 1620 9a – 5b = 570 --------------- (5) By equation (4) — equation (5): 9a - b - 9a + 5b = 1410 - 570 b = 210 From equation (4): a = 180 From equation (3): p = 15 Quantity l: Difference between the lengths of both the trains = 210- 180 = 30 m Quantity II: Speed of train P = 180 m Speed of train P = 15 + 9 = 24 m/s So, the time, in which train P will cross the 540m long bridge: (180+540)/24 = 30 seconds Hence, Quantity I = Quantity II
26.11% of ? – 521.02 = 648.51
(36.35 × 14.89) ÷ 8.78 = ? – 59.98
(129.98% of 8460) + (119.899% of 8640) = (130.009% of 15820) + ?
(3/8) × 479.84 + (2/5) × 449.67 = ? × 12.25
185.92 ÷ 5.98 - (4.002 2 + 114.03 of 5.03 ÷ 18.99 of 6.04 = 5.01 of 2.99 + ? ÷ 12.02
24.99 × 32.05 + ? - 27.01 × 19.97 = 29.99 × 27.98
(14.66)2 + (343.84 ÷ 3.88 - 55.87) = ? + 91.23
159.98% of 4820 + 90.33% of 2840 = ? + 114.99% of 1980
47.87% of 749.76 + 35.11% of 399.76 = √? + 23.15 × 20.87
(3/7 of 1049.88 + 44.95% of 799.79) ÷ (√168.89 + 24.77% of 400.11) = ?