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*1.(4) [As nothing can be said about the speed]**Distance = 200 meter**time = 16 seconds**Speed = distance/ time = 200/16**= 12.5 m/sec.**= 12.5 m/sec.**= 12.5 × 18/5**= 45 km/hr.*

*2.(1) Let the employee travelled x kms by taxi.**∴ Distance covered by him by his own car = (90 – x) km.**According to the question.**x × 7 + (90 – x) km.**According to the question.**x × 7 + (90 – x) × 6 = 675**= 7x + 540 – 6x = 675**x = 675 – 540 = 135**∴ Required distance = 135 km.*

*3.(1) Let the speed of train C be x kmph.**Speed of train B relative to C**= (120 – x) kmph**= [(120 – x) × 5/18] m/sec**= (600 – 5x)/18 m/sec.**Distance covered**= 100 + 200 = 300m**∴ 300/(600 – 5x)/18 = 120**= 300 = 120(600 – 5x)/18**= 10 × 9 = 2 (600 – 5x)**= 90 = 1200 – 10x**= 10x = 1200 – 90**= x = 1110/10 = 111**Hence, the speed of train C is 111 kmph.*

*4.(4) When a train crosses a platform it covers a**distance equal to the sum of lengths of platform and the**train itself. If the length of train be x meters, then**= (x + 300)/38 m/sec. ……….. (i)**When the train crosses a signal post it covers its own**length.**∴ Speed of train**= x / 18 m/sec. ……….. (ii)**From equati0ns (i) and (ii)**= (x + 300)/38 = x/18**= 38x – 18x = 300 × 18**= 20x = 300 × 18**= x= (300 × 18)/20**= x = 270 meters**∴ Speed of train = 270/18**= 15 m/sec.**= 15 × 18/5 = 54 kmph*

*5.(2) Average speed = 2xy/(x + y)**(when the same distances are covered)**= (2 × 24 × 36)/(24 + 36) kmph**= (2 × 24 × 36)/60 = 28.8 kmph*

*6.(3) 2 kmph = (2 ×5)/18 meter/sec.**= 10/9 meter/sec.**Let the length of the train be x meter and its speed be y**meter/sec.**Then,**x/(y – 5/9) = 9**= 9y – 5 = x**∴ 9y – x = 5 ……………. (i)**and x(9y- 10) = 9x**= 10 (9y – 10) = 9x**= 90 y – 9x = 100 ……….. (ii)**By equation (i) × 10 – equation (ii)**we have**90y – 10x = 50**90y – 9x = 100**– + .**– x = – 50**= Length of train = 50 m*

*7.(2) Let the length of train B =x**∴ Length of train A = 3x/4 meter**∴ Required ratio = 3x/4 × 33 : x/55**= 5 : 4*

*8.(3) If the distance be x km, then**x/(3/2) + x/(9/2) = 6**= (2x/3 + 2x/9) = 6**(6x + 2x) /9 = 6**= 8x = 9 × 6**= x = 54/8 = 27/4 = 27/4 km*

*9.(4) C = k x2**When x = 16 kmph**C = Rs. 64**∴64 = k × 162 = k = ¼**∴ C = ¼ x2**Total expenditure per hour**= ¼ x2 + 400**When speed = 40 kmph, the expenditure will be**minimum.*

*10.(2) Total expenditure**= ¼ × 40 × 40 × 10 + 400 × 10**= Rs. 8000*

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