*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*