Experts answer No.3,4

Experts answer No.3,4 EXERCISE 12.1 1. A distance aso in 35 min. Find the speed of the truck in and distance be covered at a speed of S tn;s? A distance of two stations in 105 min. the: (b) time taken to cover a distance of 630 km A person travels a total distance of km. He covers the first halfat a speed of the other half at 45 km'h_ Find; (a) the total time taken to complete the entire journey S A truck travels some distance at a speed of 35 knvh in 120 min and the remaining distance a speed of 48 km/h in 90 min. Determine the: (b) average speed over the entire Journey (a) Iota/ distance covered 6 Rahul covered a journey of 720 km in 420 minutes. If one-third of the journey is covered a speed of 120 km/h and half of the remaining journey at a speed of 80 km/h, at what speed did he complete his remaining journey? 7. Ramesh travels a total of 740 km, out of which he drives a car for 156 km at 52 travels 146 km in the bus at the speed of 36.5 knvh and covers the remaining distance by train in hours. Determine the: (a) speed at which the train moves (b) average speed for the whole journey

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Q 3 . D i s \tan c e t r a v e l l e d b y t r a i n = 325 k m T i m e t a k e n b y t r a i n t o t r a v e l 325 k m i s 105 m i n u t e s i . e ., \frac{105}{60} h o u r s = \frac{7}{4} h r s . (a) S p e e d o f t r a i n = \frac{D i s \tan c e}{T i m e} = \frac{325}{\frac{7}{4}} = 325 \times \frac{4}{7} = \frac{1300}{7} K m / h (b) t i m e t a k e n t o c o v e r 650 k m = \frac{d i s \tan c e}{S p e e d} = \frac{650}{325 \times \frac{4}{7}} = \frac{650 \times 7}{325 \times 4} = \frac{7}{2} = 3.5 h o u r s Q 4 . D i s \tan c e t r a v e l l e d b y p e r s o n = 360 k m S p e e d f o r f i r s t 180 k m i s 75 k m / h a n d s p e e d f o r n e x t 180 k m i s 45 k m / h (a) t i m e t a k e n t o c o v e r f i r s t 180 k m = \frac{180}{75} = 2.4 h o u r s t i m e t a k e n t o c o v e r n e x t 180 k m = \frac{180}{45} = 4 h o u r s t o t a l t i m e = 2.4 + 4 = 6.4 h o u r s (b) a v e r a g e s p e e d = \frac{360}{6.4} = 56.25 k m / h

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