CBSE Sample Paper March 2019 for Class10 CBSE Sample Paper will give you an idea about the latest examination pattern and the level of questions which can be asked in the forthcoming CBSE Board Examination. CBSE Sample Paper March 2019 for Class10 with marking scheme is available for download in PDF format. With this article students can download the complete Sample Paper along with Marking Scheme and hints. Students are advised to solve all the Sample Paper and refer their answers to the Marking Scheme to assess their level of preparation for CBSE Board Examination. All these questions are very important for forthcoming CBSE Board Examination.
10th Model Sample paper for CBSE Board Exam March 2019 latest
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Solution: Let the speed of train = x m/s Distance travel by train to cross pole = Length of train , time = 15 sec => 15x = Length of train Distance travel by train to cross platform = Length of train + Length of platform = Length of train + 100,time = 25 sec 25x = 15 x + 100 => 25x – 15x = 100 => 10x = 100 => x = 100/10 = 10m/s Length of train = 15x =15 x 10 = 150m Do yourself: A train passes an electric pole in 10 seconds and a 130 m long platform in 20 seconds. How should I calculate the length of the train? Q.2. A train 50 metres long passes a platform 100 metres long in 10 seconds. The speed of the train? Solution: Distance travel by train to cross platform = Length of train + Length of platform = 50 + 100 = 150m,,time = 10 sec Speed = d/t = 150/10= 15m/s Q.3. A train moves with a speed of 108 km/h. Find Its speed in metres per second? Solution: Speed = 108 x (1000m/3600sec) = 30m/s Q.4. In what time will a train 100 metres long cross an electric pole, if its speed be 144 km/hr? Solution: Speed = 144 x (1000m/3600sec) = 40m/s Distance travel by train to cross pole = Length of train = 100m Time =distance / time = 100/40 = 2.5sec Q.5. A train covers a distance of 12 km in 10 minutes. If it takes 6 seconds to pass a telegraph post, Find the length of the train ? Solution: Time = 10min = 10 x 60 = 600sec , distance = 12km = 12000m Speed of train = 12000m/600sec = 20m/s Length of train = Distance travel by train to cross a telegraph post = speed x time = 20 x 6 = 120m Q.6. A train 110 metres long is running with a speed of 60 km/h. In what time will it pass a man who is running at 6 km/h in the direction opposite to that in which the train is going? Solution: Speed of train relative to man = (60 + 6) km/hr = 66 km/hr = 66 x (5/18) m/s = 55/3m/s Time taken to pass the man = Distance/speed = 110 m/ (55/3)m/s = (110x3)/55 = 6 sec Q.7.Two trains 200 m and 150 m long are running on parallel rails at the rate of 40 km/h and 45 km/h respectively. In how much time will they cross each other, if they are running in the same direction? Solution: Relative speed of trains = (45  40) km/h = 5 km/h = 5 x(5/18 )m/s = 25/18m/s Total distance covered = Sum of lengths of trains = 200 + 150 = 350 m. Time taken = distance /speed = 350 /(25/18) = (350 x 18) /5 = 252 sec = 4 min 12sec th Mole concept numerical problems solved9
Download File 9th Mole concept numerical problems solved10 Download File Class 9 Atoms and molecules solved CBSE Papers New links More Problems based on mole concept 9th Atoms and Molecules  1. Mole Concept and Problems based on mole concept 2. Numerical Problems based on mole concept 9th Chemistry 3. Numerical based on mole concept by ChemistryAdda 
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February 2019
