Following are the typical problems asked in NTSE, Oympiad Maths, and other Mathematical based aptitude tests
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Q1: Ramesh walks from his home to Railway station at the rate of 5 kmph, he misses the train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance of the station from his home.
Answer: Let 'd' be the distance between Ramesh's house to Station.
We know that distance = speed x time.
If he walks at speed 5 kmph, let time taken be t1
⇒ t1 = d/5
Let the time t2 is taken when he walks with a speed of 6 kmph.
⇒ t2 = d/6
Since the time gap between two cases is 12 mins = 1/5 hrs
⇒ t1 - t2 = d/5 - d/6 = 1/5 hrs
⇒ 6d - 5d = 6
⇒ d = 6km
Q2: Roma travels from place A to B and then from B to C.
From A to B Distance is 55 km which she covers with an average speed of 22 km per hour
It takes her 1 hour 30 minutes moving from B to C with an average speed 30 km per hour
Calculate her average speed over the whole of her journey from A to C.
Answer: We know that $ Time = Distance ÷ Speed
∴ Time taken from A to B = 55/22 = 2.5 hours = 2 Hrs and 30 mins
Distance from A to B = Speed x Time = 20 x 1.5 = 45 km. (note time taken from B to C is given as 1 Hrs 30 minutes i.e. 1.5 Hrs)
Average Speed = Total Distance ÷ Total Time
∴ = (55+45) ÷ (1.5+2.5) = 100 ÷ 4 = 25 kmph
(best viewed in Chrome or Firefox).
Q1: Ramesh walks from his home to Railway station at the rate of 5 kmph, he misses the train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance of the station from his home.
Answer: Let 'd' be the distance between Ramesh's house to Station.
We know that distance = speed x time.
If he walks at speed 5 kmph, let time taken be t1
⇒ t1 = d/5
Let the time t2 is taken when he walks with a speed of 6 kmph.
⇒ t2 = d/6
Since the time gap between two cases is 12 mins = 1/5 hrs
⇒ t1 - t2 = d/5 - d/6 = 1/5 hrs
⇒ 6d - 5d = 6
⇒ d = 6km
Q2: Roma travels from place A to B and then from B to C.
From A to B Distance is 55 km which she covers with an average speed of 22 km per hour
It takes her 1 hour 30 minutes moving from B to C with an average speed 30 km per hour
Calculate her average speed over the whole of her journey from A to C.
Answer: We know that $ Time = Distance ÷ Speed
∴ Time taken from A to B = 55/22 = 2.5 hours = 2 Hrs and 30 mins
Distance from A to B = Speed x Time = 20 x 1.5 = 45 km. (note time taken from B to C is given as 1 Hrs 30 minutes i.e. 1.5 Hrs)
Average Speed = Total Distance ÷ Total Time
∴ = (55+45) ÷ (1.5+2.5) = 100 ÷ 4 = 25 kmph