TIME AND DISTANCE SHORTCUTS FOR QUANTITATIVE APTITUDE

                        TIME AND DISTANCE SHORTCUTS FOR QUANTITATIVE APTITUDE

Formula:

                                              Distance = Speed x Time
                                              Time = Distance / Speed
                                               Speed = Distance / Time

Point to care:

 To convert speed in kmph to m/sec, multiply it with 5/18
 To convert speed in m/sec to kmph , multiply it with 18/5.

Example
1.  If a men  travels from point A to point B with a speed of 'a' and back to point A (from point B) with a speed of b, then the average speed of the body is:

                                                 2ab/(a+b).

A car covers a certain distance at a speed of 90 km/hr while going and returns to the starting point at a speed of 60 km/hr. Find the average speed of the car for the whole journey?
Ans: Average speed = (2 x 90 x 60)/ (60+90)
                               = 72 km/hr
.
2. If a car does a journey in 'T' hrs, the first half at 'a' km/hr and the second half at 'b' km/hr. The total distance covered by the car:

                                        (2 x Time x a x b ) / (a + b).

 A motorcar does a journey in 10 hrs, the first half at 21 kmph and the second half at 24 kmph. Find the distance?
Ans: Distance = (2 x 10 x 21 x 24) / (21+24)
                      = 10080 / 45
                      = 224 km.


3.If a person goes from 'A' to 'B' at a speed of 'a' kmph and returns at a speed of 'b' kmph and takes 'T' hours in all, then the distance between the A and B:

 Total time taken x (Product of the two Speeds / Addition of the two speeds)

 A boy goes to school at a speed of 3 kmph and returns to the village at a speed of 2 kmph. If he takes 5 hrs in all, what is the distance between the village and the school?

Ans: Let the required distance be x km.
Then time taken during the first journey = x/3 hr.
and time taken during the second journey = x/2 hr.
x/3 + x/2 = 5 => (2x + 3x) / 6 = 5
=> 5x = 30.
=> x = 6
Required distance = 6 km.

4: Walking ¾ of his speed, a person is 10 min late to his office. Find his usual time to cover the distance?
Ans: Usual time = Late time / {1/ (3/4) - 1)
= 10 / (4/3 -1 )
= 10 / (1/3)
= 30 minutes.
5: Running 4/3 of his usual speed, a person improves his timing by 10 minutes. Find his usual timing by 10 minutes. Find his usual time to cover the distance?
Ans: Usual time = Improved time / { 1 - (1/ (3/4)}
= 10 / { 1- (3/4) }
= 40 minutes.

6 A train travelling 25 kmph leaves Delhi at 9 a.m. and another train travelling 35 kmph starts at 2 p.m. in the same direction. How many km from will they be together ?
Ans: Meeting point's distance from the starting point = [25 x 35 x (2p.m. - 9 a.m)] / (35 -25)
= (25 x 35x 5) / 10
= 4375 / 10
= 437.5 km .