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Mar 20, 2017

TIME AND DISTANCE

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Quantitative Aptitude Study Notes 
 Bank & SSC Exam

TIME AND DISTANCE
You know that quantitative aptitude section is most important in bank exams in PO and Clerk and for other competitive exams because if you want good score in bank exam then you have to score good in maths. In competitive exams the most important thing is time management, if you know how to manage your time then you can do well in Bank Exams. That’s where maths shortcut tricks and formula are comes into action. So continuously we are providing shortcut tricks on different maths topics.
TIME AND DISTANCE
 Today’s topic is TIME AND DISTANCE. This is the one of the most important topic in quantitative aptitude section in bank and ssc exam. You should know how to calculate time and distance questions and answers in very short time for bank exam. From this chapter around 1-2 questions are given in the SBI and IBPS exams. For this here we are providing shortcut tricks and quicker method to calculate time and distance in very short time.
If we want to solve time and distance questions or any other type of questions, then the first thing that we need that is Formulas about that topic. So here is the list of formulas that is used in time and distance quantitative topic.



TIME AND DISTANCE FORMULAS:
        i.            SPEED = DISTANCE / TIME
     ii.            TIME = DISTANCE / SPEED
   iii.            DISTANCE = SPEED × TIME
   iv.            If the speed of a body is changed in the ratio a:b, then the ratio of the time taken changes in the ratio b:a
      v.            x km/hr = (x × 5/18) m/sec.
   vi.            X meters/sec = (x × 18/5) km/hr.
EXAMPLE 1: Express a speed of 18 km/hr in meters per second.
Solution: 18 km/hr = ( 18 × 5/18) m/sec = 5 meters/sec.
EXAPMLE 2: Express 10 m/sec in km/hr.
Solution: 10m/sec = (10 × 18/5) km/hr = 36 km/hr.
Theorem: if a certain distance is covered at x km/hr and the same distance is covered at y km/hr then the average speed during the whole journey is 2xy/x+y km/hr.
Proof: let the distance be A km. Time taken to travel the distance at a speed of x km/hr = A/x hrs.
Time taken to travel the distance at a speed of y km/hr = A/y hrs.
Thus, we see that the total distance of 2A km is travelled in (A/x + A/y) hrs.
AVERAGE SPEED =
TIME AND DISTANCE
 
EXAMPLE 3: A man covers a certain distance by car driving by car driving at 70 km/hr and he returns back to the starting point riding on a scooter at 55 km/hr. Find his average speed for the whole journey.
Solution: Average speed = 2 * 70 * 55 / 70 + 55 km/hr
= 61.6 km/hr.
EXAMPLE 4:  A man covers a certain distance between his house and office on scooter. Having an average speed of 30 km/hr, he reaches his office 5 min earlier. Find the distance between his house and office.
Solution: let the distance be x km.
Time taken to cover x km at 30 km/hr = x/30 hrs.
Time taken to cover x km at 40 km/hr = x /40 hrs.
Difference between the time taken = 15 min = ¼ hr.
X/30 – x/40 = ¼
Or
 4x – 3x = 30
Or
x = 30
hence, the required distance is 30 km.
DIRECT FORMULA:
TIME AND DISTANCE

Thus , in this case, the required distance
= 30*40/40-30  ×  10+5/60
= 30 km
Note: 10 minutes late and 5 minutes earlier make a difference of 10 + 5 = 15 minutes. As the other unites are in km/hr, the difference in time should also be changed into hours.
EXAMPLE 5: A man walking with a speed of 5 km/hr reaches his target 5 minutes late. If he walks at a speed of 6 km/hr, he reaches on time. Find the distance of his target from his house.

Solution: This is similar to Ex. 4. Here the difference in time is 5 minutes only.
Thus, required distance = 5*6 / 6-5 * 5/60
= 5 / 2 km
=2.5 km
EXAMPLE 6: A boy goes to school at a speed of 3 km/hr and returns to the village at a speed of 2 km/hr. If he takes 5 hrs in all, what is the distance between the village and the school?
Solution: 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
DIRECT FORMULA:
TIME AND DISTANCE

= 5 ×  3*2 / 3+2
= 6 km
EXAMPLE 7: A motor car does a journey in 10 hrs, the first half at 21 km/hr and the second half at 24 km/hr. Find the distance.
Solution: this question is similar to Example 6, but we can’t use the direct formla in this case. If we use the above formula, we get half of the distance. See the detailed method first.
Let the distance be x km.
Then, x/2 km is travelled at a speed of 21 km/hr and x/2 km at a speed of 24 km/hr.
Speed of 24 km/hr.
Then time taken to travel the whole journey
= x/2*21 + x/2*24
= 10 hrs.
So, x = 2*10*21*24 / (21+24)
= 224 km
Direct formula:
Distance = 2 * time * S1 * S2 / S1 + S2
Where , s1 = speed during first half and
S2 = speed during second half of journey

Distance = 2*10*21*24 / 21+24
= 224 km
EXAMPLE 7: A man takes 8 hrs to walk to a certain place and ride back. However, he could have gained 2 hrs, if he had covered both ways by riding. How long would he take to walk both ways?
Solution: walking time + riding time = 8 hrs ........... (1)
2 * riding time = 8 – 2 = 6 hrs ........................................  (2)
Performing 2 * (1) – (2) gives the result
2 * walking time = 2 * 8 – 6 = 10 hrs.
Both ways walking will take 10 hrs.
DIRECT FORMULA: Both ways walking
=one way walking and one way riding time + Gain in time
= 8 + 2 = 10 hrs.


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