**Math Shortcut Tricks for Time & Distances questions - Math Shortcuts to solve Time and Distance of Train, bus, car etc. for Bank, Railway exams: **It is found that at
least 2-3 questions are given to calculate TIME AND DISTANCE based questions in
SSC IBPS or other exam format. These questions are very easy to solve by using SIMPLE
TRICKS in 2-3 Seconds after reading statement of question.

##
**Math Shortcut Tricks / formulae for ****Time & Distances questions**

These SUPER FAST TRICKS helps
in getting CUT-OFF score in exams. Use these tricks to solve NUMERICAL ABILIY
section of SSC IBPS or other any other competitive exam.

**SUPER FAST TRICKS are described as follows:**

**1.
**Use LCM Method for
calculation.

**2.
**Late & Late will be
subtracted.

**3.
**Early & Early will
be subtracted.

**4.
**Late & Early will be
added.

**5.
**Stopped time/hr =
Difference of speeds/ Speed without Stoppage

**6.
**When a person moves with
two different speeds and takes the same time as more or less, then

Original Speed = (2 x
Increase in speed x Decrease in Speed)/ difference between two speeds.

_{7.
}S_{A}/S_{B}
= √T_{B}/T_{A}

**-------------------------------**

**Question 1:**

A man goes to his office
at a speed of 15 Km/hr & he is late by 20 minutes. Next day he goes at a
speed of 10 KM/hr and still late by 30 minutes. Find the distance from the
office to his home.

**Solution: By Using Super Fast Trick (Rule 1-4)**

**Home to Office (Speed) LCM Early Late**

Day1 15 Km/hr 2hr --- 20 minutes

30
KM

Next Day 10 Km/hr 3hr ---- 30 minutes

3-2=1hrs
Total = 30-20 =
**10 Minutes**
=60 minutes

With two different
speeds the time difference is 1hr i.e. 60 Minutes-----1/6------------10minutes

Thus the distance from
house to office will be 30KM--------------------1/6-----------**5KM Ans**

**----------------------------------------**

**Question 2:**

A man goes to his office
at a speed of 10 Km/hr & he is late by 25 minutes. Next day he goes at a
speed of 12 KM/hr and reach early by 5 minutes. Find the distance from the
office to his home.

**Solution: By Using Super Fast Trick (Rule 1-4)**

**Home to Office (Speed) Efficiency LCM Early Late**

Day1 10 Km/hr 6hr --- 25 minutes

60
KM

Next Day 12 Km/hr 5hr 5
minutes ------

6-5=1hrs
Total = 25+5 = 30
Minutes
=60 minutes

With two different
speeds the time difference is 1hr i.e. 60 Minutes-----1/2------------30minutes

Thus the distance from
house to office will be 60KM--------------------1/2-----------30 **KM Ans**

-----------------------------------------

**Question 3:**

A Person travels a
certain distance at an average speed of 80 Km/hr without any stoppage and
covers the same distance at an average speed with stoppage. How many minutes
per hour does he stop?

**Solution: By Using Super Fast Tricks: (Rule 5)**

Stopped time/hr = Difference
of speeds/ Speed without Stoppage

** = **(80-60)/80

= 20/80 = ¼ hours

= ¼ x 60 = 15minutes Ans

-----------------------------------------

**Question 4:**

A man covers a certain
distance. When he moves 3 Km/hr faster then he would take 40 minutes less and
when he moves 2 Km/hr slower then he would take 40 minutes more. Find the
distance in Km & Find the original speed.

**Solution: By Using Super Fast Trick:**

**Original Speed** = (2 x Increase in
speed x Decrease in Speed)/ difference between two speeds

= (2 x 3 x 2) / (3-2)

= 12/1= 12 Km/hr

**Move (Speed) LCM Early Late**

Move1 15 Km/hr 2hr --- 40 minutes

30
KM

Move2 10
Km/hr 3hr 40 minutes -----

3-2=1hrs
Total = 40+40 =
**80 Minutes**
=60 minutes

With two different
speeds the time difference is 1hr i.e. 60 Minutes-----4/3------------80minutes

Thus the distance from
house to office will be 30KM--------------------4/3-----------**40 KM Ans**

**-----------------------------------------------**

**Question 5**

Two trains A & B
start from X and Y towards Y & X respectively. After meeting 1^{st}
train covers the remaining distance in 4hrs and 2^{nd} covers the
remaining distance in 9hrs. If the speed of 1^{st} is 9Km/hr then find
the speed of 2^{nd} train.

**Solution: By using Super Fast Tricks**

Let S_{A} is the
speed of 1^{st} train & S_{B} is the speed of 2^{nd}
train.

Similarly T_{A}
is the speed of 1^{st} train & T_{B }is the speed of 2^{nd}
train.

Thus

S_{A}/S_{B}
= √T_{B}/T_{A}

9 / S_{B} = √9/4
= 3/2

S_{B} = **6 Km/hr Ans**

Do the practice till
everyone can able to solve such questions in 2 seconds after reading statement
of questions. –BEST OF LUCK-

Click
here to download Super Fast Reasoning Tricks