# Shortcut Rules to Solve Problems on Time and Distance

## Effective for IBPS PO - SBI PO Exam

Here we will start a series of Quantitative Aptitude Shortcut Tricks for your upcoming SBI - IBPS - SSC and Other Government Competitive Exams. We will try to cover up all topics of the quantitative Aptitude Sections from which question was generally asked.

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Trick - 1

• If a certain distance is covered at x km/hr and the same distance is covered by y km/hr, then the average speed during the whole journey is
$\frac{2xy}{x+y}km/hr$
Trick - 2
• A person walking at a speed of x km/hr reaches his destination ${{x}_{1}}$ hrs late. Next time he increases his speed by y km/hr, but still he is late by  ${{y}_{1}}$ hrs. The distance of his destination from his house is given by
$\left[ \left( {{x}_{1}}-{{y}_{1}} \right)\left( x+y \right)\frac{x}{y} \right]km$
Trick - 3
• If a person does a journey in T hours, and the first half at ${{S}_{1}}$ km/hr and the second half at  ${{S}_{2}}$ km/hr, then the distance
$=\frac{2\times time\times {{S}_{1}}\times {{S}_{2}}}{{{S}_{1}}+{{S}_{2}}}$
where, ${{S}_{1}}$ = speed during first half and
${{S}_{2}}$ = speed during second half of journey
Trick - 4
• The distance between two stations, A and B is D km. A train starts from A and moves towards B at an average speed of x km/hr. If an another train starts from B, t hours earlier than the train at A, and moves towards A at an average speed of y km/hr, then the distance from A, where the two trains meet is

$\left[ \left( D-ty \right)\left( \frac{x}{x+y} \right) \right]km$
Trick - 5
• If a train travelling x km an hour leaves a place and t hours later  another train travelling y km an hour, where y>x, in the same direction, then they will be together after travelling $\left[ \frac{t\left( xy \right)}{y-x} \right]km$ from the starting place.

Trick - 6
• If the new speed of a person is $\frac{a}{b}$ of the usual speed, then the change in the time taken to cover the same distance is $\left( \frac{b}{a}-1 \right)\times$ usual time or, usual time is given by

$\left[ \frac{change\operatorname{in time}}{\left( \frac{b}{a}-1 \right)} \right]hrs$

Questions for Practice

Q1.  A man covers a certain distance 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.

Q2. A boy walking at a speed of 10 km/hr reaches his school 15 minutes late. Next time he increases his speed 2 km/hr, but still he is late by 5 minutes. Find the distance of his school from his house.

Q3. A mother 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.

Q4. The distance between two stations, Delhi and Amritsar is 450 km. A train starts at 4 pm from Delhi  and moves towards Amritsar at an average speed of 60 km/hr. Another train starts from Amritsar at 3.20 pm and moves towards Delhi at an average speed of 80 km/hr. How far from Delhi will the two trains meet and at what time ?

Q5. A train travelling 25 km an hour leaves Delhi at 9 a.m. and another train travelling 35 km an hour starts at 2 p.m. in the same direction. How many km from Delhi will they be together ?

Q6. Walking $\frac{3}{4}$ of his usual speed, a person is 10 min late to his office. Find his usual time to cover the distance.

Answer 5. $437\frac{1}{2}km$ 