Aptitude Questions and Answers

- 1. Important Formulae
- 2. Distance is Directly Proportional to Velocity When Time is Constant
- 3. Relative Speed
- 4. Train Problems
- 5. Boats and Streams
- 6. Clock

i) $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$

ii) $\text{Time} = \dfrac{\text{Distance}}{\text{speed}}$

iii) $\text{Distance = speed} \times \text{time}$

iv) $1 \text{km}/\text{hr} = \dfrac{5}{18} \text{m}/\text{s}$

v) $1 \text{m}/\text{s} = \dfrac{18}{5} \text{km}/\text{hr}$

vi) If the ratio of the speed of A and B is $a:b$,then the ratio of the time taken by them to cover the same distance is $\dfrac{1}{a} : \dfrac{1}{b}$ or $b:a$

vii) Suppose a man covers a distance at $x$ kmph and an equal distance at $y$ kmph, then the ** AVERAGE SPEED **during the whole journey is $\dfrac{2xy}{x+y}$ kmph

Out of time, speed and distance we can compute any one of the quantities when we happen to know the other two. For example, suppose we drive for 2 hours at 30 miles per hour, for a total of 60 miles.

If we know the time and the speed, we can find the $\text{distance:}$

$\dfrac{\text{2 hour * 30 miles}}{\text{hour}} = \text{60 miles}$

If we know the time and the distance, we can find the $\text{speed:}$

$\dfrac{\text{60 miles}}{\text{2 hours}} = 30 \dfrac{\text{miles}}{\text{hour}}$

A car travels at 30km/hr for the first 2 hrs & then 40km/hr for the next 2hrs. Find the ratio of distance travelled

$S_1/S_2=V_1/V_2=3/4$

Two cars leave simultaneously from points A & B (100km apart) & they meet at a point 40 km from A. What is $V_a/V_b$?

Time is constant so $V_1/V_2=S_1/S_2=40/60=4/6$

A train meets with an accident and moves at $(3/4)^{th}$ its original speed. Due to this, it is 20 min late. Find the original time for the journey beyond the point of accident?

Think about 2 diff. situations, 1st with accident and another w/o accident. As distance in both the cases is constant

So $\dfrac{V_1}{V_2}=\dfrac{T_2}{T_1}$

=>$\dfrac{V_1}{[(3/4)*V_1]}= \dfrac{T_1+20}{T_1}$

=> $\dfrac{4}{3}=\dfrac{T_1+20}{T_1}$

=>$T_1=60$

Velocity decreases by 25% (3/4 of original speed => decrement by 1/4) so time will increase by 33.3% (4/3 of original time => increment by 1/3)

now, $33.3\%=20$ min

=>$100\%=60$ min

Two bodies *are moving in opposite directions* at speed $V_1$ & $V_2$ respectively. The relative speed is defined as $V_r = V_1 + V_2$

Two bodies are moving *in same directions* at speed $V_1$ & $V_2$ respectively. The relative speed is defined as $V_r = |V_1 - V_2|$

The basic equation in train problem is the same $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$

The following things need to be kept in mind while solving the train related problems.

- When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object.
- The
*distance to be covered when crossing an object*, whenever trains crosses an object will be equal to*: Length of the train + Length of the object*

Let,

$U=$ Velocity of the boat in still water

$V=$ Velocity of the stream.

While moving in ** upstream**, $\text{distance covered, }$ $S= (U-V) T$

In case of ** downstream**, $\text{distance covered ,}$ $S= (U+V) T$

For clock problems consider the clock as a circular track of 60km.

Min. hand moves at the speed of 60km/hr (think min. hand as a point on the track) and hour hand moves at 5km/hr and second hand at the speed of 3600 km/hr.

*Relative speed between HOUR hand and MINUTE hand = 55*