d Question 27 & 28 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Question 27 & 28

Question 27: A train is 360 m long and is running with speed of 45 km /hr. In how much time it will cross a bridge of length 140 meter?

Solution: 

The train will  enter bridge from one end and will cover a distance equal to length of bridge and will reach other end of bridge. Then it will cross bridge completely after covering distance equal to its own length- and then its last boggy will be out from other end !

So, train has to cover total distance equal to its length + length of bridge = 360+140=500 meters;

Distance = S= 500 m

speed = V = 45km/hr

speed = V = 45\times \frac{1000m}{3600s}

speed = V = \frac{450m}{36s}

\Rightarrow speed = V = \frac{12.5m}{s}

Using formula;

 S=Vt

\Rightarrow t=\frac{S}{V}

\Rightarrow t=\frac{500}{12.5}

\Rightarrow t= 40 s

So, the train will cross the bridge in 40 seconds

Question 28: There are 5 children. Every next child is 3 years older than the previous younger one. If sum of their ages is equal to 50 years find the age of the youngest child.

Solution:

 

Let age of youngest child is ‘x’; every next child is 3 years elder than previous one – so;

Age Of Youngest Child = x

Age Of 2nd Child = x+3

Age Of 3rd Child = x+3+3=x+6

Age Of 4th Child = x+6+3=x+9

Age Of 5th Child = x+9+3=x+12

There ages sum;

Sum = x+(x+3)+(x+6)+(x+9)+(x+12)

Sum = 5x+30

It is given equal to 50 i-e;

\Rightarrow 5x+30=50

\Rightarrow 5x=50-30

\Rightarrow 5x=20

\Rightarrow x=4

So, the age of the youngest child is 4 years

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