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

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?


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;


\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.



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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