d Question 5 &6 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Question 5 &6

Question 5: A car covers 42 kms in 1 hr 40 min 48 sec with its actual speed. Find the car’s actual speed.



This is the question involving speed, distance and time. We know distance can be calculated by taking product of time and speed. But before taking product both time and speed should be taken into SI (International System of Units) units. We know SI unit for time is ‘second’ i-e if time is given in hours, we will convert it into seconds. SI unit for  speed is meter per second. Relation for Distance ‘S’, time ‘t’ and speed ‘v’ is:


We are given ;

S= 42 kms

But in 1 km there are 1000 meters, so

S= 42×1000=42000 meters

Time = 1 hour 40 minutes 48 second

Now since, there are 3600 seconds in one hour and 60 seconds in one minute;

T= 3600 seconde +40×60 seconds+45 second = 3600+2400+45 = 6045 seconds

Put values in the relation

42000=v \times 6045

\Rightarrow v= \frac{42000}{6045} = 6.94 m/s

\Rightarrow v= 7 m/s (Aproximately)

Now ,

1 meter = \frac{1}{1000}= 0.001 km


1 second = \frac{1}{3600}= 0.0002 hours

Put 0.001 in place pf ‘m’ and 0.0002 in place of ‘s’, we get;

\Rightarrow v= 7 \times \frac{0.001}{0.0002}

\Rightarrow v= \frac{0.007}{0.0002} =35 km /hour

Question 6: In 9 hours a farmer traveled 61 kms. He traveled on foot partly at the rate of 4 km/hr and partly on bicycle at the rate of 9 km/hr. The distance he walked is ?

FPSC 2018


This question seems to take  distance, speed and time related solution, but here we cant use S= VxT formula  directly for each individual case, because of ambiguity of partial distances and times taken.

We have to start with assumption !

Let the man walks ‘x’ kms on foot. Then because total distance is 61 kms, so, the distance traveled on bicycle will be 61-x (smaller distance to be subtracted from bigger distance).

Now we have two cases for using S= VxT

Case 1: On foot:

S1= x kms

V1 = 4 km/hr


Re-write S1=V1xT1

Put values;

==> x = 4 x T1

==> T1= x/4 seconds

Case2: On bicycle:

S2= 61-x kms

V2= 9 km/hr

T2 ?

Re-write S2=V2xT2

Put values of S2, V 2and T2;

==> 61-x =9 x T2

==> T2 = 61-x/9

But total time to travel all distance of  61 km is 9 hours

i-e T1 + T2 = 9 hours

Put values of T1 & T2

T1+T2=\frac{x}{4}+ \frac{61-x}{9}=9

\Rightarrow \frac{ 9x+4(61-x)}{36}=9

\Rightarrow 9x+244-4x =9\times 36

\Rightarrow 5x+244 =324

\Rightarrow 5x=324-244

\Rightarrow 5x=80

\Rightarrow x=16

So, the distance on foot is 16 kms

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