Latest Elitmus Aptitude Question SOLUTION: There is a match of 50 ovrs and each over has 6 balls. The run rate of a match is defined as number of runs per over..a team batting first played 50 ovrs with run rate of 7.8.. So

This question is not as easy to understand as it looks and at the same time very important for ELITMUS as it is repeated with different values of run-rates.
Before solving this let us first understand the concept of run-rate:
Run-rate is the number of runs scored by number of balls bowled multiplied by 6
RR = (Runs scored * 6)/(Balls bowled)

In this case,
7.8 = (Runs scored * 6) / Balls bowled
First, eliminate 6 out of the equation.
If we take 6 to the other side of the equation, it will divide 7.8 and hence give a result of 1.3

1.3 = Runs scored/ Balls bowled
If we take (Balls bowled) to the other side,
1.3 * (Balls bowled) = Runs scored

Up to this step, the process will be the same irrespective of the value given for the run-rate.

Now the question is what is the maximum possible number of instances in the match when the run-rate is exactly 7.8.

We understood that if the run-rate has to be 7.8, it has to satisfy the above equation that is
1.3 * (Balls bowled) = Runs scored

Another important point is the number of runs scored must always be an integer, so 1.3 * (Balls bowled) must be an integer.

If we go on substituting the values in the place of (Number of balls bowled), the output will be an integer for the first time when the number of balls bowled is 10.
The first instance is when the runs scored are 13 and the number of balls bowled is 10.
After 10, the other values which give integer as the output are 20, 30, 40, ……………..300

So, there will be a total of 30 instances where the run-rate can be 7.8.

For example, runs scored = 104 and balls bowled is 80.

Answer is 30.

options given above are wrong, the correct options given in the exam are 1)60 2)30 3)10 4)50

50 times since its talking about average.

Answer is 10 as in 50 overs match..in 5 over 39 den 10 ovr 78 as the whole no.will come on multiples of 5 and 10.

Can smbody make it More clear.. i felt The answer is 30 as i felt Mr.Mahesh's explanation as correct..

Sorry there is a mistake in option.. 2nd option is 30….

the answer is 6--option (c)
because to get an integral value from 7.8 we multiply 5 to get 39........i.e; if the batsman scores 39 runs in every 5 overs like (1-5),(6-10),(11-15)...then the average would be 7.8....so 7.8 occurs after every 5th over hence total no .of times =5th + 10th + ........50th=6 times.

ans. will be Option 4) 50

sorry for the previous solution ...itz wrong....admin should provide an option to delete.....