Aptitude - Average - Discussion

The average increase for one person is 0.5

Total increase is 20

Therefore Total no of pupil =20/0.5 = 40.

What if the condition is reverse entry is recorded wrong 39 instead of 63. In this case, the average marks of class reduced by 1/2? Can anyone tell me the answer?

It's easy as by simply making equations.

Consider a number of students=N.

As we know the marks of one of the students makes the number of left students= N-1.

So let the average where marks for the student is right to suppose its X.

That makes ( (N-1) +63) /N=X ---------> 1.

NOW CONSIDER WHEN MARKS ARE GIVEN WRONG THAT INCREASES THE AVERAGE BY HALF MARKS.

This means X is increased as X + 0.5.

That makes ( (N-1) +83) /N=X+0.5 --------> 2.

Here, there are two equations, two variables, Now solve for N and we can get our answer.

Ans of Savinder question
Let total no of cand-x
Dec in aver- 50-45=5
Total dec in numbers= 5x

then, 5x= 100(90-60)
x=600 ( Ans)

Simply:

x = num of pupils.
y = total average.

Therefore,
Total marks of class=xy
Total marks of class due to wrong entry = xy-63 + 83.

New average of class due to wrong entry = (xy-63 + 83)/x
Also, (according to question) new average = y+0.5 (or y+1/2)
Thus, we get:

(xy-63+83)/x = y + 0.5,
x = 40.

According to the property of Average, if each quantity is increased by certain valu "X" then the new average is increase by X.

Now in the question, it is given that average has been increased by 1/2 due to the marks entered wrongly. Therefore we can say that the marks of each n every student should increase by 1/2 as per the property.

Let x be no of students.
Therefore (x *1/2) = x/2.
Now, increase value = 83 - 63 = 20.
Therefore x/2 = 20.
x = 40.

Let us consider p be the no. of pupil, x be the average.

Case1: Original marks 63/p=x -----> average of case 1.
Case2: New marks 83/p=x+1/2 ----> average of case 2.

2-1 => (83/p-63/p) = x+1/2-x.
=> 20/p = 1/2,
=> p = 40.

Let us consider p be the no. of pupil, x be the average.

Case1: Original marks 63/p=x -----> average of case 1.
Case2: New marks 83/p=x+1/2 ----> average of case 2.

2-1 => (83/p-63/p) = x+1/2-x.
=> 20/p = 1/2,
=> p = 40.

Thank you for explaining the answer.

I didn't understand the logic. Please explain.