Aptitude - Average - Discussion

Here the simple solution;

A pupil's marks -> one student marks entered wrong.
Here 's1' means student mark.

Let (S1+S2+...63) = S total marks, 63 is the one student marks. So we don't know about remain students marks.thats why I take it as S1,S2.
Like this;
Marks entered wrong here that's 83.

(S1+S2+....83) = (S1+S2+....63) + 20 we can write like this isn't it.
(S1+S2+....83) = S +20 we can also write this.

Avg = total marks/total students.

Avg = S/X -------> (1) when correct marks(63).
Avg + 1/2 = S+20/X ----> (2) when 83 marks.

Substitute (1) in (2).

S/X + 1/2 = (S+20)/X.
( 2(S) +X ) / 2X = ( S+20 )/X

Now cross multiple.
X ( 2(S) + X ) = 2X ( S + 20).
Now cancel one 'x' from both sides.

2(S) + X = 2 ( S + 20 ) => 2S + X = 2S + 40.

Now, 2S will cancel.
Then, X = 40.

Hope it helps.

Sum1 = sum of (n-1) students + 83.
Sum2 = sum of (n-1) students + 63.
avg1 = Sum1/n;
avg2 = Sum2/n.

So, avg1-avg2 = 0.5.
Or, (83-63)/n = 0.5.
Then n=40.

Increased average = 0.5,
Increased marks = 20,
Average = sum/n.
0.5 = 20/n,
n = 40.

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.

1/2 means 0.5.

Because of difference of 20 this avg 0.5 is increased.
So distributing 0.5 equally to get total 20.

We divide 20/0.5 will give no.of student.
i.e. 20/0.5 = 40 student.

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.