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.

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.

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.

I didn't understand the logic. Please explain.

Please explain.

Can you explain it briefly please ?