Aptitude - Average - Discussion

It is actually easy if you get the proper steps, actually I kind of struggled a bit for myself to understand with the limited steps given in answer but I kind of figured out the steps.

Note : I am just an aspiring student and so I don't know if my explanation is correct but its just method I found quite easy to understand

Do it using following steps:

Let the total marks without 1 pupils mark be = y.
Let the total no. of pupil be = x.

It is given that 1 pupil marks entered wrong as 83 and thus increases avg. marks by 1/2.

Therefore,

(y+83)/x = avg.marks + 1/2 -->(i).

But,

The pupil's actual mark is 63.

(y+63)/x = avg.marks --> (ii).

Sub (ii) for avg.marks in (i), then,

(y+83)/x = (y+63)/x +1/2.
=> (y+83)/x - (y+63)/x =1/2.
=> (y + 83 - y - 63)/x = 1/2 [ i have put in brackets to show whole divided by 'x' ].
=> (83- 63)/x =1/2.
=> 20/x = 1/2.
=> x = 40.

Therefore, there are 40 pupils in class.

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.

I hope this helps you all. It's quite easy if u just go step by step according to the question.

Let no.of pupils be x.
Let total marks be y.

Earlier
y/x = Avg...eq(1).

Later
Marks increased by 20 (as 83-63=20) and the average increases by 1/2
Hence,
(y+20)/x = (1/2+Avg)....eq(2).

Now clearly we see the relation in the above 2 equations so hence we try to divide them and let's see what do we get,

Dividing eq(1)/eq(2).

(y/x)*(x/y+20) = (Avg)/(1/2+Avg).

On solving. We get,
y=40*Avg.....eq(3).

No substitute this value of 'y' eq(3) in eq(1).
i.e.,
y/x=Avg....eq(1).

So after putting value of y it becomes,
(40*Avg)/x=Avg.
Hence, x=40.

This is the easiest method I have found so far...

Its simple,

Let us assume that the total number of pupils in class be x+1,

Now Let the sum of the marks of x students (whose marks have been entered correctly be S) , and let original average be A, then (S+63)/(x+1) = A; Equation 1.

And (S+83)/(x+1) = (A+1/2); Equation 2.

Divide Equation 1 by Equation 2;

(S+63)/(S+83) = (2*A)/(2*A+1) ;

Solving it, we get: S+63=40*A; Equation 3.

Now, put value of Equation 3 in Equation 1, we get:

(40*A)/(x+1) = A;

Solve it, we get: x+1=40.

And we have assumed the total students to be x+1. Hence total students = x+1 = 40.

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.

Let considered no students be - n.
and average be - a.
And total marks be - x.

So, if correction was applied then average marks obtained will be;
a= x/n i.e n=x/a ------------> eqn(1).

Now correction is not applied then average will increase by 0.5.
And marks will also increase by 20 (83-63).

Then average will be;
a+0.5=x+20/n i.e n=x+20/a+0.5 --------> eqn(2)
From eqn 1 and 2,
x/a=x+20/a+0.5,

By doing little aligabra
You get x=40a --------------------------> eqn(3)
Substitution eqn 3 in 1.

n = 40a/a
Then, n = 40.

Let's take the actual avg marks = X and the number of pupil as Y.

Given, 63 is the actual mark obtained by Y pupil.
Which gives us, X = 63/Y (This should have been the actual avg)
So, XY = 63 ---------> (1).

But instead, due to the error, it was replaced to 83 which increased the actual avg by half
So, the new eqn for the error should be,

X + 1/2 = 83/ Y (Since the number of pupil remains the same)
Solve: (2X+1)/2 = 83/Y.
2X+1 = 166/Y.
2XY+Y = 166 -------> (2)
(1) in (2).
2(63)+Y = 166.
Y = 40 (Number of pupil).

Let suppose average marks A.
Number of students be X.

If the marks entered correctly i.e. 63 then.

A = (Total_1) / X -----------> (1).

Total_1= A * X -------------> (2).

If the marks entered is wrong, ie 83 then,

A+ (1/2) = (Total_2) /X -----------> (3).
Total_2 = [A + (1/2)] * X-------------> (4).
Total_2 = A * X + X/2 -----------> (5).

And the difference between Total_1 and total_2 is (83 - 63 = 20).
Therefore Eq (5) - (2).

A * X + X/2 - A * X = 20.
X/2 = 20.
X = 40.

Once read the question, it's stated that. Instead of 83, it's wrongly entered as 63.

Now. As we know sum/num=avg ------->1 (when mark entered is correct).
83 - 63 = 20. This is the new difference when the mark is entered wrongly.

So, sum+20/num=avg+avg (1/2) ------->2 (when mark entered is the wrong Avg is increased by 1/2).

Sum + 20/num = avg (1 + 1/2).
Sum + 20/num = avg (3/2).

Substituting 1 in 2;

Sum + 20/num = sum/num (3/2).
Sum + 20 = 3/2sum.
Sum = 40.

Take average except for the person who's marks wrongly entered = x.
LET, total no of students = N.
AND, initial average =A.

So, CORRECT AVERAGE should be ----> (x+63)/N = A { EQ. 1}
and, WRONG AVERAGE calculated is----> (x+83)/N = A+(1/2) {EQ. 2} --->(i.e increase of the average marks by 1/2).


NOW, put the value of A from EQ.1 in EQ.2 and solve for N.
i.e, (x+83)/N=(x+63)/N + 1/2,
(x+83-x-63)/N=1/2,
20/N=1/2,
N = 40.