Aptitude - Average - Discussion

Hi. I have come up with one more simple solution:

1st case:

Lets say avg is X when marks are 63. So X = 63/n.

2nd case:

When marks were noted as 83, the avg is increased by 0.5 so,

X+0.5 = 83/n.

Now if you subtract 2nd case - 1st case then (83-63) /n = X+0.5-X.

So, 20/n = 0.5 and n = 20/0.5 so n = 40.

Have a great day dudes. !

Let total number be x.
Then it is given that average for 63 is A.
i.e, sum/x = A.

When it is wrongly entered as 83 then.
(sum-63+83) /x = A + 1/2.

Then (sum- 20) /x =A + 1/2.
(sum/x) - (20/x) =A+1/2.

Now pick from above that sum/x=A.
Now A- (20/x) =A + 1/2.

Thus, it brings x=40.

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.

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...

let x be the no.of students.
When it is 83, the avg is increased by 1/2.

So we get,
x(avg+1/2) = 83 --(1).

Also given that 63 is the correct mark so we get an equ like,
x(avg) = 63 --(2).

Now (1)-(2)we get,
x(avg)+x(1/2)-x(avg) = 83-63.

Therefore, x/2 = 20.

x = 40.

Its very simple given difference of avg is half hence,
(83/x)-(63/x) = 0.5.

Hence by solve x = 40.

@Midhun J.

You are right. I can so much relate with your answer.

Average of the total as it is increased by 20 can be written as:

= s+20/n.

= s+20/n = s/n*1/2 (as the original avg is increased by 1/2).

= 2(s+20) = s.

Hence s = 40.

Let the total marks be x,

Let the total no of students be x, and let the average marks of all the students be x.

Increase in the total marks = 83-63 = 20.

So, total marks/total students = avg marks.

=> x+40/x = x+1/2.

x gets cancelled on both sides of the equation.

Then the remaining terms are,

=> 20/x = 1/2.

=> x = 40.

Let x be the no. of pupil, y is the avg.

63/x = y,

83/x = y+1/2,

By dividing these 2 equations you will get y value. Substitute y in 1st equation, you get x as 40.

Simple.