Aptitude - Area - Discussion

Suppose side = 10, Error excess = 2%.
i.e. 10 + 2% = 10.2,
A 1 = 10 * 10 = 100,
A2 = 10.2 *10.2 = 104.4,
A2 - A1 = 104.4 - 100 = 4.04%.

Suppose,

The Original Length of one side is 10m, Then area for this is 100m² &
Length when 2% error is excess is 10+0.2, it means now the length of a side is 10.2m and the area for this is 104.04m².
Now,
The area due to error excess - Original area = 104.04 - 100.
= 4.04.
Also, we can assume the length of a side is 100m or whatever we want.

Here, it is mentioned that the error excess of 2% so for that reason we add the value of the error on the original side instead of subtracting.

It is simple;

Error is 2% so, take it 1.02.
Now one side of a square is 'a'.
Area = (a*1.02)*(a*1.02),
= (a*a*1.0404).

So, the total error in area = 4.04%.

We can also solve this is in 404/10.

Why is this method wrong?

A = a^2;
dA = 2*a*da;
dA/A = 2*(da/a).

We know da/a = 2%. Hence, dA/A = 4%.

I don't seem to understand the reason why this is wrong? Can someone explain?

It can be also done as.

Assume side of the square. Let S = 100 excess error is 2% i.e., 102.

Now Area of the Square A = S*S.

A = 100*100 = 10,000 cm2 (with out error) 1.

A = 102-102 = 10,404 cm2 (with 2% excess error) 2.

Difference b/w 1 and 2 is 404.

i.e, 404/100 = 4.04%.

Hope it will help you.

The formula X + Y + (X*Y/100) is the best for square and rectangle.

Percentage error=
( Experimental value - theoretical value )/theoretical value *100.

Therefore,
Here percentage error= [(102)^2 - (100)^2]/(100)^2,
= 404*100/(100)^2.

You are right, Thanks @Harsh.

How? Explain the answer please.