Aptitude - Area - Discussion
Discussion Forum : Area - General Questions (Q.No. 2)
2.
An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:
Answer: Option
Explanation:
100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
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![]() |
404 | x 100 | ![]() |
= 4.04% |
100 x 100 |
Discussion:
40 comments Page 1 of 4.
Akash Mane said:
2 years ago
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%.
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%.
(18)
Pawar Sagar Dayanand said:
2 years ago
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.
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.
(6)
Akil said:
3 years ago
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%.
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%.
(3)
Rohith said:
6 years ago
We can also solve this is in 404/10.
(2)
Harsh said:
10 years ago
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?
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?
(1)
Sohel said:
1 decade ago
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.
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.
(1)
Pranjal said:
9 years ago
The formula X + Y + (X*Y/100) is the best for square and rectangle.
(1)
Jobin said:
7 years ago
Percentage error=
( Experimental value - theoretical value )/theoretical value *100.
Therefore,
Here percentage error= [(102)^2 - (100)^2]/(100)^2,
= 404*100/(100)^2.
( Experimental value - theoretical value )/theoretical value *100.
Therefore,
Here percentage error= [(102)^2 - (100)^2]/(100)^2,
= 404*100/(100)^2.
(1)
HARSHAVARDHAN REDDY said:
6 years ago
Use the formula {A+B+(AB/100)}.
Here A=2 for side.
For area A=B.
2+2+2*2/100=4.04.
Here A=2 for side.
For area A=B.
2+2+2*2/100=4.04.
(1)
Dwaipayan said:
6 years ago
You are right, Thanks @Harsh.
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