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.
| Percentage error = |
 |
404 | x 100 |  % |
= 4.04% |
| 100 x 100 |
Discussion:
40 comments Page 3 of 4.
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)
Engr. Md. Easir Arafat said:
1 decade ago
Let's the side to be 1.
Here calculated (1+(2/100)) = 1.02.
Hence the area becomes (1.02*1.02) = 1.0404 but the actual area was supposed to be (1*1) = 1.
The difference = 1.0404-1.00 = 0.0404 = 4.04/100 = 4.04%.
Note: Although taking only two decimals is conventional but in such types of small quantities we should take more to minimize the affect of error as much possible.
Here calculated (1+(2/100)) = 1.02.
Hence the area becomes (1.02*1.02) = 1.0404 but the actual area was supposed to be (1*1) = 1.
The difference = 1.0404-1.00 = 0.0404 = 4.04/100 = 4.04%.
Note: Although taking only two decimals is conventional but in such types of small quantities we should take more to minimize the affect of error as much possible.
Kasinath @Hyd said:
1 decade ago
Correct value is 100*100.
Measured value is 102*102.
Therefore ERROR Value is (102*102)-(100*100) = 404.
Error% = (error value/ true value)*100.
=> (404/100*100)*100 = 4.04.
Measured value is 102*102.
Therefore ERROR Value is (102*102)-(100*100) = 404.
Error% = (error value/ true value)*100.
=> (404/100*100)*100 = 4.04.
Lipu said:
1 decade ago
Error % = (error/true value)*100.
So (404/100*100)*100 = 4.04
So (404/100*100)*100 = 4.04
Nitin said:
1 decade ago
%error = a+b+ab/100.
a and b are +ve and -ve as per increment and decrement respectively.
a and b are +ve and -ve as per increment and decrement respectively.
Aarti said:
1 decade ago
Solution:
x+y+x*y/100 = 2+2+2*2/100 = 4.04.
x+y+x*y/100 = 2+2+2*2/100 = 4.04.
Aditi said:
1 decade ago
Why are we taking 100 for this sum when it is not mentioned in the question. Please explain?
Bhavana said:
1 decade ago
Why are we taking 100 for this sum when it is not mentioned in the question. Please explain?
User said:
1 decade ago
@Sreekanth you are right. Thanks for sharing with us. And this seems to be very easy method for solving these kind of problem.
Srikanth said:
1 decade ago
Remember this formula x+y+(xy/100).
For rectangle x and y are different, I mean length and breadth are different. Here is square so, only single element we can consider x and y as side of the square.
So (+ for increase and "-" for decrease) 2+2+(2*2/100) = 404/100 = 4.04%.
Lets check an example for rectangle if length increased 5% and breadth decreased 4% so the result is
5-4+(5*(-4))/100 = 0.8% over all decrease.
For rectangle x and y are different, I mean length and breadth are different. Here is square so, only single element we can consider x and y as side of the square.
So (+ for increase and "-" for decrease) 2+2+(2*2/100) = 404/100 = 4.04%.
Lets check an example for rectangle if length increased 5% and breadth decreased 4% so the result is
5-4+(5*(-4))/100 = 0.8% over all decrease.
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