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:
2%
2.02%
4%
4.04%
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 4 of 4.

Hasini said:   1 decade ago
Correct area = (100)^2.

Measured area = (102)^2.

So to get the excess area we have to subtract the incorrectly measured area by the correct area.

(102)^2-(100)^2.

(102+100)(102-100) .

202*2.

= 404.

So, 404/100.

= 4.4%.

Ramesh said:   1 decade ago
Sir Percentage error is not understanding. Please solve the answer.

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.

Bhavana said:   1 decade ago
Why are we taking 100 for this sum when it is not mentioned in the question. Please explain?

Aditi said:   1 decade ago
Why are we taking 100 for this sum when it is not mentioned in the question. Please explain?

Aarti said:   1 decade ago
Solution:

x+y+x*y/100 = 2+2+2*2/100 = 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.

Lipu said:   1 decade ago
Error % = (error/true value)*100.

So (404/100*100)*100 = 4.04

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.

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.


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