Help with using binomial theory to calculate error

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corn_she

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Hi All,

I am struggling with the question below. Does anyone know how to solve it?

A square garden slab was measured 2% too large. This measurement was used to calculate the amount of slabs needed for a square patio. The patio is required, by the customer, to be exactly 6 m2 and the slabs are 0.36 m2. What would the resulting error in the area be? What are your conclusions with respect to laying the patio?

Any help would be appreciated.
 
corn_she said:
Hi All,

I am struggling with the question below. Does anyone know how to solve it?

A square garden slab was measured 2% too large. This measurement was used to calculate the amount of slabs needed for a square patio. The patio is required, by the customer, to be exactly 6 m2 and the slabs are 0.36 m2. What would the resulting error in the area be? What are your conclusions with respect to laying the patio?

Any help would be appreciated.
Click to expand...
When you say "2% too large" is that 2% of the length or 2% of the area? You say "the slabs are 0.3 m2". Is that what they are supposed to be or the area with the 2% error0?

If you mean the error is 2% of the area of each slab, and each slab was supposed to be 0.3 m[sup2[/sup], then there should have been \(\displaystyle \frac{6}{.3}= 20\) slabs. If each slab is actually \(\displaystyle 0.3+ (.02)(0.3)= (1.02)(0.3)= 0.306\) m2 then 20 slabs would have an area (0.306)(20)= 6.12 m2. There is an error of 0.12 out of a base value of 6.
 
Is that "2% too large" 2% of a length or the area? And are we to assume the patio and slabs are square?
 
The patio is 6 square meters. If the slabs were the correct 0.36 square meters, then you would require \(\displaystyle \frac{6}{0.36}\)= 16 and 2/3 slabs. If the slabs are 2% too large then they are, in fact, 0.02(0.36)= 0.0072 square meters too large or 0.36+ 0.0072= 0.3672 square meters. 16 and 2/3 slabs would be \(\displaystyle \frac{50}{3}(0.3672)= 6.12\) square meters so 0.12 square meters too large.
 
Has anyone else got any answers? I've worked it out and get different answers.
 
corn_she said:
Has anyone else got any answers? I've worked it out and get different answers.
Click to expand...
If you were to show your work, we might be able to tell where we differ in some assumptions or where possible mistakes were made.
 
Here is my working


The area of one slab is 0.36 sq m, however the was 0.02 too large.
Patio Area needs to exactly 6 sq m.
No. of Slabs = 6/0.36=16.66667
However No of slabs including error = 6/0.34= 17.64706
Resulting Error = ((6/0.34)*0.36)-(6/0.36*0.36) =0.352941 sq m
 
corn_she said:
Here is my working


The area of one slab is 0.36 sq m, however the was 0.02 too large.
Patio Area needs to exactly 6 sq m.
No. of Slabs = 6/0.36=16.66667
However No of slabs including error = 6/0.34= 17.64706
Resulting Error = ((6/0.34)*0.36)-(6/0.36*0.36) =0.352941 sq m
Click to expand...
The difference between what you have and what is presented by HallsofIvy for example, is that you have assumed the 0.36 m2 tile is the one measured incorrectly and HallsofIvy has assumed, as I would have, that the 0.36 m2 tile is the one measured correctly. Thus you would be dividing the 0.36 by 1.02 [BTW: that 0.34 is incorrect in that case] instead of multiplying.

Edit: Looking at the title of the post: Even if you used the binomial theorem to get
1/1.02 ~ 1 - .02 + .0004
the 0.34 you have is still incorrect
 
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Ishuda said:
The difference between what you have and what is presented by HallsofIvy for example, is that you have assumed the 0.36 m2 tile is the one measured incorrectly and HallsofIvy has assumed, as I would have, that the 0.36 m2 tile is the one measured correctly. Thus you would be dividing the 0.36 by 1.02 [BTW: that 0.34 is incorrect in that case] instead of multiplying.

Edit: Looking at the title of the post: Even if you used the binomial theorem to get
1/1.02 ~ 1 - .02 + .0004
the 0.34 you have is still incorrect
Click to expand...

So HallsofIvy is correct? What would the answer be if 0.36m2 was measured incorrectly?
 
corn_she said:
So HallsofIvy is correct? What would the answer be if 0.36m2 was measured incorrectly?
Click to expand...

HallsofIvy is correct under his assumption. If it were the 0.36 m2 which was measured incorrectly, and that 0.36 m2 area was t0o large by 2%, the area of the correctly sized tile would be 0.36/1.02 ~ 0.353 m2.
 
corn_she said:
Could someone show me how to work the answer out under this assumption?
Click to expand...

If one of the measurements is too large by, say, 2% then the ratio between the larger and smaller is 1.02. That is the large measurement (ML) divided by the smaller measurement (MS) is 1.02. Thus
ML / MS = 1.02
or
ML = 1.02 MS

Halls & I assumed the 0.36 was the smaller one (MS). So the larger measurement (ML) would be
ML = 1.02 MS = 1.02 * 0.36 = 0.3672

You think the problem says the 0.36 was the larger one (ML). So
ML = 1.02 * MS = 0.36
or
MS = ML/1.02 = 0.36/1.02 ~ 0.3529

Notice that if we use the binomial theorem to approximate the reciprocal of 1+x we have, with x = .02,
1/1.02 ~ 1 - .02 = 0.98
and
MS = ML/1.02 ~ 0.36 * 0.98 ~ 0.3528
Pretty close
 
Has Halls worked the answer out using binomial theory? How do you come to answer using the other assumption like he did?
 
corn_she said:
Has Halls worked the answer out using binomial theory? How do you come to answer using the other assumption like he did?
Click to expand...
If x is small (and 0.02 is considered small) then from the binomial theorem,
1/(1+x) ~ 1 - x

Thus if
ML = MS * (1 + x)
and x is small, then
MS ~ ML * (1 - x)
If MS is given the binomial theorem isn't needed.
 
Ishuda said:
If x is small (and 0.02 is considered small) then from the binomial theorem,
1/(1+x) ~ 1 - x

Thus if
ML = MS * (1 + x)
and x is small, then
MS ~ ML * (1 - x)
If MS is given the binomial theorem isn't needed.
Click to expand...

Is hallsanswer correct using binomail theorem?
 
corn_she said:
Is hallsanswer correct using binomail theorem?
Click to expand...
Halls answer is correct if the 0.36 m2 is the smaller (and assumed correct) measurement independent of whether the binomial theorem is used or not. So, yes, HallsofIvy's answer is correct using the binomial theorem if the smaller measurement is the 0.36 m2
 
Ishuda said:
Halls answer is correct if the 0.36 m2 is the smaller (and assumed correct) measurement independent of whether the binomial theorem is used or not. So, yes, HallsofIvy's answer is correct using the binomial theorem if the smaller measurement is the 0.36 m2
Click to expand...

ishuda can you show me the workings like HallsofIvy has for the other scenario?
 
corn_she said:
ishuda can you show me the workings like HallsofIvy has for the other scenario?
Click to expand...
Copying his answer and making a few changes we have:
The patio is supposed to be 6 square meters. If the incorrect slabs are 0.36 square meters and are 2% too large then they are, in fact, 1.02 times the correct size and the correct size is (using the binomial theorem) about 2% less or .02*(0.36)=0.0072 square meters smaller large or about 0.36 - 0.0072 = 0.3528 square meters. The number of slabs required, using the correct measurement, is \(\displaystyle \dfrac{6}{0.3528} = 17.00 \)tiles. 17 tiles at 0.36 square meters each is 17*0.36 = 6.12 square meters or, using the incorrect tiles, the slab will be 0.12 square meters larger.
 
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