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percentage difference in practical

question 1 (b) where it talks about the percentage diiference I am just wondering what expression to use when calculating the percentage difference for n number of paper clips which im assuming is derived fron L= nc - 2d (n-1).

the mark scheme mentions about the expression to use for calculating percentage difference - 100 x ( 2d/c - 2d/nc) not sure how they achieved that?

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Avatar for Phichi
Phichi
11
Off the top of my head the most logical way of thinking about it would be this:

The student suggests that L≈ncL \approx nc, thus Lnc≈1\dfrac{L}{nc} \approx 1

100Lnc100\dfrac{L}{nc} represents the percentage, i.e, if nc=1.2mnc = 1.2m and L=1mL = 1m, nc is 120% L. I'm assuming that nc≥Lnc \geq L, as from the equation, L is equal to nc minus some value.

From the equation:

Lnc=1−2dnc(n−1)\dfrac{L}{nc} = 1 - \dfrac{2d}{nc}(n-1)

Converting this into a percentage as before:

100Lnc=100−100(2d)nc(n−1)100\dfrac{L}{nc} = 100 - \dfrac{100(2d)}{nc}(n-1)

From this it is clear that 100(2d)nc(n−1)\dfrac{100(2d)}{nc}(n-1) is the percentage difference.

If this isn't clear to you, if the percentage difference was 0, then we would have L=ncL = nc, or Lnc=1   ⇒   100Lnc=100\dfrac{L}{nc} = 1 \, \, \, \Rightarrow \, \, \, 100\dfrac{L}{nc} = 100

The percentage difference being 0 infers 100(2d)nc(n−1)=0\dfrac{100(2d)}{nc}(n-1) = 0

Remember 100(2d)nc(n−1)≡100×(2dc−2dnc)\dfrac{100(2d)}{nc}(n-1) \equiv 100 \times \left(\dfrac{2d}{c} - \dfrac{2d}{nc}\right) as you posted, to avoid any confusion.
(edited 9 years ago)
Reply 2
Avatar for MSB47
MSB47
OP
2
Original post by Phichi
Off the top of my head the most logical way of thinking about it would be this:

The student suggests that L≈ncL \approx nc, thus Lnc≈1\dfrac{L}{nc} \approx 1

100Lnc100\dfrac{L}{nc} represents the percentage, i.e, if nc=1.2mnc = 1.2m and L=1mL = 1m, nc is 120% L. I'm assuming that nc≥Lnc \geq L, as from the equation, L is equal to nc minus some value.

From the equation:

Lnc=1−2dnc(n−1)\dfrac{L}{nc} = 1 - \dfrac{2d}{nc}(n-1)

Converting this into a percentage as before:

100Lnc=100−100(2d)nc(n−1)100\dfrac{L}{nc} = 100 - \dfrac{100(2d)}{nc}(n-1)

From this it is clear that 100(2d)nc(n−1)\dfrac{100(2d)}{nc}(n-1) is the percentage difference.

If this isn't clear to you, if the percentage difference was 0, then we would have L=ncL = nc, or Lnc=1   ⇒   100Lnc=100\dfrac{L}{nc} = 1 \, \, \, \Rightarrow \, \, \, 100\dfrac{L}{nc} = 100

The percentage difference being 0 infers 100(2d)nc(n−1)=0\dfrac{100(2d)}{nc}(n-1) = 0

Remember 100(2d)nc(n−1)≡100×(2dc−2dnc)\dfrac{100(2d)}{nc}(n-1) \equiv 100 \times \left(\dfrac{2d}{c} - \dfrac{2d}{nc}\right) as you posted, to avoid any confusion.


That makes sense, thanks a lot for your help :smile:

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