Percentage Diff Calculation

Gator

Can somebody solve a quick maths q for me?

with the left hand i picked up 14 coins in 30 seconds. with the right i picked up 12 in 30 seconds. how much quicker am i with left? (percentage difference)

????

Michael Collins

Gator wrote: »

Can somebody solve a quick maths q for me?

with the left hand i picked up 14 coins in 30 seconds. with the right i picked up 12 in 30 seconds. how much quicker am i with left? (percentage difference)

????

Rate with Left Hand = 14/30 = 7/15 coins/sec
Rate with Right Hand = 12/30 = 6/15 coins/sec

Rate difference = 7/15 - 6/15 = 1/15 coins/sec

Fraction left is faster than right (1/15) / (6/15) = 1/6 = 16.6...%

Gator

Michael Collins wrote: »

Rate with Left Hand = 14/30 = 7/15 coins/sec
Rate with Right Hand = 12/30 = 6/15 coins/sec

Rate difference = 7/15 - 6/15 = 1/15 coins/sec

Fraction left is faster than right (1/15) / (6/15) = 1/6 = 16.6...%

thank you:)

MathsManiac

Michael Collins wrote: »

Rate with Left Hand = 14/30 = 7/15 coins/sec
Rate with Right Hand = 12/30 = 6/15 coins/sec

Rate difference = 7/15 - 6/15 = 1/15 coins/sec

Fraction left is faster than right (1/15) / (6/15) = 1/6 = 16.6...%

I would think that this is basing the percentage on the wrong rate. I would be inclined to say that the right is 16.6...% slower than the left, but the left is 14.285...% faster than the right. (i.e. (1/15) / (7/15).)

MathsManiac

MathsManiac wrote: »

I would think that this is basing the percentage on the wrong rate. I would be inclined to say that the right is 16.6...% slower than the left, but the left is 14.285...% faster than the right. (i.e. (1/15) / (7/15).)

With your left hand, you pick up 16.666% more coins in the same period of time, or you pick up the same number of coins in 14.285% less time.

Hmmmmmm.

Michael Collins

MathsManiac wrote: »

I would think that this is basing the percentage on the wrong rate. I would be inclined to say that the right is 16.6...% slower than the left, but the left is 14.285...% faster than the right. (i.e. (1/15) / (7/15).)

Hmm would that not be basing it on the wrong rate? I did wonder about which I should base it on. In the end, I decided that if the left hand were twice as fast as the right, so say for demonstrative purposes let the situation now be

Rate with Left Hand = 12/15 coins/sec
Rate with Right Hand = 6/15 coins/sec

the left now being twice as fast, intuitivaly I would consisder this as the left hand being 100% faster than the right. Would you agree?

Calc 1: [(12-6)/15]/[6/15] = 1 = 100%

Or with your way this would be the right hand is 100% slower than the left, but left is 50% faster than the right. To me this seems backwards, but there can be a certain level of ambiguity or choice, I guess.

Calc 2: [(12-6)/15]/[12/15] = 0.5 = 50%

MathsManiac

Michael Collins wrote: »

Hmm would that not be basing it on the wrong rate? I did wonder about which I should base it on. In the end, I decided that if the left hand were twice as fast as the right, so say for demonstrative purposes let the situation now be

Rate with Left Hand = 12/15 coins/sec
Rate with Right Hand = 6/15 coins/sec

the left now being twice as fast, intuitivaly I would consisder this as the left hand being 100% faster than the right. Would you agree?

Calc 1: [(12-6)/15]/[6/15] = 1 = 100%

Or with your way this would be the right hand is 100% slower than the left, but left is 50% faster than the right. To me this seems backwards, but there can be a certain level of ambiguity or choice, I guess.

Calc 2: [(12-6)/15]/[12/15] = 0.5 = 50%

Sorry! Yes, you're absolutely right and I was wrong. Don't know what I was thinking. One should base the percentage on the thing that comes after the "than". (You're very kind to suggest that there might be an ambiguity - I actually don't think there is.)