Percentage help

[petey]

I have TWO graphics cards in my computer. There is a new card coming out that is said to be 90% more powerful than BOTH cards COMBINED. How much more powerful would that be over ONE card.

Can I take 90% of 2, which is 1.8, than say 1.8 - 1 = .8, so the card is .8 more powerful than one card so .8/1 * 100 = 80% more powerful than ONE card?

Thanks

 

[DrPhil]

I have TWO graphics cards in my computer. There is a new card coming out that is said to be 90% more powerful than BOTH cards COMBINED. How much more powerful would that be over ONE card.

Can I take 90% of 2, which is 1.8, than say 1.8 - 1 = .8, so the card is .8 more powerful than one card so .8/1 * 100 = 80% more powerful than ONE card?

Thanks

90% more is a factor of 1.9. If each of your present cards is equal to x, then the power of the new card is ....
Go from there...

 

[JeffM]

I have TWO graphics cards in my computer. There is a new card coming out that is said to be 90% more powerful than BOTH cards COMBINED. How much more powerful would that be over ONE card.

Can I take 90% of 2, which is 1.8, than say 1.8 - 1 = .8, so the card is .8 more powerful than one card so .8/1 * 100 = 80% more powerful than ONE card?

Thanks

If the new card is 90% stronger than both cards combined and the new card is only 80% more powerful than one of the cards alone, then you can get 5.6% more power by throwing away one of your current cards and forgetting about the new card.

Let's say the power of the two old cards combined is 100 units.

So the power of the new card is 190 units.

But if the new card is only 80% better than a single old card, then 1.8 times the power of the old card = 190 units,
which means that the old card's power is 190 units / 1.8 = 105.6 units. The two old cards are interfering with each other.

 

[petey]

90% more is a factor of 1.9. If each of your present cards is equal to x, then the power of the new card is ....
Go from there...

But it is not 90% more than one card, that is clearly a factor of 1.9, it is 90% faster than BOTH COMBINED, so 90% of 2

 

[JeffM]

But it is not 90% more than one card, that is clearly a factor of 1.9, it is 90% faster than BOTH COMBINED, so 90% of 2

If the power of two old cards combined is 100 units and they do not degrade each other, what is the power of one old card? 50 units, right?

And if the power of the new card is 90% better than the power of the two cards combined, which is 100 units, then the new card has a power of 190 units. And 190 / 50 = 3.8. And 3.8 - 1 = 2.8 = 280% better than one of the old cards (assuming that combining old cards results in no degradation of power.)

 

[DrPhil]

But it is not 90% more than one card, that is clearly a factor of 1.9, it is 90% faster than BOTH COMBINED, so 90% of 2

190% of 2

 

[petey]

ACK I just read my original post again and I made a HUGE mistake. I cannot believe it as I read it three times before posting

Here is how it SHOULD read

I have TWO graphics cards in my computer. There is a new card coming out that is said to be 90% LESS (NOT more) powerful than BOTH cards COMBINED. How much more powerful would that be over ONE card

Looks like I was correct after all. I did the math for this problem, not the one I posted

 

[JeffM]

ACK I just read my original post again and I made a HUGE mistake. I cannot believe it as I read it three times before posting

Here is how it SHOULD read

I have TWO graphics cards in my computer. There is a new card coming out that is said to be 90% LESS (NOT more) powerful than BOTH cards COMBINED. How much more powerful would that be over ONE card

Looks like I was correct after all. I did the math for this problem, not the one I posted

If the two old cards combined have a power of 100 units and there is no interference, then each old card has a power of 50 units and the new card has a power of 10 units. The new card is 80% LESS powerful than just one of the old cards.

 

[petey]

If the two old cards combined have a power of 100 units and there is no interference, then each old card has a power of 50 units and the new card has a power of 10 units. The new card is 80% LESS powerful than just one of the old cards.

No, if the power of the new card is 90% less than the power of the two cards combined, which is 100 units, then the new card has a power of 90 units. And 90 / 50 = 1.8. And 1.8 - 1 = .8 = 80% better than one of the old cards

Which makes sense, if one card is close to two cards in performance it needs to be faster than one card alone

 

[JeffM]

No, if the power of the new card is 90% less than the power of the two cards combined, which is 100 units, then the new card has a power of 90 units. And 90 / 50 = 1.8. And 1.8 - 1 = .8 = 80% better than one of the old cards

Which makes sense, if one card is close to two cards in performance it needs to be faster than one card alone

90 units is not 90% less than 100 units. It is 10% less.

Of course if they say that the new card has 90% of the power of two combined old cards, that is a different problem.

 

[petey]

90 units is not 90% less than 100 units. It is 10% less.

Of course if they say that the new card has 90% of the power of two combined old cards, that is a different problem.

I suppose my English skills are worse than my math skills. That is what they meant and I expressed it poorly. It all started by saying it was 90% more, than switching more to less, when I should have said it was 90% AS powerfull as the new card. I had it correct in my head which is why I said the new card was 80% faster than my old card, I just screwed up the question <sigh> At least my math was correct all along