Millicent Bystander
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« on: January 20, 2009, 01:16:40 PM » |
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I need to work out some windows prices for work, as I am pretty clueless when it comes to Excel.
What I need to do is reduce the manufacturer's price by 62%
Add 50%
Add the glass price
Then do that 20,000 odd times.
Can anyone help?
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Chaos
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« Reply #1 on: January 20, 2009, 01:20:46 PM » |
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add 50%: =sum("cell"/100*150)
add glass price: =sum("cell"+"glassprice")
to do it more than once, click the little box in the bottom right of the cell and drag it down till you have it right....
hope that helped
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mark
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« Reply #2 on: January 20, 2009, 01:23:18 PM » |
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Didn't you ask this once before?
What Chaos said should work.
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Junta
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« Reply #3 on: January 20, 2009, 01:38:39 PM » |
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Just do "=(((cell * 0.38) *1.5)) + Glassprice) ? Sorry - I didn't read the original question properly, so that's why I've edited this again 
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« Last Edit: January 20, 2009, 01:42:51 PM by Junta »
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Millicent Bystander
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« Reply #4 on: January 20, 2009, 01:45:43 PM » |
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Had to do something similar but less complicated
Edit: Thanks Junta, works a treat, now to see if my OCR reader is accurate or I'll have alot of tedious typing to do
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« Last Edit: January 20, 2009, 01:51:42 PM by Millicent Bystander »
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« Reply #5 on: January 20, 2009, 01:45:58 PM » |
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add 50%: =sum("cell"/100*150)
Just do "=(((cell * 0.38) + ("cell" *1.5)) + Glassprice) ? I think you should devide that into two cells: one with the =cell*.38 and one with: =(newcell*1.5)+glassprice =cell*.38 reduces the price to 38% of the price (should equal to retracting 62%)
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Junta
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« Reply #6 on: January 20, 2009, 01:47:36 PM » |
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add 50%: =sum("cell"/100*150)
Just do "=(((cell * 0.38) + ("cell" *1.5)) + Glassprice) ? I think you should devide that into two cells: one with the =cell*.38 and one with: =(newcell*1.5)+glassprice =cell*.38 reduces the price to 38% of the price (should equal to retracting 62%) Yes, it is more readable in two cells
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Terror
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« Reply #7 on: January 20, 2009, 02:23:22 PM » |
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Should it not be:
=[CELL]*0.57+[GLASS]
I'm assuming it's an extra 50% after the reduction, not of the original value?
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r007
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« Reply #8 on: January 20, 2009, 03:00:35 PM » |
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If you want to get really fancy have the 62% reduction and 50% increase factors in seperate cells and reference them appropriately, so you can play with the values and not have to change 20.000 cells again. But whatever works I guess. 
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Millicent Bystander
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« Reply #9 on: January 20, 2009, 03:13:55 PM » |
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Junta's first way works fine, thanks for the help though, getting a page done in the time it took to do 1 line.
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Terror
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« Reply #10 on: January 20, 2009, 03:23:44 PM » |
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Yeah, I missed that. Mine was the same as John's, just I combined the 0.38 and the 1.5.
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Junta
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« Reply #11 on: January 20, 2009, 03:38:03 PM » |
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I combined the 0.38 and the 1.5.
Feel free to correct me, but I'd have thought that was not possible as two separate % values required for calculation? I'm probably missing your point lol 
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Terror
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« Reply #12 on: January 20, 2009, 03:52:52 PM » |
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I combined the 0.38 and the 1.5.
Feel free to correct me, but I'd have thought that was not possible as two separate % values required for calculation? I'm probably missing your point lol  (x*0.38)*1.5≡x*(0.38*1.5) Therefore: (x*0.38)*1.5≡x*0.57 I don't see why that would be wrong, but I did (nearly) fail maths last year.
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claws
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« Reply #13 on: January 20, 2009, 03:56:35 PM » |
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It's associative that way, but I guess Junta thought he meant 0.38x + 0.50x which is two operations unless you simplify it to 0.88x 
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Junta
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« Reply #14 on: January 20, 2009, 06:05:01 PM » |
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I combined the 0.38 and the 1.5.
Feel free to correct me, but I'd have thought that was not possible as two separate % values required for calculation? I'm probably missing your point lol  (x*0.38)*1.5≡x*(0.38*1.5) Therefore: (x*0.38)*1.5≡x*0.57 I don't see why that would be wrong, but I did (nearly) fail maths last year. Nothing wrong - a perfect explanation  ty Me being thick and not considering that as an option - I guess my work usually leads me to separate variables that might change.
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« Last Edit: January 20, 2009, 06:07:14 PM by Junta »
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