Pls help. How do you calculate the time difference between 2 cells? Eg A1=08:23, B1=13:40, C1=formula??  Also if I need cell C1 to be 'blank' if A1 & B2 are blank, what would the formula be?
199403
Put this in C1:\n\n=(12-A1)-(12-B1)\n\nHave the cell formated to time.
32283
TIME(HOUR(A1)+16, MINUTE(A1), SECOND(A1)) \n\nworked fine for me when I had A1 as\n\n="04:00"\n\n(Note the equal sign and quotation marks.)
505545
Work in 24 hour times.\n\n10:18-16:23\n\n16-10 is 6 hours.\n23-18 is 5 minutes.\n\n6hr 5mins.\n\nFor an automatic way of doing it, visit: \nhttp://www.timeanddate.com/date/timeduration.html\n\nJust enter the hours and minutes, leave the days and months blank. It will accept 4pm or 16 as an entry for hours.\n\nYet another way, and good if you want to do them in batch. Use Microsoft Excel. Enter all the start times in 24 hours format in one column, and all the end times in 24 hour format in the next column. In a 3rd column, in the first cell type = then click the top cell in the 2nd column. now type - and click the top cell in the first column and hit enter. This calculates the first row. Now click on the result cell again, and you'll notice a box in the bottom right corner, hold and drag this (should get a + symbol) drag down to the bottom row, and it auto fills in all your formulas.
27573
According to my Excel Spreadsheet.\n\n03-20-08 minus 01-11-06 equals 799 Days\n\nIt is very easy. You must configure the first two cells as dates.\n\nBut then you have to format the resulting cell as a number
270648
An alternative method:\nThe times taken by A, B and C to complete one lap around the track would be 10/7, 10/8 and 10/9 mins, respectively. Now we just need the LCM of these three to find out when they will all catch up with each other. \n\nThis turns out to be 10 mins.
389252
The easiest way if you use the default sheet names Sheet1, Sheet2 .... Sheet30 \n\n=SUM(Sheet1:Sheet30!A1) \n\nIf you use custom sheet names: \n\nInsert a new sheet before the first sheet and name it START. Insert another new sheet after the last sheet and name it END. \n\n=SUM(Start:End!A1) \n\nIf you will need to create any new sheets that should also be used in this calculation just insert them between START and END.
469570
Let D = distance between home and school.\n\nUsing the formula : distance = speed * time, we have :\n\nFrom home to school : D = x * T1\nFrom school to home : D = kx * T2\n\nTherefore, x * T1 = kx * T2\nDividing through by x gives :\nT1 = k * T2\n\nWe know that T1 + T2 = 1 (total time travelled)\nTherefore, T2 = 1 - T1.\n\nSubstituting this into T1 = k * T2 gives :\nT1 = k (1 - T1)\n\nFrom this, we find that :\nT1 = k / (k + 1)\n\nSubstituting into 'home to school equation' gives :\nD = x * k / (k + 1)\n\nTherefore total distance covered\n= 2D\n= 2 * x * k / (k + 1) miles.
37166
That's a tough 8th grade question; I've seen similar ones in algebra 2.  You have to figure the amount of the job that gets done in one hour. If x is the total time for the faster machine, 2x is the total time for the slower machine.  Together it takes 3 hours, so together they can do 1/3 of the job in one hour.  The slow machine can do 1/x of the job in an hour; the faster, 1/(2x).  This gives the equation:\n\n1/x + 1/(2x) = 1/3\n\nmultiply both sides of this equation by the LCD (6x) as if it were a giant distributive property example and get:\n\n6 + 3 = 2x       (the denominators all cancel out)\n\nso 9 = 2x (the slower machine takes 9 hrs)\n4.5 = x (the faster one takes 4 1/2 hours)
257824
Type "=B1-C1+D1*A1" without the quotes inside another cell such as E1.  This is assuming that your equation is the correct algebraic form and solved for one variable.  Your text is a bit jumbled so I didn't parse all of it.
673313
Tim is right.  i just want to explain it further so you will understand.\n\nso the problem is we are looking for the time spent in both trips. lets call it "X". one is 1 hour longer than the other so we can call it "X + 1". \n\nif trip1 with a distance of 180 miles is taken in X with an additional 1 hour we can call this equation as 180 per X+1 or in mathematical form its 180/(X+1)\n\nif trip 2 with a distance of 135 miles is taken in X we can call it 135 per X or 135/X\n\nwe have now 2 equations 180/X+1 and 135/X\n\nsince the rates are the same we can say that the rate for 180/(X+1) is equal to rate for 135/X.  or in equational form it is\n180/(X+1)=135/X.\n\nWe now have a working equation.  If you notice, we are dealing with fractions and to equate it we need to transpose it using cross multiplication.  Cross multiplication is done by pairing the numerator of one side to the denominator of the other side and then proceed to multiplication.\n\nlooking at the working equation, we will pair 180 and X or 180X and another pair which is 135 and (X+1) or 135(X+1)\n\nnow we have a new working equation 135(X+1) = 180X\n\nto simplify 135(X+1) we multiply 135 with X and with 1. 135 and X is equal to 135X and 135 and 1 is equal to 135\n\nnow we have simplified the equation to 135X + 135 = 180X\n\nto get the value of X, we combine similar terms.  they are 135X and 180X.  We combine this by transposition.  Remember that in order not to change the value of the equation we must change the sign of what is moved to another side.  In this case i will move 180X to pair it with 135X and move 135 to the other side. so now we get 135X - 180X = -135\n\nwe simplify the like terms and we get -45X.\n\nnow we have -45X = -135\n\nbut we are only looking for X so we must eliminate -45 from the X.  identify the numerical coefficient of X which is -45 and devide it from the original. what you do to one side, you should also do it to the other.\n\n-45X/-45 = -135/-45 with this equation we can simplify it into\nX=3\n\nnow we have the value for X\n\nso one trip will be 3 and since another is 1 hour longer then it must 3 + 1 = 4\n\nheres the solution\n\n180/(X+1)    =      135/X\n135(X+1)     =      180X\n135X + 135   =    180X\n135X - 180X   =    -135  \n            -45X   =    -135\n         -45X/-45=    -135/-45\n                   X =  3\n          \n             X + 1 =  trip 2\n             3+1 = 4
534536
If t is the time that has been measured in the clock's reference frame, then t/2 is the time that has been measured by the observer.\n\nNow t = (t/2)[1-(v^2)/(c^2)]^(1/2)\n\nthe t's cancel and you are left with 1 = (1/2)[1-(v^2)/(c^2)]^(1/2)\nor [1-(v^2)/(c^2)]^(1/2) = 1/2\n\nsquare both sides to get 1 - (v^2)/(c^2) = 1/4\n\nsubtract one from each side to get -(v^2)/(c^2) = -3/4\n\nor (v^2)/(c^2) = 3/4\n\nthis means that v^2 = (3/4)(c^2)\n\ntaking the square root of both sides gives the final answer:\n\nv = [3^(1/2)]c/2.\n\nNot the most intuitive answer in the world.
560635
A cell can either contain data or a formula, not both.\n\nYou will need to create a macro and assign it to the onChange event for that cell.
81422
Try this:\n=IF(A1>0,A1-(A1*0.1),A1) \nin cell C1. \n\nAlso, you didn't specify what happens when A1=0. If C1 = A1, then the above will work, if not, try this:\n=IF(A1>0,A1-(A1*0.1),IF(A1<0,A1,Zero Value))
41587
I assume both Worksheets are in the same file...\n\nIf so, you simply do the following:\nIn the second worksheet in the cell you want to display something from the first worksheet, type:\n\n=SheetName!CellName\n\ni.e. if your info is on a Worksheet called Sheet1 and the cell is B1\n\n=Sheet1!B1
9934
If you're talking about Excel from Microsoft, heres a couple pointers\n\n1) every formula starts with an equals sign '='\n2) there are countless formulas you can use, your best bet would be to click on the '=' in the upper left of the window and a box will pop up to help you.
246498
Ok, first you want to see where they meet. To make this easier, advance train A for 5/6ths of an hour and see the problem in terms of when train B starts out. 65 x 5/6= 325/6\nkilometers. So the problem is the same only they start out at the same time over a distance of 600- 325/6 kilometer =\n545 5/6 kilometers.\n\nNow, they travel in the proportion of 65/105 kilometers per hour. So the distance till they meet will be in the same proportion. That is 545 5/6 x 65/105 = 212875/630 or about 338 kilometers for train A to travel. Since it travels at 65 kph, this means it travels for 5.2 hours from the new start time equal to B's which was 6:20 AM. Ends up being 11:32AM.
358403
Tokyo Japan time\nhttp://www.timeanddate.com/worldclock/city.html?n=248\n\nCairo Egypt time\nhttp://www.timeanddate.com/worldclock/city.html?n=53\n\nIdeally there is a 7 hour time difference between them wrt GMT. However, since you see Egypt is running at 1 hour daylight saving time, the time difference now is 6 hours!\n\nHope this helps :-)