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Default conversion question

is there a formula to convert 5.08 into years/months? I know 5 = 5 years,
but can't figure out how to determine what .08 is. Thanks
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Maybe
=INT(A1)&" Years "&ROUND(MOD(A1,1)*12,2)&" Months"
where A1 = 5.08

which is 0.96. If you have specific start/end dates, you may want to look
at DateDif. Whether or not 0.08 of a year is one month depends on which
month you're talking about.


"Tammi P" wrote:

is there a formula to convert 5.08 into years/months? I know 5 = 5 years,
but can't figure out how to determine what .08 is. Thanks

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Hi,

Assuming your data is in A1 and that 0.08 is a decimal representation

B1 = DATE(2000+FLOOR(A1,1),1,MOD(A1,1)*365)

You could use the data at this point by custom formatting it,

or you could add two more cells with
=MONTH(B1)
=YEAR(B1)

and custom format them. Either way, it will work.


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Default conversion question

Thank you! I will see if there are specific start/end dates - although I'm
not sure how to use the DateDif - so I may post again.

"JMB" wrote:

Maybe
=INT(A1)&" Years "&ROUND(MOD(A1,1)*12,2)&" Months"
where A1 = 5.08

which is 0.96. If you have specific start/end dates, you may want to look
at DateDif. Whether or not 0.08 of a year is one month depends on which
month you're talking about.


"Tammi P" wrote:

is there a formula to convert 5.08 into years/months? I know 5 = 5 years,
but can't figure out how to determine what .08 is. Thanks

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JMB JMB is offline
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Default conversion question

You're welcome.

"Tammi P" wrote:

Thank you! I will see if there are specific start/end dates - although I'm
not sure how to use the DateDif - so I may post again.

"JMB" wrote:

Maybe
=INT(A1)&" Years "&ROUND(MOD(A1,1)*12,2)&" Months"
where A1 = 5.08

which is 0.96. If you have specific start/end dates, you may want to look
at DateDif. Whether or not 0.08 of a year is one month depends on which
month you're talking about.


"Tammi P" wrote:

is there a formula to convert 5.08 into years/months? I know 5 = 5 years,
but can't figure out how to determine what .08 is. Thanks

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