joan and jane are twin sisters and right now she is 16 years old..and mon is 28.....2 years ack the respective ages were 12 and 24....mathematicaly correct question with a poor logic from the asker for biologically its very difficult for a 11 year old to conceive....very very rare phenomenon....
or i think its 28 24 and 12
dont waste time on this question....
2007-03-21 19:25:12
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answer #1
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answered by Anonymous
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Here are the equations:
(1) Jean = Jane - 12
(2) Joan = Jean * 2
(3) Joan - 4 = Jane
(4) Jane - 4 = Jean * 2
Solve the two Jean/Jane equations:
(4) becomes: Jane = (Jean * 2) +4
and then (1) becomes
Jean = ((Jean * 2) +4) - 12
or
Jean = 8
Checking the math:
(1) Jean = Jane - 12
so Jane = 20
(2) Joan = Jean * 2
so Joan = 16
(3) Joan - 4 = Jane
(correct)
(4) Jane - 4 = Jean * 2
(correct)
2007-03-21 19:26:34
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answer #2
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answered by Captain Al 2
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Jean is 16 years old now. Present Age of...
Joan =32 Years
Jane =28 Years
Jean =16 Years
Any way, a foolish way to pass the time!!!
2007-03-21 19:38:54
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answer #3
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answered by suryakant d 2
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Jean is 16 years old now.
so joan is 32 now and jane is 28. 4 years ago joan was 28 (as jane now) and jane was 24 (twice of then jean's age).
2007-03-21 19:27:20
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answer #4
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answered by dew 1
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Jene is 16 Years now.
Jane is 28 Years now.
Joan is 32 Years now.
4 Years ago
Jene was 12 Years
Jane was 24 Years
Joan was 28 Years.
(Yeeeeeeeeeeeeeeeeek)
2007-03-21 21:01:36
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answer #5
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answered by Rajasekaran M 1
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Jean is 16 years old now.
2007-03-21 19:31:46
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answer #6
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answered by Satish K 1
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Thats impossible to answer. It all depends on what age Joan had her daughter...and that can vary.
2007-03-21 19:19:39
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answer #7
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answered by Anonymous
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Oh My God, it's the SAT's all over again!!!! This must be a nightmare because I never knew these answers when I was in school and I don't know them now!!!!!!!
2007-03-22 04:10:14
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answer #8
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answered by moviegirl 6
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sorry dear, this quetion could have been answered through Linear Equations in 1 variable, but you have not provided any constant!!
2007-03-21 19:22:50
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answer #9
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answered by Anonymous
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