2006-11-08 06:25:43 · 33 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Definitely 49
2006-11-08 06:33:51 · answer #1 · answered by Anonymous · 1⤊ 0⤋
35 years at 5 days a week = x years at 7 days a week
35/5 = x/7
x = 7 x 7
x = 49
Her real age is 49 years old.
For anyone trying to work this using 2 extra days per week there are two flaws in the logic. First there aren't exactly 52 weeks in a year. Second, you need to account for 2 extra days for *49* years (her real age) not her imagined age. You can see how this would be hard because you are trying to figure her real age and don't know it yet...
35 is 5/7 of her real age, so her real age is 35 * 7/5 = 49 years. It's that easy.
2006-11-08 06:36:02 · answer #2 · answered by Puzzling 7 · 3⤊ 0⤋
There are 7 days in each week. 2 of those days are Saturday and Sunday. this suggests that 2/7 of each week includes Saturday and Sunday. To count quantity one's age, all days of the week ought to be secure. subsequently, in elementary terms 5/7 is counted in to make 35 years. We ought to discover the genuine age. Make a proportion: 35:? ; 5:7 or 35 / ? ; 5 / 7 (manufactured from the extremes/outer numbers is comparable to the manufactured from the means/inner numbers) 245 = 5? Divide the two sides via 5 to isolate (?) ? = 40 9 the lady's genuine age is 40 9.
2016-12-14 03:46:31 · answer #3 · answered by tollefson 4 · 0⤊ 0⤋
Every year she lied about 104 days (52 weeks x 2 days)
In 35 years that would be 35 x 104 = 3540 days = 9 years, 355 days
She says she is 35, but she is actually 44, nearly 45. If today was her birthday, of course!
If her birthday was more than 10 days ago, assuming there were no saturdays or sundays, then she would definitely be 45. And a bit.
Tell her to stop lying about her age, will you?
2006-11-08 06:34:45 · answer #4 · answered by superman in disguise 4 · 0⤊ 1⤋
About 45
2006-11-08 06:42:44 · answer #5 · answered by Anonymous · 0⤊ 1⤋
She was counting only 5/7 of the days (leaving out Sat+Sun)
5/7 of age = 35
7/7 of age = 35 x 7/5
=49
7/7 (whole ) of age = 49
2006-11-08 09:28:16 · answer #6 · answered by rosie recipe 7 · 0⤊ 0⤋
Divide by 5, multiply by 7, she's 49 in that cruel world that doesn't let a lady count the way she likes! I was 27 for three years (26/27/28 in that other world) just cuz I liked the number... then jumped to 29, another nice number.
If you follow her system however, she's 35 on weekdays, and 14 on the weekend.
2006-11-08 06:29:45 · answer #7 · answered by Holly 3 · 1⤊ 0⤋
Intuitively I'd say (7/5)*35 = 49.
She counts a week as 5 days rather than 7 so she undercounts by the factor 5/7. So after 5 of her weeks she'd actually aged 7 normal weeks, a factor of 7/5.
I feel confindent about my ratio of 7/5...
Notice how much more difficult it would be to work this problem by trying to add 2 days for every week in 35 years... yuck. Recognizing that it's a ration is the important part of solving this problem.
2006-11-08 06:30:38 · answer #8 · answered by modulo_function 7 · 1⤊ 1⤋
Sat & Sun are 2/7 of all days so her age is
35/(1-2/7)=35/(5/7)=35*7/5=49
2006-11-08 06:33:04 · answer #9 · answered by yupchagee 7 · 1⤊ 0⤋
She only counted 5 days out of every 7, meaning she reported 5/7 of her true age. Multiply what she said by 7/5 to get the real number. 35*(7/5) = 49.
2006-11-08 06:29:26 · answer #10 · answered by DavidK93 7 · 1⤊ 1⤋