Image by JND90745, distributed under CCL.
Your answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.
Friday, March 16, 2012
Celebrity Age Puzzle
Image by JND90745, distributed under CCL.
Your answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.
13 comments:
He is 24 and she is 21. I let x be her earliest age mentioned in the problem. So he was 2x. When she was 2x, he was 3x. Now he's 4 times as old as her 2x, so he is 8x. Since it is their birthday, his age must be an integral multiple of 8. The only such number in the 20's is 24. Thus he is 24 and x is 3.
His age is 24 and her age is 21.
Solution:
Assume her age is x when his age was twice as her's.
So her age was 2x when his age was 3x.
Hence his current age is 8x and her current age is 7x. Also both are in their twenties. So only possible value for x is 3.
That's a great thing to do to the Papparazzi! They might be good with cameras, but a complext sentence like that one, maybe not so good.
Start with their youngest ages.
The possibilities are
2 and 4;;;;;3 and 6;;;;;4 and 8 Look below.
4 and 6;;;;;6 and 9;;;;;8 and 12
14 and16;;;;22and24;;;;;28and 32
The last one is too large (she is in her 20'd he is 32)
The First one is too small (she is 14 he is 16. The only answer is the middle one.
So let's follow it through. The first condition is really the last one. He must be twice as old as she is at some point.
She's 3; he's 6.
The middle step is he is three times her years when she's three. So he's 9 and she's three years younger which is six. The years between them does not change.
The last fact is that he is currently 4 times her age when he was 9 which means that he is 4*6 which is 24. Then she must be 21. That 3 year difference is still true.
He is 24 and she is 21.
Fun puzzle!
I think that he is 24 now, and that she is 21.
He is four times as old (24) as she was (6) when he was three times as old (9) as she was (3) when he was twice as old (6) as she was (3).
He is 24, she is 23. He was going to be 20, 24, or 28, since those are divisible by 4. Of those, only 24/4 gives you a number divisible by 3 (6), which is convent lay divisible by 2. So when he was 2, she was 1, and they are one year apart.
He is 24 and she is 21.
When he was twice her age she was x yrs. old and he was 2x yrs. old. When he was 3 times as old as she was when he was twice her age he was 3x. x years passed between the time he was 2x and 3x. So when he was 3x she was 2x.
His present age is 4(2x - her age when he was 3x) = 8x
Since they are both in their 20's, 20 < 8x </=29. The only value (integer) for x that fits the parameters is 3. Thus he is 24. She is 21 since she is 3 yrs. younger than he ( 2x =6, x =3, and 3x=9, 2x=6).
He is 24 and she is 21.
I found this puzzle easier to work backwards. There are three time references: 1) now, 2) when he was twice as old as she, and 3) when he was three times as old as she was at the earliest time.
So I picked some numbers for their earliest ages starting with 10 and 5, which would make them 40 and 35 now. Then I tried 4 and 2 which would make them in their teens. Six and 3 were the magic numbers, leading to 9 and 6 and finally 24 and 21.
He is 24 and she is 21.
I found this puzzle easier to work backwards. There are three time references: 1) now, 2) when he was twice as old as she, and 3) when he was three times as old as she was at the earliest time.
So I picked some numbers for their earliest ages starting with 10 and 5, which would make them 40 and 35 now. Then I tried 4 and 2 which would make them in their teens. Six and 3 were the magic numbers, leading to 9 and 6 and finally 24 and 21.
From the description of the problem, there are three moments in time:
T1: when he was twice as old as she was
T2: when he was three times as old as she was at time T1.
T3 (now): when he is four times as old as she was at time T2.
Let's assign unknowns to be their ages at these respective times: H stands for him and S stands for her, for six unknowns total - H1, S1, H2, S2, H3, S3 . From the conditions we can write out the following equations:
H3=S2*4
H2=S1*3
H1=S1*2
H2-H1=S2-S1
H3-H2=S3-S2
Only five equations for six unknowns, are we in trouble here? Seemingly, yes, but let's see what can be simplified.
From the 2nd, 3rd and 4th, we get 2*S1 = S2. Feed that into the first and we get H3=8*S1. Luckily, the problem stated that they are both in their twenties, and there is only one multiple of 8 in the twenties, and that is 24. So from that we solve the rest of the unknowns, leading eventually to S3 = 21. So his age is 24 and hers is 21.
I wonder how soon AI would be able to solve problems of this sort. E.g. would the Jeopardy champion Watson be able to solve it today?
Bravo to all of you!
The answer is: 24 (he) and 21 (she).
All provided terrific explanations, but some (aka Ilya) made is a bit more complex than it is. I am synthesizing Ilya's and Annie's explanations to make it easy for those who couldn't solve it:
From the description of the problem, there are three moments in time:
T1: when he was twice as old as she was
T2: when he was three times as old as she was at time T1.
T3 (now): when he is four times as old as she was at time T2.
Assume her age at time T1 is X.
Then at T1: she is X, he is 2X, age diff=X.
At T2 he is three times as old as she was at time T1. So at T2: he is 3X, age diff is still X, therefore she is 2X.
At T3 (that is now) he is four times as old as she was at time T2. T3: he is 4 x 2X = 8X, age diff is still X, therefore she is 7X.
he is 8X, she is 7X, both in their 20th
X=3
she is 3 x 7 = 21, he is 4 x 7 = 28
Bean - you were very close but mixed up a bit with the dependencies.
A puzzle point for everyone.
Maria, are you sure you've listed the answer correctly? It makes everyone wrong.
Yes, sorry. You are absolutely right Jerome. I made a mistake in the last line of my explanation.
he is 8X, she is 7X, both in their 20th.
X=3
she is 7 x 3 = 21, he is 8 x 3 = 24.
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