Question

The oldest child in a family is 5 years older than her brother. The mother is 1 year younger than twice the sum of her children's ages, and the father equals the mother's age plus the son's age. If the sum of all four ages is three times a prime number and the mother was younger than 50 when she had her second child, how old is the daughter?

Answer 1

Father, Mother, Daughter, Son
Let their respective ages be: F, M, D and S.
D = S + 5
M = 2 * (D + S) - 1
F = M + S
F + M + D + S = 3 * prime number
M - S < 50

Solve for D

Answer 2

So the daughter would be 9?

Answer 3

I haven't done it. But if you show the work I can see if it is correct.

Answer 4

d=s+5 i subtracted over and got. d-5=s
I plugged this to m= 2*(D+S)-1 and simplified to get m=4d-11
I moved over to f=m+s and used the first equation of d-5=s and ended with f=m+d-5
Using the last two equations I found. I made f=4d-11+d-5 and simplified to f=5d-16
With the F+M+D+S= 3*prime# I made a equation of d+d-5+5d-16+4d-9=3(prime#)
Simplified to get 11d-30=3(prime#) and after a couple of tries found 23 as the prime # works. I plugged it into the 11d-30=69 (at this point) and solved for her age to be 9.

Answer 5

actually if you add all their ages (9,4,25, and 29) you get 67. Not the 3*23 which it should be.

Answer 6

That means there is a mistake in the calculation somewhere. Let me see if I can locate it.

Answer 7

These are the numbers I am getting: son = 8, daughter = 13, mother = 41 and father = 49. The prime number is 37. It satisfies all the equations given in the problem.

Answer 8

so the prime number just needed to be different is all?

Answer 9

If I plug 37 into your equation it does not work. There is an error somewhere.

Answer 10

Aha, the mistake is in: F+M+D+S. Do that part again.

Answer 11

F = 5D - 16
M = 4D - 11
D = D
S = D - 5

F+M+D+S = 11D - 32

Answer 12

i caught it. there was a number from one of the first times i attempted and i didn't change it with the correct #. The math works out correctly now. Thank you. :) I have another question. And i'll go type that in. I don't know what kind of math you do (since the other question has to do with triangles and consecutive integers). But if you could help me with that one as well i'd appreciate it.

Answer 13

Go ahead.

Answer 14

Okay. On the date 11/12/13, that date was a ratio of the angles of a triangle that result in all three angles being integers. Determine the next date consisting of three consecutive integers that can also be used as the continued ratio of integral angle measures for a triangle.

Answer 15

Let the integers be N, N+1 and N+2
The three angles are in the ratio: N:N+1:N+2
Let k be the constant of proportionality
Then the three angles are: k*N, k*(N+1) and k*(N+2)
Three angles of a triangle add up to 180 degrees.
k * (N+N+1+N+2) = 180
k * (3N + 3) = 180
k * 3(N+1) = 180
k * (N+1) = 60

Come up with a suitable values for k and N

Answer 16

60 can be:
1 x 60
2 x 30
3 x 20
4 x 15
5 x 12
6 x 10
etc.

Answer 17

so should N be 12?

Answer 18

It wouldn't be 11 since 11/12/13 is what it's starting with. So next would be 12/13/14 which would still fall in the calendar. 12 being N. then 12+1=13 and so on getting 12/13/14. Or am I completely off?

Answer 19

Yes, it has to be 12/13/14 because they want the NEXT date. Didn't see that before.

Answer 20

But we didn't need any of the above calculations to conclude that because 180 degrees can always be split into the ratio of many three consecutive numbers.

Answer 21

So really the only way to "determine" the next set of consecutive integers is to know that 12/13/14 is the only set of consecutive numbers that don't fall outside of the calendar year?

Answer 22

Oh, wait a minute. They want all ANGLES to be integers. That won't work for 12/13/14. So the above calculations didn't go to waste after all.

Answer 23

So both k and N has to be integers.

Answer 24

I don't think by next date they mean the date after 11/12/13. They are just asking for another date where this will work.
12/13/14 - no
11/12/13 - done already
10/11/12 - k non-integer
9/10/11 - works

Answer 25

04/05/06 works too.

Answer 26

03/04/05 is good too.

Answer 27

01/02/03 works. And that is it.

Answer 28

so for the work i put down what should i label "k" as? You said before it was a constant of proportionality. but what does it really do? what is it?

Answer 29

using k you can find the actual angles which will all be INTEGERS and still have ratio of three consecutive integers.

Answer 30

Take 09/10/11 for example.
N = 9
Put it in k * (N+1) = 60
k * (9+1) = 60
k = 6
The three consecutive integers are: 9, 10, 11
The three angles are: 9*6, 10*6, 11*6 = 54, 60, 66
54+60+66 = 180 which satisfies the condition that all three angles of a triangle add to 180 degrees.
The angles are in the ratio - 54:60:66
reduce the ratio to the lowest form by dividing by 6
9:10:11
three consecutive numbers as they wanted.

Answer 31

Wow. That is beautiful. Thank you so much once again! I really had no clue where to even start with that last one. But thank you for sticking with me!

Answer 32

You are welcome.