[Guest]

Jill is 5 years older than 3 times as old as Ian. In ten years, Jill will be 4 times as old as Ian. How old are they both now?

[Guest]

15 and 50

[Guest]

Jill is 5 years older than 3 times as old as Ian. In ten years, Jill will be 4 times as old as Ian. How old are they both now?

Shouldn't it be as Ian is now? Or else it would be impossible.

But 15 and 50

[Guest]

Shouldn't it be as Ian is now? Or else it would be impossible.

But 15 and 50

I agree with you, zee.

[Guest]

this one is killing me- in high school we would have ones like this- ugh- we had powerful calculators as well....

[Guest]

Jill would be -70, and Ian -25 years old. In 10 years, Jill will be -60 and Ian -15.

If Ian's age is constant, Ian is 15, so Jill would be 50 and then 60.

Jill is 5 years older than 3 times as old as Ian. In ten years, Jill will be 4 times as old as Ian. How old are they both now?

[Guest]

According to your logic....Ian's age=-25yrs

JIll's age=-70yrs.

[Guest]

Ummmmmm. . . I got

Ian=5

Jill=20

I guess I could have messed up, but here's my math:

J=Jill E=Ian (so it won't look like ones)

J=3E+5

J+10=4E+10 (+10 Because of "in ten years") Then substitute

3E+5+10=4E+10 Then combine like terms

3E+15=4E+10

-10 -10

3E+5=4E

-3E -3E

5=Ian's Age Then substitute again

J=3(5)+5

J=15+5

J=20

Edited March 10, 2008 by Sharpie357

[Guest]

the problem with you math comes when you look at the statement - in ten years Jill will be four times Ian's age

If Ian = 5 and Jill = 20 then in ten years

Jill = 30 and that is not 4 times Ian's 15

The second equation should have been

Jill + 10 = 4 (Ian + 10) and then you have to distribute and then

Ian = -25 and Jill = -70 which is 5 more than three times Ian's age

and in 10 years

Ian = -15 and Jill = -60 which is four times Ian's age

[Guest]

Oh. . . . oops. . . . . . . .