Question

There are 5 people standing in a library. Ricky is 5 times the age of Mickey who is half the age of Laura. Eddie is 30 years younger than double Laura and Mickey's combined ages. Dan is 79 years younger than Ricky. The sum of their ages is 271. Dan's age?

HELP!!!!!!!!!

Answer 1

This is a fun simultaneous equations problem. The solution is that Dan is `21` years old.

Explanation:

Let's use the first letter of each person's name as a pronumeral to represent their age, so Dan would be `D` years old.

Using this method we can turn words into equations:

Ricky is 5 times the age of Mickey who is half the age of Laura.

`R=5M` (Equation1)

`M=L/2` (Equation 2)

Eddie is 30 years younger than double Laura and Mickey's combined ages.

`E=2(L+M)-30` (Equation 3)

Dan is 79 years younger than Ricky.

`D=R-79` (Equation 4)

The sum of their ages is 271.

`R+M+L+E+D=271` (Equation 5)

Now we have five equations in five unknowns, so we're in good shape to use simultaneous equations to find out everyone's age.

Let's multiply Equation 2 by 2 (I hate fractions!) so

`2M=L`

If we substitute in `2M` where we see `L` in Equation 3, it gets simpler:

`E=2(2M+M)-30`
`E=2(3M)-30=6M-30`

Now we have values for both `E` and `L` in terms of `M`.

In Equation 1 we also have a value for `R` in terms of `M`. If we use that in Equation 4 we can create a value for `D` in terms of `M` too:

`D=R-79=5M-79`

Just to make it super clear, let me line them all up:

`R=5M`
`L=2M`
`E=6M-30`
`D=5M-79`
And, of course: `M=M`!

Now we can substitute all of these values into Equation 5, and we'll have an equation that is only in terms of one variable, and we know how to solve those:

`5M+M+2M+(6M-30)+(5M-79)=271`

Collect like terms:

`19M=380`

Divide both sides by 19:

`M=20`

Great! We know Mickey's age! But we were asked for Dan's age in the question. Fortunately, we already have an equation for Dan's age (`D`) in terms of Mickey's age (`M`):

`D=5M-79=5(20)-79=100-79=21`

And we're done!