The sum of four numbers is 1,320. The sum of three of the numbers is 40% more than the other number, x. What is the value of x?

550

Step-by-step explanation:

We aren't given that they are consecutive so lets just pretend the numbers are x, y, z, and r.

We are given x+y+z+r=1320

We are also given that y+z+r=.4x+x

Simplifying my second equation gives us y+z+r=1.4x. I'm going to plug this into first equation giving me

x+1.4x=1320

2.4x=1320

So x=1320/2.4=550

Since the numbers are not consecutive natural numbers, how about if I let the 4 numbers be a, b, c, and d?

The sum is given to be 1320.

a + b + c + d = 1320 ... I will call this the first equation.

Can I then say that the second equation is b + c + d = 0.4a + a? Yes, I can do this.

Simplifying my second equation,

I get b + c + d = 1.4a.

Plugging this into my first equation, I get

a + 1.4a = 1320

2.4a = 1320

So, a = 1320 / (2.4) = 550.

Answer: 550