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
Answer:
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
