Find the numbers?

Question

Three numbers added together total 68. The second number is four more than the first, and the third number is twice the first. Find the numbers?

Answer 1

answer
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x+x+4+2x=68
==>4x=64
==>x=16;

Numbers are 16,20,32.

Answer 2

Thank you for posting your question ... this kind of problem stumps a LOT of students, although I hope you soon see it's simple.

First, it's important to make up an expression for each of the unknown quantities ... we should try to convert from "written words" to "mathematical statements."

When I read: "Three numbers ... added ... total 68"
I think: "A + B + C = 68"

Notice how the 2nd and 3rd numbers are each compared to the 1st one? That's a hint that I can label the 1st number "plain old x." The others will be represented by terms containing x:

1st number: x
2nd number: x+4 "four ... more than"
3rd number: 2x "twice ... x"

Let's add them up!

(x) + (x+4) + (2x) = 68

Collect like terms:

4x + 4 = 68

Solve:

4x = 64
x = 16

Don't forget to state the complete solution:
"The numbers are 16, 20 and 32."

You can check your work simply by adding up those answers ... and, yes, they do add up to 68! Congratulations.

Hope this helps!

Answer 3

the first number is a

the second is 4a

the third is 2a

so the sum is 7a = 68

a = 68/7 , b = 272/7 , c = 134/7

Answer 4

let x=first number, y=second number, z=third number

x+y+z=68
y=x+4
z=2x

substitute y and z into the first equation and solve for x

x+(x+4)+2x=68
4x+4=68
4x=64
x=16

Now solve for y and z

y=x+4 = 16+4 = 20
z=2x = 2*16 = 32

Verify
16 +20 +32 = 68
68 = 68

Answer 5

they r : X , X+4 and 2X

X + X + 4 + 2X = 68
4X = 64
X = 16

i.e : numbers are : 16 , 20 and 32


piece of cake huh? salam...:)

Answer 6

a+b+c=68
b=a+4
c=2a
a+a+4+2a=68
4a=64
a=16
b=20
c=32
16+20+32=68

Answer 7

x.1+x.2+x.3=68
x.2=4+x.1
x.3=2x.1

x.1+x.2+x.3=68
x.1+4+x.1+2x.1=68
4x.1=64
x.1=64/4=16
x.2=20
x.3=32