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Question 253475: The first of three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the numbers is 54, find the numbers.
Found 2 solutions by palanisamy, vksarvepalli:
Answer by palanisamy(496)   (Show Source):
You can put this solution on YOUR website! Let the three numbers be x,y and z.
Given,The first of three numbers exceeds twice the second number by 4
x = 2y+4 ...(1)
The third number is twice the first.
z = 2x
2x-z= 0 ...(2)
The sum of the numbers is 54
x+y+z = 54 ...(3)
(2)+(3)=> 3x+y = 54 ...(4)
Put x = 2y+4 in (4)
3(2y+4)+y = 54
6y+12+y = 54
7y = 54-12
7y = 42
y = 42/7
y = 6
Substituting in (1), we get
x = 2*6+4
x = 12+4
x = 16
Substituting in (2), we get
2*16-z = 0
32 = z
Thereforethe solution is x=16, y=6 and z = 32
Answer by vksarvepalli(154)  (Show Source):
You can put this solution on YOUR website! let the numbers be a,b,c
given,
a=2b+4 ------------- 1
c=2a ------------- 2
a+b+c=54------------- 3
from 2, a+b+2a=54
=> 3a+b=54 ------4
using 1 in 4
6b+12+b=54
=>7b=42
=>b=6
so, a=16
c=32
ans: a=16, b=6, c=32
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