What 4 consecutive even integers whose sum is 140?
回答 (10)
2008-07-23 5:33 pm
✔ 最佳答案
The first integer will be n. The second integer = n + 2
The third integer = n + 4
The fourth integer = n + 6
The sum of all the integers
n + n + 2 + n + 4 + n + 6 = 140
combine like terms
4n + 12 = 140
Subtract 12 from both sides
4n = 128
n = 32
n + 2 = 34
n + 4 = 36
n + 6 = 38
check it: 32 + 34 + 36 + 38 = 140
and so your integers are 32, 34, 36, and 38.
2008-07-24 12:30 am
x + x+2 + x+4 + x+6 = 140
4x = 128
x = 32
32, 34, 36 and 38
4x = 128
x = 32
32, 34, 36 and 38
2008-07-24 6:05 pm
Dividing by 4,
140/4=35.
The four parts are 35, 35, 35 and 35.
As we have to find out even integers, the four numbers can be rewritten as 34, 34, 36, 36. Still the sum is 140.As we have to find out consecutive even integers, we can transform the numbers as 32, 34, 36, 38.
140/4=35.
The four parts are 35, 35, 35 and 35.
As we have to find out even integers, the four numbers can be rewritten as 34, 34, 36, 36. Still the sum is 140.As we have to find out consecutive even integers, we can transform the numbers as 32, 34, 36, 38.
2008-07-24 12:10 pm
let 1st no. = a
therefore 2nd no.= a+2
3rd no. = a+4
4th no. = a+6
according to question
a + a +2 + a + 4 + a +6 = 140
4a + 12 =140
4a = 128
a = 32
therefore numbers r
32, 34, 36, 38
therefore 2nd no.= a+2
3rd no. = a+4
4th no. = a+6
according to question
a + a +2 + a + 4 + a +6 = 140
4a + 12 =140
4a = 128
a = 32
therefore numbers r
32, 34, 36, 38
2008-07-24 8:51 am
Consider the first no. to be x
then as they are consecutive even nos. the other nos. will be x+2,x+4,x+6.
By data
x+x+2+x+4+x+6=140
Add the variables and constants seperately.
4x+12=140
Taking 12 that side it becomes -12
4x=140-12.
x=128/4
x=32.
therefore the nos are
32,34,36,38.
then as they are consecutive even nos. the other nos. will be x+2,x+4,x+6.
By data
x+x+2+x+4+x+6=140
Add the variables and constants seperately.
4x+12=140
Taking 12 that side it becomes -12
4x=140-12.
x=128/4
x=32.
therefore the nos are
32,34,36,38.
2008-07-24 8:49 am
w > x > y > z (solve by using substitution)
w - 2 = x
x - 2 = y
y - 2 = z
w - 2 = x
w = x + 2
w + x + y + z = 140
(x + 2) + x + (x - 2) + (y - 2) = 140
x + 2 + x + x - 2 + y - 2 = 140
x + 2x + y = 140 - 2 + 2 + 2
3x + y = 142
x - 2 = y
x - y = 2
x = 2 + y
3x + y = 142
3(2 + y) + y = 142
6 + 3y + y = 142
3y + y = 142 - 6
4y = 136
y = 136/4
y = 34
y - 2 = z
34 - 2 = z
z = 32
x - 2 = y
x - 2 = 34
x = 34 + 2
x = 36
w - 2 = x
w - 2 = 36
w = 36 + 2
w = 38
â´ w = 38 , x = 36 , y = 34 , z = 32
w - 2 = x
x - 2 = y
y - 2 = z
w - 2 = x
w = x + 2
w + x + y + z = 140
(x + 2) + x + (x - 2) + (y - 2) = 140
x + 2 + x + x - 2 + y - 2 = 140
x + 2x + y = 140 - 2 + 2 + 2
3x + y = 142
x - 2 = y
x - y = 2
x = 2 + y
3x + y = 142
3(2 + y) + y = 142
6 + 3y + y = 142
3y + y = 142 - 6
4y = 136
y = 136/4
y = 34
y - 2 = z
34 - 2 = z
z = 32
x - 2 = y
x - 2 = 34
x = 34 + 2
x = 36
w - 2 = x
w - 2 = 36
w = 36 + 2
w = 38
â´ w = 38 , x = 36 , y = 34 , z = 32
2008-07-24 2:08 am
32,34,36,38
2008-07-24 1:25 am
as the numbers are even they can be represented as x,x+2,x+4
and x+6.
heir sum=x+(x+2)+(x+4)+(x+6)=140
or 4x+12=140
or 4x=140-12
or x=128/4=32
thus the numbers are 32,(32+2),(32+4),(32+6) that is32,34,36,38
and x+6.
heir sum=x+(x+2)+(x+4)+(x+6)=140
or 4x+12=140
or 4x=140-12
or x=128/4=32
thus the numbers are 32,(32+2),(32+4),(32+6) that is32,34,36,38
2008-07-24 12:33 am
x +(x +2) +(x +4) +(x +6) = 140
4x +12 = 140
x = 32
x +2 = 34
x +4 = 36
x +6 = 38
32, 34, 36, 38.
4x +12 = 140
x = 32
x +2 = 34
x +4 = 36
x +6 = 38
32, 34, 36, 38.
2008-07-24 12:31 am
Represent the first even number as 2n.
The next even numbers are 2n+2, 2n+4, 2n+6.
The sum of all four numbers is 4(2n) + 2 + 4 + 6 = 8n + 12
8n + 12 = 140
8n = 128
n = 16
The numbers are 32, 34, 36, 38
The next even numbers are 2n+2, 2n+4, 2n+6.
The sum of all four numbers is 4(2n) + 2 + 4 + 6 = 8n + 12
8n + 12 = 140
8n = 128
n = 16
The numbers are 32, 34, 36, 38
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