Question

Find four integers whose sum is 400 such that the first integer is equal to twice the second integer, three times the third integer and four time the fourth integer.

Answer 1

a=192

b=96

c=64

d=48

Explanation:

hope this helps

Answer 2

a=192

b=96

c=64

d=48

Explanation:

So we have
`a+b+c+d=400` where
`a,b,c,` and
`d` are integers.

We also have
`a=2b`and
`a=3c`and
`a=4d.`

`a=2b` means
`a/2=b`

`a=3c` means
`a/3=c`

`a=4d` means
`a/4=d`

Let's plug those in:

`a+b+c+d=400`

`a+(a)/(2)+(a)/(3)+(a)/(4)=400`

Multiply both sides by 4(3)=12 to clear the fractions:

`12a+6a+4a+3a=4800`

Combine like terms:

`25a=4800`

Divide both sides by 25:

`a=(4800)/(25)`

Simplify:

`a=192`.

Let's go back and find
`b,c,d` now.

b is half of a so half of 192 is 96 which means b=96

c is a third of a so a third of 192 is 64 which means c=64

d is a fourth of a so a fourth of 192 is 48 which means d=48

So

a=192

b=96

c=64

d=48