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# divide 116 among four part such that 5 added to first,4 subtracted from the second,third multiplied by 3 and f?

divide 116 among four part such that 5 added to first,4 subtracted from the second,third multiplied by 3 and fourth divided by 2 give the same result.

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- Anonymous8 years agoFavorite Answer
A + B + C + D = 116

A + 5 = B - 4 = 3C = D/2

B = A + 9

C = (A + 5)/3

D = 2(A + 5)

Using the first equation: A + (A + 9) + (A + 5)/3 + 2(A + 5) = 116

4A + A/3 = 116 - 9 - 10 - 5/3

13A = 286

A = 22

You take it from there.

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- 8 years ago
a+5 = b -4 = 3*c = d/2

a=3c-5

b=3c+4

d=6c

a+b+c+d=116

3c-5 + 3c +4 +c +6c =116

13c = 117

c=9

a=22

b=31

d=54

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- 8 years ago
5+a=b-4=c*3=d/2=x

a=x-5

b=x+4

c=x/3

d=x*2

add all these

so 4x+x/3-1=116

13x/3=117

x=27

numbers are 22,31,9,54

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