Divide 116 among four parts such that 5 added to first, 4 subtracted from second, third multiples by 3 and fourth divided by 2 give the same result

Question

Plzz help in me solving some maths problems and few issues that r difficult and solve all solution

Answer 1

Call the 4 numbers a,b,c,d

 

a + b + c + d = 116

 

a+5 = b-4 = 3c = d/2

 

*********************

Let's put everything in terms of a single varialbe... Let's use "a"

 

a+5 = b- 4

b = a + 9

 

a + 5 = 3c

c = (a+5)/3

 

a+5 = d/2

d = 2a + 10

 

Substituting back into a+b+c+d = 116

 

a + a + 9 + (a+5)/3 + 2a + 10 = 116

4a + (a+5)/3 + 19 = 116

4a + (a+5)/3 = 97

12a + a+5 = 291

13a = 286

a = 22

 

Now just use

b = a+9

c = (a+5)/3

d + 2a + 10 

to find the other three numbers