Question

divide 1896 into three parts such that first part be double of that of second part and second part be ⅓ of the third part​

Answer 1

Step-by-step explanation:

Let the three parts be x,y,z

x = 2y (given)------(1)

y = 1/3z (given)------(2)

Putting the value of y in (1) we get,

x = 2/3z

Now,

x + y + z = 1896

2/3z + 1/3z + z = 1896

6z = 5688

z = 948

x = 2/3z = 2/3 * 948 = 632

y = 1/3z = 1/3 * 948 = 316

Hope it will help you.

Answer 2

Answer:

Let the three parts be x,y,z x = 2y (given)----(1) y = 1/3z (given)--2) Putting the value of y in (1) we get, x = 2/3z Now, x +y +z = 1896 2/3z + 1/3z + z = 1896 6z = 5688 %3D z = 948 x = 2/3z = 2/3 * 948 = 632 y = 1/3z = 1/3 * 948 = 316