Question

1
3. Divide 400 into two parts such that
of the first part is 40 less than the other part. Find the two parts​

Answer 1

Question:

Divide 400 into two parts such that  $\frac{1}{3}$ of the first part is 40 less than the other part. Find the two parts​

Given:

• $\frac{1}{3x }= y-40$

• $y= \frac{1}{3x} + 40$

To Find:

• x

• y

Solution:

Let the first part be x and the second part be y.

Hence, $\frac{1}{3x }= y-40$

And,$y= \frac{1}{3x} + 40$

Therefore,$( \frac{1}{3x} + 40)+ x = 400$

$\frac{1}{3x} + x = 360\\\\\\\frac{4}{3x} = 360\\\\\\\frac{4}{3x} = 360\\\\\\x=360\times\frac{3}{4}\\\\\\x=90\times3\\\\\\x=270\\\\\\y=\frac{1}{3}x+40\\\\\\y=\frac{1}{3}\times270+40\\\\\\y=1\times90+40\\\\\\y=90+40\\\\\\y=130$

Answer:

• x = 270

• y=130