If I have 2 steel rods, one with 17% purity and one with 25% purity. How much should I take (in kgs) from each bar?

250 kg of 17% and 150 kg of 25%

Step-by-step explanation:

For answering this, we do it by proportions.

take the difference

17%  ===  20% === 25%

         3              5

Then ratio

17% :25% = 5:3

For 400 kg,

17% = 5/(5+3)  * 400 = 250 kg

25% = 3/(5+3) * 400 = 150 kg.

Above is a mental method, no calculation or pencil need.

I will edit to give an algebraic method.

Algebraic method:

Given:

17% rods and 25% rods.

Need to make 400 kg of 20% rods.

Let  

A=weight of 17% rods

B=weight of 25% rods

to make 20% rods, we need  

( 0.17A + 0.25B ) / (A+B) = 0.20  (1)

But we know that  

A+B=400 (2)

substitute

B=400-A in (1)

( 0.17A + 0.25 (400-A) ) / 400 = 0.20

Cross multiply and simplify

(0.17-0.25)A +100 = 80

-0.08A = -20

A = -20 / (-0.08)

A = 250 (3)  (17% rods)

substitute (3) in (2)

B = 400-250 = 150   (25% rods)

Same answer as mental calculations

Even simpler albebra, same answer!

Given:

17% rods and 25% rods.

Need to make 400 kg of 20% rods.

Let  

W=weight of 25% rods

400-W = weight of 17% rods

To make 20% purity rods, we need to have 400*0.2=80 kg of pure metal, which is

0.25W + 0.17(400-W) = 80

Expand and simplify

0.25W + 68 - 0.17W = 80

0.08W = 80-68 = 12

W = 12/0.08 = 150  (25% rods)

400-W = 250 kg (17% rods)

the barter