0
$\begingroup$

Two cubes of Bronze have their total weight equivalent to 60Kg The first piece contains 10kg of Pure Zinc and the second piece contains 8kg of pure Zinc.what is the percentage of Zinc in the first piece of bronze ,if the second piece contains 15% more zinc than first?

I have tried :

first piece be X and Second piece be 60-X

for first piece Zinc Percentage is

10/X *100

Second Piece Zinc Percentage is 

8/60-X*100

According to Question:

(8/60-X *100) - (10/x*100) =15

I have calculated X= 40 percentage

but when i am calculating it takes so Much of steps as X^2 like that , Is there any alternative or Shortcut Method for this sum,Guide me for the Answer

1 Answers 1

1

Let's call $x$ and $y$ the weight of each cube, then:

$$x+y=60 \quad (1)$$

In the first cube we have $k\%$ of zinc, so:

$$\frac{k}{100}\cdot x=10\to x=\frac{1000}{k}\quad (2)$$

In the second cube we have $15\%$ more zinc which is equivalent to $8kg$ so:

$$1,15\cdot\frac{k}{100}\cdot y=8\to y=\frac{800}{1,15\cdot k}\quad(3)$$

Now replace $(2)$ and $(3)$ in $(1)$ and get $k$.

  • 0
    for second equation it is K+15 /100 *y = 8 – 2017-02-11
  • 0
    How you formed second equation please explain – 2017-02-11
  • 0
    Once there is $15\%$ more in the second then it is $(100\%+15\%)\cdot k\%=1,15\cdot k\%$. – 2017-02-11