A chemist needs 500mL of 20% acid and 80% water mix for an experiment. He adds x mL of a 10% acid and 90% water and y mal of a 30% acid and 70% water mix to the 20% acid and 80% water mix. Find the amounts x and y

  x = y = 250 mL

Step-by-step explanation:

The desired concentration of acid (20%) is exactly halfway between the concentrations of the available supplies (10%, 30%). So, the mix will be equal parts of each of those. x and y are both half the total quantity required:

  x = y = (500 mL)/2 = 250 mL

If you need an equation to solve this, you can let x = 500 -y and write the equation for the acid volume in the final mix:

  10%(500 -y) +30%(y) = 20%(500)

  50 -0.1y +0.3y = 100 . . . . . eliminate parentheses

  0.2y = 50 . . . . . . . . . . . . . . subtract 50, collect terms

  y = 250 . . . . . . . . . . . . . . . multiply by 5 (equivalently, divide by 0.2)

  x = 500 -250 = 250

what is the original number? you forgot to add it.

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

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the measure of angle xoy is equal to

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