Question 790009


Stuck need help with this problem:

A chemist needs a 110 mL of 32% solution, but only has 29% & 40% solutions available. How much of each solution does he need?

Can you please walk me thru this to help me get the answer?


Let amount of 29% to be used be T
Then amount of 40% to be used = 110 – T
Therefore, .29T + .4(110 – T) = .32(T + 110 – T)
.29T + 44 - .4T = .32T + 35.2 - .32T
.29T - .4T = 35.2 – 44 
- .11T = - 8.8
T, amount of 29% solution to be used = {{{(- 8.8)/-.11}}}, or {{{880/11}}}, or {{{highlight_green(80)}}} mL


Amount of 40% solution to be used = 110 - 80, or {{{highlight_green(30)}}} mL


You can do the check!! 


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