Mixture Problem
Algebra
Level
pending
There are two alcohol solutions with different strengths. The first one is 40% alcohol solution while the second one is 45% alcohol solution. How many cubic cm of each must be used so as to make mixture of 35 cm³ which contain 42% alcohol?
15 and 21
14 and 21
14 and 20
15 and 22
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1 solution
1stly there was no need to even calculate.. Addition of the to solution x1 being 1st solutions volume and x2 being 2nd solutions volume had to be 35
x1+x2=35 -------eq1
Whose only ans was 14 and 21
But for those who want mathematical way, 1st solutions alcohol contain is 40% i.e 40/100 2nd solutions alcohol contain is 45% i.e 45/100 And the mixture of both contain 42% i.e 42/100
Vol of alcohol in 1st sol+Vol of alcohol in 2nd sol=Vol of alcohol in final sol (40/100)x1+(45/100)x2=(42/100)*35----eq2
Solving equation 1 & 2 gives the solution..