when we have a mixture problem, we solve it similarly.Whats different in this case?
e.g: Mixture Problem : Can A has 5 liter of milk with 50% protein and Can B has 20% protein in milk.What amount of B should be added in a solution to have 35% protein if all the milk in Can A is added?
5 * 50/100 + x * 20/100 = 35/100 * (5+x)
The equation above translated into words:
5 liters of 50% protein + x liters of 20% protein = (5+x) liters of 35% protein.
This equation accurately reflects the mixture described in the prompt.
prachi18oct wrote:Hi GMATGuruNY,
Can you please explain whats wrong in below?
70*60/100 + x * 40/100 = 80/100 * (70+x)
The equation above translated into words:
70 3-exam units with a 60% score + x final-exam units with a 40% score = (70+x) course units with an 80% score.
This equation does not reflect the mixture described in the prompt.
In the prompt, the 3 exams represent 60% of the course grade, while the final exam represents 40% of the course grade.
Thus, if the course is composed of 100 units, we get:
60 3-exam units with a 70% score + 40 final-exam units with an x% score = 100 course units with an 80% score.
Translated into math:
60(70/100) + 40(x/100) = 100(80/100)
4200 + 40x = 8000
420 + 4x = 800
4x = 380
x = 95.
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