Hi Swati:

Let F be the grade required in the final exam to get to 80% overall.
Then we set up our equation.
For the first three exams, the weight is 60%=0.6 and value is 70
For the final exam, the weight is 40% = 0.4 and value is F.

Then 80 = 0.6*70+0.4*F
=> F = 38/0.4 = 95.

Let me know if this helps

leonswati wrote:Katie has taken three exams and received an average of 70% on them .Her score on these three exams will determine 60% of her course grade while her score in the final exam will determine 40% of her course grade.What grade does she need to receive on her final exam in order to achieve an average of 80% in her course?

Can someone help me solve it...

Step 1: Assign the appropriate WEIGHT to each part of the course.
Let the course grade = 100 units.
Since the 3 exams account for 60% of the course grade, the 3 exams = 60 units.
The final exam = 40 units.

Step 2: Calculate the SUM for each part of the course.
Sum for the course grade = (units)(average) = 100*80 = 8000.
Sum for the 3 exams = (units)(average) = 60*70 = 4200.
Thus, the sum needed on the final = 8000-4200 = 3800.
Average needed on the final = sum/units = 3800/40 = 95.

Hi GMATGuruNY,

Can you please explain whats wrong in below?

70*60/100 + x * 40/100 = 80/100 * (70+x)

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)

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.

Katie has taken three exams and received an average of 70% on them .Her score on these three exams will determine 60% of her course grade while her score in the final exam will determine 40% of her course grade.What grade does she need to receive on her final exam in order to achieve an average of 80% in her course?

The weightage of First three exams is 60% which should be considered that 60% of (average) exam score (of first three scores) should be considered as contributing marks in the Consolidated grading system

The weightage of Second three exams is 40% which should be considered that 40% of (final) exam score should be considered as contributing marks in the Consolidated grading system

Now 80% in the last line refers to the total score requirement as percentage of total marks i.e.100

so equation should be

[(60/100) x (Average of first three scores)] + [(40/100) x (Final exam scores)] = [(80/100) x 100]

i.e. 0.6 x 70 + 0.4 x X = 0.8 x 100
i.e. 42 + 0.4X = 80
i.e. X = 38/0.4 = 95