both equations combined give us 70 who scored atleaast 6 points or higher in final exams compared to mid term. But we don't know many scored let's say 2 points or 1 point higher. How does both equations combined solve that?
Bunuel
The students in a certain college class took a midterm exam and a final exam, both of which were scored out of a total of 100 points. What percent of the students earned a higher score on the final than on the midterm?
(1) Of the students in the class, 28% scored at least 6 points higher on the final than on the midterm.
(2) Of the students who scored higher on the final than on the midterm, 40% scored at least 6 points higher on the final.
IMO-Option-CSay, x be the total number of students and y be the number of students scoring higher in final term.
Asked: \(\frac{y}{x}*100\)
St-1:Of the students in the class, 28% scored at least 6 points higher on the final than on the midterm.
No. of students scoring atleast six points higher = 0.28x
No information of students other with 1 to 5 points higher. Thus y can't be determined.
InsufficientSt-2:Of the students who scored higher on the final than on the midterm, 40% scored at least 6 points higher on the final.
No. of students scoring atleast six points higher = 0.4y
We do not have any info about x
Insufficient(1) + (2)0.28x = 0.4y ==> \(\frac{y}{x}*100\) = \(\frac{0.28}{0.40}*100\) = 70%
Sufficient