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Bunuel
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the approach used is to find 40 percent of the 28 percent and one can arrive at the answer
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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?


Aryaa03

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-C

Say, 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.
Insufficient

St-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


­
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vektr
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?


Aryaa03

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-C

Say, 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.
Insufficient

St-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


­

Say 100 students took the tests.

(1) says 28 students scored at least 6 points higher on the final.
(2) says these 28 students make up 40% of ALL students who scored higher on the final.

So, if 40% corresponds to 28 students, then 100% of the students who scored higher on the final is 28/40 * 100 = 70 students.

Hope it's clear.
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