Bunuel
In a state bar exam with two sections, 35% of candidates failed one section, 42% failed the other section, and 15% failed both sections. If 5000 candidates took the exam, how many passed one section but not both?
A. 650
B. 2150
C. 2350
D. 4150
E. 4500
Attachment:
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I used a Venn diagram but took a different route (no percents after initial calculation).
I focused on calculating "failed only one."
From percentages given, find the numbers of
candidates who failed first, second, or both sections
With those three sets of numbers, we have "captured" the population
in whom we are interested, because
Passed both are excluded entirely (we want pass
one)
Passed neither (passed 0 = failed both) get subtracted (we want pass
one)
"Failed only one" is equivalent to "passed only one."
We are down to two sets that mirror one another.
Members of either set failed one section only.
Those members simultaneously also passed one section only.
If you failed one section but not the other, then
you also passed one section but not the otherMETHOD
Calculate the actual numbers of candidates who
• failed Section #1
• failed Section #2
• failed BOTH sections
NUMBER of candidates who failed one or two sections
Section #1, 35% failed: (.35*5,000) =
1,750 TOTALSection #2, 42% failed: (.42*5,000) =
2,100 TOTALBoth sections, 15% failed: (.15*5,000) =
750 STEPS
Draw two intersecting circles
Write
totals for S #1 and S #2
outsideNext steps:
1) Inside first. Put 750 in the gray overlap (failed both #1 and #2) = 750
2)
S #1 ONLY? Pink shading
S #1 ONLY = (Section 1 TOTAL - overlap). S #1 ONLY: (1,750 - 750) =
1,0003)
S #2 ONLY? Green shading. (2,100 - 750) =
1,350S #2 ONLY = (Section 2 TOTAL - overlap)
4) Add the ONLY parts: (1,000 + 1,350) =
2,350 candidates
failed one section but not the other, and thus passed one section but not the other
Answer C
EDIT:
AweG , of the great username, I meant to give you kudos.
By "different route" I mean that once I figured out the number who
failed one only, I did not extrapolate percentages as you did. +1