A city volunteer program recorded whether each of its 480 volunteers

Question

A city volunteer program recorded whether each of its 480 volunteers held each of three certifications: logistics, first aid, and radio operation. Every volunteer with first-aid certification also had logistics certification, and no volunteer held all three certifications. If 90 volunteers held only the radio-operation certification, how many volunteers held exactly two of the three certifications?

(1) 70 volunteers held only the logistics certification.
(2) 30 volunteers held none of the three certifications.

Bunuel · Accepted Answer

GMAT Club Official Solution:

A city volunteer program recorded whether each of its 480 volunteers held each of three certifications: logistics, first aid, and radio operation. Every volunteer with first-aid certification also had logistics certification, and no volunteer held all three certifications. If 90 volunteers held only the radio-operation certification, how many volunteers held exactly two of the three certifications?

Check the diagram below:

https://gmatclub.com/forum/download/file.php?id=88389

Since every volunteer with First Aid also had Logistics, the green First Aid circle must lie entirely inside the red Logistics circle.

Since no volunteer held all three certifications, the green First Aid circle cannot overlap the blue Radio Operation circle. Otherwise, anyone in that overlap would have First Aid, Radio Operation, and since First Aid is inside Logistics, that person would also have Logistics, meaning all three, which is not allowed.

So the volunteers who held exactly two certifications are  the sum of the two yellow regions :

the green circle, which represents Logistics and First Aid  
 the overlap of the red and blue circles, which represents Logistics and Radio Operation

(1) 70 volunteers held only the logistics certification.

This tells us that the red-only region, meaning volunteers with only Logistics, is 70:

https://gmatclub.com/forum/download/file.php?id=88390

But we still do not know how many volunteers are outside all three circles.

So we cannot yet determine the total in the yellow regions.

Not sufficient.

(2) 30 volunteers held none of the three certifications.

This tells us that 30 volunteers are outside all three circles: 
 https://gmatclub.com/forum/download/file.php?id=88390

But we still do not know how many volunteers are in the red-only region.

So we still cannot determine the total in the yellow regions.

Not sufficient.

(1)+(2) Now we know:
blue-only region = 90
red-only region = 70
outside all circles = 30
total volunteers = 480

https://gmatclub.com/forum/download/file.php?id=88391

So the yellow regions together must contain:

480 - 90 - 70 - 30 = 290

Therefore, 290 volunteers held exactly two of the three certifications.

Sufficient.

Answer: C.

ankushsambare · Answer

Let:
L = logistics
F = first aid
R = radio operation
Given:
 
 Every F volunteer also has L, so F is a subset of L. 
 No one has all three certifications. 
 90 volunteers have only R. 
 Total volunteers = 480.  
We need the number who held exactly two certifications.
Because F ⊂ L and nobody has all three, the only possible “exactly two” groups are:
 
 L and F only 
 L and R only  
Let:
a = only L
b = L and F only
c = L and R only
d = only F (impossible since every F has L)
e = only R = 90
f = none
So total:
a + b + c + 90 + f = 480
We want:
b + c
Statement (1): 70 held only L
So:
70 + b + c + 90 + f = 480
b + c + f = 320
Not enough because f unknown.
Statement (2): 30 held none
So:
a + b + c + 90 + 30 = 480
a + b + c = 360
Not enough because a unknown.
Combine (1) and (2):
70 + b + c + 90 + 30 = 480
b + c = 290
Unique value obtained.
Therefore BOTH statements together are sufficient, but neither alone is sufficient.
 Answer: C

BongBideshini · Answer

Let:
 
 L = logistics 
 F = first aid 
 R = radio operation  
Given:
 
 Total volunteers = 480 
 Every F also has L → F is a subset of L 
 No one has all three certifications 
 90 held only R

Since every F holder also has L, the only possible “exactly two certifications” group is:
 
 L and F only 
 L and R only  
Let:
 
 a = only L 
 b = L and F only 
 c = L and R only 
 d = only R = 90 
 e = none  
Total:
a + b + c + d + e = 480
So:
a + b + c + e = 390
We need:
b + c
 Statement (1):  
a = 70
Then:
70 + b + c + e = 390
> b + c + e = 320
We still don’t know " e "
 Not sufficient.

Statement (2):

e = 30
Then:
a + b + c = 360
But we don’t know  "a" 
 Not sufficient.

Combine (1) and (2) 
From (1):
a = 70
From (2):
e = 30
Using:
a + b + c + e = 390
70 + b + c + 30 = 390
⇒ b + c = 290

Both together are sufficient.
 Answer: C.

pappal · Answer

given total=480, E3=0 & all of F's are included in L and R only=90
so need to find E2=?
since E3=0 so there are no intersection values between R&F, R&L possible.
1. L only=70, not sure of n(none) -----insufficient
2. n(none)=30 , but not sure of other requisites----insufficient
together
union value=480-30=450=E1+E2 (since E3=0)
so E2=450-E1 
here E1=90+70=160 as F only=0
hence E2=290 sufficient
C

SS990 · Answer

Answer is C

what we have from Q stem = 480=90+La+e2+n... La=L Alone, e2 is exactly 2 whch we want and n=none

1) 480=90+70+e2+n....2 unknowns 1 equation..insufficient
2) 480= 90+ La+e2+30 insufficent

Together we know La and n value so e2 can be found. Hence C

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