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.

ankushsambare · Answer

Let L = logistics, F = first aid, R = radio. Total volunteers = 480. F is always a subset of L, and no one has all three. Only R = 90.
We classify:
Only L = x, Only R = 90, L+F only = a, L+R only = b.
Total gives: x + 90 + a + b = 480 so x + a + b = 390.
We need a + b.
(1) Only L = 70, so x = 70.
70 + a + b = 390 so a + b = 320 → sufficient.
(2) None = 30, so inside groups = 450.
x + a + b + 90 = 450 so x + a + b = 360.
Cannot find a + b alone → insufficient.
 Answer: A (Statement 1 alone is sufficient)

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

A city volunteer program recorded whether each of its 480 volunteers

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