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22 Dec 2025, 23:33
Kudos
Bookmarks
data analyst only - x
pm only - y
nt only - z
da and pm only - a
pm and nt only - b
da and nt only - c
all 3 - m
x = 6z
y = 12c
z=1/4 y
y+a+b+m = 220
x+y+z+a+b+c+m = 308
x+z+c = 308-220
x+z+c = 88
6z+ z + z/3 = 88
(22/3)z = 88
z = 12
y = 48
x = 72
x+y+z = 132
ans: option D
pm only - y
nt only - z
da and pm only - a
pm and nt only - b
da and nt only - c
all 3 - m
x = 6z
y = 12c
z=1/4 y
y+a+b+m = 220
x+y+z+a+b+c+m = 308
x+z+c = 308-220
x+z+c = 88
6z+ z + z/3 = 88
(22/3)z = 88
z = 12
y = 48
x = 72
x+y+z = 132
ans: option D
23 Dec 2025, 00:50
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Having drawn a three-part Venn diagram (attached), we can write down a bunch of equations:
The question stands for \(a+b+c=?\)
Subtracting PM attendees from total, we get \(a+x+c=308-220=88\)
Then, \(a+x+c=6c+c/3+c =22/3*c =88\), so \(c=12\)
Therefore, \( a = 12*6=72\) and \(b=4c=48\)
Finally, \(a+b+c=12+72+48=132,\) and the answer is D.
- \(a+b+c+x+y+z+k=308\) for total participants
- \(b+y+z+k=220\) for PM participants
- \(a=6c\)
- \(b=12x=4c\), so \(x=c/3\)
The question stands for \(a+b+c=?\)
Subtracting PM attendees from total, we get \(a+x+c=308-220=88\)
Then, \(a+x+c=6c+c/3+c =22/3*c =88\), so \(c=12\)
Therefore, \( a = 12*6=72\) and \(b=4c=48\)
Finally, \(a+b+c=12+72+48=132,\) and the answer is D.
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23 Dec 2025, 00:54
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Let those that attended DA be a, pm be b and Negotiation be c, D&P be d, D&N be e, P&N be g and g be for those attending all the three.
a=6c
b=12e
c=1/4b
Now a+b+c+d+e+f+g+h= 308
Given b+d+f+g=220. Substituting c=b/4, a= 6xb/4, and e=b/12
3/2b+b+b/4+d+b/12+f+g=308
Given b+d+f+g=220
So 17/b-b=88 b =48
a= 3/2*48=72
c=48/4= 12
e=48/12=12
Exactly one is thus 72+48+12= 132
a=6c
b=12e
c=1/4b
Now a+b+c+d+e+f+g+h= 308
Given b+d+f+g=220. Substituting c=b/4, a= 6xb/4, and e=b/12
3/2b+b+b/4+d+b/12+f+g=308
Given b+d+f+g=220
So 17/b-b=88 b =48
a= 3/2*48=72
c=48/4= 12
e=48/12=12
Exactly one is thus 72+48+12= 132
Bunuel
