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14 Jul 2026, 10:56
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Morning track Afternoon Track Total
Data Visualization 2/5X( statement-1) k 30
Non-data visualization 3/5X( statement-1) Y-k 45
Total X Y 75( combined total)
( given in question that total are 75 and each participants will have exactly one track and each track will have one participant minimum, 30 participants in total data visualization participants so remaining 45 out of 75 will be non-data visualization participant
Total participants will constitute following categories as tabulated above: morning track data visualization, morning track non data visualization and afternoon data visualization and afternoon non-data visualization( 4 categories)
From Table as per statement-1
0.4X+k=30
0.6X+Y-k=45 we want to know if we can know (Y-k/Y)
Y-k=45-0.6X (45-0.6X)/(75-X)
Take 0.6 common from nominator 0.6*(75-x)/(75-x)=0.6 so we can determine 0.6 fraction or 60% of afternoon track are non-data visualization track.
Statement-1 is sufficient
Statement-2: Lets assume , afternoon track non-data visualization participant are a,
then data visualization participants in afternoon track is 2/3a.
Total participants in afternoon track is a+2/3a= 5/3 a.
Ratio we are interested in is afternoon non-data visualization track participants/ total afternoon track participants= (a)/(5/3a)= 3/5 again same 60% or 0.6 fraction. answerable.
Statement-2 is also sufficient
Each statement alone is sufficient
Data Visualization 2/5X( statement-1) k 30
Non-data visualization 3/5X( statement-1) Y-k 45
Total X Y 75( combined total)
( given in question that total are 75 and each participants will have exactly one track and each track will have one participant minimum, 30 participants in total data visualization participants so remaining 45 out of 75 will be non-data visualization participant
Total participants will constitute following categories as tabulated above: morning track data visualization, morning track non data visualization and afternoon data visualization and afternoon non-data visualization( 4 categories)
From Table as per statement-1
0.4X+k=30
0.6X+Y-k=45 we want to know if we can know (Y-k/Y)
Y-k=45-0.6X (45-0.6X)/(75-X)
Take 0.6 common from nominator 0.6*(75-x)/(75-x)=0.6 so we can determine 0.6 fraction or 60% of afternoon track are non-data visualization track.
Statement-1 is sufficient
Statement-2: Lets assume , afternoon track non-data visualization participant are a,
then data visualization participants in afternoon track is 2/3a.
Total participants in afternoon track is a+2/3a= 5/3 a.
Ratio we are interested in is afternoon non-data visualization track participants/ total afternoon track participants= (a)/(5/3a)= 3/5 again same 60% or 0.6 fraction. answerable.
Statement-2 is also sufficient
Each statement alone is sufficient
14 Jul 2026, 10:59
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Need to find percentage of afternoon track people that did NOT register for the data visualization session
Total participants 75
Total registered for data visualization 30
Total no registered for data 45
Let M be morning group
Let A be afternoon group
than M + A = 75
Statement 1
Morning people 2/5 registered for data
2/5M
Since there are 30 total registered for data
Afternoon registrants is therefore 30- (2/5M)
Afternoon non data registrants
A - (30 - 2/5M)
Since afternoon people A equals 75 - M
then
75 - M - (30 - 2/5M) = 75 - M - 30 + 2/5M
=45 - 3/5M
Percentage is
(45 - 3/5M) / (75 - M) = (75 - M) 3/5 over 75 -M
remove (75 - M)
you get 3/5
Statement 1 is sufficent
Statement 2
Among the afternoon participants, the number who registered is 2/3 the number not registered
Let R be afternoon registrants
Let N be afternoon non registrants
Then
R = 2/3 N
So R to N is 2 to 3 and total participants is 5
There for non registrants to total is
3 to 5 or 3/5. Sufficient
Therefore both statements are sufficient
Answer D
Total participants 75
Total registered for data visualization 30
Total no registered for data 45
Let M be morning group
Let A be afternoon group
than M + A = 75
Statement 1
Morning people 2/5 registered for data
2/5M
Since there are 30 total registered for data
Afternoon registrants is therefore 30- (2/5M)
Afternoon non data registrants
A - (30 - 2/5M)
Since afternoon people A equals 75 - M
then
75 - M - (30 - 2/5M) = 75 - M - 30 + 2/5M
=45 - 3/5M
Percentage is
(45 - 3/5M) / (75 - M) = (75 - M) 3/5 over 75 -M
remove (75 - M)
you get 3/5
Statement 1 is sufficent
Statement 2
Among the afternoon participants, the number who registered is 2/3 the number not registered
Let R be afternoon registrants
Let N be afternoon non registrants
Then
R = 2/3 N
So R to N is 2 to 3 and total participants is 5
There for non registrants to total is
3 to 5 or 3/5. Sufficient
Therefore both statements are sufficient
Answer D
14 Jul 2026, 11:12
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| Morning | Afternoon | ||
| Data Vis. | 2/5a | 30 | |
| No data Vis. | 3/5a | ||
| a | 75 |
Total is 75
St1.
If x is morning count of participants. 2/5 of "a" are registered for data visualisation.
30=Morning registered + Afternoon registered
30=2/5"a" + Afternoon registered
We do not know "a" so we hit a wall.
Not sufficient
St2. for every 2 registered , 3 did not registered.
What percentage of the afternoon track participants did not register= 3x/5x=3/5=60%
Sufficient
Answer is B.
