Comprehension
Help Distress (HD) - Volunteer Project Information
Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has 37 volunteers. They are involved in three projects:
- Tsunami Relief (TR) in Tamil Nadu
- Flood Relief (FR) in Maharashtra
- Earthquake Relief (ER) in Gujarat
Each volunteer working with Help Distress has to be involved in at least one relief work project.
Given Information:
- A maximum number of volunteers are involved in the FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project.
- The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects.
- 17 volunteers are involved in the TR project.
- The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER project alone.
- 10 volunteers involved in the TR project are also involved in at least one more project.
Question: 1
Based on the information given above, the minimum number of volunteers involved in both FR and TR projects, but not in the ER project is:
Based on the information given above, the minimum number of volunteers involved in both FR and TR projects, but not in the ER project is:
Show Hint
Fill the seven regions of a three-circle Venn diagram; everything is fixed except the split of six volunteers between 'FR and TR only' and 'TR and ER only', which is limited by FR having the most volunteers.
Updated On: Nov 4, 2026
- 1
- 4
- 3
- 5
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Set up a Venn diagram:
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers (more than ER and TR), so \(16+d>22-d\), giving \(d>3\), i.e. \(d\ge 4\) (and \(d\le 6\)).
Step 4: Minimum:
The number of volunteers in FR and TR but not ER is \(d\), and \(d\ge 4\). With \(d=4\): FR \(=20\), ER \(=18\), TR \(=17\), which is consistent.
Final Answer:
The correct option is (2): 4
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers (more than ER and TR), so \(16+d>22-d\), giving \(d>3\), i.e. \(d\ge 4\) (and \(d\le 6\)).
Step 4: Minimum:
The number of volunteers in FR and TR but not ER is \(d\), and \(d\ge 4\). With \(d=4\): FR \(=20\), ER \(=18\), TR \(=17\), which is consistent.
Final Answer:
The correct option is (2): 4
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Question: 2
Which of the following additional information would enable to find the exact number of volunteers involved in various projects?
Which of the following additional information would enable to find the exact number of volunteers involved in various projects?
Show Hint
After using the given conditions, only the split of six volunteers between two regions remains unknown. See which option fixes it.
Updated On: Nov 4, 2026
- Twenty three volunteers are involved in exactly one project.
- Four volunteers are involved in all the three projects.
- Twenty volunteers are involved in FR.
- No need for any additional information.
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Set up a Venn diagram:
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers (more than ER and TR), so \(16+d>22-d\), giving \(d>3\), i.e. \(d\ge 4\) (and \(d\le 6\)).
Step 4: What remains unknown:
Only \(d\) (and then \(f=6-d\)) is undetermined, with \(d\in\{4,5,6\}\).
Step 5: Test each option:
(3) FR \(=20\Rightarrow 16+d=20\Rightarrow d=4\). This determines everything.
(2) Four volunteers in all three: already derived (\(g=4\)), so no new information.
(1) Exactly one project: \(a+b+c=8+7+8=23\) always, so no new information.
(4) Not sufficient, as \(d\) can be 4, 5 or 6.
Final Answer:
The correct option is (3): Twenty volunteers are involved in FR.
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers (more than ER and TR), so \(16+d>22-d\), giving \(d>3\), i.e. \(d\ge 4\) (and \(d\le 6\)).
Step 4: What remains unknown:
Only \(d\) (and then \(f=6-d\)) is undetermined, with \(d\in\{4,5,6\}\).
Step 5: Test each option:
(3) FR \(=20\Rightarrow 16+d=20\Rightarrow d=4\). This determines everything.
(2) Four volunteers in all three: already derived (\(g=4\)), so no new information.
(1) Exactly one project: \(a+b+c=8+7+8=23\) always, so no new information.
(4) Not sufficient, as \(d\) can be 4, 5 or 6.
Final Answer:
The correct option is (3): Twenty volunteers are involved in FR.
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Question: 3
After some time, the volunteers who were involved in all the three projects were asked to withdraw from one project. As a result, one of the volunteers opted out of the TR project, and one opted out of the ER project, while the remaining ones involved in all the three projects opted out of the FR project. Which of the following statements, then, necessarily follows?
After some time, the volunteers who were involved in all the three projects were asked to withdraw from one project. As a result, one of the volunteers opted out of the TR project, and one opted out of the ER project, while the remaining ones involved in all the three projects opted out of the FR project. Which of the following statements, then, necessarily follows?
Show Hint
Find the new totals of FR, TR and ER in terms of the single unknown d, and test each statement for all admissible values of d.
Updated On: Nov 4, 2026
- The number of volunteers in the ER Project is even.
- More volunteers are now in TR project as compared to ER project.
- More volunteers are now in FR project as compared to ER project.
- The number of volunteers in the FR Project is even.
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Set up a Venn diagram:
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers (more than ER and TR), so \(16+d>22-d\), giving \(d>3\), i.e. \(d\ge 4\) (and \(d\le 6\)).
Step 4: After the withdrawal:
The 4 volunteers in all three projects each leave one project: one leaves TR, one leaves ER, and the other two leave FR.
FR \(=16+d-2=14+d\); TR \(=17-1=16\); ER \(=22-d-1=21-d\).
Step 5: Test the statements for \(d=4,5,6\):
FR: 18, 19, 20. ER: 17, 16, 15. TR: 16.
(1) ER even: false for \(d=4\) (17). Not necessary.
(3) FR \(>\) ER: \(14+d>21-d\iff d\ge 4\). True always.
(2) TR \(>\) ER: \(16>21-d\iff d>5\). False for \(d=4\), so not necessary.
(4) FR even: false for \(d=5\) (19). Not necessary.
Final Answer:
The correct option is (3): More volunteers are now in FR project as compared to ER project.
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers (more than ER and TR), so \(16+d>22-d\), giving \(d>3\), i.e. \(d\ge 4\) (and \(d\le 6\)).
Step 4: After the withdrawal:
The 4 volunteers in all three projects each leave one project: one leaves TR, one leaves ER, and the other two leave FR.
FR \(=16+d-2=14+d\); TR \(=17-1=16\); ER \(=22-d-1=21-d\).
Step 5: Test the statements for \(d=4,5,6\):
FR: 18, 19, 20. ER: 17, 16, 15. TR: 16.
(1) ER even: false for \(d=4\) (17). Not necessary.
(3) FR \(>\) ER: \(14+d>21-d\iff d\ge 4\). True always.
(2) TR \(>\) ER: \(16>21-d\iff d>5\). False for \(d=4\), so not necessary.
(4) FR even: false for \(d=5\) (19). Not necessary.
Final Answer:
The correct option is (3): More volunteers are now in FR project as compared to ER project.
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Question: 4
After the withdrawal of volunteers, as indicated in the previous question, some new volunteers joined the NGO. Each one of them was allotted only one project in a manner such that, the number of volunteers working in one project alone for each of the three projects became identical. At that point, it was also found that the number of volunteers involved in FR and ER projects was the same as the number of volunteers involved in TR and ER projects. Which of the projects now has the highest number of volunteers?
After the withdrawal of volunteers, as indicated in the previous question, some new volunteers joined the NGO. Each one of them was allotted only one project in a manner such that, the number of volunteers working in one project alone for each of the three projects became identical. At that point, it was also found that the number of volunteers involved in FR and ER projects was the same as the number of volunteers involved in TR and ER projects. Which of the projects now has the highest number of volunteers?
Show Hint
After the withdrawal nobody is in all three projects, and the new volunteers raise the 'alone' groups to a common size, so the comparison depends only on the two-project groups; the 'FR and ER' versus 'TR and ER' statement fixes them.
Updated On: Nov 4, 2026
- FR
- ER
- TR
- Cannot be determined
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Set up a Venn diagram:
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers, so \(16+d\ge 22-d\), i.e. \(d\ge 3\) (and \(d\le 6\)); the final condition of this question will decide \(d\).
Step 4: After the withdrawal (nobody is in all three):
FR and TR only \(=d+1\); FR and ER only \(=4+1=5\); TR and ER only \(=f+2=8-d\).
Step 5: After the new volunteers join:
The 'alone' groups (FR only, TR only, ER only) all become equal, say \(N\) each. The two-project groups do not change.
FR \(=N+(d+1)+5\); TR \(=N+(d+1)+(8-d)=N+9\); ER \(=N+5+(8-d)\).
Step 6: Use the extra condition:
The number in both FR and ER (5) equals the number in both TR and ER \((8-d)\), so \(d=3\). Then FR and TR \(=4\), FR and ER \(=5\), TR and ER \(=5\).
FR \(=N+9\); TR \(=N+9\); ER \(=N+10\).
ER has the highest number of volunteers.
Final Answer:
The correct option is (2): ER
Let FR only \(=a\), TR only \(=b\), ER only \(=c\), FR and TR only \(=d\), FR and ER only \(=e\), TR and ER only \(=f\), all three \(=g\). Total \(a+b+c+d+e+f+g=37\).
Step 2: Translate the conditions:
ER only is double the all-three: \(c=2g\).
TR only is one less than ER only: \(b=c-1=2g-1\).
Ten TR volunteers are also in another project: \(d+f+g=10\); TR total is 17, so \(b=17-10=7\). Hence \(2g-1=7\), \(g=4\), \(c=8\).
FR only equals the FR volunteers who are also in ER: \(a=e+g=e+4\).
Total: \(a+7+8+d+e+f+4=37\Rightarrow a+e+(d+f)=18\). Since \(d+f=10-4=6\), \(a+e=12\), so \(e=4\) and \(a=8\).
Step 3: Totals in each project:
FR \(=a+d+e+g=16+d\); TR \(=17\); ER \(=c+e+f+g=16+f=22-d\) (using \(f=6-d\)).
FR has the maximum number of volunteers, so \(16+d\ge 22-d\), i.e. \(d\ge 3\) (and \(d\le 6\)); the final condition of this question will decide \(d\).
Step 4: After the withdrawal (nobody is in all three):
FR and TR only \(=d+1\); FR and ER only \(=4+1=5\); TR and ER only \(=f+2=8-d\).
Step 5: After the new volunteers join:
The 'alone' groups (FR only, TR only, ER only) all become equal, say \(N\) each. The two-project groups do not change.
FR \(=N+(d+1)+5\); TR \(=N+(d+1)+(8-d)=N+9\); ER \(=N+5+(8-d)\).
Step 6: Use the extra condition:
The number in both FR and ER (5) equals the number in both TR and ER \((8-d)\), so \(d=3\). Then FR and TR \(=4\), FR and ER \(=5\), TR and ER \(=5\).
FR \(=N+9\); TR \(=N+9\); ER \(=N+10\).
ER has the highest number of volunteers.
Final Answer:
The correct option is (2): ER
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