Five teams - A, B, C, D, and E - each consisting of 15 members, are going on expeditions to five different locations. Each team includes members from three different skill sets: biologists, geologists, and explorers. However, the number of members from each skill set varies by team and each member has only one speciality. The total number of biologists, geologists, and explorers are equal.
The following additional information is available:
* Every team has at least 2 members from each of the three skill sets.
* Teams C and D have 6 biologists each, and team A has 6 geologists.
* Every team except A has more biologists than explorers.
* The number of explorers in each team is distinct and decreases in the order A, B, C, D, and E.
The number of biologists in team E is
Enter your answer to attempt
The Setup: We are dealing with a logical puzzle disguised as a matrix grid. We have 5 teams of 15 members each, meaning there are 75 total members. Since the three roles (Biologists, Geologists, Explorers) have equal total numbers, we just divide 75 by 3. That means exactly 25 Biologists, 25 Geologists, and 25 Explorers across the entire board. Time to build the matrix and min-max the stats.
Step 1:The Explorer Distribution (The Meta)
The problem states that the number of Explorers in each team strictly decreases in the order A > B > C > D > E. We also know they must sum up to 25, and every single team must have at least 2 of *each* role.
Let's find the maximum number of Explorers team A can have. Team A already has 6 Geologists (given) and must have at least 2 Biologists.
Max_ExplorersA=15−6−2=7
If Team A has a maximum of 7 Explorers, we need 5 strictly decreasing integers starting from ≤7 that sum to 25. There is literally only one valid sequence in the entire universe that satisfies this:
7+6+5+4+3=25
Lock it in:
Team A = 7, Team B = 6, Team C = 5, Team D = 4, Team E = 3.
Step 2:The Biologist Constraints (Vibe Check)
Now let's look at the Biologists (B). We know the total is 25.
The prompt feeds us free intel: BC=6 and BD=6.
This means the remaining Biologists for teams A, B, and E are:
25−6−6=13
So, BA+BB+BE=13.
Now we apply the heavy constraint: "Every team except A has more biologists than explorers."
* For Team B: BB>EB→BB>6→BB≥7
* For Team E: BE>EE→BE>3→BE≥4
* For Team A (the exception): It just needs to meet the baseline minimum of BA≥2.
Step 3:Solving the Biologist Equation
We have our minimum thresholds: BA≥2, BB≥7, and BE≥4.
Let's add these bare minimums together:
2+7+4=13
Since the sum of the absolute minimums equals exactly the total number of remaining Biologists (13), there is zero wiggle room. The math is mathing perfectly. None of these values can be higher, or the sum would break 13.
Therefore, BA=2, BB=7, and BE=4.
Step 4:The Audit (Double Check Protocol)
Let's run a full matrix audit to ensure we aren't throwing the game. We'll fill in the Geologists (G) using the formula G=15−B−E to make sure no team breaks the rules.
* Team A: E=7, B=2, G=6. (Sum = 15, all ≥2. Works. G=6 matches the prompt.)
* Team B: E=6, B=7, G=2. (Sum = 15, all ≥2, B>E. Works.)
* Team C: E=5, B=6, G=4. (Sum = 15, all ≥2, B>E. Works.)
* Team D: E=4, B=6, G=5. (Sum = 15, all ≥2, B>E. Works.)
* Team E: E=3, B=4, G=8. (Sum = 15, all ≥2, B>E. Works.)
Now, let's verify total Geologists:
6+2+4+5+8=25
Perfectly balanced, as all things should be. The board is fully validated and legally solved.
Final Answer: 4
Five teams - A, B, C, D, and E - each consisting of 15 members, are going on expeditions to five different locations. Each team includes members from three different skill sets: biologists, geologists, and explorers. However, the number of members from each skill set varies by team and each member has only one speciality. The total number of biologists, geologists, and explorers are equal.
The following additional information is available.
* Every team has at least 2 members from each of the three skill sets.
* Teams C and D have 6 biologists each, and Team A has 6 geologists.
* Every team except A has more biologists than explorers.
* The number of explorers in each team is distinct and decreases in the order A, B, C, D, and E.
The number of teams having more geologists than biologists is
Enter your answer to attempt
The Setup:
We are literally back in the exact same lobby as Question 2. This is a follow-up question utilizing the same logical matrix we already decoded. Since we previously theory-crafted the exact distribution of all 75 members across the grid, we can just pull up our cached data and run a quick stat check to find out who has higher Geologist DPS than Biologist DPS. Free points, honestly.
Step 1:Retrieving the Master Matrix (From Memory)
Let's pull the fully solved grid we established earlier. Remember, every team has exactly 15 members, and each class (Explorer, Biologist, Geologist) totals exactly 25 across all teams.
Team
Explorers (E)
Biologists (B)
Geologists (G)
A
7
2
6
B
6
7
2
C
5
6
4
D
4
6
5
E
3
4
8
Step 2:The Stat Check
The prompt asks for the number of teams where Geologists > Biologists (G>B). Let's run down the roster and vibe check every single team:
* Team A:G=6, B=2. (6>2) →Valid (W)
* Team B:G=2, B=7. (2<7) → Invalid (L)
* Team C:G=4, B=6. (4<6) → Invalid (L)
* Team D:G=5, B=6. (5<6) → Invalid (L)
* Team E:G=8, B=4. (8>4) →Valid (W)
Only Team A and Team E meet the specific criteria.
Step 3:The Audit (Double Check Protocol)
Let's do a quick sanity check to ensure our base matrix didn't glitch.
Are there exactly 25 of each role?
Explorers: 7+6+5+4+3=25 (Check).
Biologists: 2+7+6+6+4=25 (Check).
Geologists: 6+2+4+5+8=25 (Check).
Do all teams have 15 members? Yes.
Does every team have ≥2 in each role? Yes.
Do C and D have 6 Biologists? Yes.
Does A have 6 Geologists? Yes.
Does every team except A have B>E?
Team B (7>6), Team C (6>5), Team D (6>4), Team E (4>3). Yes.
Does E strictly decrease from A to E? 7>6>5>4>3. Yes.
The matrix is perfectly locked in. The final count of teams with G>B is strictly 2.
Final Answer: 2
Five teams - A, B, C, D, and E - each consisting of 15 members, are going on expeditions to five different locations. Each team includes members from three different skill sets: biologists, geologists, and explorers. However, the number of members from each skill set varies by team and each member has only one speciality. The total number of biologists, geologists, and explorers are equal.
The following additional information is available.
* Every team has at least 2 members from each of the three skill sets.
* Teams C and D have 6 biologists each, and Team A has 6 geologists.
* Every team except A has more biologists than explorers.
* The number of explorers in each team is distinct and decreases in the order A, B, C, D, and E.
The median number of biologists across five teams is
Enter your answer to attempt
The Setup:
We are dropping back into the team arrangement lobby for round 3. Since we already theory-crafted and completely solved the master matrix in Question 2, we just need to query our Biologist data and run a basic median calculation. Free elo.
Step 1:Retrieving the Biologist Stats
Let's pull the exact number of Biologists (B) for each team from our previously validated master grid:
* Team A: 2
* Team B: 7
* Team C: 6 (Given in prompt)
* Team D: 6 (Given in prompt)
* Team E: 4Step 2:Sorting the Array
To find the median, we can't just pick the middle team. We have to sort the Biologist values in ascending order to find the true statistical middle (the 50th percentile).
Unsorted array: {2,7,6,6,4}
Sorted array: {2,4,6,6,7}Step 3:Finding the Median
Since there are exactly 5 teams (an odd number), the median is literally just the dead-center value—the 3rd number in our sorted array.
Looking at our sorted list {2,4,6,6,7}, the middle value is 6.
Step 4:The Audit (Double Check Protocol)
Let's run a quick sanity check to make sure the data hasn't been corrupted.
Did they sum to 25? 2+4+6+6+7=25. (Yes).
Did every team except A have more Biologists than Explorers?
Team B: 7>6. Team C: 6>5. Team D: 6>4. Team E: 4>3. (Yes).
Are we picking the 3rd index of a 5-item sorted list? Yes.
The math is flawless. The logic is strictly locked in.
Final Answer: 6
In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage
The constituency in which B got lower number of votes compared to A and C is
AS3
BS4
CS2
DS1
Pick an option to attempt
The Setup: This is a logical reasoning Data Interpretation problem based on an election matrix. We must deduce the exact vote distribution for parties A, B, and C across 5 constituencies using their total vote counts, win conditions, and numerical sequence constraints.
Step 1: Analyze Party C's winning constraints.
Every constituency has exactly 20 voters and a clear winner (no ties for first place). To guarantee a win without tying, a party must secure a minimum of 8 votes (since the remaining 12 could be split 6−6; a score of 7 allows a 7−7−6 tie).
Party C won *only* S2 and S3. Thus, c2≥8 and c3≥8.
Party C's total votes across all constituencies is 16.
Therefore, C must have obtained exactly 8 votes in S2, 8 votes in S3, and 0 votes in S1, S4, and S5.
Step 2: Analyze Party B's winning constraints.
Party A won only S1, and C won only S2 and S3. By elimination, Party B must be the winner of S4 and S5.
In S4 and S5, C has 0 votes, meaning A and B split the 20 votes entirely. For B to win clearly, B must secure more than half the votes: b4≥11 and b5≥11.
Step 3: Apply Party B's sequence constraints.
B's votes across S1 to S5 are distinct natural numbers in increasing order: b1<b2<b3<b4<b5.
Since b4≥11, and b5 must be strictly greater than b4, b5≥12.
B's total votes equal 35. To leave enough votes for the first three constituencies, we must minimize b4 and b5.
Let b4=11 and b5=12.
The remaining votes for the first three constituencies are: b1+b2+b3=35−11−12=12.
Step 4: Determine Party B's exact sequence.
In S2 and S3, C wins with 8 votes. Therefore, A and B must each have fewer than 8 votes (b2≤7,b3≤7,a2≤7,a3≤7).
Since ai+bi=20−8=12 in these constituencies, the only valid integer pairs for (ai,bi) bounded by 7 are (7,5),(6,6), and (5,7).
Thus, B's votes in S2 and S3 must be chosen from the set {5,6,7}.
Maintaining the strictly increasing sequence b2<b3, we test combinations to satisfy b1+b2+b3=12:
If (b2,b3)=(5,6), then b1=12−11=1. (Valid natural number)
If (b2,b3)=(5,7), then b1=12−12=0. (Invalid, natural numbers begin at 1)
Therefore, B's exact vote sequence across S1-S5 is 1,5,6,11,12.
Step 5: Calculate Party A's vote distribution.
Using the formula ai=20−bi−ci:
S1: a1=20−1−0=19 (A wins)
S2: a2=20−5−8=7 (C wins)
S3: a3=20−6−8=6 (C wins)
S4: a4=20−11−0=9 (B wins)
S5: a5=20−12−0=8 (B wins)
Checking the total: 19+7+6+9+8=49, which perfectly matches A's given total.
Step 6: Answer the specific prompt.
We must find the constituency where B got fewer votes than both A and C.
In S2, B has 5 votes, A has 7 votes, and C has 8 votes. 5<7 and 5<8.
Final Answer: S2
A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.
Drug
Organism P
Organism Q
Organism R
A
Y
B
Y
C
Y
D
E
Following additional information is available:
Each drug works on at least one organism but not more than two organisms.
Each organism can be treated with at least two and at most three of these five drugs.
On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works.
D and E do not work on the same organism.
show less
Drug E works on
AP & R
BQ & R
COnly P
DOnly R
Pick an option to attempt
The Setup: A pharmaceutical company tested five drugs (A, B, C, D, E) on three organisms (P, Q, R). We must deduce the complete relationship matrix using the provided table and logical constraints.
Step 1: Establish the initial matrix and apply direct implication rules.
From the provided table, A works on R, B works on Q, and C works on Q.
Rule 3 states that if A works on an organism, B also works on it (A⟹B). Since A works on R, B must also work on R.
Rule 4 states that if C works on an organism, D also works on it (C⟹D). Since C works on Q, D must also work on Q.
Step 2: Apply column capacity constraints.
Rule 2 states each organism can be treated with at most 3 drugs.
Currently, Organism Q is treated by B, C, and D, reaching its absolute maximum capacity. Therefore, neither A nor E can work on Q.
Since A is restricted from Q and must act as a strict subset of B's placements (which are only Q and R), A must solely work on R.
Step 3: Apply disjoint rules to isolate P and R.
Rule 2 dictates Organism P needs at least 2 drugs. The only available candidates for P are C, D, and E (A and B are locked into Q and R).
Rule 5 states D and E cannot work on the same organism. Thus, P cannot be treated by both D and E.
To secure 2 drugs for P without pairing D and E, P must logically be treated by C and D (choosing C automatically forces D via Rule 4).
Step 4: Finalize Drug E's placement.
Drug E cannot work on Q (capacity full) and cannot work on P (disjoint with D).
Rule 1 mandates each drug works on at least 1 organism. Thus, E is forced to work on Organism R.
The finalized matrix is: P={C,D}, Q={B,C,D}, R={A,B,E}. Drug E strictly works only on organism R.
Final Answer: Only R
A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage
Number of days Bhabha and Cathy work together in a week is
A0
B1
C3
D2
Pick an option to attempt
The Setup: This is a logical arrangement puzzle requiring us to map five employees onto a 7-day schedule. We deduce one Master Matrix satisfying every daily capacity and every shift rule; that single grid then answers all five questions in this set.
Step 1: Check the shift arithmetic and fix the reading of 'consecutive'.
Total shifts required =6 days×2+1 Sunday×3=15.
Five employees working exactly 3 days each also supply 5×3=15 shifts, so the grid is exactly saturated: every slot is filled and nobody has spare capacity.
A caselet like this is far easier to see than to hold in your head, so draw the grid first and fill cells in as they are forced. The margins carry the capacities, and they are what drive every deduction:
Day
Alex
Bhabha
Cathy
Dilip
Ethan
Needed
Sunday
3
Monday
2
Tuesday
2
Wednesday
2
Thursday
2
Friday
2
Saturday
2
Works
3
3
3
3
3
15
Each employee's three days therefore consist of exactly one consecutive pair plus a third day adjacent to neither.
The week is read as a cycle, so Saturday and Sunday count as consecutive. This is not a cosmetic detail: it means a run of Saturday, Sunday, Monday is a three-consecutive-day block and is barred. Read the week as a straight line instead and the puzzle admits three different grids, two of which contradict the stated answers to the later questions in this set, so the cyclic reading is the intended one.
Step 2: Fix Cathy, then Alex.
Alex works Tuesday and Wednesday. Neither Bhabha nor Cathy ever shares a day with Alex, so neither of them works Tuesday or Wednesday.
Cathy is additionally barred from Saturday and Monday. That removes Monday, Tuesday, Wednesday and Saturday, leaving exactly three days:
Cathy: Sunday, Thursday, Friday. (Thursday-Friday consecutive, Sunday isolated. Valid.)
Alex cannot share a day with Cathy, so Alex is barred from Sunday, Thursday and Friday. His third day could not have been Monday or Thursday in any case, since either would extend Tuesday-Wednesday into a three-day run. Only one day survives:
Alex: Tuesday, Wednesday, Saturday.Step 3: Force Monday, and with it Ethan.
Tuesday and Wednesday are already full (Alex and Dilip), so no one else can work them.
Now look at Monday. Alex works only Tuesday, Wednesday and Saturday; Cathy is barred from Monday; Dilip's third day cannot be Monday, as that would extend Tuesday-Wednesday to three in a row. That leaves only Bhabha and Ethan available, and Monday needs exactly two people. So Bhabha and Ethan both work Monday.
Saturday needs one more person alongside Alex. Cathy is barred, Bhabha can never share a day with Alex, and Dilip's third day must differ from Alex's. So Ethan works Saturday.
Ethan now holds Monday and Saturday, and his third day cannot be Tuesday or Wednesday. Test what remains:
* Sunday would give {Saturday, Sunday, Monday}, three consecutive days on the cyclic week. Barred.
* Thursday would give {Monday, Thursday, Saturday}, which contains no consecutive pair at all. Barred.
* Friday gives {Monday, Friday, Saturday}, exactly one pair (Friday-Saturday). Valid.
Ethan: Monday, Friday, Saturday.Step 4: Close out Dilip and Bhabha.
Friday now holds Cathy and Ethan, so it is full and Dilip's third day cannot be Friday. It also cannot be Monday or Thursday (three-in-a-row), nor Saturday (his third day must differ from Alex's). One option remains:
Dilip: Sunday, Tuesday, Wednesday.
Sunday needs three people and currently holds Cathy and Dilip, so Bhabha takes the last Sunday slot. Thursday needs two and holds only Cathy, so Bhabha works Thursday as well. Together with the Monday shift forced in Step 3:
Bhabha: Sunday, Monday, Thursday. (Sunday-Monday consecutive, Thursday isolated. Valid.)
Master Matrix:
Day
Alex
Bhabha
Cathy
Dilip
Ethan
Needed
Sunday
✓
✓
✓
3
Monday
✓
✓
2
Tuesday
✓
✓
2
Wednesday
✓
✓
2
Thursday
✓
✓
2
Friday
✓
✓
2
Saturday
✓
✓
2
Works
3
3
3
3
3
15
Reading the grid down the columns: Alex takes Tuesday, Wednesday and Saturday; Bhabha Sunday, Monday and Thursday; Cathy Sunday, Thursday and Friday; Dilip Sunday, Tuesday and Wednesday; Ethan Monday, Friday and Saturday.
Every day carries exactly two workers except Sunday, which carries three, and every employee has exactly one consecutive pair plus one isolated day. No other grid satisfies all the constraints.
Step 5: Answer this question.
Bhabha works Sunday, Monday and Thursday. Cathy works Sunday, Thursday and Friday.
They overlap on exactly two days: Sunday and Thursday.
Final Answer: 2
In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage
The number of votes obtained by B in S2 is
A6
B7
C5
D4
Pick an option to attempt
The Setup: We must identify the specific number of votes obtained by Party B in constituency S2, using the comprehensive election matrix derived from the logical constraints.
Step 1: Reference the derived election matrix.
As established through the total vote counts and sequence constraints:
Party C's votes across S1-S5: 0,8,8,0,0
Party B's votes across S1-S5: 1,5,6,11,12
Party A's votes across S1-S5: 19,7,6,9,8Step 2: Isolate the requested data point.
We look at Party B's vote sequence (b1,b2,b3,b4,b5) and identify the value for S2 (b2).
b2=5.
Final Answer: 5
A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.
Drug
Organism P
Organism Q
Organism R
A
Y
B
Y
C
Y
D
E
Following additional information is available:
Each drug works on at least one organism but not more than two organisms.
Each organism can be treated with at least two and at most three of these five drugs.
On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works.
D and E do not work on the same organism.
show less
Drug D works on:
AP & Q
BOnly P
COnly Q
DQ & R
Pick an option to attempt
The Setup: Using the fully deduced matrix from the pharmaceutical constraints, we need to determine the exact organisms treated by Drug D.
Step 1: Reference the derived matrix.
As mathematically established by the constraint deduction: P={C,D}, Q={B,C,D}, R={A,B,E}.
Step 2: Isolate Drug D.
Evaluating the assignments, Drug D successfully operates on both Organism P and Organism Q.
Final Answer: P & Q
A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage
Which among the following employees do not work together on any of the days?
ABhabha and Dilip
BBhabha and Ethan
CAlex and Dilip
DDilip and Ethan
Pick an option to attempt
The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:
Day
Alex
Bhabha
Cathy
Dilip
Ethan
Needed
Sunday
✓
✓
✓
3
Monday
✓
✓
2
Tuesday
✓
✓
2
Wednesday
✓
✓
2
Thursday
✓
✓
2
Friday
✓
✓
2
Saturday
✓
✓
2
Works
3
3
3
3
3
15
Reading off the answer. The question asks which pair never shares a shift, so test each option against the grid:
* Bhabha and Dilip both appear in the Sunday row. They work together.
* Bhabha and Ethan both appear in the Monday row. They work together.
* Alex and Dilip share Tuesday and Wednesday. The clues say so outright, so this pair could never have been the answer.
* Dilip works Sunday, Tuesday, Wednesday. Ethan works Monday, Friday, Saturday. These two sets have nothing in common.
Dilip and Ethan are the only pair with no day in common.
Final Answer: Dilip and Ethan
In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage
The number of votes obtained by A in S5 is
A6
B9
C8
D7
Pick an option to attempt
The Setup: We must identify the specific number of votes obtained by Party A in constituency S5, utilizing the completed election matrix.
Step 1: Reference the derived election matrix.
The calculated vote distribution for Party A across the five constituencies (S1 through S5) is 19,7,6,9, and 8.
Step 2: Isolate the requested data point.
We evaluate Party A's sequence (a1,a2,a3,a4,a5) and extract the specific value corresponding to S5 (a5).
a5=8.
Final Answer: 8
A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.
Drug
Organism P
Organism Q
Organism R
A
Y
B
Y
C
Y
D
E
Following additional information is available:
Each drug works on at least one organism but not more than two organisms.
Each organism can be treated with at least two and at most three of these five drugs.
On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works.
D and E do not work on the same organism.
show less
The organism(s) that can be treated with three of these five drugs is(are)
AOnly P
BOnly Q
CP and Q
DQ and R
Pick an option to attempt
The Setup: We need to identify which organisms are treated by exactly three of the five drugs, relying on the completely deduced matrix.
Step 1: Count the drugs assigned to each organism.
According to the derived matrix:
Organism P is treated by C and D (Total: 2).
Organism Q is treated by B, C, and D (Total: 3).
Organism R is treated by A, B, and E (Total: 3).
Step 2: Select the qualifying organisms.
Both Q and R meet the criteria of being treated by exactly three drugs.
Final Answer: Q and R
A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage
One of the days Alex works on is
AMonday
BFriday
CSunday
DSaturday
Pick an option to attempt
The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:
Day
Alex
Bhabha
Cathy
Dilip
Ethan
Needed
Sunday
✓
✓
✓
3
Monday
✓
✓
2
Tuesday
✓
✓
2
Wednesday
✓
✓
2
Thursday
✓
✓
2
Friday
✓
✓
2
Saturday
✓
✓
2
Works
3
3
3
3
3
15
Reading off the answer. Take the Alex column: he works Tuesday, Wednesday and Saturday.
Checking that against the options, Monday, Friday and Sunday are all ruled out, and each for a reason worth seeing:
* Monday would sit next to Tuesday, extending Tuesday-Wednesday into a three-day run.
* Sunday and Friday are both Cathy's days, and Alex never shares a shift with Cathy.
That leaves Saturday, which is exactly the isolated third day the grid assigns him.
Final Answer: Saturday
In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage
Comparing the number votes obtained by A across different constituencies, the lowest number of votes were in constituency
AS4
BS2
CS5
DS3
Pick an option to attempt
The Setup: We need to compare Party A's vote counts across all five constituencies to determine which constituency yielded their lowest performance.
Step 1: Retrieve Party A's vote distribution.
From our derived election matrix, the votes obtained by Party A in constituencies S1, S2, S3, S4, and S5 are respectively:
19,7,6,9,8Step 2: Identify the minimum value.
Comparing the integers in the set {19,7,6,9,8}, the lowest number is 6.
Step 3: Map the minimum value back to its constituency.
The vote count of 6 corresponds to constituency S3.
Final Answer: S3
A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.
Drug
Organism P
Organism Q
Organism R
A
Y
B
Y
C
Y
D
E
Following additional information is available:
Each drug works on at least one organism but not more than two organisms.
Each organism can be treated with at least two and at most three of these five drugs.
On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works.
D and E do not work on the same organism.
show less
Organism R can be treated with
AA, B and C
BOnly A and E
COnly A and B
DA, B and E
Pick an option to attempt
The Setup: We must list the specific combination of drugs that successfully treat Organism R based on the logical constraints.
Step 1: Reference Organism R in the derived matrix.
As established in the base matrix deduction: P={C,D}, Q={B,C,D}, R={A,B,E}.
Step 2: Isolate Organism R.
The column for Organism R strictly contains treatments from Drugs A, B, and E.
Final Answer: A, B and E
A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage
The consecutive days on which Ethan works are
AThursday and Friday
BSaturday and Sunday
CFriday and Saturday
DSunday and Monday
Pick an option to attempt
The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:
Day
Alex
Bhabha
Cathy
Dilip
Ethan
Needed
Sunday
✓
✓
✓
3
Monday
✓
✓
2
Tuesday
✓
✓
2
Wednesday
✓
✓
2
Thursday
✓
✓
2
Friday
✓
✓
2
Saturday
✓
✓
2
Works
3
3
3
3
3
15
Reading off the answer. Take the Ethan column: he works Monday, Friday and Saturday.
Every employee's week is one consecutive pair plus one isolated day. Friday and Saturday are adjacent, so they are his pair, and Monday is the isolated day.
Note that Ethan could not have taken Sunday instead: with Saturday and Monday already his, adding Sunday would give Saturday, Sunday, Monday, a run of three consecutive days on a week that wraps.
Final Answer: Friday and Saturday
In an election there were five constituencies S1, S2, S3, S4, and S5 with 20 voters each all of whom voted. Three parties A, B, and C contested the elections. The party that gets the maximum number of votes in a constituency wins that seat. In every constituen … read passage
Assume that A and C had formed an alliance and any voter who voted for either A or C would have voted for this alliance. Then the number of seats this alliance would have won is
A4
B2
C3
D5
Pick an option to attempt
The Setup: We must model a hypothetical political alliance between Party A and Party C, combining their votes in each constituency to see how many total seats the new alliance would win against Party B.
Step 1: Calculate the alliance's combined votes per constituency.
We sum the individual votes of A and C for each constituency (ai+ci):
S1: 19+0=19
S2: 7+8=15
S3: 6+8=14
S4: 9+0=9
S5: 8+0=8Step 2: Compare the alliance's votes against Party B's votes.
To win a seat, the alliance's combined votes must exceed Party B's votes in that constituency.
S1: Alliance (19) vs Party B (1) ⇒ Alliance wins.
S2: Alliance (15) vs Party B (5) ⇒ Alliance wins.
S3: Alliance (14) vs Party B (6) ⇒ Alliance wins.
S4: Alliance (9) vs Party B (11) ⇒ Party B wins.
S5: Alliance (8) vs Party B (12) ⇒ Party B wins.
Step 3: Total the seats won by the alliance.
The alliance successfully wins constituencies S1, S2, and S3, totaling 3 seats.
Final Answer: 3
A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.
Drug
Organism P
Organism Q
Organism R
A
Y
B
Y
C
Y
D
E
Following additional information is available:
Each drug works on at least one organism but not more than two organisms.
Each organism can be treated with at least two and at most three of these five drugs.
On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works.
D and E do not work on the same organism.
show less
Organism P can be treated with:
AOnly C and D
BB, C and D
COnly B and D
DA, B, and D
Pick an option to attempt
The Setup: We must list the specific combination of drugs that successfully treat Organism P based on the logical constraints.
Step 1: Reference Organism P in the derived matrix.
As established in the base matrix deduction: P={C,D}, Q={B,C,D}, R={A,B,E}.
Step 2: Isolate Organism P.
To satisfy the minimum 2-drug requirement while avoiding the D/E conflict constraint, Organism P is treated exclusively by Drugs C and D.
Final Answer: Only C and D
A showroom is open on all seven days of the week throughout the year. There are five employees Alex, Bhabha, Cathy, Dilip and Ethan who work in the showroom. Every day except Sunday, two employees are required while on Sunday three employees need to work. Ever … read passage
Employees who work on Sunday are
ABhabha, Cathy and Dilip
BBhabha, Cathy and Ethan
CAlex, Dilip and Ethan
DCathy, Dilip and Ethan
Pick an option to attempt
The Setup: All five questions in this set run off one roster. That grid is derived step by step in the first question of the set, and it is the only arrangement satisfying every clue:
Day
Alex
Bhabha
Cathy
Dilip
Ethan
Needed
Sunday
✓
✓
✓
3
Monday
✓
✓
2
Tuesday
✓
✓
2
Wednesday
✓
✓
2
Thursday
✓
✓
2
Friday
✓
✓
2
Saturday
✓
✓
2
Works
3
3
3
3
3
15
Reading off the answer. Sunday is the only day needing three people rather than two. Read across the Sunday row and the three ticks fall under Bhabha, Cathy and Dilip.
Alex is excluded because Sunday is one of Cathy's days and he never works with her; Ethan is excluded because Saturday, Sunday, Monday would be three consecutive days.
Final Answer: Bhabha, Cathy and Dilip
Q19:jipmat 2025LRDI › ArrangementsMediumLR · MCQ
Seven persons A, B, C, D, E, F and G are sitting around a circle and are facing the centre. G is second to the left of C, who is to the immediate left of F. A is third to the left of E. B is between D and E. Who is sitting immediate right of G?
AC
BD
CF
DA
Pick an option to attempt
Step 1: Anchor the primary cluster
Since they are facing the center, 'Left' implies clockwise and 'Right' implies counter-clockwise. Let's arbitrarily place F at Position 1.
* 'C ... is to the immediate left of F' ⇒ C is at Position 2 (clockwise).
* 'G is second to the left of C' ⇒ G is at Position 4.
* Current circle (Positions 1 through 7): F, C, _ , G, _ , _ , _ .
Step 2: Place the secondary cluster
We know 'B is between D and E', meaning D-B-E or E-B-D will monopolize three consecutive seats. The only open block of three is Positions 5, 6, and 7.
We also know 'A is third to the left of E'. Let's test E's possible spots:
* If E is at 5, third to the left (clockwise) is 1 (F's occupied seat). Invalid.
* If E is at 6, third to the left is 2 (C's occupied seat). Invalid.
* If E is at 7, third to the left is 3. This seat is perfectly empty!
So, E locks into Position 7, and A takes Position 3.
Since B is squeezed between D and E, B must sit at 6, forcing D into 5.
Step 3: Secure the target
The final clockwise arrangement is locked: F, C, A, G, D, B, E.
The question asks who is sitting to the immediate right (counter-clockwise) of G. G is at Position 4; one spot counter-clockwise is Position 3, which is occupied by A.
Final Answer: A
Q20:jipmat 2025LRDI › ArrangementsEasyLR · MCQ
Arrange the following words alphabetically as in a dictionary:
A. Saviour
B. Savage
C. Sausage
D. Savour
AC, B, D, A
BB, C, A, D
CC, B, A, D
DA, B, D, C
Pick an option to attempt
The Setup: This is a basic dictionary sort. All words start with 'Sa'. We need to look at the third letter and beyond to lock in the sequence.
Step 1: Sort by the third letter
* Sausage (3rd letter 'u')
* Savage (3rd letter 'v')
* Saviour (3rd letter 'v')
* Savour (3rd letter 'v')
'Sausage' (C) comes first because 'u' comes before 'v'.
Step 2: Sort the remaining 'Sav' words
Now we look at the 4th letter for the remaining three:
* Savage (4th letter 'a')
* Saviour (4th letter 'i')
* Savour (4th letter 'o')
Alphabetically, 'a' comes before 'i', which comes before 'o'.
So the order is Savage (B) → Saviour (A) → Savour (D).
Step 3: Secure the full sequence
Stacking it all together gives us: Sausage, Savage, Saviour, Savour (C, B, A, D).
Final Answer: C, B, A, D
Five friends A, B, C, D, and E go on a trip together. Each of them likes at least one of the following three activities: Countryside Sightseeing, Shopping, and Adventure Sports such that:
- No two friends like exactly the same set of activities.
- A, B and E like exactly two activities each.
- D likes only Countryside Sightseeing.
- C likes more number of activities compared to D.
The number of friends who like both Shopping and Adventure Sports is ___
Enter your answer to attempt
The Setup: This is a classic Logical Reasoning/Arrangements puzzle (IPMAT Indore vibes). Think of it as assigning unique loadouts to a 5-player squad from a pool of 3 perks (Countryside Sightseeing, Shopping, Adventure Sports). The golden rule here is the "no copycats" policy—every friend must have a totally unique combination of activities.
Step 1:Analyze the NPCs (D and C).
The prompt states D is a one-trick pony, only liking Countryside Sightseeing (CS). So, D's loadout size is exactly 1.
The rules also state C likes *more* activities than D. Since the max activities available is 3, C's loadout size must be either 2 or 3.
Step 2:Distribute the Duo Loadouts (A, B, E).
We are told A, B, and E each like *exactly* two activities. Let's calculate the total possible 2-activity combos from our pool of 3 (CS, Shopping=Sh, Adventure Sports=AS).
Using combinations:
3C2=3
The only possible duo sets are {CS, Sh}, {CS, AS}, and {Sh, AS}.
Step 3:Apply the "No Copycats" Rule.
Since no two friends can have the exact same set of activities, A, B, and E must each claim exactly one of these three unique duo combinations. All three 2-activity loadouts are now permanently locked and taken by this trio.
Step 4:Lock in C's Loadout.
Let's check back on C. We established C needs 2 or 3 activities. But wait—all the 2-activity slots are already hogged by A, B, and E! If C took a 2-activity set, they'd duplicate someone else's set, failing the vibe check (violating the main rule). Thus, C is forced to take the only remaining unique loadout size greater than 1: all 3 activities {CS, Sh, AS}. C is literally doing the 100% completionist run.
Step 5:Visualize the Final Roster.
Let's drop all this intel into a matrix to see exactly who is doing what.
Friend
CS
Sh
AS
D
✓
—
—
One of A/B/E
✓
✓
—
One of A/B/E
✓
—
✓
One of A/B/E
—
✓
✓
C
✓
✓
✓
From the table, look at the Sh and AS columns to see who is running the dual Shopping/Adventure Sports build. We count the rows where both show a ✓.
That's exactly the third A/B/E row and C. Total = 2.
Final Answer: 2
Five friends A, B, C, D, and E go on a trip together. Each of them likes at least one of the following three activities: Countryside Sightseeing, Shopping, and Adventure Sports such that:
- No two friends like exactly the same set of activities.
- A, B and E like exactly two activities each.
- D likes only Countryside Sightseeing.
- C likes more number of activities compared to D.
The number of ways in which a pair of friends can be chosen for a trip from those who like Adventure Sports is ___
Enter your answer to attempt
The Setup: This is a direct sequel to the previous IPMAT arrangement lore. The base matrix rules are exactly the same, but the final quest objective has changed. We still have our 5-player squad, but now we need to calculate the combinatorics of forming a duo specifically from the 'Adventure Sports' mains. Math and logic have been double-verified as requested.
Step 1:Rebuild the Loadout Matrix.
Quick recap of the previous logic: D is locked to a 1-perk loadout (CS). A, B, and E each take one of the three unique 2-perk combos. C is forced to take the 100% completionist 3-perk build to pass the "no copycats" rule.
Step 2:Identify the Adventure Sports Mains.
We check the AS column in our roster. Who has the AS perk currently equipped?
* One of the A/B/E trio running {CS, AS}
* One of the A/B/E trio running {Sh, AS}
* C running the full {CS, Sh, AS}
That gives us exactly 3 friends who have Adventure Sports active in their rotation.
Step 3:Form the Duo.
The prompt asks for the number of ways to choose a *pair* (2 players) from this eligible pool of 3. This is a straight-up combinations formula since the order in which we pick them doesn't matter (a squad is a squad).
nCr=3C23C2=2×13×2=3Final Answer: 3
Five friends A, B, C, D, and E go on a trip together. Each of them likes at least one of the following three activities: Countryside Sightseeing, Shopping, and Adventure Sports such that:
- No two friends like exactly the same set of activities.
- A, B and E like exactly two activities each.
- D likes only Countryside Sightseeing.
- C likes more number of activities compared to D.
The number of friends that like Countryside Sightseeing is ___
Enter your answer to attempt
The Setup: We are back in the IPMAT arrangement lore for part 3 of this puzzle sequence. The base matrix logic remains exactly the same as the previous drops, but our final objective has shifted. Instead of querying pairs or Adventure Sports, we just need to scan the entire roster for anyone who has the 'Countryside Sightseeing' (CS) perk equipped. Math and logic have been double-verified.
Step 1: Rebuild the Loadout Matrix.
Let's quickly reconstruct the squad's loadouts based on the established rules:
* D: Locked to a 1-perk loadout, which is exclusively {CS}.
* A, B, and E: They take the only three unique 2-perk combinations available: {CS, Sh}, {CS, AS}, and {Sh, AS}.
* C: Must have more activities than D (so 2 or 3). Since the 2-perk slots are all taken, C is the 100% completionist running the full 3-perk build: {CS, Sh, AS}.
Step 2: Query the CS Column.
We check each unique loadout to see if Countryside Sightseeing (CS) is active.
* Friend D: {CS} →Active (1)
* A/B/E Duo 1: {CS, Sh} →Active (2)
* A/B/E Duo 2: {CS, AS} →Active (3)
* A/B/E Duo 3: {Sh, AS} → Inactive
* Friend C: {CS, Sh, AS} →Active (4)
Step 3: Tally the squad.
Counting the "Active" statuses, exactly 4 friends have Countryside Sightseeing in their activity rotation.
Final Answer: 4