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Alia, Badal, Clive, Dilshan, and Ehsaan played a game in which each asks a unique question to all the others and they respond by tapping their feet, either once or twice or thrice. One tap means “Yes”, two taps mean “No”, and three taps mean “Maybe”.

A total of 40 taps were heard across the five questions. Each question received at least one “Yes”, one “No”, and one “Maybe.”

The following information is known.

  1. Alia tapped a total of 6 times and received 9 taps to her question. She responded “Yes” to the questions asked by both Clive and Dilshan.
  2. Dilshan and Ehsaan tapped a total of 11 and 9 times respectively. Dilshan responded “No” to Badal.
  3. Badal, Dilshan, and Ehsaan received equal number of taps to their respective questions.
  4. No one responded “Yes” more than twice.
  5. No one’s answer to Alia’s question matched the answer that Alia gave to that person’s question. This was also true for Ehsaan.
  6. Clive tapped more times in total than Badal.

What was Clive’s response to Ehsaan’s question?

Solution

✅ Correct Option: 3
Slide 1/11

Each of the five people asks exactly one question, answered by the other four. A response is 1 tap (Yes), 2 taps (No), or 3 taps (Maybe).

The rows are the responders; the columns are the five questions, where QA is Alia's question, QB is Badal's, QC is Clive's, QD is Dilshan's, QE is Ehsaan's. A cell holds the taps the row-person gives to that question. The diagonal is blank — no one answers their own question.

Each row total is that person's total taps; each column total (the Received row) is the taps that question collected. Both add to 40.

ResponderQAQBQCQDQETapped
Alia—6
Badal—
Clive—
Dilshan—11
Ehsaan—9
Received40

From clues 1 and 2, the totals Alia =6=6, Dilshan =11=11, Ehsaan =9=9 are placed.


Each question gets four answers and must include at least one Yes(1), one No(2), one Maybe(3).

The minimum is 1+2+3=61+2+3=6, plus a fourth answer of 11 to 33.

So every question receives 77, 88, or 99 taps.

ResponderQAQBQCQDQETapped
Alia—6
Badal—
Clive—
Dilshan—11
Ehsaan—9
Received9878840

The Received row is filled.

Alia's question received 99 taps (clue 1).

By clue 3, Badal's, Dilshan's and Ehsaan's questions each received the same amount tt; let Clive's be cc.

9+3t+c=40⇒3t+c=319 + 3t + c = 40 \Rightarrow 3t + c = 31, with t,c∈{7,8,9}t, c \in \{7,8,9\}.

Only t=8, c=7t = 8,\ c = 7 works.

So QB == QD == QE =8= 8 and QC =7= 7.


For the row totals:

6+B+C+11+9=40⇒B+C=146 + B + C + 11 + 9 = 40 \Rightarrow B + C = 14, where BB and CC are Badal's and Clive's total taps.

ResponderQAQBQCQDQETapped
Alia—116
Badal—
Clive—
Dilshan2—11
Ehsaan—9
Received9878840

From clue 1, Alia answered Yes to Clive and Dilshan:

(A,QC)=1(A,QC)=1

(A,QD)=1(A,QD)=1

From clue 2, Dilshan answered No to Badal:

(D,QB)=2(D,QB)=2

ResponderQAQBQCQDQETapped
Alia—21126
Badal—
Clive—
Dilshan2—11
Ehsaan—9
Received9878840

Alia's row total is 66, and (A,QC)+(A,QD)=2(A,QC)+(A,QD)=2, so

(A,QB)+(A,QE)=4(A,QB)+(A,QE)=4.

Clue 4 caps Yes responses at two per person, and Alia already gave Yes to Clive and Dilshan.

So neither remaining answer can be 11.

Two values each at least 22 summing to 44 forces

(A,QB)=2(A,QB)=2

(A,QE)=2(A,QE)=2

ResponderQAQBQCQDQETapped
Alia—21126
Badal—
Clive—
Dilshan323—311
Ehsaan—9
Received9878840

Dilshan's row total is 1111 with (D,QB)=2(D,QB)=2, so

(D,QA)+(D,QC)+(D,QE)=9(D,QA)+(D,QC)+(D,QE)=9.

Three answers, each at most 33, summing to 99 must all be 33:

(D,QA)=3(D,QA)=3

(D,QC)=3(D,QC)=3

(D,QE)=3(D,QE)=3

Dilshan's row is complete.

ResponderQAQBQCQDQETapped
Alia—21126
Badal—
Clive2—
Dilshan323—311
Ehsaan—9
Received9878840

Clue 5 (Alia's question): each person's answer to QA differs from Alia's answer to their own question.

(A,QB)=2⇒(B,QA)∈{1,3}(A,QB)=2 \Rightarrow (B,QA)\in\{1,3\}

(A,QC)=1⇒(C,QA)∈{2,3}(A,QC)=1 \Rightarrow (C,QA)\in\{2,3\}

(A,QE)=2⇒(E,QA)∈{1,3}(A,QE)=2 \Rightarrow (E,QA)\in\{1,3\}


Column QA sums to 99 with (D,QA)=3(D,QA)=3, so

(B,QA)+(C,QA)+(E,QA)=6(B,QA)+(C,QA)+(E,QA)=6.

If (C,QA)=3(C,QA)=3 then (B,QA)+(E,QA)=3(B,QA)+(E,QA)=3, impossible with both in {1,3}\{1,3\}.

So (C,QA)=2(C,QA)=2, and {(B,QA),(E,QA)}={1,3}\{(B,QA),(E,QA)\}=\{1,3\}.

ResponderQAQBQCQDQETapped
Alia—21126
Badal1—
Clive2—
Dilshan323—311
Ehsaan3—9
Received9878840

Two options remain for column QA: either (B,QA)=1,(E,QA)=3(B,QA)=1,(E,QA)=3 or (B,QA)=3,(E,QA)=1(B,QA)=3,(E,QA)=1.

Limits on Ehsaan's other answers:

Column QB sum 88 with (A,QB)=(D,QB)=2(A,QB)=(D,QB)=2 gives (C,QB)+(E,QB)=4(C,QB)+(E,QB)=4; to keep a Yes and a Maybe these are {1,3}\{1,3\}, so (E,QB)∈{1,3}(E,QB)\in\{1,3\}.

Column QC sum 77 with (A,QC)=1,(D,QC)=3(A,QC)=1,(D,QC)=3 gives (B,QC)+(E,QC)=3(B,QC)+(E,QC)=3, so (E,QC)∈{1,2}(E,QC)\in\{1,2\}.

Clue 5 (Ehsaan): (D,QE)=3⇒(E,QD)∈{1,2}(D,QE)=3 \Rightarrow (E,QD)\in\{1,2\}.


If (E,QA)=1(E,QA)=1, Ehsaan's other three answers must total 88, but their maximum is 3+2+2=73+2+2=7. Impossible.

So (B,QA)=1(B,QA)=1 and (E,QA)=3(E,QA)=3.

ResponderQAQBQCQDQETapped
Alia—21126
Badal1—
Clive21—
Dilshan323—311
Ehsaan33—9
Received9878840

Ehsaan's total is 99 with (E,QA)=3(E,QA)=3, so

(E,QB)+(E,QC)+(E,QD)=6(E,QB)+(E,QC)+(E,QD)=6.

Since (E,QC)≤2(E,QC)\le 2 and (E,QD)≤2(E,QD)\le 2, their sum is at most 44, forcing (E,QB)≥2(E,QB)\ge 2.

As (E,QB)∈{1,3}(E,QB)\in\{1,3\}, this gives (E,QB)=3(E,QB)=3.

Column QB then gives (C,QB)=4−3=1(C,QB)=4-3=1.

ResponderQAQBQCQDQETapped
Alia—21126
Badal1—
Clive21—
Dilshan323—311
Ehsaan3312—9
Received9878840

Now (E,QC)+(E,QD)=3(E,QC)+(E,QD)=3, each in {1,2}\{1,2\}.

Test (E,QC)=2,(E,QD)=1(E,QC)=2,(E,QD)=1:

column QC gives (B,QC)=1(B,QC)=1; clue 5 (Ehsaan) needs (C,QE)≠(E,QC)(C,QE)\neq(E,QC), forcing (C,QE)=1(C,QE)=1 and so (B,QE)=2(B,QE)=2.

Column QD then needs (B,QD)+(C,QD)=6(B,QD)+(C,QD)=6; Badal already has two Yes (QA, QC), so (B,QD)≠1(B,QD)\neq 1, forcing (B,QD)=3,(C,QD)=3(B,QD)=3,(C,QD)=3.

This makes Badal == Clive =7= 7, violating clue 6 (C>BC>B).

So (E,QC)=1(E,QC)=1 and (E,QD)=2(E,QD)=2. Ehsaan's row is complete.

ResponderQAQBQCQDQETapped
Alia—21126
Badal1—21
Clive21—2
Dilshan323—311
Ehsaan3312—9
Received9878840

Column QC: (B,QC)=3−(E,QC)=2(B,QC)=3-(E,QC)=2.

Clue 5 (Ehsaan): (C,QE)≠(E,QC)=1(C,QE)\neq(E,QC)=1, and (C,QE)∈{1,2}(C,QE)\in\{1,2\}, so (C,QE)=2(C,QE)=2.

Column QE: (B,QE)=3−(C,QE)=1(B,QE)=3-(C,QE)=1.

ResponderQAQBQCQDQETapped
Alia—21126
Badal1—2216
Clive21—328
Dilshan323—311
Ehsaan3312—9
Received9878840

Column QD: (B,QD)+(C,QD)+(E,QD)=7(B,QD)+(C,QD)+(E,QD)=7 with (E,QD)=2(E,QD)=2, so (B,QD)+(C,QD)=5(B,QD)+(C,QD)=5.

Badal already has two Yes (QA, QE), so (B,QD)≠1(B,QD)\neq 1.

(B,QD)=2⇒(C,QD)=3(B,QD)=2 \Rightarrow (C,QD)=3: Badal 66, Clive 88 — satisfies C>BC>B.

(B,QD)=3⇒(C,QD)=2(B,QD)=3 \Rightarrow (C,QD)=2: Badal 77, Clive 77 — fails C>BC>B.

So (B,QD)=2(B,QD)=2 and (C,QD)=3(C,QD)=3. The grid is complete and unique.


Verification:

Row totals 6,6,8,11,96,6,8,11,9 sum to 4040.

Column totals 9,8,7,8,89,8,7,8,8 sum to 4040.

Every question contains a 11, a 22, and a 33.

Yes counts: Alia 22, Badal 22, Clive 11, Dilshan 00, Ehsaan 11 — none exceeds two.

Clive (88) is greater than Badal (66), and B+C=14B+C=14.

Clue 5 holds for both Alia's and Ehsaan's questions.


Tracing the grid, Clive's response to Ehsaan's question is (C,QE)=2(C,QE)=2, which means No. (It was fixed as 22 because clue 5 for Ehsaan forces (C,QE)≠(E,QC)=1(C,QE)\neq(E,QC)=1, leaving (C,QE)=2(C,QE)=2.) Answer: option 3 (No).

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