Token Queue at the Single-Window Counter
On 12 August a branch in Nagpur runs a single-window token system between 10:00 a.m. and 1:00 p.m. Six customers - Asha, Bipin, Chanchal, Dev, Esha and Farid - hold tokens 1 to 6 and are served strictly in that order. Each came for exactly one service: cash deposit, cash withdrawal, KYC update, cheque-book request, demand draft issuance, or passbook printing; no two came for the same service. The counter register records the following. (i) Chanchal held token 3 and her work was the KYC update. (ii) Dev was served immediately before Asha. (iii) Asha's work was a cash deposit, and Bipin was served exactly three tokens after her. (iv) Farid was served immediately before Bipin, and Farid's work was a cash withdrawal. (v) Dev's work was demand draft issuance. (vi) Esha was served last. (vii) Bipin did not come for a cheque-book request. The counter clerk must reconstruct the full sequence before the day's report is signed.
1. Reconstruct the complete token order, naming the customer and the service at each of the six positions.
Read the worked answer
Step 1 - anchor the fixed seats. Chanchal is token 3 (clue i) and Esha is token 6 (clue vi). Free seats are 1, 2, 4 and 5.
Step 2 - place the two blocks. Dev–Asha must be adjacent (clue ii) and Farid–Bipin must be adjacent (clue iv). The only adjacent pairs available among 1, 2, 4, 5 are (1, 2) and (4, 5). So Dev = 1, Asha = 2, Farid = 4, Bipin = 5.
Step 3 - verify with clue (iii). Bipin at 5 is exactly three tokens after Asha at 2. Consistent, so the placement is forced and unique.
Step 4 - assign services. Chanchal = KYC, Dev = draft, Asha = deposit, Farid = withdrawal. Cheque book and passbook printing remain for Bipin (5) and Esha (6). Clue (vii) bars Bipin from the cheque book, so Bipin = passbook printing and Esha = cheque book.
Answer: 1 Dev–draft; 2 Asha–cash deposit; 3 Chanchal–KYC update; 4 Farid–cash withdrawal; 5 Bipin–passbook printing; 6 Esha–cheque-book request.
2. The clerk later finds that Bipin actually came for a cheque-book request, not passbook printing. State exactly what changes and whether the arrangement remains uniquely determined.
Read the worked answer
Only the service column changes; the token order is untouched, because every clue that fixes positions (i, ii, iii, iv, vi) is independent of Bipin's service. Clue (vii) said Bipin did not want a cheque book, so it is now false and must be discarded.
With Bipin = cheque book, the only unassigned service is passbook printing, and the only customer without a service is Esha at token 6. Esha must therefore take passbook printing. So the revised answer is: 1 Dev–draft; 2 Asha–cash deposit; 3 Chanchal–KYC; 4 Farid–withdrawal; 5 Bipin–cheque book; 6 Esha–passbook printing.
The arrangement stays unique, but note why: uniqueness now comes from exhaustion of the service list, not from clue (vii). If the six services were not a closed list, the solution would collapse to two possibilities (Bipin cheque book/Esha printing, or the reverse), which is the trap to flag.
3. A trainee offers two possibilities: (a) Esha's service was passbook printing; (b) Farid held token 2. Test each against the clues and say whether it can be true.
Read the worked answer
(a) Esha = passbook printing - false on the original clues. Esha is at token 6. Passbook printing is the service left over for Bipin, because cheque book and printing are the two services not yet fixed, and clue (vii) removes cheque book from Bipin. If Esha were printing, Bipin would have to hold the cheque book, directly contradicting clue (vii). So (a) contradicts the record, and would become true only under the revised fact in the previous task.
(b) Farid = token 2 - false. Token 2 belongs to Asha. Clue (ii) puts Dev immediately before Asha, and clue (iv) puts Farid immediately before Bipin. Among the free seats 1, 2, 4 and 5, the only two adjacent pairs are (1, 2) and (4, 5); placing Asha anywhere other than 2 would leave Dev with no adjacent free seat. Farid is therefore locked at token 4, not 2.
Lesson: in linear arrangements, fix the block pairs first and only then test individual claims.