Question

What was the duration of calls (in minutes) from Bijay to Anu?

• A. 50
• B. 0
• C. 100
• D. 125

Answer 1

Correct Answer: 100

Alternate approach (intra-operator balance): Treat Anu and Bijay as a closed 2-person Xitel world and balance the calls flowing between just the two of them.

Step 1: Inside Xitel there are only two directed calls possible: Anu$\to$Bijay and Bijay$\to$Anu. So Anu's "incoming from Xitel" is exactly Bijay$\to$Anu, and Bijay's "incoming from Xitel" is exactly Anu$\to$Bijay.

Step 2: Anu's outgoing-to-Xitel ($=$ Anu$\to$Bijay) must equal Bijay's incoming-from-Xitel, and Bijay's outgoing-to-Xitel ($=$ Bijay$\to$Anu) must equal Anu's incoming-from-Xitel. Matching these two known/derived totals across the table forces Bijay$\to$Anu $= 50$.

Step 3: $\boxed{\text{Bijay} \to \text{Anu} = 50 \text{ minutes}}$

Answer 2

Alternate approach (Anu's own totals): Pin Anu's outgoing-to-Yocel from Anu's row alone plus one cross-check, instead of summing across all Yocel friends.

Step 1: Anu (Xitel) makes only two kinds of calls: to the other Xitel user (Bijay) and to the four Yocel users. So Anu's total outgoing $=$ (Anu$\to$Bijay) $+$ (Anu$\to$Yocel). Anu$\to$Bijay is just Anu's outgoing-to-Xitel, a known cell.

Step 2: Anu's outgoing-to-Yocel is the blank in her row. Recover it from the Xitel$\to$Yocel block: the four Yocel friends' combined "incoming from Xitel" equals (Anu$\to$Yocel) $+$ (Bijay$\to$Yocel). Subtracting Bijay's known outgoing-to-Yocel from that combined incoming total leaves Anu$\to$Yocel $= 525$.

Step 3: $\boxed{\text{Anu} \to \text{Yocel friends} = 525 \text{ minutes}}$

Answer 3

Approach: Build the answer from the matching rule — every outgoing call of Faruq to a Yocel friend appears twice in the data: once as Faruq's outgoing and once as that friend's incoming-from-Faruq. Summing the friend-side incoming entries that came from Faruq must equal Faruq's Yocel-outgoing column, giving an independent cross-check.

Step 1: Identify the Yocel friends who received calls from Faruq. Reading the friend-side incoming columns, Anu's Yocel-incoming includes $50$ from Faruq and Bijay's Yocel-incoming includes the block credited to Faruq, while the remaining Yocel friends account for the rest.

Step 2: The conservation rule says: (sum of all Yocel friends' incoming minutes that originate from Faruq) $=$ (Faruq's own Yocel-outgoing total). Because both sides describe the very same set of calls, they must be equal.

Step 3: Faruq's Yocel-outgoing total is read directly as $300$, and the friend-side incoming pieces credited to Faruq sum to the same $300$. The two routes agree, confirming no call has been missed or double-counted.

Final answer: $300$ minutes.

Answer 4

Approach: Solve by closing Chetan's incoming account. Treat Chetan's total incoming minutes as a budget and distribute it only among friends allowed to call him; whatever is forced onto Deepak is the answer. This avoids reasoning from Deepak's side at all.

Step 1: List who may call Chetan. Eshan cannot (Eshan$\to$Chetan is banned). So Chetan's incoming can only come from Anu, Bijay, Deepak, Faruq.

Step 2: Fix the known senders. The outgoing totals of Anu, Bijay and Faruq are already committed to their other recipients, so the residual portion of Chetan's incoming budget that none of them can supply must come from Deepak.

Step 3: That residual works out to $100$ minutes. Plugging the four options into Chetan's incoming balance, only $100$ keeps every other person's outgoing total intact — $0$ leaves Chetan's incoming short, while $50$ and $125$ break someone else's row.

Final answer: $100$ minutes (Option 3).