CAT 2024 Slot 1 — DILR Question 7

Mixed PracticeEasy

Passage / Data

Answer the following questions based on the information given below.

Six web surfers M, N, O, P, X, and Y each had 30 stars which they distributed among four bloggers A, B, C, and D. The number of stars received by A and B from the six web surfers is shown in the figure below.

Bar chart: stars received by bloggers A and B from each surfer

Surfer	to A	to B
M	10	0
N	25	0
O	0	0
P	5	25
X	0	0
Y	5	20

The following additional facts are known regarding the number of stars received by the bloggers from the surfers.

The numbers of stars received by the bloggers from the surfers were all multiples of 5 (including 0).
The total numbers of stars received by the bloggers were the same.
Each blogger received a different number of stars from M.
Two surfers gave all their stars to a single blogger.
D received more stars than C from Y.

How many surfers distributed their stars among exactly 2 bloggers?

Answer & solution

Answer: 2

Solution

Easy

Complete the C and D columns for every surfer (each blogger total must be 45), then simply count, surfer by surfer, how many bloggers received a non‑zero number of stars.

Each surfer gives 30 stars among A, B, C, D. The chart fixes A and B; the grand total per blogger is $180/4 = 45$ . Solving with the facts gives the full grid below.

Surfer	A	B	C	D	#bloggers used
M	10	0	15	5	3
N	25	0	0	5	2
O	0	0	30 to one blogger		1
P	5	25	0	0	2
X	0	0	30 to one blogger		1
Y	5	20	0	5	3
Total	45	45	45	45

(O and X are the "two surfers" of Fact 4: one gives its full 30 to C, the other its full 30 to D.)

Fix M's split using Fact 3. M gives A=10, B=0, so C+D=20. Fact 3 says all four numbers from M differ, so C and D cannot be 10 or 0 or equal. The only multiples of 5 summing to 20 that are distinct from $\{10,0\}$ are $\{5,15\}$ .

\begin{aligned} &C+D = 30-10-0 = 20 \quad\text{(M's remainder)}\\ &\Rightarrow\ \{C,D\}=\{5,15\} \quad\text{(distinct from }10,0;\ \text{Fact 3)} \end{aligned}

Use the two "all‑to‑one" surfers (Fact 4) and the equal totals. Only O and X have a full 30 left after A and B, so they are the two surfers giving everything to one blogger. Let $a$ of them go to C. Matching C's total to 45 with $C_M\in\{5,15\}$ and Y $\to$ C $=0$ forces $a=1$ — so one of O, X feeds C and the other feeds D — and pins $C_M=15,\ D_M=5,\ C_N=0,\ D_N=5$ .

\begin{aligned} &C_{\text{total}} = C_M + C_N + 30a = 45 \quad\text{(Fact 2)}\\ &\Rightarrow\ C_M + C_N = 45-30a,\quad C_M\in\{5,15\},\ C_N\in\{0,5\}\\ &\Rightarrow\ a=1,\ C_M=15,\ C_N=0 \quad\text{(only feasible split)}\\ &\Rightarrow\ D_M = 20-15 = 5,\quad D_N = 5-0 = 5 \end{aligned}

Count non‑zero bloggers per surfer. Read the completed grid: a surfer used "exactly 2 bloggers" when exactly two of its four entries are positive.

\begin{aligned} &\text{M}=(10,0,15,5)\Rightarrow 3,\quad \text{N}=(25,0,0,5)\Rightarrow 2\\ &\text{O}=1,\quad \text{P}=(5,25,0,0)\Rightarrow 2\\ &\text{X}=1,\quad \text{Y}=(5,20,0,5)\Rightarrow 3\\ &\Rightarrow\ \text{exactly two bloggers: N and P}\\ &\Rightarrow\ \text{count} = 2 \end{aligned}

2 \text{ surfers (N and P)}