CAT 2024 Slot 1 — DILR Question 8

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.

Which of the following can be determined with certainty?

I. The number of stars received by C from M
II. The number of stars received by D from O

Answer & solution

Only I
B
Neither I nor II
C
Both I and II
D
Only II

Solution

Hard

Fix each blogger’s total at 45, then use Facts 3–5 to nail down the C and D entries. Statement I (C from M) is forced; Statement II (D from O) is not, because O’s “all-to-one” donation could go to either C or D.

Set-up (C+D per surfer = 30−A−B):

Surfer	A	B	C+D
M	10	0	20
N	25	0	5
O	0	0	30
P	5	25	0
X	0	0	30
Y	5	20	5

Each blogger total = 45 (from Fact 2 and 180/4).

Forced single-blogger rows. P gives C+D=0, so C and D get nothing from P. From Y, C+D=5 with D > C (Fact 5) and multiples of 5, so D gets 5 and C gets 0.

\begin{aligned} &C_P=D_P=0,\qquad C_Y=0,\ D_Y=5 \quad\text{(Fact 5)} \end{aligned}

Use Fact 3 on M. M gives A=10, B=0, and C, D the remaining 20, all four amounts distinct multiples of 5. So $\{C_M,D_M\}=\{5,15\}$ .

\begin{aligned} &C_M+D_M=20,\quad C_M\ne D_M\ \Rightarrow\ \{C_M,D_M\}=\{5,15\} \quad\text{(Fact 3)} \end{aligned}

Use Fact 4 on O and X. Only O and X have C+D=30; they are the two surfers who give all 30 to one blogger. They cannot both feed C (that would give C at least 60 > 45) nor both feed D — so one of them gives 30 to C and the other 30 to D.

\begin{aligned} &\{O,X\}\ \text{give } 30\text{ to } C\ \text{and } 30\text{ to } D\ \text{(one each)} \quad\text{(Fact 4)} \end{aligned}

Close C’s total to find C from M. C gets 0 from P and Y, 30 from one of O/X, and $C_N$ from N (where $C_N\le5$ ). With C total = 45:

\begin{aligned} &C_M+C_N+30=45\ \Rightarrow\ C_M+C_N=15 \quad\text{(C total = 45)}\\ &\Rightarrow\ C_M=15,\ C_N=0 \quad\text{(since } C_M\in\{5,15\},\ C_N\le5) \end{aligned}

So Statement I is fixed: C received 15 stars from M.

Test Statement II (D from O). From step 3 the 30-star gift to D comes from O or X — nothing distinguishes them. So D from O is either 30 or 0; it is not determinable.

\begin{aligned} &D_O\in\{0,\ 30\}\ \text{(O could be the C-donor or the D-donor)}\ \Rightarrow\ \text{not certain} \end{aligned}

\textbf{Only I}

Need a hint?

The symmetry between O and X is the whole point: any fact that fixes one of them would fix the other. None does, so D-from-O stays ambiguous while C-from-M is pinned by the totals.