CAT 2024 Slot 1 — DILR Question 6

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.

What was the number of stars received by D from Y?

Answer & solution

A
10
B
0
5
D
Cannot be determined

Solution

Easy

Each surfer hands out exactly 30 stars among A, B, C, D. The chart fixes the columns for A and B, so each surfer's C+D is just 30 − (to A) − (to B). Use the equal-total rule and the five facts to fill the C and D columns; the answer falls straight out of Fact 5 for surfer Y.

Each of the six surfers M, N, O, P, X, Y gives out 30 stars among bloggers A, B, C, D. Stars to A and B are read off the bar chart; C+D is the remainder. All entries are multiples of 5, and each blogger's grand total is the same — since the grand total is $6\times 30=180$ , every blogger total is $180/4=45$ .

Surfer	to A	to B	C+D (=30−A−B)
M	10	0	20
N	25	0	5
O	0	0	30
P	5	25	0
X	0	0	30
Y	5	20	5
Total	45	45	90

Facts: (1) all entries are multiples of 5 incl. 0; (2) the four blogger totals are equal (=45 each); (3) M gives a different number to each blogger; (4) two surfers give all 30 stars to a single blogger; (5) D received more stars than C from Y.

Y splits its leftover 5 between C and D. From the table, Y gives A=5, B=20, so C+D = 5. Both must be multiples of 5, so the only options are $(C,D)=(5,0)$ or $(0,5)$ .

\begin{aligned} &\text{to A}=5,\ \text{to B}=20\\ &\Rightarrow\ C+D = 30-5-20 = 5 \quad\text{(remaining stars)}\\ &\Rightarrow\ (C,D)\in\{(5,0),\,(0,5)\} \quad\text{(multiples of 5, Fact 1)} \end{aligned}

Fact 5 breaks the tie. D received more stars than C from Y, so $D>C$ . With $C+D=5$ that forces $C=0,\ D=5$ .

\begin{aligned} &D > C \quad\text{(Fact 5)}\\ &\Rightarrow\ (C,D)=(0,5) \quad\text{(from step 1)}\\ &\Rightarrow\ \text{Y}\to\text{D} = 5 \end{aligned}

\text{Stars from Y to D} = 5