CAT 2023 Slot 1 DILR Set 2: Tables (5 Questions) with Solutions

Totals: restaurant sums are R1 17, R2 11, R3 19, R4 14, R5 17. Worker sums are Ullas 11, Vasu 19, Waman 17, Xavier 18, Yusuf 13.

Ullas: R1 gave 1 and the range is 3, so the maximum is 4. Sorted order is $1, a, 2, b, 4$ with $a+b=4$. Only $1,2,2,2,4$ has mode 2, so Ullas has {1, 2, 2, 2, 4}.

Vasu: R5 gave 5 and the range is 3, so the minimum is 2. Sorted order is $2, a, 4, b, 5$ with $a+b=8$. Mode 4 forces {2, 4, 4, 4, 5}.

Waman: R2 gave 1, and R1 and R3 gave 5. Sorted order is $1, a, 4, b, 5$ with $a+b=7$. Mode 5 forces $b=5$, so Waman has {1, 2, 4, 5, 5}.

Xavier: sorted order is $1, a, 4, 5, 5$ with sum 18, so $a=3$. Xavier has {1, 3, 4, 5, 5}.

Yusuf: R2 and R3 gave 1 and the range is 3, so the maximum is 4. Modes 1 and 4 with median 3 and sum 13 give {1, 1, 3, 4, 4}.

R2's row: it gave Waman 1, Xavier 5 and Yusuf 1, so Ullas + Vasu $=11-7=4$. That gives 2 and 2.

R3's row: it gave Waman 5, Xavier 5 and Yusuf 1, so Ullas + Vasu $=8$. The leftover values force 4 and 4.

R4 and R1: Ullas's last two 2s go to R4 and R5. Vasu's remaining 4s go to R1 and R4. The row sums then fix every remaining cell: R1 gave Xavier 3 and Yusuf 4; R4 gave Waman 4, Xavier 1 and Yusuf 3; R5 gave Waman 2, Xavier 4 and Yusuf 4.

R2's mean is 2.2, so its ratings total 11.

R2 gave Waman 1, Yusuf 1 and Xavier 5, which total 7. Ullas + Vasu $=4$.

Ullas's ratings are {1, 2, 2, 2, 4}, and R1 already gave him the 1. Vasu's ratings are {2, 4, 4, 4, 5}, and R5 gave him the 5.

The only way for the two ratings to sum to 4 is $2+2$. A 4 from either worker would push the sum above 4.

R2's ratings are therefore 2, 2, 1, 5, 1. Check: $2+2+1+5+1=11$.

Xavier has {1, 3, 4, 5, 5}. R2 and R3 gave the 5s, so R1, R4 and R5 share 1, 3 and 4.

Yusuf has {1, 1, 3, 4, 4}. R2 and R3 gave the 1s, so R1, R4 and R5 share 3, 4 and 4.

R1's total is 17. It gave Ullas 1, Waman 5 and Vasu 4, because the 2s and 4s of Ullas and Vasu are fixed by R2 and R3. So Xavier + Yusuf $=17-10=7$.

Yusuf must be 3 or 4 and Xavier must be 1, 3 or 4. The pairs summing to 7 are (3, 4) and (4, 3).

Test R4 and R5 for $X=4, Y=3$ at R1. R4 would have to take Xavier 1 or 3, and R5 would have to take the other, with Yusuf 4 and 4. R4's total of 14 then requires Waman + Xavier + Yusuf $=8$, and R5's total of 17 requires 10. Checking Waman's 2 and 4 against both rows fails.

With $X=3, Y=4$ at R1, R4 gets Waman 4, Xavier 1 and Yusuf 3, which sum to 8. R5 gets Waman 2, Xavier 4 and Yusuf 4, which sum to 10. Both rows match.

R3's mean is 3.8, so its ratings total 19.

R3 gave Waman 5, Xavier 5 and Yusuf 1, which total 11. Ullas + Vasu $=8$.

R2 uses Ullas's and Vasu's 2s (their row needs $U+V=4$). R1 has Ullas 1, and R5 has Vasu 5. The leftover values for R3 are Ullas 2 or 4, and Vasu 4.

A sum of 8 forces Ullas 4 and Vasu 4.

R3's ratings sorted are 1, 4, 4, 5, 5. The middle value is 4.

The full table, in order Ullas, Vasu, Waman, Xavier, Yusuf, is:

R1 sorted is 1, 3, 4, 4, 5, with median 4. Two workers got 4.

R2 sorted is 1, 1, 2, 2, 5, with median 2. Two workers got 2.

R3 sorted is 1, 4, 4, 5, 5, with median 4. Two workers got 4.

R4 sorted is 1, 2, 3, 4, 4, with median 3. Only Yusuf got 3.

R5 sorted is 2, 2, 4, 4, 5, with median 4. Two workers got 4.