CAT 2024 Slot 1 DILR Set 4: Bar Graphs (4 Questions) with Solutions

The chart gives A: $10+25+0+5+0+5=45$.

The chart gives B: $0+0+0+25+0+20=45$.

Six surfers give $6\times 30=180$ stars in all. Condition 2 makes the four totals equal, so each blogger gets $180/4=45$.

D is one of the four bloggers, so D received 45.

Y gives A 5 and B 20, which is 25 stars. So C and D together get $30-25=5$ from Y.

Condition 1 makes every amount a multiple of 5, so the split is $(C,D)=(0,5)$ or $(5,0)$.

Condition 5 says D received more than C from Y. That rules out $(5,0)$ and leaves D = 5 and C = 0.

P gives A 5 and B 25, which is 30 stars. C and D get 0, so P uses exactly 2 bloggers.

Y gives A 5, B 20 and D 5 (from condition 5), so Y uses 3 bloggers.

M gives A 10 and B 0. Condition 3 needs four different amounts, so C and D must be different and neither can be 0 or 10. With $C+D=20$, the split is 5 and 15, so M uses 3 bloggers.

N gives A 25 and B 0. The remaining 5 goes to one of C or D, so N uses exactly 2 bloggers.

Condition 4 says two surfers gave all 30 stars to one blogger. M, N, P and Y cannot be those two, so they are O and X. Each gives all 30 to C or to D, so each uses 1 blogger.

The surfers using exactly 2 bloggers are N and P, which is 2.

From the chart, A receives $10+25+0+5+0+5=45$ and B receives $0+0+0+25+0+20=45$. Clue 2 makes every blogger's total 45, so the grand total is $4\times 45=180$ (6 surfers × 30). C and D therefore receive 45 each.

The chart leaves C+D stars for each surfer: M gets 20, N gets 5, O gets 30, P gets 0, X gets 30 and Y gets 5.

P: A and B already take 5+25=30, so C = D = 0 from P.

Y: C+D = 5. Clue 5 (D > C) with multiples of 5 forces C = 0 and D = 5.

M: Clue 3 says all four bloggers get different amounts from M. A gets 10 and B gets 0, so C and D cannot be 10 or 0. With C+D = 20, they must be 5 and 15 in some order.

Clue 4: two surfers give all 30 to one blogger. M, N, P and Y each give stars to at least two bloggers, so the two surfers are O and X. Each gives 30 to C or D. If both went to the same blogger, that blogger would have 60 > 45. So one gives 30 to C and the other gives 30 to D.

C's total: $30+M_C+N_C=45$, so $M_C+N_C=15$. N has only 5 stars for C and D, so $N_C\le 5$. If $M_C=5$, then $N_C=10$, which is impossible. So $M_C=15$ and $N_C=0$.

Check D: $M_D=5$, $N_D=5$, $Y_D=5$ and 30 from one of O or X give $45$. ✓

C receives 15 from M, which is certain.

D's 30 stars come from either O or X. Nothing in the clues separates O from X, so D's stars from O could be 0 or 30. This cannot be determined.