A ball makes a ping on a hoop exactly when its diameter is not larger than the hoop's diameter, i.e. $ball \le hoop$. A ball gets stuck (no ping) when $ball > hoop$. We build a grid with the six balls as rows and the four hoops as columns. In each cell we write $P$ for a ping ($ball \le hoop$) and $X$ for no ping ($ball > hoop$). Every $P$ gives an inequality $ball \le hoop$ and every $X$ gives $ball > hoop$; these inequalities are exactly what let us order the ball sizes and the hoop sizes. | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | | | | | | B2 | | | | | | B3 | | | | | | B4 | | | | | | B5 | | | | | | B6 | | | | | —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | | | | | B2 | P | | | | | B3 | X | | | | | B4 | P | | | | | B5 | P | | | | | B6 | P | | | | From clue (3), every ball except B3 pings on H1, so the H1 column is $P$ for B1, B2, B4, B5, B6 and $X$ for B3.

Each ping means $ball \le H1$, and B3's miss means $B3 > H1$. Since $B1, B2, B4, B5, B6 \le H1$ while $B3 > H1$, B3 is larger than every other ball. So B3 is the largest ball. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | | | | B2 | P | P | | | | B3 | X | X | | | | B4 | P | X | | | | B5 | P | X | | | | B6 | P | X | | | From clue (4), only B2 pings on H2, so the H2 column is $P$ for B2 and $X$ for B1, B3, B4, B5, B6.

Here $B2 \le H2$ while every other ball is $> H2$. So B2 is smaller than every other ball. So B2 is the smallest ball. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | X | | | B2 | P | P | | | | B3 | X | X | | | | B4 | P | X | P | | | B5 | P | X | | | | B6 | P | X | | | From clue (2), B4 pings on H3 but B1 does not, so H3 is $P$ for B4 and $X$ for B1.

This gives $B4 \le H3$ and $B1 > H3$, so $B4 \le H3 < B1$. Therefore $B4 < B1$. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | X | P | | B2 | P | P | | | | B3 | X | X | | | | B4 | P | X | P | | | B5 | P | X | | X | | B6 | P | X | | P | From clue (1), B1 and B6 ping on H4 but B5 does not, so H4 is $P$ for B1 and B6 and $X$ for B5.

This gives $B1 \le H4$, $B6 \le H4$ and $B5 > H4$, so $B1 \le H4 < B5$ $B6 \le H4 < B5$. Therefore $B1 < B5$ and $B6 < B5$. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | X | P | | B2 | P | P | | | | B3 | X | X | X | X | | B4 | P | X | P | | | B5 | P | X | | X | | B6 | P | X | | P | The hoops are now ordered using the marks already placed: $H2 < H1$, since B1 misses H2 but pings H1: $H2 < B1 \le H1$. $H4 < H1$, since B5 misses H4 but pings H1: $H4 < B5 \le H1$. $H3 < H4$, since B1 misses H3 but pings H4: $H3 < B1 \le H4$. $H2 < H3$, since B4 misses H2 but pings H3: $H2 < B4 \le H3$. Combining these gives $H2 < H3 < H4 < H1$.

H1 is the largest hoop, and $B3 > H1$, so B3 is larger than every hoop. B3 therefore misses every hoop, so its H3 and H4 cells are $X$. B3 makes $0$ pings in total. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | X | P | | B2 | P | P | P | P | | B3 | X | X | X | X | | B4 | P | X | P | | | B5 | P | X | | X | | B6 | P | X | | P | B2 is the smallest ball with $B2 \le H2$, and from the hoop order $H2 < H3 < H4$. So $B2 < H3$ and $B2 < H4$.

B2 therefore pings H3 and H4, so both cells are $P$. B2 pings all four hoops. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | X | P | | B2 | P | P | P | P | | B3 | X | X | X | X | | B4 | P | X | P | P | | B5 | P | X | | X | | B6 | P | X | | P | B4 pings H3, so $B4 \le H3$, and from the hoop order $H3 < H4$. So $B4 < H4$, meaning B4 pings H4 and that cell is $P$. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | X | P | | B2 | P | P | P | P | | B3 | X | X | X | X | | B4 | P | X | P | P | | B5 | P | X | X | X | | B6 | P | X | | P | B5 misses H4, so $B5 > H4$, and from the hoop order $H3 < H4$. So $B5 > H4 > H3$, meaning B5 misses H3 and that cell is $X$. —-- | Ball | H1 | H2 | H3 | H4 | |------|----|----|----|----| | B1 | P | X | X | P | | B2 | P | P | P | P | | B3 | X | X | X | X | | B4 | P | X | P | P | | B5 | P | X | X | X | | B6 | P | X | P/X| P | For B6 on H3: B6 pings H4 so $B6 \le H4$, and B6 misses H2 so $B6 > H2$. Thus $H2 < B6 \le H4$, with H3 lying in between, and no clue fixes whether $B6 \le H3$ or $B6 > H3$. So B6 on H3 is undetermined, marked $P/X$.

The grid is complete; the only undetermined cell is B6 on H3. Hoop order: $H2 < H3 < H4 < H1$. Ball facts: B2 is smallest, B3 is largest, with $B4 < B1 < B5$ and $B6 < B5$; the place of B6 relative to B4 and B1 is not fixed. Pings per hoop: $H1 = 5$, $H2 = 1$, $H3 = 2$ or $3$, $H4 = 4$, giving a total of $12$ or $13$ pings.

CAT (DILR) PYQs

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