CAT 2022 DILR Questions — Slot 3

The trick is to follow each starting group as one moving bar. The study starts in 1980 and each figure is exactly $10$ years later, so every group of $250$ slides up one age band each decade. Reading the four figures and filling back $1980=250$ for each, the number still alive in each starting group is:

A group's current age is its 1980 age plus the years elapsed — e.g. the "20–30 in 1980" group is $40$–$50$ by 2000 and $50$–$60$ by 2010.

This question: in 2000, the ratio of dead males to dead females. Add the 2000 column for each sex; the dead are $1000$ minus the living.

• Males alive $=180+205+160+100=645$, so dead males $=1000-645=355$.
• Females alive $=210+175+150+120=655$, so dead females $=1000-655=345$.

$355:345=71:69.$

Answer: 71: 69

The trick is to follow each starting group as one moving bar. The study starts in 1980 and each figure is exactly $10$ years later, so every group of $250$ slides up one age band each decade. Reading the four figures and filling back $1980=250$ for each, the number still alive in each starting group is:

A group's current age is its 1980 age plus the years elapsed — e.g. the "20–30 in 1980" group is $40$–$50$ by 2000 and $50$–$60$ by 2010.

This question: those aged $30$–$40$ in 1980 (both sexes) who were still alive in 2010. That is the Males 30–40 and Females 30–40 rows, 2010 column: $90+100=190.$

Answer: 190

The trick is to follow each starting group as one moving bar. The study starts in 1980 and each figure is exactly $10$ years later, so every group of $250$ slides up one age band each decade. Reading the four figures and filling back $1980=250$ for each, the number still alive in each starting group is:

A group's current age is its 1980 age plus the years elapsed — e.g. the "20–30 in 1980" group is $40$–$50$ by 2000 and $50$–$60$ by 2010.

This question: those under $30$ in 1980 — the $10$–$20$ and $20$–$30$ groups of both sexes — still alive in 2020 (the 2020 column):

• Males: $140+125=265$.
• Females: $100+105=205$.

$265+205=470.$

Answer: 470

The trick is to follow each starting group as one moving bar. The study starts in 1980 and each figure is exactly $10$ years later, so every group of $250$ slides up one age band each decade. Reading the four figures and filling back $1980=250$ for each, the number still alive in each starting group is:

A group's current age is its 1980 age plus the years elapsed — e.g. the "20–30 in 1980" group is $40$–$50$ by 2000 and $50$–$60$ by 2010.

This question: males aged $20$–$30$ in 1980 who died during 2000–2010. Take that single row: $205$ alive in 2000, $165$ alive in 2010. The drop is the deaths in that decade: $205-165=40.$

Answer: 40

The trick is to follow each starting group as one moving bar. The study starts in 1980 and each figure is exactly $10$ years later, so every group of $250$ slides up one age band each decade. Reading the four figures and filling back $1980=250$ for each, the number still alive in each starting group is:

A group's current age is its 1980 age plus the years elapsed — e.g. the "20–30 in 1980" group is $40$–$50$ by 2000 and $50$–$60$ by 2010.

This question: females aged $20$–$30$ in 1980 who died between the ages of 50 and 60. That group is $40$–$50$ in 2000 and $50$–$60$ in 2010, so they pass through the $50$–$60$ band exactly across the 2000–2010 decade. The fall in that row over those years is the deaths in that age band: $175-145=30.$

💡 Teacher tip: "Died between ages 50 and 60" isn't a fixed year — first find which decade this cohort is aged 50–60, then take the drop across it.

Answer: 30

Every first-year CS student takes both AI and ML; a non-CS student takes exactly one of them. Failing gives an F; passing gives A, B or C. Let us build the two grade tables from the clues.

Counting the students.

1. Let the non-CS students in AI and ML be $2x$ and $5x$ (ratio 2: 5). Non-CS students take only one course, so there are $2x+5x = 7x$ non-CS students, and this equals the number of CS students: $\text{CS} = 7x$.
2. No CS student fails AI. Let CS failures in ML be $g$, and let each of the two non-CS failure counts be $f$ (the two are equal).
3. The ratio of total non-CS failures to CS failures is 3: 1, so $\dfrac{2f}{g} = 3$, i.e. $2f = 3g$.
4. A total of 30 students fail ML: $f + g = 30$. Combined with $2f = 3g$ we get $g = 12$ and $f = 18$.
5. CS students passing ML split A: B: C = 4: 5: 2, so the C-grade count is $\frac{2}{11}(\text{CS} - g)$. This equals $2f = 36$: solving $36 = \frac{2}{11}(\text{CS} - 12)$ gives $\text{CS} = 210$. Hence $x = 30$, so non-CS in AI $= 60$ and non-CS in ML $= 150$.

Filling the grades.

• AI has $210 + 60 = 270$ students, of whom 18 (all non-CS) fail, leaving 252 passers. Half of the passers get B, so B = 126; the A and C counts are equal, so A = C = 63. CS grades in AI split 3: 5: 2 of 210, i.e. 63, 105, 42. No non-CS student gets an A in AI, so the 42 non-CS passers give B = 21 and C = 21.
• ML has $210 + 150 = 360$ students, 30 fail, 330 pass. B = 165; the A: C ratio is 3: 2 with A + C = 165, so A = 99, C = 66. CS ML passers (198) split 4: 5: 2 into 72, 90, 36. The rest are non-CS: A = 27, B = 75, C = 30, F = 18.

AI course (270 students)

ML course (360 students)

This question. AI is taken by all 210 CS students plus 60 non-CS students: $210 + 60 = 270$.

💡 Teacher tip: The knot unties at "30 fail ML" combined with the 3: 1 failure ratio — that pins the failure counts, which then unlock the total CS strength.

Answer: 270

Every first-year CS student takes both AI and ML; a non-CS student takes exactly one of them. Failing gives an F; passing gives A, B or C. Let us build the two grade tables from the clues.

Counting the students.

1. Let the non-CS students in AI and ML be $2x$ and $5x$ (ratio 2: 5). Non-CS students take only one course, so there are $2x+5x = 7x$ non-CS students, and this equals the number of CS students: $\text{CS} = 7x$.
2. No CS student fails AI. Let CS failures in ML be $g$, and let each of the two non-CS failure counts be $f$ (the two are equal).
3. The ratio of total non-CS failures to CS failures is 3: 1, so $\dfrac{2f}{g} = 3$, i.e. $2f = 3g$.
4. A total of 30 students fail ML: $f + g = 30$. Combined with $2f = 3g$ we get $g = 12$ and $f = 18$.
5. CS students passing ML split A: B: C = 4: 5: 2, so the C-grade count is $\frac{2}{11}(\text{CS} - g)$. This equals $2f = 36$: solving $36 = \frac{2}{11}(\text{CS} - 12)$ gives $\text{CS} = 210$. Hence $x = 30$, so non-CS in AI $= 60$ and non-CS in ML $= 150$.

Filling the grades.

• AI has $210 + 60 = 270$ students, of whom 18 (all non-CS) fail, leaving 252 passers. Half of the passers get B, so B = 126; the A and C counts are equal, so A = C = 63. CS grades in AI split 3: 5: 2 of 210, i.e. 63, 105, 42. No non-CS student gets an A in AI, so the 42 non-CS passers give B = 21 and C = 21.
• ML has $210 + 150 = 360$ students, 30 fail, 330 pass. B = 165; the A: C ratio is 3: 2 with A + C = 165, so A = 99, C = 66. CS ML passers (198) split 4: 5: 2 into 72, 90, 36. The rest are non-CS: A = 27, B = 75, C = 30, F = 18.

AI course (270 students)

ML course (360 students)

This question. From the failure analysis, CS failures in ML are $g = 12$ (see the F row of the ML table).

Answer: 12

Every first-year CS student takes both AI and ML; a non-CS student takes exactly one of them. Failing gives an F; passing gives A, B or C. Let us build the two grade tables from the clues.

Counting the students.

1. Let the non-CS students in AI and ML be $2x$ and $5x$ (ratio 2: 5). Non-CS students take only one course, so there are $2x+5x = 7x$ non-CS students, and this equals the number of CS students: $\text{CS} = 7x$.
2. No CS student fails AI. Let CS failures in ML be $g$, and let each of the two non-CS failure counts be $f$ (the two are equal).
3. The ratio of total non-CS failures to CS failures is 3: 1, so $\dfrac{2f}{g} = 3$, i.e. $2f = 3g$.
4. A total of 30 students fail ML: $f + g = 30$. Combined with $2f = 3g$ we get $g = 12$ and $f = 18$.
5. CS students passing ML split A: B: C = 4: 5: 2, so the C-grade count is $\frac{2}{11}(\text{CS} - g)$. This equals $2f = 36$: solving $36 = \frac{2}{11}(\text{CS} - 12)$ gives $\text{CS} = 210$. Hence $x = 30$, so non-CS in AI $= 60$ and non-CS in ML $= 150$.

Filling the grades.

• AI has $210 + 60 = 270$ students, of whom 18 (all non-CS) fail, leaving 252 passers. Half of the passers get B, so B = 126; the A and C counts are equal, so A = C = 63. CS grades in AI split 3: 5: 2 of 210, i.e. 63, 105, 42. No non-CS student gets an A in AI, so the 42 non-CS passers give B = 21 and C = 21.
• ML has $210 + 150 = 360$ students, 30 fail, 330 pass. B = 165; the A: C ratio is 3: 2 with A + C = 165, so A = 99, C = 66. CS ML passers (198) split 4: 5: 2 into 72, 90, 36. The rest are non-CS: A = 27, B = 75, C = 30, F = 18.

AI course (270 students)

ML course (360 students)

This question. In ML, the total A grades number 99, of which CS students earned 72. So non-CS A grades are $99 - 72 = 27$.

Answer: 27

Every first-year CS student takes both AI and ML; a non-CS student takes exactly one of them. Failing gives an F; passing gives A, B or C. Let us build the two grade tables from the clues.

Counting the students.

1. Let the non-CS students in AI and ML be $2x$ and $5x$ (ratio 2: 5). Non-CS students take only one course, so there are $2x+5x = 7x$ non-CS students, and this equals the number of CS students: $\text{CS} = 7x$.
2. No CS student fails AI. Let CS failures in ML be $g$, and let each of the two non-CS failure counts be $f$ (the two are equal).
3. The ratio of total non-CS failures to CS failures is 3: 1, so $\dfrac{2f}{g} = 3$, i.e. $2f = 3g$.
4. A total of 30 students fail ML: $f + g = 30$. Combined with $2f = 3g$ we get $g = 12$ and $f = 18$.
5. CS students passing ML split A: B: C = 4: 5: 2, so the C-grade count is $\frac{2}{11}(\text{CS} - g)$. This equals $2f = 36$: solving $36 = \frac{2}{11}(\text{CS} - 12)$ gives $\text{CS} = 210$. Hence $x = 30$, so non-CS in AI $= 60$ and non-CS in ML $= 150$.

Filling the grades.

• AI has $210 + 60 = 270$ students, of whom 18 (all non-CS) fail, leaving 252 passers. Half of the passers get B, so B = 126; the A and C counts are equal, so A = C = 63. CS grades in AI split 3: 5: 2 of 210, i.e. 63, 105, 42. No non-CS student gets an A in AI, so the 42 non-CS passers give B = 21 and C = 21.
• ML has $210 + 150 = 360$ students, 30 fail, 330 pass. B = 165; the A: C ratio is 3: 2 with A + C = 165, so A = 99, C = 66. CS ML passers (198) split 4: 5: 2 into 72, 90, 36. The rest are non-CS: A = 27, B = 75, C = 30, F = 18.

AI course (270 students)

ML course (360 students)

This question. In AI, A equals C and B is half of the 252 passers, so B = 126 and A = C = 63. Every A grade in AI belongs to a CS student. So the number of A grades in AI is 63.

Answer: 63

Every first-year CS student takes both AI and ML; a non-CS student takes exactly one of them. Failing gives an F; passing gives A, B or C. Let us build the two grade tables from the clues.

Counting the students.

1. Let the non-CS students in AI and ML be $2x$ and $5x$ (ratio 2: 5). Non-CS students take only one course, so there are $2x+5x = 7x$ non-CS students, and this equals the number of CS students: $\text{CS} = 7x$.
2. No CS student fails AI. Let CS failures in ML be $g$, and let each of the two non-CS failure counts be $f$ (the two are equal).
3. The ratio of total non-CS failures to CS failures is 3: 1, so $\dfrac{2f}{g} = 3$, i.e. $2f = 3g$.
4. A total of 30 students fail ML: $f + g = 30$. Combined with $2f = 3g$ we get $g = 12$ and $f = 18$.
5. CS students passing ML split A: B: C = 4: 5: 2, so the C-grade count is $\frac{2}{11}(\text{CS} - g)$. This equals $2f = 36$: solving $36 = \frac{2}{11}(\text{CS} - 12)$ gives $\text{CS} = 210$. Hence $x = 30$, so non-CS in AI $= 60$ and non-CS in ML $= 150$.

Filling the grades.

• AI has $210 + 60 = 270$ students, of whom 18 (all non-CS) fail, leaving 252 passers. Half of the passers get B, so B = 126; the A and C counts are equal, so A = C = 63. CS grades in AI split 3: 5: 2 of 210, i.e. 63, 105, 42. No non-CS student gets an A in AI, so the 42 non-CS passers give B = 21 and C = 21.
• ML has $210 + 150 = 360$ students, 30 fail, 330 pass. B = 165; the A: C ratio is 3: 2 with A + C = 165, so A = 99, C = 66. CS ML passers (198) split 4: 5: 2 into 72, 90, 36. The rest are non-CS: A = 27, B = 75, C = 30, F = 18.

AI course (270 students)

ML course (360 students)

This question. In ML the total B grades number 165, of which CS students earned 90. So non-CS B grades are $165 - 90 = 75$.

Answer: 75

Four players — Pulak, Qasim, Ritesh, Suresh — start with Rs.10 each, so the four always hold Rs.40 between them. Each round they split into two pairs (pairs must change from the previous round), and each pair either sits out or bets Re.1 or Rs.2, the winner taking that amount. Because the total is always 40, any row of the table can be completed once three of its four entries are known.

Reconstructing the amounts.

1. Conservation fills gaps: Pulak on Round 1 $= 40-8-10-10 = 12$; Ritesh on Round 2 $= 9$ and on Round 5 $= 7$. Round 7 gives Pulak + Suresh $= 24$; Round 8 gives Qasim + Ritesh $= 17$.
2. The disclosure rule (every time a player is at 10, plus each player's own highest and lowest) narrows each player's undisclosed days: Pulak is 11 or 12 except where shown; Qasim is 9 or 11 except where shown, with a high of 12 in Round 7; Ritesh sits between 5 and 9 in the middle days; Suresh is 9, 11 or 12 except where shown.
3. Round 7: Pulak is 11 or 12 and Suresh is 9, 11 or 12, and they sum to 24 — only Pulak = 12, Suresh = 12 works.
4. Round 8: a player's amount changes by at most 2 each round, so Ritesh (4 in Round 7) is 5 or 6; with Qasim being 9 or 11 and Qasim + Ritesh = 17, only Ritesh = 6, Qasim = 11 fits.
5. Every player's round-to-round change is $0, \pm1$ or $\pm2$. Tracking Ritesh $(10, 9, \cdot, \cdot, 7, \cdot, 4, 6)$ with steps of at most 2 forces Round 6 = 5.
6. Pulak and Qasim held equal amounts at the end of Round 4; combined with the step rule this pins Rounds 3 and 4 at Pulak 11 / Qasim 11, with Ritesh 8 then 6 and Suresh 10 then 12.

Solved table — money (Rs.) at the end of each round

Who paired with whom, and the winner each round

This question. Round 8 forces Qasim + Ritesh = 17. Qasim is restricted to 9 or 11, and Ritesh is within 2 of his Round-7 value of 4, so Ritesh is 5 or 6. The only fit is Ritesh = 6, Qasim = 11.

💡 Teacher tip: The "changes by at most 2 each round" rule is the engine of this set — it turns a nearly blank table into a chain you can pin down end to end.

Answer: Exactly Rs.6

Four players — Pulak, Qasim, Ritesh, Suresh — start with Rs.10 each, so the four always hold Rs.40 between them. Each round they split into two pairs (pairs must change from the previous round), and each pair either sits out or bets Re.1 or Rs.2, the winner taking that amount. Because the total is always 40, any row of the table can be completed once three of its four entries are known.

Reconstructing the amounts.

1. Conservation fills gaps: Pulak on Round 1 $= 40-8-10-10 = 12$; Ritesh on Round 2 $= 9$ and on Round 5 $= 7$. Round 7 gives Pulak + Suresh $= 24$; Round 8 gives Qasim + Ritesh $= 17$.
2. The disclosure rule (every time a player is at 10, plus each player's own highest and lowest) narrows each player's undisclosed days: Pulak is 11 or 12 except where shown; Qasim is 9 or 11 except where shown, with a high of 12 in Round 7; Ritesh sits between 5 and 9 in the middle days; Suresh is 9, 11 or 12 except where shown.
3. Round 7: Pulak is 11 or 12 and Suresh is 9, 11 or 12, and they sum to 24 — only Pulak = 12, Suresh = 12 works.
4. Round 8: a player's amount changes by at most 2 each round, so Ritesh (4 in Round 7) is 5 or 6; with Qasim being 9 or 11 and Qasim + Ritesh = 17, only Ritesh = 6, Qasim = 11 fits.
5. Every player's round-to-round change is $0, \pm1$ or $\pm2$. Tracking Ritesh $(10, 9, \cdot, \cdot, 7, \cdot, 4, 6)$ with steps of at most 2 forces Round 6 = 5.
6. Pulak and Qasim held equal amounts at the end of Round 4; combined with the step rule this pins Rounds 3 and 4 at Pulak 11 / Qasim 11, with Ritesh 8 then 6 and Suresh 10 then 12.

Solved table — money (Rs.) at the end of each round

Who paired with whom, and the winner each round

This question. Pulak goes from Rs.10 at the end of Round 5 to Rs.12 at the end of Round 7. With steps of at most 2 and the pairings shown, his Round 6 amount is forced to exactly Rs.12 (a Rs.2 win over Ritesh).

Answer: Exactly Rs.12

Four players — Pulak, Qasim, Ritesh, Suresh — start with Rs.10 each, so the four always hold Rs.40 between them. Each round they split into two pairs (pairs must change from the previous round), and each pair either sits out or bets Re.1 or Rs.2, the winner taking that amount. Because the total is always 40, any row of the table can be completed once three of its four entries are known.

Reconstructing the amounts.

1. Conservation fills gaps: Pulak on Round 1 $= 40-8-10-10 = 12$; Ritesh on Round 2 $= 9$ and on Round 5 $= 7$. Round 7 gives Pulak + Suresh $= 24$; Round 8 gives Qasim + Ritesh $= 17$.
2. The disclosure rule (every time a player is at 10, plus each player's own highest and lowest) narrows each player's undisclosed days: Pulak is 11 or 12 except where shown; Qasim is 9 or 11 except where shown, with a high of 12 in Round 7; Ritesh sits between 5 and 9 in the middle days; Suresh is 9, 11 or 12 except where shown.
3. Round 7: Pulak is 11 or 12 and Suresh is 9, 11 or 12, and they sum to 24 — only Pulak = 12, Suresh = 12 works.
4. Round 8: a player's amount changes by at most 2 each round, so Ritesh (4 in Round 7) is 5 or 6; with Qasim being 9 or 11 and Qasim + Ritesh = 17, only Ritesh = 6, Qasim = 11 fits.
5. Every player's round-to-round change is $0, \pm1$ or $\pm2$. Tracking Ritesh $(10, 9, \cdot, \cdot, 7, \cdot, 4, 6)$ with steps of at most 2 forces Round 6 = 5.
6. Pulak and Qasim held equal amounts at the end of Round 4; combined with the step rule this pins Rounds 3 and 4 at Pulak 11 / Qasim 11, with Ritesh 8 then 6 and Suresh 10 then 12.

Solved table — money (Rs.) at the end of each round

Who paired with whom, and the winner each round

This question. Reading straight from the solved table, Ritesh held Rs.6 at the end of Round 4.

Answer: 6

Four players — Pulak, Qasim, Ritesh, Suresh — start with Rs.10 each, so the four always hold Rs.40 between them. Each round they split into two pairs (pairs must change from the previous round), and each pair either sits out or bets Re.1 or Rs.2, the winner taking that amount. Because the total is always 40, any row of the table can be completed once three of its four entries are known.

Reconstructing the amounts.

1. Conservation fills gaps: Pulak on Round 1 $= 40-8-10-10 = 12$; Ritesh on Round 2 $= 9$ and on Round 5 $= 7$. Round 7 gives Pulak + Suresh $= 24$; Round 8 gives Qasim + Ritesh $= 17$.
2. The disclosure rule (every time a player is at 10, plus each player's own highest and lowest) narrows each player's undisclosed days: Pulak is 11 or 12 except where shown; Qasim is 9 or 11 except where shown, with a high of 12 in Round 7; Ritesh sits between 5 and 9 in the middle days; Suresh is 9, 11 or 12 except where shown.
3. Round 7: Pulak is 11 or 12 and Suresh is 9, 11 or 12, and they sum to 24 — only Pulak = 12, Suresh = 12 works.
4. Round 8: a player's amount changes by at most 2 each round, so Ritesh (4 in Round 7) is 5 or 6; with Qasim being 9 or 11 and Qasim + Ritesh = 17, only Ritesh = 6, Qasim = 11 fits.
5. Every player's round-to-round change is $0, \pm1$ or $\pm2$. Tracking Ritesh $(10, 9, \cdot, \cdot, 7, \cdot, 4, 6)$ with steps of at most 2 forces Round 6 = 5.
6. Pulak and Qasim held equal amounts at the end of Round 4; combined with the step rule this pins Rounds 3 and 4 at Pulak 11 / Qasim 11, with Ritesh 8 then 6 and Suresh 10 then 12.

Solved table — money (Rs.) at the end of each round

Who paired with whom, and the winner each round

This question. Go through the pairings and pick out the Rs.2 games: Round 1 Pulak–Qasim, Round 2 Qasim–Suresh, Round 3 Pulak–Suresh, Round 4 Ritesh–Suresh, Round 6 Pulak–Ritesh, and Round 8 Ritesh–Suresh. That is 6 games played for a Rs.2 stake.

Answer: 6

Four players — Pulak, Qasim, Ritesh, Suresh — start with Rs.10 each, so the four always hold Rs.40 between them. Each round they split into two pairs (pairs must change from the previous round), and each pair either sits out or bets Re.1 or Rs.2, the winner taking that amount. Because the total is always 40, any row of the table can be completed once three of its four entries are known.

Reconstructing the amounts.

1. Conservation fills gaps: Pulak on Round 1 $= 40-8-10-10 = 12$; Ritesh on Round 2 $= 9$ and on Round 5 $= 7$. Round 7 gives Pulak + Suresh $= 24$; Round 8 gives Qasim + Ritesh $= 17$.
2. The disclosure rule (every time a player is at 10, plus each player's own highest and lowest) narrows each player's undisclosed days: Pulak is 11 or 12 except where shown; Qasim is 9 or 11 except where shown, with a high of 12 in Round 7; Ritesh sits between 5 and 9 in the middle days; Suresh is 9, 11 or 12 except where shown.
3. Round 7: Pulak is 11 or 12 and Suresh is 9, 11 or 12, and they sum to 24 — only Pulak = 12, Suresh = 12 works.
4. Round 8: a player's amount changes by at most 2 each round, so Ritesh (4 in Round 7) is 5 or 6; with Qasim being 9 or 11 and Qasim + Ritesh = 17, only Ritesh = 6, Qasim = 11 fits.
5. Every player's round-to-round change is $0, \pm1$ or $\pm2$. Tracking Ritesh $(10, 9, \cdot, \cdot, 7, \cdot, 4, 6)$ with steps of at most 2 forces Round 6 = 5.
6. Pulak and Qasim held equal amounts at the end of Round 4; combined with the step rule this pins Rounds 3 and 4 at Pulak 11 / Qasim 11, with Ritesh 8 then 6 and Suresh 10 then 12.

Solved table — money (Rs.) at the end of each round

Who paired with whom, and the winner each round

This question. In Round 5 the two pairs were Qasim–Ritesh and Pulak–Suresh. Of the options listed, the pairing that actually occurred is Pulak and Suresh.

Answer: Pulak and Suresh

We are tracking new cases in four neighbourhoods over five days. Each day a neighbourhood records 0, 1, 2 or 3 new cases. The five-day totals are fixed: Levmisto 12, Tyhrmisto 12, Pesmisto 5, Kitmisto 14 — so the whole city records $12+12+5+14 = 43$ cases.

Let us pin the grid down one neighbourhood at a time.

1. Pesmisto totals 5, its highest single-day figure is 2 and that 2 happens exactly once; every other day is 0 or 1. So its five days are the numbers $\{2,1,1,1,0\}$ in some order.
2. Kitmisto totals 14 with at most 3 per day, which is only possible as four 3's and one 2. We are told Kitmisto is the only neighbourhood with 3 cases on Day 2, so Day 2 is one of its 3's.
3. The city's daily total keeps rising, and Day 3 is exactly one more than Day 2. The most a single day can reach is $3+3+2+3 = 11$. The only rising five-day sequence that sums to 43 with Day 3 one more than Day 2 is 5, 8, 9, 10, 11.
4. Day 1 totals 5 with every neighbourhood having at least one case, so it is $1+1+1+2$. Since Kitmisto always exceeds Pesmisto, Kitmisto cannot be the 1 — so Kitmisto = 2 on Day 1 and the other three are 1 each. Hence Kitmisto is 3 on Days 2 to 5.
5. Day 5 totals 11, achievable only as $3+3+2+3$, so Pesmisto = 2 here (its single 2) and Levmisto = Tyhrmisto = 3.
6. On Day 2 only Kitmisto may be 3, so Levmisto and Tyhrmisto are at most 2 that day. Each of them totals 12 and is 1 on Day 1 and 3 on Day 5, so their Days 2–4 must add to 8 with Day 2 $\le 2$: that forces 2, 3, 3.
7. Pesmisto then follows from each day's city total: Day 2 needs Pesmisto = 1, Day 3 needs 0, Day 4 needs 1.

Every clue checks out and the grid is unique:

This question. The Day 2 column adds to $2+2+1+3 = 8$. Because the grid is completely forced, there is no other possibility.

💡 Teacher tip: The rising-totals clue does the heavy lifting — once you spot that 5, 8, 9, 10, 11 is the only valid daily sequence, every column is locked.

Answer: Exactly 8

We are tracking new cases in four neighbourhoods over five days. Each day a neighbourhood records 0, 1, 2 or 3 new cases. The five-day totals are fixed: Levmisto 12, Tyhrmisto 12, Pesmisto 5, Kitmisto 14 — so the whole city records $12+12+5+14 = 43$ cases.

Let us pin the grid down one neighbourhood at a time.

1. Pesmisto totals 5, its highest single-day figure is 2 and that 2 happens exactly once; every other day is 0 or 1. So its five days are the numbers $\{2,1,1,1,0\}$ in some order.
2. Kitmisto totals 14 with at most 3 per day, which is only possible as four 3's and one 2. We are told Kitmisto is the only neighbourhood with 3 cases on Day 2, so Day 2 is one of its 3's.
3. The city's daily total keeps rising, and Day 3 is exactly one more than Day 2. The most a single day can reach is $3+3+2+3 = 11$. The only rising five-day sequence that sums to 43 with Day 3 one more than Day 2 is 5, 8, 9, 10, 11.
4. Day 1 totals 5 with every neighbourhood having at least one case, so it is $1+1+1+2$. Since Kitmisto always exceeds Pesmisto, Kitmisto cannot be the 1 — so Kitmisto = 2 on Day 1 and the other three are 1 each. Hence Kitmisto is 3 on Days 2 to 5.
5. Day 5 totals 11, achievable only as $3+3+2+3$, so Pesmisto = 2 here (its single 2) and Levmisto = Tyhrmisto = 3.
6. On Day 2 only Kitmisto may be 3, so Levmisto and Tyhrmisto are at most 2 that day. Each of them totals 12 and is 1 on Day 1 and 3 on Day 5, so their Days 2–4 must add to 8 with Day 2 $\le 2$: that forces 2, 3, 3.
7. Pesmisto then follows from each day's city total: Day 2 needs Pesmisto = 1, Day 3 needs 0, Day 4 needs 1.

Every clue checks out and the grid is unique:

This question. Reading Levmisto's row, Day 3 is 3, and this value is forced by the grid.

Answer: Exactly 3

We are tracking new cases in four neighbourhoods over five days. Each day a neighbourhood records 0, 1, 2 or 3 new cases. The five-day totals are fixed: Levmisto 12, Tyhrmisto 12, Pesmisto 5, Kitmisto 14 — so the whole city records $12+12+5+14 = 43$ cases.

Let us pin the grid down one neighbourhood at a time.

1. Pesmisto totals 5, its highest single-day figure is 2 and that 2 happens exactly once; every other day is 0 or 1. So its five days are the numbers $\{2,1,1,1,0\}$ in some order.
2. Kitmisto totals 14 with at most 3 per day, which is only possible as four 3's and one 2. We are told Kitmisto is the only neighbourhood with 3 cases on Day 2, so Day 2 is one of its 3's.
3. The city's daily total keeps rising, and Day 3 is exactly one more than Day 2. The most a single day can reach is $3+3+2+3 = 11$. The only rising five-day sequence that sums to 43 with Day 3 one more than Day 2 is 5, 8, 9, 10, 11.
4. Day 1 totals 5 with every neighbourhood having at least one case, so it is $1+1+1+2$. Since Kitmisto always exceeds Pesmisto, Kitmisto cannot be the 1 — so Kitmisto = 2 on Day 1 and the other three are 1 each. Hence Kitmisto is 3 on Days 2 to 5.
5. Day 5 totals 11, achievable only as $3+3+2+3$, so Pesmisto = 2 here (its single 2) and Levmisto = Tyhrmisto = 3.
6. On Day 2 only Kitmisto may be 3, so Levmisto and Tyhrmisto are at most 2 that day. Each of them totals 12 and is 1 on Day 1 and 3 on Day 5, so their Days 2–4 must add to 8 with Day 2 $\le 2$: that forces 2, 3, 3.
7. Pesmisto then follows from each day's city total: Day 2 needs Pesmisto = 1, Day 3 needs 0, Day 4 needs 1.

Every clue checks out and the grid is unique:

This question. Pesmisto's row is $1, 1, 0, 1, 2$. Its only zero sits on Day 3.

Answer: Only Day 3

We are tracking new cases in four neighbourhoods over five days. Each day a neighbourhood records 0, 1, 2 or 3 new cases. The five-day totals are fixed: Levmisto 12, Tyhrmisto 12, Pesmisto 5, Kitmisto 14 — so the whole city records $12+12+5+14 = 43$ cases.

Let us pin the grid down one neighbourhood at a time.

1. Pesmisto totals 5, its highest single-day figure is 2 and that 2 happens exactly once; every other day is 0 or 1. So its five days are the numbers $\{2,1,1,1,0\}$ in some order.
2. Kitmisto totals 14 with at most 3 per day, which is only possible as four 3's and one 2. We are told Kitmisto is the only neighbourhood with 3 cases on Day 2, so Day 2 is one of its 3's.
3. The city's daily total keeps rising, and Day 3 is exactly one more than Day 2. The most a single day can reach is $3+3+2+3 = 11$. The only rising five-day sequence that sums to 43 with Day 3 one more than Day 2 is 5, 8, 9, 10, 11.
4. Day 1 totals 5 with every neighbourhood having at least one case, so it is $1+1+1+2$. Since Kitmisto always exceeds Pesmisto, Kitmisto cannot be the 1 — so Kitmisto = 2 on Day 1 and the other three are 1 each. Hence Kitmisto is 3 on Days 2 to 5.
5. Day 5 totals 11, achievable only as $3+3+2+3$, so Pesmisto = 2 here (its single 2) and Levmisto = Tyhrmisto = 3.
6. On Day 2 only Kitmisto may be 3, so Levmisto and Tyhrmisto are at most 2 that day. Each of them totals 12 and is 1 on Day 1 and 3 on Day 5, so their Days 2–4 must add to 8 with Day 2 $\le 2$: that forces 2, 3, 3.
7. Pesmisto then follows from each day's city total: Day 2 needs Pesmisto = 1, Day 3 needs 0, Day 4 needs 1.

Every clue checks out and the grid is unique:

This question. Check each statement against the forced grid.

• Statement A: "2 new cases in Tyhrmisto on Day 3." Tyhrmisto on Day 3 is 3, not 2 — so A is false.
• Statement B: "No new cases in Pesmisto on Day 2." Pesmisto on Day 2 is 1, not 0 — so B is false.

Both statements are necessarily false.

Answer: Both Statement A and Statement B

We are tracking new cases in four neighbourhoods over five days. Each day a neighbourhood records 0, 1, 2 or 3 new cases. The five-day totals are fixed: Levmisto 12, Tyhrmisto 12, Pesmisto 5, Kitmisto 14 — so the whole city records $12+12+5+14 = 43$ cases.

Let us pin the grid down one neighbourhood at a time.

1. Pesmisto totals 5, its highest single-day figure is 2 and that 2 happens exactly once; every other day is 0 or 1. So its five days are the numbers $\{2,1,1,1,0\}$ in some order.
2. Kitmisto totals 14 with at most 3 per day, which is only possible as four 3's and one 2. We are told Kitmisto is the only neighbourhood with 3 cases on Day 2, so Day 2 is one of its 3's.
3. The city's daily total keeps rising, and Day 3 is exactly one more than Day 2. The most a single day can reach is $3+3+2+3 = 11$. The only rising five-day sequence that sums to 43 with Day 3 one more than Day 2 is 5, 8, 9, 10, 11.
4. Day 1 totals 5 with every neighbourhood having at least one case, so it is $1+1+1+2$. Since Kitmisto always exceeds Pesmisto, Kitmisto cannot be the 1 — so Kitmisto = 2 on Day 1 and the other three are 1 each. Hence Kitmisto is 3 on Days 2 to 5.
5. Day 5 totals 11, achievable only as $3+3+2+3$, so Pesmisto = 2 here (its single 2) and Levmisto = Tyhrmisto = 3.
6. On Day 2 only Kitmisto may be 3, so Levmisto and Tyhrmisto are at most 2 that day. Each of them totals 12 and is 1 on Day 1 and 3 on Day 5, so their Days 2–4 must add to 8 with Day 2 $\le 2$: that forces 2, 3, 3.
7. Pesmisto then follows from each day's city total: Day 2 needs Pesmisto = 1, Day 3 needs 0, Day 4 needs 1.

Every clue checks out and the grid is unique:

This question. Levmisto's row is $(1,2,3,3,3)$ and Tyhrmisto's row is also $(1,2,3,3,3)$ — identical on every one of the five days.

Answer: 5