Arithmetical reasoning covers short number puzzles: ages, heads and legs, notes, fences, cuts, schedules and simple equations. ASO asks 3 to 7 of them every year. Most take one line of algebra once you set up the unknown correctly.
Age problems
- Ratio type: take the present ages as ax and bx, write the condition for "n years ago" or "after n years", and solve for x. Example (ASO 2025): 3 : 5 now and 13 : 20 after 5 years → (3x + 5)/(5x + 5) = 13/20 → x = 7, so B = 35.
- "Times as old" type: father = k × son now and m × son n years ago → k·s − n = m(s − n).
- Remember: the difference between two people's ages never changes.
- Average type: turn averages into sums first. avg(A, B, C) = 45 → A + B + C = 135.
Heads and legs
Pretend every animal has 2 legs. The extra legs ÷ 2 = the number of four-legged animals. 80 heads and 200 legs → 200 − 160 = 40 extra → 20 deer and 60 peacocks (ASO 2021).
Counting rules
- Pieces and cuts: n pieces need n − 1 cuts (ASO 2018: 10 pieces = 9 cuts).
- Trees or poles in a line: number of gaps + 1.
- Poles round a square with n on each side: 4n − 4 (ASO 2015: 27 on each side → 104).
- Handshakes: n(n − 1) ÷ 2.
- Digit count 1–100: each digit 1–9 appears 20 times (10 in the units place, 10 in the tens place).
Equal notes
If the counts are equal, add the note values once and divide the total by that. ₹480 in ₹1, ₹5 and ₹10 → 480 ÷ 16 = 30 of each → 90 notes (ASO 2015).
Two equations
When the coefficients are swapped (2B + 3C and 3B + 2C), add the equations to get B + C and subtract them to get B − C (ASO 2018).
Working backwards
When you know the end result (eggs left, money left), undo each step in reverse. Undo "half plus half an egg" with x → 2x + 1: 3 → 7 → 15 → 31 (ASO 2021).