Averages, Ages & Number System

Snapshot

• Averages, Ages & Number System bundles three quick-scoring families: the average (mean), age problems (linear equations in disguise), and number-system facts (divisibility, unit digits, HCF/LCM, remainders).
• Each question is one or two steps once it is set up correctly, so the marks go to whoever sets up fast and avoids the careless slip — and to whoever has the divisibility and cyclicity rules memorised.
• This guide covers every common average pattern, the two things that make age problems trip students, and the number-system toolkit the General Test reuses constantly.
• Exam reality: +5 / −1 — set the equation, solve, sanity-check the size.

Part 1 — Averages

Average = (sum of values) ÷ (number of values), and rearranged, sum = average × count — the form you actually use.

• Missing value. Five tests average 50 ⇒ total 250; if four add to 190, the fifth is 60 (the bar above the dashed line).
• New entrant / departure. Work with the change in total. If the average of 10 numbers rises by 2 when one is replaced, the new number is 10 × 2 = 20 more than the old.
• Average of consecutive numbers = the middle one (or the mean of the two middle ones) — the average of 1…99 is just 50.
• Weighted average (unequal groups): (n₁a₁ + n₂a₂) ÷ (n₁ + n₂). Classes of 40 @ 60 and 60 @ 70 average (2400 + 4200) ÷ 100 = 66, not 65.
• Average speed for equal distances is the harmonic mean, 2ab ÷ (a + b) — not (a + b) ÷ 2. (For equal times, the ordinary mean is correct.)

Part 2 — Ages

Translate words into one variable. "A is 5 years older than B" → A = B + 5. "Their ages are in ratio 3 : 4" → 3x and 4x. The two killers:

• Past/future shifts apply to everyone equally — "5 years ago" subtracts 5 from every age in the problem.
• A present ratio plus a past/future ratio gives two equations — solve them together.

Part 3 — Number System

Divisibility: by 3 if the digit-sum is divisible by 3; by 4 if the last two digits are; by 8 if the last three are; by 9 if the digit-sum is; by 11 if (sum of odd-place − sum of even-place digits) is 0 or a multiple of 11.

Unit digit of powers cycles in 4. The unit of 2ⁿ runs 2, 4, 8, 6; of 3ⁿ runs 3, 9, 7, 1; of 7ⁿ runs 7, 9, 3, 1. Only the exponent's remainder ÷ 4 matters (a remainder of 0 means the last digit in the cycle).

HCF & LCM: for two numbers, HCF × LCM = product of the numbers. HCF ≤ each number ≤ LCM — a fast sanity check.

Number of factors: write N = pᵃ · qᵇ; the count of factors is (a + 1)(b + 1). So 36 = 2²·3² has (2+1)(2+1) = 9 factors.

Remainders: for "remainder when … divided by", reduce each part by its own remainder first (modular arithmetic) instead of computing the giant number.

Part 4 — Speed techniques

1. Work with the total, not the average, whenever items join or leave.
2. Assume the average as a baseline and track only the deviations ("+5, −3, …") for fast mental means.
3. Let the smallest age be x to keep age equations clean.
4. Test divisibility by 3 and 9 with the digit-sum, by 4 and 8 with the last digits.
5. Read unit digits from the cycle of four — never compute the full power.
6. Use HCF × LCM = product to recover a missing number instantly.

Part 5 — Worked examples

1. Average, departure. The average of 6 numbers is 30; removing one makes it 32. Removed number? old total 180, new total 32×5 = 160, removed = 20.

2. Weighted. 30 boys average 45 kg, 20 girls average 40 kg. Class average? (30×45 + 20×40) ÷ 50 = (1350 + 800) ÷ 50 = 43 kg.

3. Average speed. A car goes to a town at 40 km/h and returns at 60 km/h. Average speed? 2×40×60 ÷ (40+60) = 4800 ÷ 100 = 48 km/h (not 50).

4. Ages. A father is 3× his son. In 12 years he will be 2×. Present ages? 3x + 12 = 2(x + 12) → x = 12 → son 12, father 36.

5. Ages ratio. Present ratio 5 : 7; after 6 years it is 3 : 4. Ages? (5x+6)/(7x+6) = 3/4 → 20x + 24 = 21x + 18 → x = 6 → 30 and 42.

6. Unit digit. Unit of 3⁵⁰? cycle 3,9,7,1; 50 ÷ 4 leaves remainder 2 → 2nd term → 9.

7. Factors. How many factors does 72 have? 72 = 2³·3² → (3+1)(2+1) = 12.

8. HCF/LCM. HCF of two numbers is 6, LCM 60, one number 12. Other? (6×60)/12 = 30.

Part 6 — Common traps

• Average is not the middle value for non-consecutive sets — compute it.
• Average speed is harmonic for equal distances — do not just average the speeds.
• Time shifts hit every age — don't age only one person.
• Unit-digit remainder 0 = last in the cycle, not the first.
• HCF ≤ number ≤ LCM — use it to reject impossible options.

Part 7 — How to use this page

Lock the divisibility rules, the cyclicity table and "sum = average × count", re-solve the eight examples closed-book, then attempt the practice set and the timed test.

One-line revision: sum = average × count, equal-distance average speed is 2ab/(a+b), time shifts hit every age equally, test 3 and 9 by digit-sum, and read the unit digit from the power's cycle of four.