Eleven kinds of stories, forty-eight worked examples, and one good habit: draw the picture first.
11 topics48 problemsHints + full solutionsVisual bar models
Every problem here can be drawn before it's solved. The drawing — usually a
bar model — turns a tangle of words into a picture you can count.
Most of the work is in choosing what to draw. Once the picture is right,
the arithmetic almost does itself.
Each topic opens with the idea, then three problems — an easy one,
a medium one, and a stretch. Try the problem. Open the hint if you're stuck.
Open the solution to check.
01
Chapter One
Sum, Difference & Multiple
The whole family of "two amounts, one relationship" problems.
The three pieces
Almost every problem about two (or three) people sharing something can be described with three pieces of information:
Sum — how much in total. Difference — how much more one has than the other. Multiple — how many times one is the size of the other.
If you have any two of these three, you can find everything. Bar models show you which two you've got.
§ 1.1
Sum–Difference Problems
You know the total and you know the gap. Cover the gap with your finger and the two bars become equal — that's the trick.
Q1 Classic warm-up
Some boys and girls plant a total of 128 trees, with boys planting 20 more trees than girls. How many trees do boys plant?
If you "remove" the extra 20 trees from boys' bar, both bars become the same length. What's the new total? Then how big is each bar?
Sum = 128 trees. Difference = 20 trees more for boys.
01Remove the extra 20 from the total: 128 − 20 = 108. Now both bars are equal, summing to 108.
02One bar (Girls) is half of 108: 108 ÷ 2 = 54.
03Boys' bar is Girls + 20: 54 + 20 = 74.
Boys planted 74 trees.
Q2 The "after the swap" trick
There are a total of 120 pears in two baskets. If 10 pears are taken from the first basket and placed in the second basket, then the number of pears in both baskets is equal. How many pears were in each basket before?
Should you draw the situation before the swap or after? Draw the one where you have more concrete information. Does moving pears between baskets change the total?
Moving pears between two baskets does not change the sum — it's still 120. We draw the model after the swap, because that's the moment we know the relationship (the two are equal).
01After the swap, the two baskets are equal and still sum to 120, so each holds 120 ÷ 2 = 60 pears.
02Walk the swap backwards. The first basket lost 10, so before the swap it had 60 + 10 = 70 pears.
03The second basket gained 10, so before the swap it had 60 − 10 = 50 pears.
First basket had 70 pears; second basket had 50 pears.
Q32023 GEP · three people
Alex, Ben, and Cindy have a total of 44 marbles. Alex bought 10 marbles and Ben sold 6. The three of them now have an equal number of marbles. How many marbles did Cindy have at the beginning?
Unlike Q2, this time the sum does change. Why? Then notice: Cindy didn't buy or sell anything, so her "before" equals her "after."
Alex bought 10 (so 10 more marbles enter); Ben sold 6 (so 6 leave). The new sum is 44 + 10 − 6 = 48.
01All three are equal: 48 ÷ 3 = 16 marbles each.
02Cindy never changed, so her "before" = her "after."
Cindy started with 16 marbles.
§ 1.2
Sum–Multiple Problems
You know the total and you know that one is so many times the other. Always draw the multiple first: small bar = 1 unit, big bar = however many units.
Q12024 GEP · the basic version
Amy has 5 times as many books as Ben. Together they have 60 books. How many books does Ben have?
Ben is 1 unit. Amy is 5 units. Add them up — how many units altogether?
01Total units: 1 + 5 = 6 units = 60 books.
02One unit: 60 ÷ 6 = 10 books.
Ben has 10 books.
Q2 Sum + multiple + extra piece
Alex and Ben moved 200 books together. Alex moved 20 books more than twice as many as Ben. How many books did Ben move?
Draw the multiple part first (Ben = 1 unit, Alex = 2 units). Then tack on the extra "+20" piece for Alex. If you remove the +20 from the total, what's left in whole units?
01Remove the extra 20 from Alex (and from the total): 200 − 20 = 180.
02Now we have whole units: 1 + 2 = 3 units = 180. So 1 unit = 180 ÷ 3 = 60.
03Alex's actual total: 2 × 60 + 20 = 140 books.
Ben moved 60 books; Alex moved 140 books.
Q3 When the sum itself changes
There were 400 passengers on two ships A and B. After arriving, 40 got off ship A and 30 got on ship B. Now ship A has exactly 5 times the passengers of ship B. How many were on each ship originally?
The "5 times" relationship is true after the change, not before. Find the new sum first, then solve, then walk each ship back to "before."
The multiple holds after the change. New sum: 400 − 40 + 30 = 390.
01Total units: 1 + 5 = 6 units = 390 after. So 1 unit = 390 ÷ 6 = 65.
02After: B has 65, A has 65 × 5 = 325.
03Walk back. A lost 40 → before: 325 + 40 = 365.
04B gained 30 → before: 65 − 30 = 35.
Check: 365 + 35 = 400 ✓
Originally, A had 365 passengers; B had 35 passengers.
§ 1.3
Difference–Multiple Problems
No sum this time. Instead, you're told the gap and the multiple. The gap occupies a specific number of units — that's how you find one unit.
Q12023 GEP
A toy car costs $30 more than a soft toy, and the toy car equals the cost of 3 soft toys. How much does the soft toy cost?
Soft toy = 1 unit, toy car = 3 units. The $30 gap is how many extra units?
01The gap of $30 covers 2 units (the toy car has 2 more units than the soft toy).
02One unit: 30 ÷ 2 = $15.
The soft toy costs $15.
Q22023 GEP · with a twist
Alex has 2 fewer cards than Cindy. Alex receives 14 more cards. Now Alex has twice as many cards as Cindy. How many cards did Alex have at first?
Find the new difference between Alex and Cindy after the 14 cards. Alex was 2 behind, then gained 14 — how far ahead is he now? That's where the multiple kicks in.
New difference (after): Alex was 2 behind, then gained 14 → he is now 14 − 2 = 12 ahead of Cindy. And Alex is now twice Cindy. Draw the model after the change:
01The new gap = 1 unit = 12 cards.
02Alex (after) = 2 units = 12 × 2 = 24 cards.
03Walk back: Alex gained 14, so originally he had 24 − 14 = 10.
Alex had 10 cards at first.
Q3 Geometry in disguise
Take a square. One side increases by 10 cm; the other decreases by 5 cm. The new rectangle's length is four times its width. How long is the width?
The square's two sides were equal. One grew by 10 and the other shrank by 5 — how big is the gap between them now?
Both sides started equal. One increased by 10, the other decreased by 5, so the new gap is 10 + 5 = 15 cm. And length = 4 × width.
01The 15 cm gap covers 3 units (length is 3 units longer than width).
02One unit: 15 ÷ 3 = 5 cm.
The width of the rectangle is 5 cm.
§ 1.4
Age Problems
The Age Rule
If A is 10 years older than B today, A will still be 10 years older in 20 years, in 50 years, forever. The age difference never changes. When the sum or multiple is given for "X years later," translate the relationship using this rule.
Q12023 GEP
Ahmad is 10 years older than Tom. 2 years later, Ahmad will be twice as old as Tom. What are their ages now?
Draw the model for "2 years later" because that's when the multiple is true. The gap (10 years) is the same then as now.
2 years later, Tom = 1 unit, Ahmad = 2 units. The gap of 10 years is also 1 unit.
011 unit = 10 years. Two years later: Tom = 10, Ahmad = 20.
02Walk back 2 years: today Tom = 8, Ahmad = 18.
Tom is 8 and Ahmad is 18 today.
Q2 The sneaky version
The sum of the ages of an older sister and a younger sister in 3 years is 33. The younger sister's age this year equals the age difference between them. How old is the older sister this year?
Two big ideas: (1) sums of ages change with time — if they sum to 33 in 3 years, what's their sum today? (2) If younger's age = the difference, then older = 2 × younger.
Three years from now both sisters will be 3 years older, so their sum grows by 6: today's sum is 33 − 3 − 3 = 27.
If younger's age = (older − younger), then older = 2 × younger. So today, older = 2 units and younger = 1 unit, summing to 27:
011 + 2 = 3 units = 27. So 1 unit = 9.
02Older = 2 × 9 = 18; younger = 9.
The older sister is 18 years old this year.
Q3Extension · Teacher Tian
Alex asks Teacher Tian his age. Teacher Tian says: "When I was as old as you, my age was 10 times yours. When you become as old as I am now, I will be 56." How old is Teacher Tian now?
Draw an age ruler. Place four marks on it: Alex-then (in the past), Alex-now, Tian-now, and 56 (Tian's future). The age gap Tian−Alex is fixed forever, so the three jumps Alex-then → Alex-now, Alex-now → Tian-now, and Tian-now → 56 are all equal to that gap.
Key idea: the age gap between two people never changes. So if at some past moment Alex was a and Tian was a + g, then today Tian = Alexnow + g, and any future age difference is still g.
"When I was your age" puts us at a past moment where Tian's age then = Alex's age now. At that moment Tian was 10 × Alexthen. Call Alexthen = 1 unit; then Tianthen = 10 units, and the fixed gap is 10 − 1 = 9 units.
01Three equal jumps of 9 units each, plus the 1-unit head start from 0 to Alex-then, span all of 56 years: 1 + 9 + 9 + 9 = 28 units = 56.
02So 1 unit = 56 ÷ 28 = 2 years.
03Tian-now is 9 units short of 56: 56 − 9 × 2 = 38. (Check: Alex-now = 10 × 2 = 20; gap = 18; Tian-now = 20 + 18 = 38 ✓)
Teacher Tian is 38 years old now.
02
Chapter Two
Grouping Problems
When two kinds of things come in a fixed ratio, glue them together into one bundle and count bundles.
The idea
If the problem tells you something like "twice as many big packs as small packs" or "A and B use stickers in lengths that match up," you can group the items into a repeating bundle whose total cost / size / count you can compute. Then the question becomes: how many bundles?
Q12023 GEP
Alex bought some files and pencil cases for $42. Each file costs $4, each pencil case costs $2. There were 3 more files than pencil cases. How many files did Alex buy?
First strip away the 3 extra files (and the money they cost) so you have equal numbers of files and pencil cases. Then bundle: 1 file + 1 pencil case = ?
01Remove the 3 extra files first: 3 × $4 = $12. Remaining cost: $42 − $12 = $30.
02Now files = pencil cases. Bundle one of each: $4 + $2 = $6.
03Number of bundles: $30 ÷ $6 = 5.
04Each bundle has 1 file, plus the 3 extras: 5 + 3 = 8 files.
Alex bought 8 files.
Q22025 GEP
String A uses 2 cm stickers; String B uses 3 cm stickers. There are 45 stickers in total, and the two strings have the same length. How long is one string?
How many A-stickers add up to the same length as how many B-stickers? Find the smallest matching group, then count.
3 stickers of A = 3 × 2 = 6 cm. 2 stickers of B = 2 × 3 = 6 cm. So bundle 3 A's with 2 B's — that's 5 stickers covering equal length on both strings.
01One bundle has 5 stickers. Number of bundles: 45 ÷ 5 = 9.
02String A has 9 × 3 = 27 stickers → length = 27 × 2 = 54 cm.
Each string is 54 cm long.
Q32023 GEP · packs
A "small pack" contains 1 pen + 2 erasers. A "big pack" contains 2 pens + 4 rulers. Twice as many big packs as small packs were sold. If 60 pens were sold, how many rulers were sold?
Bundle 1 small pack + 2 big packs together. How many pens in one bundle?
Small pack
Big pack
1 pen 2 erasers
2 pens 4 rulers
01Bundle: 1 small + 2 big = 1 + 2 × 2 = 5 pens.
02Number of bundles: 60 ÷ 5 = 12.
03Big packs sold: 12 × 2 = 24. Rulers come only from big packs: 24 × 4 = 96.
96 rulers were sold.
03
Chapter Three
Excess & Shortage
Two ways to distribute, two different leftover stories — find the unknown by comparing.
The trick
You're given two distribution scenarios with the same set of items or people. Lay them out as Case 1 and Case 2. Find: (a) the change per item from one case to the other, and (b) the total extra/missing items between the cases. Divide and you've found the count.
Q12023 GEP
A teacher has some bags. If she puts 5 pieces of sugar in each, all bags are used up. If she puts 6 pieces in each instead, she runs out with 2 empty bags left over. How many bags are there?
The 2 leftover bags in Case 1 would have held 5 × 2 = 10 sugars. Where do those 10 sugars "go" in Case 2? Each filled bag in Case 2 has 1 extra sugar — so how many filled bags share those 10?
01The 2 leftover bags in Case 1 had 2 × 5 = 10 sugars in them. In Case 2 they're empty, so those 10 sugars went into other bags.
02From Case 1 to Case 2, each filled bag gets 1 extra sugar. So 10 sugars fill 10 ÷ 1 = 10 bags.
03Total bags = 10 filled in Case 2 + 2 emptied = 12.
There are 12 bags in total.
Q22024 GEP
The numbers of girls and boys are the same. Every girl gets 2 candies. Five boys are absent, so every present boy gets 3 candies. Total candies to boys equals total to girls. How many girls are there?
The 5 absent boys are "phantom" people. The candies that would have gone to them (had they all received 2 like the girls) are 10 extra candies. Now distribute those 10 extras to the present boys, who each get one more than a girl.
Imagine a Case 1 where everyone — girls and an equal number of boys — got 2 candies. Then in Case 2, boys get 3 each (1 more) but 5 boys are missing.
01The 5 absent boys would have received 5 × 2 = 10 candies if they'd shown up. Those 10 candies are "extra" now.
02Each present boy gets 1 more candy than a girl. So 10 extra candies feed 10 ÷ 1 = 10 present boys.
03Total boys = 10 present + 5 absent = 15. Girls = boys = 15.
There are 15 girls.
Q32023 GEP · balls
There are 10 more girls than boys. Each girl has 2 red balls; each boy has 3 blue balls. The total number of red balls equals the total number of blue balls. How many boys are there?
The 10 "extra" girls contribute 10 × 2 = 20 red balls. To balance the blue-ball total, those 20 extra must be matched by extra blue balls. Each boy contributes 1 more ball than a girl (3 vs 2).
01The 10 extra girls contribute 10 × 2 = 20 extra red balls compared to a matched set.
02For totals to be equal, those 20 extra reds must be matched by extra blue balls. Each boy provides 1 more ball than a girl (3 − 2 = 1), so the number of boys is 20 ÷ 1 = 20.
03Girls = 20 + 10 = 30.
There are 20 boys (and 30 girls).
04
Chapter Four
Period Problems
Whenever a pattern repeats, divide by the length of the repeat and read the remainder.
§ 4.1
Repeating Patterns
Q12023 GEP
PEARPEARPEAR… — what is the 35th letter?
The pattern repeats every 4 letters. Divide 35 by 4 and look at the remainder.
35 ÷ 4 = 8 remainder 3. So 8 full cycles of PEAR (32 letters), then 3 more: P, E, A.
The 35th letter is A.
Q22023 GEP · two-row pattern
Top row: △△ △ △△ △ … Bottom row: □○○ □○○ □○○ … When the bottom row has 60 shapes, how many triangles are in the top row?
The bottom row repeats every 3 shapes. The top row's triangle pattern repeats every 2 of those bottom-groups (3 triangles per 2 columns).
01Bottom columns: every 3 bottom shapes = 1 column. So 60 ÷ 3 = 20 columns.
02Every 2 columns has 3 triangles on top. So 20 ÷ 2 = 10 repeats.
03Triangles: 10 × 3 = 30.
30 triangles.
Q3Extension · checkers
2018 white and black checkers are arranged: ○●○●●○●●●○●○●●○●●●… (white, 1 black, white, 2 black, white, 3 black, repeats). How many black checkers are there?
One full group is: white-black-white-blackblack-white-blackblackblack. Count whites and blacks in one group. After dividing, don't forget the remainder.
Numbers arranged in a rectangle that repeats — another period problem.
Q1 Three-column zig-zag
Numbers are arranged in 3 columns, rows alternating direction: 1 2 3 / 6 5 4 / 7 8 9 / 12 11 10 / … Which column is 2025 in?
Every 2 rows (6 numbers) the pattern repeats. Divide 2025 by 6 and read the remainder.
1
2
3
6
5
4
7
8
9
Every 6 numbers, the pattern repeats. 2025 ÷ 6 = 337 R 3. So 2025 is the 3rd number in a "1 2 3"-style row → column 3.
2025 is in column 3.
Q22024 GEP · the moving box
In an 8-column grid filled 1, 2, 3, … row by row, a 2×2 box surrounds four numbers. The four numbers sum to 226. Find the sum of the largest and smallest numbers inside the box.
Inside the box, the bottom-row numbers are 8 more than the top-row numbers (since rows have 8 columns). The differences from the largest to the other three are 1, 8, 9. So sum + 18 = 4 × largest.
Label the four numbers a, a+1, a+8, a+9. The largest is a+9. The differences between the largest and the others are 9, 8, 1.
01If we boost every number up to the largest: sum + (9 + 8 + 1) = 4 × largest, i.e. 226 + 18 = 244 = 4 × largest.
02Largest = 244 ÷ 4 = 61. Smallest = 61 − 9 = 52.
Smallest + Largest = 52 + 61 = 113.
Q3 Even numbers in columns
Even numbers are arranged in columns A B C D E, alternating direction. Row 1 (left to right, starting at B): 2 4 6 8. Row 2 (right to left): 16 14 12 10. Row 3 (left to right starting at A): 18 20 22 24. … Which column contains 2024?
Convert 2024 to its position in the sequence of evens: it's the 2024 ÷ 2 = 1012th even number. The full pattern (two rows = 8 numbers) repeats.
012024 is the 2024 ÷ 2 = 1012th even number in the list.
02The arrangement repeats every 8 numbers. 1012 ÷ 8 = 126 R 4.
03So 2024 is the 4th number in the current cycle, which lands in column E.
2024 is in column E.
05
Chapter Five
Fraction Word Problems
Convert fractions to a common denominator and slice the bar into that many equal pieces.
The idea
Most fraction word problems become easy once you pick a denominator that fits all the fractions in the question. Then the whole becomes that many units, and each fraction is a clean number of units.
Q12023 GEP
Mark got some pocket money from his dad. He spent 1/3 on food and 1/9 on stationery. What fraction of his money does he have left?
Pick a common denominator that works for both 1/3 and 1/9. (Hint: 9.) Now express everything in ninths.
Common denominator: 9. So 1/3 = 3/9. Draw the bar as 9 equal units. Mark off 3 (food) + 1 (stationery) = 4.
01Spent: 3/9 + 1/9 = 4/9.
02Left: 1 − 4/9 = 5/9.
Mark has 5/9 of his money left.
Q22023 GEP · find the start
David spends 2/5 of his money on food. He then spends $6 on sweets. He now has half his original amount left. How much did he have at first?
Common denominator for 2/5 and 1/2: tenths is messy, but you can just work in fifths. Whole = 5 units. Food = 2 units. Half = 2½ units. So sweets = 5 − 2 − 2½ = ½ unit = $6.
Draw the original amount as 5 equal units. Subtract: 2 units (food) + $6 (sweets), then he has 5/2 = 2½ units left.
01Originally 5 units. After food (2 units removed): 3 units left. After sweets ($6): only 2½ units remain.
02So $6 = ½ unit, meaning 1 unit = $12.
03Original = 5 units = $60.
David had $60 at first.
Q3Extension · three fractions
Victor has some cookies. He gives 1/4 to his brother, eats 14 himself, then gives the rest to his mother — which is exactly 1/6 of the original total. How many cookies did Victor have originally?
Common denominator for 1/4 and 1/6 is 12. Make the whole = 12 units. Brother = 3 units. Mother = 2 units. The 14 cookies Victor ate are the rest.
Convert: 1/4 = 3/12 and 1/6 = 2/12. So the whole = 12 units.
01Brother: 3 units. Mother: 2 units. Victor ate the rest: 12 − 3 − 2 = 7 units.
027 units = 14 cookies, so 1 unit = 14 ÷ 7 = 2 cookies.
03Total: 12 × 2 = 24 cookies.
Victor had 24 cookies originally.
06
Chapter Six
Distance Problems
When two things move, freeze the moment they take one "step" together. Then count steps.
The idea
If A walks 3 cm "every time" B walks 2 cm, treat that pairing as one step. In a single step the gap closes by 3 + 2 = 5 cm (if moving toward each other) or A pulls ahead of B by 3 − 2 = 1 cm (if same direction). Speed problems also reduce to: distance = speed × time.
Q12023 GEP · ants
Two ants are a distance apart, facing each other. Ant A walks 3 cm every time ant B walks 2 cm. If ant B walks 12 cm before meeting ant A (A is walking too), what was the distance between them at first?
One step = both ants together close 5 cm (B walks 2 + A walks 3). B walked a total of 12 cm. How many "steps" was that for B?
01In one step: gap closes by 3 + 2 = 5 cm. B walks 2 cm per step.
02Number of steps for B: 12 ÷ 2 = 6 steps.
03Total gap closed = total starting distance = 6 × 5 = 30 cm.
They started 30 cm apart.
Q22023 GEP · walkers
A and B are 600 m apart. Every 4 m A walks, B walks 6 m. They start walking towards each other. When they meet, how far has A travelled?
One "step" = both move together, closing 4 + 6 = 10 m. How many steps to close 600 m?
01Per step, gap closes by 4 + 6 = 10 m.
02Number of steps: 600 ÷ 10 = 60.
03A walks 4 m per step, so A walked 60 × 4 = 240 m.
A travelled 240 m.
Q3Extension · A turns around
Two cyclists A and B are some distance apart. For every 3 m A rides, B rides 5 m. They start riding towards each other. After a while, A turns and rides away from B (who keeps going). From the moment A turns, it takes exactly 6 minutes for A and B to meet. In those 6 minutes, A rides 360 m and B rides 600 m. A had been riding for 5 minutes before turning. What was their starting distance?
Find each cyclist's speed (distance ÷ time). Then there are two phases: (1) before A turns — closing in; (2) after A turns — B catches up. Add the distances.
01Speed of A: 360 ÷ 6 = 60 m/min. Speed of B: 600 ÷ 6 = 100 m/min.
02After A turns: both move same direction, so B catches up at 100 − 60 = 40 m/min. In 6 minutes, B catches up 40 × 6 = 240 m. So at the moment A turned, they were 240 m apart.
03Before A turned: they were closing in at 100 + 60 = 160 m/min. For 5 minutes: gap closed = 160 × 5 = 800 m.
04Starting distance = 800 + 240 = 1040 m.
They were 1040 m apart at first.
07
Chapter Seven
Chicken & Rabbit Problems
Two things, two totals. Assume they're all one type, then swap one at a time.
The "assume all one" trick
If 14 items in total made of "Monopoly" and "UNO" cost $300, pretend all 14 are Monopoly. Compute the would-be total. The gap between that and the real $300 tells you how many to swap for UNO. Each swap shifts the total by a fixed amount — the price difference between the two items.
Q12023 GEP · board games
A Monopoly costs $24; a UNO card game costs $18. $300 buys exactly 14 of Monopoly and UNO combined. How many Monopoly games can be bought?
Pretend all 14 are Monopoly. Total would be $336. That's $36 too much. Each "swap" of Monopoly for UNO saves $24 − $18 = $6.
01Assume all 14 are Monopoly: 14 × $24 = $336. That's $336 − $300 = $36 too much.
02Each Monopoly → UNO swap saves $24 − $18 = $6.
03Number of swaps: $36 ÷ $6 = 6 UNO games.
04Monopoly games: 14 − 6 = 8.
8 Monopoly games can be bought.
Q22025 GEP · clothes & buttons
A T-shirt needs 4 buttons; a dress needs 6 buttons. A total of 10 pieces of clothing use 54 buttons. How many T-shirts are there?
Assume all 10 are T-shirts: 40 buttons. That's 14 buttons short. Each T-shirt → dress swap adds 2 buttons.
01Assume all 10 are T-shirts: 10 × 4 = 40 buttons. We need 54 − 40 = 14 more.
A mall held a lucky draw with three prizes: 1st prize $1000, 2nd prize $250, 3rd prize $50. 100 people won prizes; the total prize money was $9500. How many people won the 2nd prize?
Try guessing the number of 1st prizes. If you fix the 1st-prize count, the rest is a 2-type chicken-and-rabbit between 2nd and 3rd. Try 1 first-prize, then 2.
With three types, fix one count and reduce to a 2-type problem.
01Try 1 first-prize: $9500 − $1000 = $8500 left for 99 winners with 2nd and 3rd prizes. Assume all 99 are 3rd prize: 99 × $50 = $4950. Need $8500 − $4950 = $3550 more. Each swap adds $250 − $50 = $200. But $3550 ÷ $200 doesn't divide evenly. Bad guess.
02Try 2 first-prizes: $9500 − $2000 = $7500 left for 98 winners. Assume all 98 are 3rd: 98 × $50 = $4900. Need $7500 − $4900 = $2600 more. Swap value $200 — $2600 ÷ $200 = 13 swaps.
03So 2 first, 13 second, 85 third.
13 people won the 2nd prize.
08
Chapter Eight
Interval Problems
Posts and gaps. The trick is remembering the +1 (or −1) on the count.
The post-and-gap rule
If you plant n trees along a straight road with equal spacing, there are only n − 1gaps between them (the first tree has no gap behind it). Distance = (n − 1) × gap-length. The "off-by-one" trip-wire here catches everyone — always pause and count gaps, not posts.
Q12025 GEP
Distance between every 2 flowers is 20 m. There are 50 flowers along the road. How long is the road?
50 flowers means 49 gaps. Multiply 49 by 20.
Gaps: 50 − 1 = 49. Length: 49 × 20 = 980 m.
The road is 980 m long.
Q2 Around a pond
40 trees are planted at equal distances around a circular pond. The distance between neighbouring trees is 6 m. What is the distance around the pond?
On a straight road, n trees give n − 1 gaps. But on a circle, the last tree's gap closes back to the first tree — so the count of gaps changes.
01The circular trick: on a closed loop, every tree has exactly one gap to its right that ends at the next tree. With no "start" and "end," the number of gaps equals the number of trees.
0240 trees ⇒ 40 gaps of 6 m each. Distance around: 40 × 6 = 240 m.
The pond is 240 m around.
Compare
If those same 40 trees were on a straight road, the road would only be 39 × 6 = 234 m — one gap shorter, because the road has two ends. The +1/−1 trip-wire flips depending on whether the path closes.
Q3Extension · mixed removal
A straight road has lampposts on both sides. Originally, each side had lampposts 12 m apart, with a lamppost at both ends. Later, 3 lampposts on the right side were removed; the left side stayed the same. After removal: 59 lampposts total, and the left has 3 more than the right. How long is the road?
"Total 59, left has 3 more" is a small sum-difference problem (Chapter 1). Solve it first. Then add the 3 back to find the original right-side count.
01Sum-difference: left + right = 59, left = right + 3. So right = (59 − 3) ÷ 2 = 28 and left = 31.
02Originally right side had 28 + 3 = 31 lampposts — same as left side.
0331 lampposts means 30 gaps per side. Road length = 30 × 12 = 360 m.
The road is 360 m long.
09
Chapter Nine
Page-Number Problems
Counting digits, not numbers. Break the range into 1-digit, 2-digit, 3-digit chunks.
Two flavours
How many digits appear in writing the numbers 1 to N? Count each block (1–9 is one digit, 10–99 is two digits, …) separately.
How many times does a specific digit appear (say the digit 8 from 1 to 300)? Count by place: ones, tens, hundreds.
Q12023 GEP
Write out: 1, 2, 3, …, 21, 22. How many digits are there in total?
Split into 1-digit numbers (1–9) and 2-digit numbers (10–22). Count digits in each block separately.
01Numbers 1 to 9: each has 1 digit → 9 × 1 = 9 digits.
02Numbers 10 to 22: that's 13 numbers, each with 2 digits → 13 × 2 = 26 digits.
03Total: 9 + 26 = 35 digits.
There are 35 digits in total.
Q2 Count the 8s
A book has 300 pages numbered 1 to 300. How many times does the digit "8" appear in all the page numbers?
Count the 8s at each position: ones place, tens place, hundreds place. In each block of 100, the digit 8 appears 10 times in the ones place and 10 times in the tens place.
01Ones place: 8, 18, 28, …, 298 — every 10 numbers contains one 8 in the ones place. So 300 ÷ 10 = 30 eights.
02Tens place: 80–89 (10 of them), 180–189, 280–289. Total 10 × 3 = 30 eights.
03Hundreds place: would need 800–899, but we stop at 300. So 0 eights here.
04Total: 30 + 30 + 0 = 60.
The digit 8 appears 60 times.
Q3Extension · count the 0s
In a math textbook the digit "0" has appeared 123 times. How many pages does this book have, at least?
From 1–99, only 9 zeros (in 10, 20, …, 90). Each later century (100–199, 200–299, …) contributes 20 zeros. Find how many full centuries fit, then count what's left in the partial century.
011–99: zeros only in 10, 20, …, 90 — 9 zeros.
02100–199: in 100–109 there are 11 zeros (the 100–109 each have a zero in tens or ones, with 100 having two). Counting: 100, 101–109 each contribute 1 zero (plus 100 has an extra), totalling 11. Then 110–199 contribute 9 zeros (110, 120, …, 190). Total 20.
03Same for 200–299, etc. So after 1–99 we have 9 zeros. Need 123 − 9 = 114 more zeros from blocks of 100s.
04114 ÷ 20 = 5 R 14. So 5 complete blocks (100–599) contribute 100 zeros. We need 14 more zeros from 600 onwards.
05From 600–609: 11 zeros (600 = 2, then 601–609 = 1 each, totalling 2 + 9 = 11). Need 14 − 11 = 3 more. Each of 610, 620, 630 adds 1 zero. So we stop at page 630.
The book has at least 630 pages.
10
Chapter Ten
Worker Problems
Three quantities — workers, time, work done — reduce to two by combining workers × time.
Worker-days (or worker-hours)
If 2 workers labour for 2 hours, that's 4 worker-hours of effort. Combining workers and time into one "effort unit" turns a 3-variable problem into a familiar ratio: effort ↔ work. From there it's just proportion.
Q12022 GEP
2 workers take 2 hours to sew 3 dresses. How many hours does it take for 6 workers to sew 9 dresses?
Compute the worker-hours per dress, then scale up.
029 dresses = 3 × 3 dresses, so we need 4 × 3 = 12 worker-hours.
036 workers share 12 worker-hours: each works 12 ÷ 6 = 2 hours.
It takes 2 hours.
Q2 Two different workers
Anna alone can sort a stack of letters in 6 hours. Ben alone takes 12 hours for the same stack. If they sort the stack together, how many hours does it take?
When two workers have different speeds, a worker-hour from Anna isn't the same as a worker-hour from Ben — you have to compare in letters per hour. Imagine the stack has 12 letters (so both rates come out whole).
Trick: pretend the stack has a convenient number of letters — the LCM of 6 and 12, which is 12 letters. The rates are then whole numbers.
01Anna sorts 12 ÷ 6 = 2 letters per hour. Ben sorts 12 ÷ 12 = 1 letter per hour.
02Together they sort 2 + 1 = 3 letters per hour.
03Time for the whole stack: 12 ÷ 3 = 4 hours.
Together they take 4 hours.
Why Q1's "worker-hour" trick doesn't work here
In Q1 every worker had the same speed, so "2 workers × 2 hours = 4 worker-hours" was a fair count. Here Anna and Ben are different — one of Anna's hours is worth two of Ben's. Convert to a common unit first (letters per hour), then add the rates.
Q3Extension · road repair
Repairing a road takes 24 people exactly 12 days. After 24 people have worked for 4 days, 8 more people are added. How many days from then will it take to finish?
Total worker-days for the whole job = 24 × 12 = 288. Subtract what's done. New team size is 24 + 8 = 32.
A small bag of techniques that don't fit elsewhere — but each rewards a careful drawing.
Q12023 GEP · chairs in a grid
There are 100 chairs in a 10×10 grid. 3 rows and 3 columns are removed. How many chairs were taken away?
Remove the 3 rows first. Now only 7 rows remain. The 3 columns from those 7 rows are shorter than before — only 7 chairs each, not 10.
01Remove 3 full rows: 3 × 10 = 30 chairs.
02Now there are only 7 rows left. Each remaining column has 7 chairs. Removing 3 columns: 3 × 7 = 21 chairs.
03Total taken: 30 + 21 = 51.
51 chairs were removed.
Q22023 GEP · chairs rearranged
There are 10 rows of chairs in a hall. Some rows have 8 chairs and the rest have 9 chairs. They are rearranged into rows of 11 chairs. How many rows are there now?
Total chairs is between 80 (all 8s) and 90 (all 9s). It must be divisible by 11 for the new arrangement to work. Which multiple of 11 lies in that range?
01If all 10 rows had 8 chairs: 10 × 8 = 80. If all had 9: 10 × 9 = 90. So the total is between 80 and 90.
02For rows of 11 to work, the total must be a multiple of 11. The only multiple of 11 between 80 and 90 is 88.
03New rows: 88 ÷ 11 = 8.
There are 8 rows now.
Q32024 GEP · Florida's shopping
Florida has some money. If she buys an umbrella, she has $3 left. If she buys a ball, she has $2 left. If she buys both, she's $17 short. How much does an umbrella cost?
Draw Florida's money as a bar. Both ball and umbrella together = money + $17. After buying just the ball, $2 is left. So the umbrella covers that $2 plus the $17 short.
After buying just the ball, $2 remains. Then to buy the umbrella she still needs $17 more — meaning the umbrella costs the $2 she had left plus the $17 she's short.
01Umbrella = $2 + $17 = $19.
The umbrella costs $19.
Final Thoughts
Five habits that beat any word problem.
Read once for the story, again for the numbers. The numbers stay still while the story moves.
Find the moment you can draw. If "now" is unknown, draw "later." If "before" is unknown, draw "after."
Pick the smallest as 1 unit. Everything else is built up from it, so you stay with whole numbers.
Extra pieces ("+20", "−$5") are glued on, not units. Strip them off before counting units.
Check by retelling. Substitute your answer back. Every sentence in the problem should come out true.
Bars and arithmetic are not different math. They're the same idea wearing different costumes. Draw the picture. Then count.