This is the marking copy and it carries every answer. Her sheet is the other file — same numbering, same margin codes, so you can mark straight down the page.
A road divider is painted in repeating segments: white 2 m, black 5 m, white 2 m, black 5 m, and so on, beginning and ending with white. If the road divider is 149 m long, how many white segments are there?
[white 2m][ black 5m ][white 2m][ black 5m ][white 2m] ... |<----- one repeat = 7 m ----->| the whole divider is 149 m
answer 22
method One repeat is a white and a black together: 2 + 5 = 7 m. In 149 m: 149 ÷ 7 = 21 repeats with 2 m left over. The 21 repeats give 21 white segments, and the last 2 m is exactly one more white segment. So 21 + 1 = 22.
watch Answering 21 and ignoring the 2 m remainder — which is precisely the length of one more white segment, and is why the divider ends white.
Bob wrote letters in a repeating pattern: Z E S T Z E S T Z E S T Z E … How many letters ‘Z’ and ‘T’ are there altogether if there is a total of 87 letters in the whole series?
Z E S T Z E S T Z E S T Z E ... |<- one repeat = 4 letters ->| 87 letters altogether
answer 43
method One repeat is Z E S T — 4 letters. In 87 letters: 87 ÷ 4 = 21 repeats with 3 left over. The 21 repeats give 21 Zs and 21 Ts. The 3 leftover letters restart the pattern as Z, E, S — one more Z and no more T. So Z = 22 and T = 21, giving 22 + 21 = 43.
watch Splitting the leftovers evenly, or counting a T in the remainder. The leftover always starts from the BEGINNING of the pattern, so it reaches Z before it reaches T.
A length of sticky tape is made up of repeated designs: a 5 cm striped band, then a 3 cm checked band, then a 4 cm grid band, repeating. The sticky tape is 80 cm long. How many checked bands are there altogether?
| striped | checked | grid | striped | checked | grid | ... |<- 5 cm ->|<- 3 cm ->|<-4cm->| |<-------- one repeat = 12 cm -------->| the whole tape is 80 cm
answer 7
method One full repeat is 5 + 3 + 4 = 12 cm. In 80 cm: 80 ÷ 12 = 6 repeats with 8 cm left over. Those 6 repeats give 6 checked bands. The leftover 8 cm starts the next repeat: 5 cm of stripes, then 3 cm of checks — exactly enough for one more complete checked band. So 6 + 1 = 7.
watch Answering 6 and ignoring the leftover. Always ask what the remaining centimetres are long enough to reach.
Some circles and triangles are arranged in a repeating pattern: triangle, circle, circle, circle, triangle, circle, circle, circle, triangle, … How many circles are there if there are 122 shapes?
/\ O O O /\ O O O /\ ... |<- one repeat = 4 shapes ->| (1 triangle and 3 circles)
answer 91
method One repeat is a triangle and 3 circles — 4 shapes. In 122 shapes: 122 ÷ 4 = 30 repeats with 2 shapes left over. The 30 repeats give 30 × 3 = 90 circles. The 2 leftover shapes continue the pattern: a triangle, then a circle. So 90 + 1 = 91 circles.
watch Answering 90 and forgetting the leftovers, or counting the leftover 2 shapes as 2 circles. The pattern always restarts with a TRIANGLE.
There are 17 pupils in a class. The teacher gets each pupil to give a hi-five to every classmate once. How many hi-fives are exchanged altogether?
answer 136
method Each pupil hi-fives the other 16, which suggests 17 × 16 = 272. But a hi-five needs two people, so every one has been counted twice — once from each side. The real number is 272 ÷ 2 = 136.
watch Answering 272 and forgetting to halve. A hi-five between Amy and Ben is the SAME hi-five, however you look at it.
Jean is watching a movie in a hall. The chairs are arranged in rows with the same number, and all are occupied. There are 7 people behind Jean and 4 people in front of her. There are 3 people on her left and 4 people on her right. How many people are watching the movie?
7 behind
|
3 left --- JEAN --- 4 right
|
4 in front
(towards the screen)answer 96
method Count Jean's own column: 7 behind + 4 in front + Jean herself = 12 rows. Count her row: 3 on the left + 4 on the right + Jean herself = 8 seats across. Every seat is taken, so the hall holds 12 × 8 = 96 people.
watch Forgetting to add Jean herself to each count, giving 11 × 7 = 77. She is in her own row and her own column.
Study the staircase pattern: Figure 1 has 2 squares along the bottom and 1 above the right-hand one (3 squares). Figure 2 has 3 along the bottom, then 2, then 1 (6 squares). Figure 3 has 4, then 3, then 2, then 1 (10 squares). Find the total number of squares needed to form Figure 50.
Figure 1 Figure 2 Figure 3
[]
[] [] [][]
[][] [][] [][][]
[][][] [][][][]
3 6 10 squaresanswer 1326
method Each figure is a staircase. Figure 1 is 2 + 1 = 3, Figure 2 is 3 + 2 + 1 = 6, Figure 3 is 4 + 3 + 2 + 1 = 10. So Figure n counts down from (n + 1) to 1. Figure 50 is 51 + 50 + 49 + … + 1. Pair the ends: 51 + 1 = 52, 50 + 2 = 52, and so on — there are 51 numbers, giving 51 × 52 ÷ 2 = 1326.
watch Counting down from 50 instead of 51. Figure 1's bottom row has 2 squares, not 1, so the bottom row of Figure 50 has 51.
Sam had three coins in his wallet. They could be 10-cent coins, 20-cent coins or 50-cent coins. How many different possible amounts of money could Sam have in his wallet?
answer 10
method List the ways to choose 3 coins from the three kinds, working down in an order so none is missed: three 10s = 30c; two 10s and a 20 = 40c; two 10s and a 50 = 70c; one 10 and two 20s = 50c; one of each = 80c; one 10 and two 50s = 110c; three 20s = 60c; two 20s and a 50 = 90c; one 20 and two 50s = 120c; three 50s = 150c. That is 10 combinations, and all 10 totals are different, so there are 10 possible amounts.
watch Counting the ORDER of the coins as different (10-10-20 and 10-20-10 are the same wallet), or stopping before all ten combinations are found. A written-down order is what stops one being missed.
Mrs Bala wanted to sew 6 ribbons on each side of a square handkerchief. There was a ribbon at each corner of the handkerchief. How many ribbons did she sew in total?
R--R--R--R--R--R | | 6 ribbons along EACH side, R R one at each corner | | R--R--R--R--R--R
(A) 18 (B) 20 (C) 24 (D) 36
answer (B) — 20
method Counting 6 ribbons on each of the 4 sides gives 4 × 6 = 24 — but each corner ribbon has been counted twice, once for each side it sits on. There are 4 corners, so take 4 off: 24 − 4 = 20 ribbons.
watch Answering 24 and forgetting that a corner ribbon belongs to two sides at once. Mark the corners on a quick sketch before multiplying.
Jamie decorated a square classroom of side 6 m 40 cm. She tied balloons in a repeating pattern on a string and hung the string around the classroom once. In every 80 cm of string the pattern holds 4 small balloons (with big balloons between them). How many small balloons did she use to decorate the 4 sides of the classroom?
(BIG) o o (BIG) o o (BIG) |<--------- 80 cm --------->| o = small balloon · 4 small balloons in every 80 cm
answer 128
method Find how much string is needed: the classroom is a square of side 6 m 40 cm = 640 cm, so once round is 4 × 640 = 2560 cm. The pattern repeats every 80 cm, so it repeats 2560 ÷ 80 = 32 times. Each repeat carries 4 small balloons, so she used 32 × 4 = 128 small balloons.
watch Working out the balloons for one side and forgetting to multiply by 4, or counting the big balloons too. The question asks only for the SMALL ones.
A farmer planted some trees in a straight row, at an equal distance apart from one another. The distance between the 2nd tree and the 5th tree was 2340 m. What was the distance between the 1st tree and the 10th tree?
T1 T2 T3 T4 T5 ... T10
|<--------------->|
2340 m
the trees are equally spaced
(count the SPACES between them, not the trees)answer 7020
method Count GAPS, not trees. From the 2nd tree to the 5th there are 5 − 2 = 3 gaps, so one gap = 2340 ÷ 3 = 780 m. From the 1st tree to the 10th there are 10 − 1 = 9 gaps, so the distance = 9 × 780 = 7020 m.
watch Counting trees instead of gaps — using 4 trees and 10 trees instead of 3 gaps and 9 gaps. Mark the gaps on a quick sketch before dividing.
John divided the corridor of a school building into equal parts of length 4 m, and placed 2 potted plants in each part. For the same corridor, he divided it into equal parts of length 6 m and hung 5 lanterns in each part. If there were 24 more lanterns than potted plants, how long was the corridor?
Figure 1 |<-- 4 m -->| 2 potted plants in each part Figure 2 |<--- 6 m --->| 5 lanterns in each part
answer 72
method The two patterns use different part lengths, so compare them over a length both divide — 12 m. In 12 m there are 12 ÷ 4 = 3 parts of plants, giving 3 × 2 = 6 plants, and 12 ÷ 6 = 2 parts of lanterns, giving 2 × 5 = 10 lanterns. That is 10 − 6 = 4 more lanterns for every 12 m. We need 24 more, so the corridor is 24 ÷ 4 = 6 lots of 12 m = 72 m. (Check: 72 m gives 18 × 2 = 36 plants and 12 × 5 = 60 lanterns, and 60 − 36 = 24 ✓)
watch Comparing 2 plants against 5 lanterns directly and using a difference of 3. The parts are different LENGTHS, so the counts can only be compared over the same distance.
How many rectangles are there in the figure below?
┌───┬───┬───┐ │ │ │ │ └───┴───┴───┘
(A) 3 (B) 4 (C) 5 (D) 6 (E) 9
answer (D) — 6
method A rectangle is made by picking 2 of the up-and-down lines and 2 of the across lines. There are 4 up-and-down lines, giving 3 + 2 + 1 = 6 pairs, and only 2 across lines, giving 1 pair. So 6 × 1 = 6. (Counting by size agrees: 3 single squares + 2 doubles + 1 triple = 6.)
How many rectangles are there in the figure below?
┌───┬───┐ │ │ │ ├───┼───┤ │ │ │ └───┴───┘
(A) 4 (B) 6 (C) 8 (D) 9 (E) 12
answer (D) — 9
method 3 up-and-down lines give 2 + 1 = 3 pairs; 3 across lines give 3 pairs. 3 × 3 = 9. The four little squares are only the start — there are also 2 wide ones, 2 tall ones, and the whole square itself.
How many ways are there from A to B, moving only up or right along the lines?
┌───┬───┐ B
│ │ │
├───┼───┤
│ │ │
A └───┴───┘(A) 4 (B) 5 (C) 6 (D) 8 (E) 12
answer (C) — 6
method Write 1 at A and fill in every dot as below + left. Bottom row: 1, 1, 1. Middle row: 1, 2, 3. Top row: 1, 3, 6. B says 6.
How many ways are there from A to B, moving only up or right along the lines?
┌───┬───┐ B
│ │ │
├───┼───┤
│ │
A └───┘(A) 4 (B) 5 (C) 6 (D) 8 (E) 10
answer (B) — 5
method Same rule, but the shape has a bump: the dot at the far right of the middle row has no line coming up from below, so that side gives 0 and it just copies the 2 from its left. Filling in: A = 1, the dot right of A = 1, up the left = 1, then 1 + 1 = 2, then 2. Top row: 1, then 1 + 2 = 3, then 3 + 2 = 5. B says 5. (Treating it as a plain 2-by-2 grid gives 6 — that is the trap.)
How many squares are there in the figure below?
┌───┬───┬───┐ │ │ │ │ ├───┼───┼───┤ │ │ │ │ ├───┼───┼───┤ │ │ │ │ └───┴───┴───┘
(A) 9 (B) 13 (C) 14 (D) 22 (E) 36
answer (C) — 14
method Count them by size: 9 small 1-by-1 squares, 4 of the 2-by-2 squares, and the whole 3-by-3 square itself. 9 + 4 + 1 = 14. (36 is the number of RECTANGLES in this figure — a different question.)
In the same 3-by-3 figure as the question above, how many rectangles are there that are NOT squares?
┌───┬───┬───┐ │ │ │ │ ├───┼───┼───┤ │ │ │ │ ├───┼───┼───┤ │ │ │ │ └───┴───┴───┘
(A) 9 (B) 14 (C) 22 (D) 27 (E) 36
answer (C) — 22
method All the rectangles first: 4 up-and-down lines give 3 + 2 + 1 = 6 pairs, and the same across, so 6 × 6 = 36. Of those, 14 are squares (9 + 4 + 1). The ones that are not squares are 36 − 14 = 22. Read the last line of the question carefully — 36 and 14 are both right answers to a different question, and both are sitting in the options.