Answer key and worked explanations · P4 · 45 questions · 100 marks · 90 minutes
Mixed provenance. Each question is tagged OFFICIAL or DERIVED — trust them differently.
Each question carries the method under Why, and the mistake to watch for under Where she’ll slip. The slip is the useful half: it is what to ask about before she starts writing.
Write the missing number in the number pattern below: 7648, 7798, 7948, __________, 8248
8098
Find the step first: 7798 − 7648 = 150, and 7948 − 7798 = 150 as well. So the pattern adds 150 each time: 7948 + 150 = 8098. (Check forwards: 8098 + 150 = 8248 ✓)
Guessing from the last gap only. Check the step between at least two pairs, then check your answer reaches the number after it.
Round the sum of 18.208 and 1.42 to the nearest whole number.
20
Add first, lining up the decimal points: 18.208 + 1.420 = 19.628. Then round to the nearest whole number — the digit after the point is 6, which is 5 or more, so round up to 20.
Rounding each number first (18 + 1 = 19). The question says round the SUM, and rounding early loses the 0.628 that pushes it over.
Find the sum of all the factors of 81.
121
List the factors in pairs: 1 × 81, 3 × 27, 9 × 9. So the factors are 1, 3, 9, 27 and 81 — the 9 appears only once because it pairs with itself. Sum = 1 + 3 + 9 + 27 + 81 = 121.
Writing 9 twice because 9 × 9 = 81. A factor is listed once however many times it appears in a pair.
David went to sleep at 21 50. He slept for 9 h 20 min. At what time did he wake up? (Give your answer in 24-hour clock format.)
0710
Add the hours first: 21 50 + 9 h = 30 50, which is past midnight, so take off 24 h to get 06 50 the next morning. Then add the 20 min: 06 50 + 20 min = 07 10.
Writing 30 50, or adding the minutes as if 50 + 20 made 70. Sixty minutes make an hour, so 50 + 20 rolls over into the next hour.
I am a 3-digit odd number. All my three digits are different. All my three digits are multiples of 3. The digit in the ‘hundreds’ place is twice the digit in the ‘ones’ place. What number am I?
693
The digits that are multiples of 3 are 3, 6 and 9. The number is odd, so the ones digit is 3 or 9. If the ones digit were 9, the hundreds digit would be 18 — not a digit. So the ones digit is 3 and the hundreds digit is 2 × 3 = 6. The tens digit must be a different multiple of 3, which leaves 9. The number is 693.
Counting 0 as a multiple of 3 and offering 603 as well. In primary work the multiples of 3 are 3, 6, 9, … — they start at 3, not 0.
Using a protractor, measure ∠ABC. BA runs horizontally to the left from B; BC rises steeply up and slightly to the right.
C
/
/
/
/
A ---------------------/ B
(angle here)
measure from BA, round to BC107
Put the protractor's centre on B with its baseline along BA, then read where BC crosses the scale. BC leans back past the upright, so the angle is obtuse — a little over 90°, not under it. The reading is 107°.
Reading the wrong scale on the protractor and getting 73° instead. The two scales run in opposite directions, so decide FIRST whether the angle is acute or obtuse and pick the reading on that side of 90°. (A reading a degree or two either way is a fair measurement; the published key gives 107.)
Mary gave 55 beads to Betty. Then she received 36 beads from Lydia. Mary had 342 beads in the end. How many beads did Mary have at first?
361
Work backwards, undoing each step. Before receiving 36 she had 342 − 36 = 306. Before giving away 55 she had 306 + 55 = 361.
Running the story forwards from 342, which adds the 55 and subtracts the 36 the wrong way round. Going backwards means every give becomes a take and every take becomes a give.
A man changed $210 into $5 and $2 notes. The number of $5 notes and $2 notes was the same. How many $5 notes did he get?
30
Because the numbers are equal, take one of each together: $5 + $2 = $7 per pair. Then $210 ÷ $7 = 30 pairs, so he got 30 five-dollar notes (and 30 two-dollar notes).
Dividing $210 by $5 alone. The pairing is what makes the two equal counts usable in one division.
The line graph shows the number of books borrowed by students in a library: March 500, April 600, May 800, June 1100, July 800. How many more books were borrowed in July than in March?
Mar Apr May Jun Jul 500 600 800 1100 800
300
Read July off the graph: 800. Read March: 500. "How many more" means subtract: 800 − 500 = 300.
Reading the peak (June, 1100) instead of July. Follow the month labels along the bottom, not the highest point.
Using the same line graph (March 500, April 600, May 800, June 1100, July 800), what was the total number of books borrowed from March to June?
Mar Apr May Jun Jul 500 600 800 1100 800 |<---- March to June ---->|
3000
Add the four months named — March, April, May and June: 500 + 600 + 800 + 1100 = 3000 books. July is not included, because the question stops at June.
READ THIS ONE BEFORE TEACHING IT. The published answer key says 3800, which is the total of all FIVE months including July. For the question as printed — March to June — the answer is 3000. Check the wording with her rather than the key: if she answers 3000 she has read the question correctly.
The box below contains a mixture of circles and stars: 4 stars and 6 circles. ?/5 of the shapes in the box are stars. What is the missing number?
+--------------------------+ | * O O * O | | O O * O * | +--------------------------+ 4 stars, 6 circles, 10 shapes in all
2
Count first: 4 stars out of 10 shapes, so the fraction is 4/10. The answer must be written in fifths, so simplify by dividing top and bottom by 2: 4/10 = 2/5. The missing number is 2.
Answering 4 — that is the number of stars, not the numerator once the fraction is written in fifths.
The figure is made up of 5 identical squares arranged as a cross of diamonds, each outer square joined to the middle one along a full side. The perimeter of the figure is 132 cm. What is the area of each square?
/\ /\
/ \ / \
/ \ / \
\ \/ /
\ /\ /
\ / \ /
\/ \/
/\ /\
/ \ / \
/ \/ \
\ /\ /
\ / \ /
\/ \/
5 identical squares · the middle one shares all 4 of its sides121
Five squares have 5 × 4 = 20 sides altogether. The middle square shares each of its 4 sides with an outer square, so 4 + 4 = 8 of those sides are tucked inside the figure and do not show. That leaves 20 − 8 = 12 sides on the outside. So 12 sides = 132 cm, one side = 132 ÷ 12 = 11 cm, and the area is 11 × 11 = 121 cm².
Dividing 132 by 20 (all the sides) or by 4 (one square). Only the sides on the OUTSIDE make up the perimeter — count how many are hidden first.
Andy bought 4 identical shirts at $12.80 each and spent the rest of the money on 8 identical pairs of socks. He gave $100 to the cashier and received 80 cents change. How much did each pair of socks cost?
6
He spent $100 − $0.80 = $99.20 in total. The shirts cost 4 × $12.80 = $51.20, so the socks cost $99.20 − $51.20 = $48. There are 8 pairs, so each pair is $48 ÷ 8 = $6.
Treating 80 cents as $80, or forgetting to take the change off the $100 before splitting the rest.
Siti is 17 years old and Julia is 5 years old. How old will Julia be when she is 3/5 as old as Siti?
18
The gap between their ages never changes: 17 − 5 = 12 years, now and forever. At the moment we want, Julia is 3 units and Siti is 5 units, so the gap is 5 − 3 = 2 units. So 2 units = 12 years, 1 unit = 6 years, and Julia (3 units) is 3 × 6 = 18.
Taking 3/5 of 17 straight away and getting 10.2. Both of them get older, so the fraction applies to their ages LATER, not to Siti's age now — the fixed gap is what pins the moment down.
The number of girls is 3/7 the number of boys at a National Kids’ Run competition. There are 960 more boys than girls. How many children took part in the competition?
2400
Draw boys as 7 units and girls as 3 units. The difference is 7 − 3 = 4 units = 960, so 1 unit = 960 ÷ 4 = 240. Altogether there are 7 + 3 = 10 units, so 10 × 240 = 2400 children.
Answering 1680 (the boys) or 720 (the girls). The question asks for everyone, which is all 10 units.
Mrs Thompson baked 175 tarts. She sold 3/5 of them. Each tart was sold for $4. How much money did she collect in total?
420
Find how many she sold: 1/5 of 175 is 35, so 3/5 is 3 × 35 = 105 tarts. Each sold for $4, so she collected 105 × $4 = $420.
Multiplying all 175 tarts by $4. She only sold three fifths of them — the unsold ones brought in nothing.
In the diagram, ABEF is a square and BCDE is a rectangle. The length of AF is twice the length of BC and AB = 56 cm. Find the length of ED.
A B C +--------------------+---------+ | | /| | square | rect / | | ABEF | / | +--------------------+---------+ F E D AB = 56 cm · AF = 2 x BC
28
ABEF is a square, so all four of its sides are equal: AF = AB = 56 cm. We are told AF is twice BC, so BC = 56 ÷ 2 = 28 cm. In rectangle BCDE, ED is the side opposite BC, and opposite sides of a rectangle are equal — so ED = 28 cm.
Doubling instead of halving, giving 112. "AF is twice BC" makes BC the SMALLER one.
Using the same diagram (ABEF a square, BCDE a rectangle, with A, B, C on one straight line and F, E, D on another), ∠ACF = 35° and ∠DCE = 18°. Find ∠FCE.
A B C
+--------------------+---------+
\ \35 /|
\ \ / | 18
\ X |
+------\---------------------\-+
F (to F) E D
at C the whole angle from CA down to CD is 90 degrees37
At corner C the angle between the top line (towards A) and the side CD is a right angle, 90°. Three angles sit inside it, side by side: ∠ACF = 35°, then ∠FCE, then ∠ECD = 18°. So 35 + ∠FCE + 18 = 90, giving ∠FCE = 90 − 53 = 37°.
Adding the two given angles and stopping at 53, or subtracting from 180 instead of 90. The corner of a rectangle is 90°.
Jeff is 12 years older than Leon now. 10 years ago, the total age of Jeff and Leon was 78 years. How old is Leon now?
43
Ten years ago both were 10 years younger, so their ages today total 78 + 10 + 10 = 98. Jeff is 12 more than Leon, so take the 12 off and split what is left equally: (98 − 12) ÷ 2 = 86 ÷ 2 = 43. Leon is 43 now.
Adding only 10 to the 78 instead of 10 for EACH of them, or answering 55 (Jeff's age).
Shirley packed 163 lollipops into plastic bags of 7 lollipops. She used the most number of plastic bags possible. She sold each plastic bag of lollipops at $6 and each remaining lollipop at $2. How much did she get after selling all the lollipops?
142
Divide to find the bags: 163 ÷ 7 = 23 bags with 2 lollipops left over. The bags bring in 23 × $6 = $138, and the 2 loose lollipops bring in 2 × $2 = $4. Total = $138 + $4 = $142.
Throwing away the remainder. The 2 left-over lollipops are still sold — at a different price, which is exactly why the remainder matters here.
Melody baked some muffins. 2/7 of the muffins were blueberry muffins and the rest were chocolate muffins. There were 72 more chocolate muffins than blueberry muffins. How many muffins did she bake altogether?
168
Blueberry is 2 units out of 7, so chocolate is the other 5 units. The difference is 5 − 2 = 3 units = 72 muffins, so 1 unit = 24. Altogether there are 7 units: 7 × 24 = 168 muffins.
Treating 72 as one of the fractions rather than as the DIFFERENCE between them, or answering 120 (the chocolate ones).
A glue stick cost $1.80. A pair of scissors cost $3.20. Thomas bought an equal number of glue sticks and scissors. He spent $25 in all. How many glue sticks and scissors did he buy in total?
10
Buy them in pairs, since the numbers are equal: one glue stick and one pair of scissors cost $1.80 + $3.20 = $5. Then $25 ÷ $5 = 5 pairs. Each pair is 2 items, so in total he bought 5 × 2 = 10 items.
Answering 5 — that is the number of PAIRS. The question asks for glue sticks and scissors added together.
Gupta boarded the bus at 11.40 a.m. The bus ride took 38 minutes. After alighting from the bus, he walked for 28 minutes from the bus stop to his school. What time did he reach his school? (Give your answer in 24-hour clock format.)
1246
Add the two journeys to 11.40 a.m. A bus ride of 38 min: 11.40 + 20 min reaches 12.00 noon, and 18 min more gives 12.18 p.m. Then walk 28 min: 12.18 + 28 min = 12.46 p.m. In 24-hour clock that is 1246.
Writing 0046 or 1146 by mishandling the crossing of 12 noon, or converting to 24-hour clock by adding 12 to a time that is already past noon.
The 8-point compass shows the houses of 8 children around point O: Perry north, Ziming north-east, Bill east, Hadi south-east, Ali south, Sam south-west, Dave west, Shawn north-west. Roy is at point O facing Perry’s house. How many degrees must he turn in a clockwise direction so that he can face Shawn’s house?
Perry
Shawn | Ziming
\ | /
Dave -------O------- Bill
/ | \
Sam | Hadi
Ali315
Each step round an 8-point compass is 360 ÷ 8 = 45°. Starting at Perry (north) and going CLOCKWISE, count the steps to Shawn (north-west): Ziming, Bill, Hadi, Ali, Sam, Dave, Shawn — that is 7 steps. So the turn is 7 × 45 = 315°.
Turning the short way and answering 45°. That is anti-clockwise; the question asks for clockwise, which is the long way round.
Tania had 156 stickers and Amelia had 114 stickers. After each of them gave away an equal number of stickers, Tania had 4 times as many stickers as Amelia. How many stickers did each of them give away?
100
They give away the same number, so the gap between them never changes: 156 − 114 = 42. At the end Tania is 4 units and Amelia is 1 unit, so the gap is 3 units = 42, giving 1 unit = 14. So Amelia ended with 14 stickers, and she gave away 114 − 14 = 100.
Answering 14 (what Amelia has LEFT) instead of what she gave away. The unchanging gap is the key — it is what makes the 42 usable.
A rectangular piece of paper is folded as shown below, with both top corners folded down. The flat top that is left is 7 cm across, each folded corner covers 6 cm, and the height of the figure is 23 cm. What is the area of the piece of paper at first?
- - - - - +-------------+ - - - - - -+ | /| 7 cm |\ | | | / | | \ | | +----/----+-------------+----\----+ | |<-6cm->| |<-6cm->| 23 cm | | | | | | | | | +---------------------------------+ -+ the dashed lines show where the corners came from
437
The dashed lines show the paper's original top edge, which runs the full width: 6 + 7 + 6 = 19 cm. The folds only turned the corners down — they did not shorten the sheet — so the height is still 23 cm. Area = 19 × 23 = 437 cm².
Using 7 cm as the width because that is the top edge you can see. The folded corners are part of the same sheet, so their 6 cm each still counts.
The table shows the boys and girls in four Primary Four classes, with some cells blank: 4A has 22 boys and 18 girls; 4B has 25 girls and 41 students in total; 4C has 13 boys; 4D has 20 girls. There are as many students in Primary 4A as in Primary 4C. How many girls are there in Primary 4C?
Class | Boys | Girls | Total ------+------+-------+------ 4A | 22 | 18 | 4B | | 25 | 41 4C | 13 | | 4D | | 20 |
27
Fill in 4A first: 22 + 18 = 40 students. We are told 4C has as many students as 4A, so 4C also has 40. Of those, 13 are boys, so the girls number 40 − 13 = 27.
Comparing 4A's BOYS with 4C's boys. The sentence says as many STUDENTS, which means the totals match, not the boys.
Using the same table, the four classes have a total of 160 students. How many students are from the class that has the least number of students?
4A = 22 + 18 = 40 4B = 41 (given) 4C = 40 (same as 4A) 4D = ? all four classes together = 160
39
Three of the classes are now known: 4A = 40, 4B = 41, 4C = 40, which is 121 students. So 4D = 160 − 121 = 39. Comparing 40, 41, 40 and 39, the smallest is 4D with 39 students.
Stopping once 4D is found without checking it really is the smallest, or picking 4C's 13 boys as "the least". The question asks about whole classes.
ABCD is a square of side 8 cm, cut out of a 28 cm by 12 cm rectangle. What is the area of the shaded part (the rectangle outside the square)?
|<----------------- 28 cm ----------------->| +###########################################+ -+ ###########+-----------+##################### | ###########| A B |##################### | ###########| 8 cm |##################### 12 cm ###########| D C |##################### | ###########+-----------+##################### | +###########################################+ -+
272
Find the whole rectangle: 28 × 12 = 336 cm². Find the square: 8 × 8 = 64 cm². The shaded part is everything except the square, so 336 − 64 = 272 cm².
Taking 8 cm as the square's AREA instead of its side, or forgetting to subtract the square at all.
The figures show a square and a rectangle. Square X has sides of 10 cm. Rectangle Y is 2 cm wide. The area of Square X is 4 times the area of Rectangle Y. Find the perimeter of Rectangle Y.
Square X Rectangle Y +----------+ +--+ | | | | | 10 cm | | | 2 cm wide | | | | +----------+ +--+
29
Square X has area 10 × 10 = 100 cm². Rectangle Y is a quarter of that: 100 ÷ 4 = 25 cm². Its width is 2 cm, so its length is 25 ÷ 2 = 12.5 cm. Perimeter = 2 × (12.5 + 2) = 2 × 14.5 = 29 cm.
Multiplying by 4 instead of dividing. Square X is the BIGGER one, so Rectangle Y's area must come out smaller.
Find the perimeter of the I-shaped figure below. All lines meet at right angles. The figure is 14 cm across and 16 cm tall; the top and bottom bars are each 5 cm deep, and the waist is set in 3 cm on the left and 4 cm on the right.
|<-------- 14 cm -------->|
+-------------------------+ -+
| | | 5 cm
+----+---------------+----+ -+
|3cm | | 4cm|
| | 16 cm
+----+---------------+----+
| | | 5 cm
+-------------------------+ -+74
For a shape like this with no overhangs, start with the rectangle that just surrounds it: 2 × (14 + 16) = 60 cm. Then each notch cut into the side adds twice its depth, because you walk in and back out again: the left notch adds 2 × 3 = 6 cm and the right notch adds 2 × 4 = 8 cm. Perimeter = 60 + 6 + 8 = 74 cm.
Assuming the notches make the perimeter SMALLER. Cutting a bite out of the side removes no edge — it replaces one straight edge with three, so the perimeter grows.
A bookshop had a total of 160 erasers and pencils. After 28 erasers and 24 pencils were sold, the number of erasers became thrice the number of pencils. How many pencils did the bookshop have at first?
51
After the sale, 160 − 28 − 24 = 108 items were left. At that point erasers were 3 units and pencils 1 unit, so 4 units = 108 and 1 unit = 27 — that is the pencils left. Before the sale the shop had 27 + 24 = 51 pencils.
Answering 27, the pencils remaining. The 24 sold pencils have to be added back to reach "at first".
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
22
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.
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.
There are 256 balls in a box. 1/4 of the balls are red and 5/8 of them are white. The rest are black. How many more white balls than black balls are there?
128
Red = 1/4 of 256 = 64. White = 5/8 of 256 = 5 × 32 = 160. Black is what is left: 256 − 64 − 160 = 32. So there are 160 − 32 = 128 more white balls than black.
Answering 160 (the white balls) instead of the difference, or adding 1/4 and 5/8 as 6/12 by adding tops and bottoms.
John spilled some paint on his report card, hiding his Mathematics and Science scores. English was 86, Chinese was 80, and the total of all four subjects was 342. He scored 12 fewer marks in Science than in Mathematics. How many marks did he score for Mathematics?
Subject | Score -------------+------- English | 86 Chinese | 80 Mathematics | ##### Science | ##### Total | 342
94
Maths and Science together = 342 − 86 − 80 = 176. Science is 12 fewer than Maths, so take that 12 off and split the rest equally: (176 − 12) ÷ 2 = 164 ÷ 2 = 82 — that is Science. Maths = 82 + 12 = 94.
Splitting the 176 evenly into 88 and 88 and forgetting the 12, or adding the 12 to the wrong subject. Science is the SMALLER one.
Charles bought 9 bags of sweets. Each bag had 48 sweets. He gave away 5/6 of all the sweets. How many sweets did he have left?
72
Find the total first: 9 × 48 = 432 sweets. He gave away 5/6, so he kept the other 1/6: 432 ÷ 6 = 72 sweets.
Working out 5/6 of 432 (= 360) and giving that as the answer. That is what he gave AWAY; the question asks what is left.
Study the pattern of hexagons made from sticks: Pattern 1 is one hexagon (6 sticks), Pattern 2 is two hexagons sharing a side (11 sticks), Pattern 3 is three in a row (16 sticks). In a particular pattern, 206 sticks are used. What is the pattern number?
Pattern 1 Pattern 2 Pattern 3
/ \ / \ / \ / \ / \ / \
| | | | | | | | |
\ / \ / \ / \ / \ / \ /
6 sticks 11 sticks 16 sticks41
The first hexagon takes 6 sticks. Every hexagon after that shares a side with the one before, so it only needs 5 more. Sticks = 6 + 5 × (pattern number − 1). Setting that to 206: 206 − 6 = 200 extra sticks, and 200 ÷ 5 = 40 more hexagons, so the pattern number is 40 + 1 = 41.
Dividing 206 by 6 because each hexagon "has 6 sides". After the first, each new hexagon adds only 5 sticks — the shared side is already there.
The figure is made up of 4 identical rectangles: two standing upright side by side (22 cm tall, 14 cm across the pair) resting centrally on two more lying end to end to form the base. Find the length of XY, the part of the base sticking out to the right of the upright pair.
|<- 14 cm ->|
+-----+-----+ -+
| | | |
| | | 22 cm
| | | |
+-----+-----+-----+----+----+ -+
| | X----Y |
+-----------+---------------+ -+15
The two uprights together are 14 cm across, so each rectangle is 14 ÷ 2 = 7 cm wide, and each is 22 cm long. The base is two of the same rectangles laid end to end, so it is 22 + 22 = 44 cm long. The upright pair sits centrally, so the base sticks out equally at both ends: (44 − 14) ÷ 2 = 30 ÷ 2 = 15 cm.
Forgetting that all four rectangles are IDENTICAL, so the base's length comes from the uprights' 22 cm. Or taking the whole 30 cm overhang instead of halving it between the two sides.
Kate and Sam had the same number of books at first. Kate gave away 2/3 of her books while Sam gave away 1/6 of his books. The number of books that Kate gave away was 63 more than Sam. Find the number of books that each of them had at first.
126
They started with the same number, so both fractions are of the same amount. Kate gave 2/3 and Sam gave 1/6. In sixths that is 4/6 and 1/6, so Kate gave 3/6 — that is half — more than Sam. If half the books is 63, the whole is 63 × 2 = 126 books each.
Subtracting the fractions as 2/3 − 1/6 = 1/3 by taking the bottoms away too. Rewrite both in sixths before subtracting.
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
43
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.
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.
The mass of a container filled with 2 identical ping pong balls was 80 g. The mass of the same container filled with 3 identical marbles was 220 g. The mass of each marble was 3 times the mass of each ping pong ball. What was the mass of 1 ping pong ball?
20
Swap the marbles for ping pong balls: each marble weighs the same as 3 ping pong balls, so 3 marbles weigh the same as 9 ping pong balls. Now both weighings hold the same container: container + 2 balls = 80 g, and container + 9 balls = 220 g. Subtracting removes the container: 7 balls = 220 − 80 = 140 g, so one ball is 140 ÷ 7 = 20 g.
Dividing 80 by 2 and forgetting the container has a mass of its own. Subtracting the two weighings is what makes the container disappear.
Mrs Sim wanted to buy 20 glass bowls but she was short of $8. She then bought 19 glass bowls and had $6 left. Each glass bowl cost the same. How much money did Mrs Sim have at first?
272
Compare the two plans. Buying one bowl fewer took her from being $8 short to having $6 spare — a swing of 8 + 6 = $14. That swing is the price of exactly one bowl, so a bowl costs $14. She bought 19 of them and had $6 left, so she started with 19 × $14 + $6 = $266 + $6 = $272.
Subtracting the 8 and the 6 instead of adding them. Being short and having spare are on opposite sides of zero, so the gap between the two plans is their sum.
Lisa and Max each had a roll of ribbon with the same length. They cut their roll of ribbon into shorter pieces. Each piece of Lisa’s ribbon was 6 cm. Each piece of Max’s ribbon was 10 cm. After cutting, Lisa had 16 more pieces of ribbon than Max. What was the length of each roll of ribbon?
240
Look at a length both cut sizes divide — 30 cm. In every 30 cm, Lisa gets 30 ÷ 6 = 5 pieces and Max gets 30 ÷ 10 = 3 pieces, so Lisa is 2 pieces ahead for every 30 cm. She is 16 pieces ahead in all, which is 16 ÷ 2 = 8 lots of 30 cm. So each roll was 8 × 30 = 240 cm.
Multiplying 16 by 6 or by 10. The 16 counts PIECES, not centimetres, so it has to be converted through a length that both cut sizes fit into.
A picture frame is made up of 4 identical rectangular pieces joined together as a pinwheel. The area of each rectangular piece is 69 cm² and its breadth is 3 cm. A square picture fits exactly in the centre of the frame. What is the area of the picture?
+--+------------------+
| | |
| +---------------+ |
| | | |
| | picture | |
| | | |
| +---------------+ |
| | |
+------------------+--+
|<- 3 cm ->| (the frame's width)400
Each frame piece is 69 cm² with breadth 3 cm, so its length is 69 ÷ 3 = 23 cm. In a pinwheel frame each piece lies along one side of the picture, overlapping the next at the corner: the picture's side is the piece's length minus its breadth, 23 − 3 = 20 cm. So the picture's area is 20 × 20 = 400 cm².
Taking the picture's side as the full 23 cm. Look at one side of the hole: the neighbouring piece's 3 cm width eats into it at one end.
Sally arranges a pattern with sticks and coins. Pattern 1 uses 6 coins and 7 sticks (13 in all), Pattern 2 uses 9 coins and 12 sticks (21 in all), Pattern 3 uses 12 coins and 17 sticks (29 in all). There is a total of 93 coins and sticks in one of the patterns. What is the Pattern Number?
Pattern | coins | sticks | total
--------+-------+--------+------
1 | 6 | 7 | 13
2 | 9 | 12 | 21
3 | 12 | 17 | 29
4 | ? | ? | ?11
Look at the totals: 13, 21, 29. Each one is 8 more than the last. So the total is 13 + 8 × (pattern number − 1). Setting that to 93: 93 − 13 = 80, and 80 ÷ 8 = 10 more steps, so the pattern number is 10 + 1 = 11.
Dividing 93 by 8 and answering 11 remainder 5 without checking, or forgetting the +1 because Pattern 1 is the starting point, not a step.