Answer key and worked explanations · P4 · 31 questions · 100 marks · 90 minutes
These answers were worked out here, not copied from an official key. The source paper prints no answers at all, so every answer below is reasoning that can be wrong. Where one looks surprising, re-read the method before telling her she is wrong — the method is printed so it can be checked.
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.
What is the value of 20 thousands and 9 tens?
(D) 20 090
Build the number place by place: 20 thousands is 20 000, and 9 tens is 90. Add them: 20 000 + 90 = 20 090.
Writing the 9 in the ones place (20 009). "Tens" means the second column from the right, so the 9 needs a 0 after it.
How many one-thirds are there in 4 wholes?
(D) 12
One whole holds 3 thirds, so 4 wholes hold 4 × 3 = 12 thirds.
Answering 3 — that is how many thirds are in ONE whole. Or dividing 3 by 4 instead of multiplying.
Which of the following is NOT an equivalent fraction of 1/4?
(C) 5/8
Multiply the top and bottom of 1/4 by the same number: × 2 gives 2/8 ✓, × 3 gives 3/12 ✓, × 25 gives 25/100 ✓. For eighths you would need 2/8 to make 1/4, so 5/8 is the odd one out.
Reading past the word NOT and picking one that IS equivalent. Underline NOT before looking at the options.
A figure folded in half along the dotted line of symmetry is shown: below a horizontal dotted line sit two half-circles, each with one dot and a single stroke leaning to the right. Which of the following figures is the symmetric figure when opened up?
- - - - - - - - - - - - <- the fold line (horizontal) | ( . / ) ( . / ) | +------------------------------+ everything shown is the BOTTOM half; opening up mirrors it UPWARDS
(B) two faces, dots stacked, each with a curve bulging right ( ) )
The fold line is HORIZONTAL, so opening up reflects everything upwards. The single dot in each half-circle gains a partner directly above it, giving two dots stacked one over the other. The leaning stroke reflects too: a stroke leaning one way, joined to its mirror leaning the other way, makes a curve that bulges out to the right — a ) shape. Only B has both.
Picking D, which mirrors the two faces LEFT to RIGHT. The fold line drawn is horizontal, so the mirror is up-and-down, and both faces must end up identical to each other.
The table shows the number of pets kept by a group of children. How many children have MORE than 1 pet?
Number of pets | 0 | 1 | 2 | 3 | 4 -------------------+----+----+---+---+--- Number of children | 10 | 17 | 5 | 8 | 2
(A) 15
"More than 1 pet" means 2, 3 or 4 pets — not 1, and not 0. Add those three columns: 5 + 8 + 2 = 15 children.
Including the 17 children who have exactly 1 pet (giving 32). "More than 1" excludes 1 itself; "1 or more" would include it.
The figure below is made up of 5 identical squares in an I-shape — two across the top, one in the middle, two across the bottom. The area of each square is 36 cm². What is the perimeter of the figure?
+------+------+
| | |
+------+------+
| |
| |
+------+------+
| | |
+------+------+
5 identical squares, 36 cm2 each(B) 72 cm
Each square has area 36 cm², so its side is 6 cm (because 6 × 6 = 36). Start with the rectangle that just surrounds the figure: it is 2 squares wide and 3 squares tall, so 12 cm by 18 cm, giving 2 × (12 + 18) = 60 cm. Then each notch cut into the waist adds twice its depth, because you walk in and back out: 2 notches, each half a square (3 cm) deep, add 2 × 3 × 2 = 12 cm. Perimeter = 60 + 12 = 72 cm.
Taking 36 cm as the SIDE of each square instead of the area, or adding the five squares' perimeters (5 × 24 = 120 cm), which counts the edges where they join.
Arrange the following from the greatest to the smallest: 9/20, 0.54, 4.05, 1/2
(D) 4.05, 0.54, 1/2, 9/20
Turn everything into decimals so they can be compared directly: 9/20 = 45/100 = 0.45, and 1/2 = 0.5. Now the list is 0.45, 0.54, 4.05, 0.5. Greatest to smallest: 4.05, then 0.54, then 0.5 (= 1/2), then 0.45 (= 9/20).
Comparing 0.54 and 0.5 by counting digits and deciding 0.54 is "longer so bigger" — true here, but it fails for 0.5 versus 0.45. Line up the tenths first, every time.
Jason has $20. He bought 2 packets of chips and 6 oranges. Chips are $2.50 a packet and oranges are 3 for $1.20. How much money did he have left?
1 packet of chips $2.50 3 oranges for $1.20
(D) $12.60
Chips: 2 × $2.50 = $5.00. Oranges: 6 is two lots of 3, so 2 × $1.20 = $2.40. He spent $5.00 + $2.40 = $7.40. Money left = $20 − $7.40 = $12.60.
Answering $7.40 — that is what he SPENT. The question asks what is left, so there is one more subtraction to do.
Kayden is standing at point K facing the park. He makes a 135° clockwise turn followed by a 315° anticlockwise turn. Where will he be facing now?
Shopping Centre North
| ^
Library | Park |
\ | /
Cinema ---------K--------- Market
/ | \
Food Centre | Playground
|
School(A) Food Centre
Combine the two turns first instead of doing them one at a time: 135° clockwise then 315° anticlockwise leaves 315 − 135 = 180° anticlockwise — which is the same as a half turn, so he ends up facing the exact opposite direction. He started facing the Park (north-east), and the opposite of north-east is south-west, which is the Food Centre.
Working the two turns separately and losing track. Also note a 180° turn needs no direction — clockwise or anticlockwise both land in the same place.
Linda had 3 litres of milk. She spilled 1/6 ℓ of milk and used 3/4 ℓ to bake some cookies. How much milk did she have left?
(B) 2 1/12 ℓ
Put everything over twelfths: 1/6 = 2/12 and 3/4 = 9/12, so she lost 2/12 + 9/12 = 11/12 ℓ altogether. Taking that from 3 ℓ: 3 − 11/12 = 2 12/12 − 11/12 = 2 1/12 ℓ.
Answering 11/12 — that is how much milk went, not how much is left. Or subtracting only one of the two amounts.
A protractor is placed with its centre at Y and its baseline along the bottom. Two lines are drawn from Y: one up to the left towards X, reading 18° above the baseline, and one up to the right towards Z, reading 38° above the baseline. Find the value of ∠XYZ.
Z
X /
\ /
\ /
-----\--/--------- (the protractor baseline)
\/
Y
X is 18 deg above the baseline on the LEFT
Z is 38 deg above the baseline on the RIGHT(C) 124°
The whole straight line along the protractor's base is 180°. The angle sits between the two drawn lines, so take off what lies outside it on each side: the 18° between YX and the baseline on the left, and the 38° between YZ and the baseline on the right. ∠XYZ = 180 − 18 − 38 = 124°.
Answering 142° by treating YX as lying ALONG the baseline (180 − 38). It does not — it is drawn 18° above it, and that 18° has to come off too.
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
(B) 20
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.
Answering 24 and forgetting that a corner ribbon belongs to two sides at once. Mark the corners on a quick sketch before multiplying.
Mariam started her piano lesson at the time shown on the clock (the minute hand points at 12 and the hour hand at 1, so 1 o’clock in the afternoon). The duration of the piano lesson is 75 minutes. However, the lesson ended 10 minutes later than that. What time did her lesson end? Give your answer in 24-hour clock format.
12
11 | 1
| /
10 | / 2
|/
9 * 3 minute hand: 12
hour hand: 1
8 4
7 6 5
the lesson starts at 13 00(D) 14 25
The clock reads 1 o’clock in the afternoon, which is 13 00. The lesson runs 75 minutes, which is 1 hour 15 minutes, so it was due to end at 13 00 + 1 h 15 min = 14 15. It actually ended 10 minutes after that: 14 15 + 10 min = 14 25.
Answering 14 15 and forgetting the extra 10 minutes, or writing 13 65 — there is no such time, because 60 minutes roll over into the next hour.
The graph shows the number of watches sold by a shop from January to April: January 38, February 22, March 52, April 16. The total number of watches sold from January to April was 4 times the number of watches sold in May. How many watches were sold in May?
Jan Feb Mar Apr 38 22 52 16
(B) 32
Add the four months: 38 + 22 + 52 + 16 = 128 watches. That total is 4 times May's figure, so May = 128 ÷ 4 = 32 watches.
Multiplying by 4 instead of dividing. The four months together are the BIG number, so May must be smaller than it.
A piece of paper is cut straight down into Square A and Rectangle B. The area of A is twice the area of B. The breadth of Rectangle B is 14 cm. What is the area of Square A?
+----------------+---------+
| | |
| A : B | the cut is straight down,
| (square) : | so A and B are the same height
| | |
+----------------+---------+
|<- 14 cm ->|(D) 784 cm²
The cut is straight down, so A and B are the same height — call it h. A is a SQUARE, so its width is h too and its area is h × h. B is h tall and 14 cm wide, so its area is 14 × h. Now use "A is twice B": h × h = 2 × 14 × h, so h = 28 cm. The area of Square A = 28 × 28 = 784 cm².
Answering 784 by luck but 28 by stopping early — 28 is the SIDE, and the question asks for the area. Or using 14 as the square's side.
A stall had a promotion: buy 1 muffin for $3 each, or buy 4 muffins at $11. Mandy wanted to buy 50 muffins. What was the least amount of money Mandy had to pay for all the 50 muffins?
Buy 1 muffin for $3 each Buy 4 muffins at $11
138
The 4-for-$11 deal works out at $2.75 a muffin, cheaper than $3, so take as many fours as possible: 50 ÷ 4 = 12 sets of four (48 muffins) with 2 left over. That is 12 × $11 = $132, plus 2 singles at $3 = $6. Total = $138. (Buying 13 sets would give 52 muffins for $143 — more muffins AND more money, so it is not cheaper.)
Buying 13 sets of four to avoid paying full price for the last two. Always price the leftover BOTH ways — here the singles win.
Ashley is 10 years old and her mother is 46 years old. In how many years’ time will Ashley’s mother be 4 times as old as Ashley?
2
The gap between their ages never changes: 46 − 10 = 36 years, now and always. At the moment we want, Ashley is 1 unit and her mother is 4 units, so the gap is 4 − 1 = 3 units = 36, giving 1 unit = 12. So Ashley will be 12, which is 12 − 10 = 2 years from now.
Answering 12 — that is Ashley's AGE at that time, not how many years away it is. Read the last line again before writing.
In the figure, ABCD is a rectangle and DEFG is a square, overlapping at D. ∠CDG = 64° and ∠EDH = 33°, where DH is a straight line from D. Find ∠ADH.
C D E
\ 64 / \ 33 /
\ / \ /
\ / G \ / H
\ / \
A
at D, the rays run in the order C, G, A, H, E
rectangle corner ADC = 90 deg · square corner GDE = 90 deg31
Use the two right angles the shapes give you. ABCD is a rectangle, so its corner at D is 90°: ∠CDA = 90. Since ∠CDG = 64, what is left is ∠GDA = 90 − 64 = 26°. DEFG is a square, so its corner at D is also 90°: ∠GDE = 90. Going from G round to E the rays pass A then H, so ∠GDA + ∠ADH + ∠HDE = 90, that is 26 + ∠ADH + 33 = 90. So ∠ADH = 90 − 59 = 31°.
Adding 64 and 33 and subtracting from 90 straight away. The 64° is measured inside the RECTANGLE's corner and the 33° inside the SQUARE's — two different right angles, so each has to be used on its own shape first.
There were 163 ribbons in Box A and 115 ribbons in Box B. Some ribbons were moved from Box A to Box B. In the end, there were 26 more ribbons in Box A than Box B. How many ribbons were there in Box A in the end?
152
Moving ribbons between the boxes does not change the TOTAL: 163 + 115 = 278 ribbons, before and after. At the end A is 26 more than B, so take that 26 off and split the rest equally: (278 − 26) ÷ 2 = 252 ÷ 2 = 126 — that is Box B. Box A = 126 + 26 = 152.
Trying to work out how many ribbons were moved first. You never need to know — the unchanged total plus the final difference is enough.
Square A and Rectangles B, C and D form the rectangle WXYZ. A sits top-left and is a square of side 7 cm, B is top-right, D is bottom-left and C is bottom-right. The area of Rectangle B is twice the area of Rectangle D. The shaded part (A, B and D together) has an area of 385 cm². Find the area of Rectangle C, in cm².
W X +--------+-----------------------+ | A | B | 7 cm |(square)| | +--------+-----------------------+ | D | C | | | | +--------+-----------------------+ Z Y A is a 7 cm square · B = 2 x D · shaded (A + B + D) = 385 cm2
512
A is a 7 cm square, so the top row is 7 cm tall and the left column is 7 cm wide. Call the right column's width w and the bottom row's height h. Then B = 7 × w, D = 7 × h, and C = w × h. Since B is twice D: 7w = 2 × 7h, so w = 2h. The shaded area is A + B + D = 49 + 7w + 7h = 385, so 7w + 7h = 336 and w + h = 48. Putting w = 2h in: 3h = 48, so h = 16 and w = 32. Area of C = 16 × 32 = 512 cm².
Trying to find C by subtracting 385 from the whole rectangle — you do not know the whole rectangle's area yet. Naming the two unknown lengths and using "B is twice D" is what unlocks it.
Julia had some beads in a jar. 1/4 of the beads were red and the rest were green. After Julia put another 318 red beads into the jar, the fraction of green beads in the jar became 3/7. What was the total number of beads in the jar at first?
424
The GREEN beads never change — only red ones were added. At first green was 3/4 of the jar; afterwards green is 3/7 of the bigger jar. Say the green beads number G. Then at first the jar held G ÷ 3 × 4 beads, and at the end it held G ÷ 3 × 7. The jar grew by exactly the 318 red beads added, so G ÷ 3 × 7 − G ÷ 3 × 4 = 318, that is G ÷ 3 × 3 = 318, so G = 318. The jar at first = 318 ÷ 3 × 4 = 424 beads.
Trying to track the red beads, which change. Spotting the quantity that STAYS THE SAME — the green ones — is what makes the two fractions comparable.
David paid $7374 for 3 laptops and 2 headphones. Kumar paid $5002 more than David for 5 laptops and 4 headphones. What was the cost of 1 headphone?
129
Kumar paid 7374 + 5002 = $12 376 for 5 laptops and 4 headphones. Now double David's order so the headphones match: 6 laptops and 4 headphones cost 2 × $7374 = $14 748. Comparing that with Kumar's 5 laptops and 4 headphones, the headphones cancel and the difference is exactly 1 laptop: 14 748 − 12 376 = $2372. So 3 laptops cost 3 × 2372 = $7116, leaving 7374 − 7116 = $258 for 2 headphones, and one headphone is $129.
Subtracting David's order from Kumar's directly (2 laptops + 2 headphones = $5002) and then guessing. Doubling one order first is what makes a pair of items cancel exactly.
The total mass of a box and a watermelon was 7 kg 34 g. When some strawberries were added into the box, the total mass became 8600 g. The watermelon was 3 times as heavy as all the strawberries added. Find the mass of the box, in grams.
2336
Work in grams: 7 kg 34 g = 7034 g. Adding the strawberries took the total from 7034 g to 8600 g, so the strawberries weigh 8600 − 7034 = 1566 g. The watermelon is 3 times that: 3 × 1566 = 4698 g. The box and watermelon together were 7034 g, so the box = 7034 − 4698 = 2336 g.
Reading 7 kg 34 g as 7340 g. It is 7000 + 34 = 7034 g — the 34 fills the ones and tens columns, not the hundreds.
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
128
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.
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.
At a fruit stall, mangoes are sold at 2 for $7 and pears are sold at 3 for $5.50. Mdm Fauziah bought the same number of mangoes and pears. She paid $160 in total. How many mangoes did she buy?
Mangoes 2 for $7 Pears 3 for $5.50
30
Choose a number of each that both deals divide into — 6 is the smallest. 6 mangoes are three lots of 2, costing 3 × $7 = $21. 6 pears are two lots of 3, costing 2 × $5.50 = $11. So 6 of each costs $21 + $11 = $32. Then $160 ÷ $32 = 5 such groups, and she bought 5 × 6 = 30 mangoes.
Working out the price of one mango and one pear. A pear costs $5.50 ÷ 3, which is not a whole number of cents — buying in sixes keeps every figure exact.
Mr Lee had an apple, a papaya and a honeydew. He weighed them in pairs: apple + papaya = 730 g, apple + honeydew = 1960 g, honeydew + papaya = 2470 g. What was the mass of the honeydew, in grams?
Setup 1 apple + papaya = 730 g Setup 2 apple + honeydew = 1960 g Setup 3 honeydew + papaya = 2470 g
1850
Add all three weighings: 730 + 1960 + 2470 = 5160 g. Every fruit has been weighed exactly twice, so the three fruits together weigh 5160 ÷ 2 = 2580 g. The honeydew is the one missing from Setup 1, so honeydew = 2580 − 730 = 1850 g.
Subtracting two of the weighings and hoping. Adding all three and halving is what turns three pair-weights into the single total, and the rest is one subtraction.
Kenny has 3 pieces of rope of length 24 cm, 60 cm and 96 cm. He wants to cut them into smaller pieces of equal length without any leftovers. Each smaller piece must be as long as possible. How many such pieces of equal length can he get?
24 cm |------------| 60 cm |------------------------------| 96 cm |------------------------------------------------|
15
The piece length must divide 24, 60 and 96 exactly, and be as long as possible — so it is the highest common factor. 24 = 12 × 2, 60 = 12 × 5, 96 = 12 × 8, and no bigger number divides all three, so each piece is 12 cm. Counting the pieces: 2 + 5 + 8 = 15 pieces.
Answering 12 — that is the LENGTH of each piece, and the paper's answer line unhelpfully prints "cm" even though the question asks how many. Read the question, not the blank.
Sam used Square X (side 5 cm), Rectangle Y (9 cm by 4 cm) and Rectangle Z (14 cm by 3 cm) — with a second copy of Square X — to form Figure A: Z lies along the bottom, one X sits on Z at the left, Y stands upright beside it, and the other X sits at the right-hand end, level with the bottom of Z. What is the perimeter of Figure A, in cm?
+---+
| |
| Y | X = 5 by 5 (square)
+-----+ | Y = 9 tall, 4 wide
| X | | Z = 14 wide, 3 tall
| | | +-----+
+-----+---+----------+ X |
| Z | |
+--------------------+-----+
Figure A
note the STEP on the right: the square stands 2 cm proud of Z66
Set the bottom-left corner of Z at 0. Z runs from 0 to 14 across and 0 to 3 up. The left X sits on Z, from 0 to 5 across and 3 to 8 up. Y stands next to it, from 5 to 9 across and 3 to 12 up. The right X sits beyond Z, from 14 to 19 across and 0 to 5 up. Now walk the outline once: up the left side 8, across the top of X 5, up Y's side 4, across Y's top 4, down Y's far side 9, along Z's top 5, up the step 2, across the right X's top 5, down its side 5, and back along the bottom 19. Adding: 8 + 5 + 4 + 4 + 9 + 5 + 2 + 5 + 5 + 19 = 66 cm.
Adding the pieces' perimeters (5 × 4 + 5 × 4 + 26 + 34 = 100 cm). Every edge where two pieces touch is inside the figure and must not be counted — walking the outline once is the only safe way.
A spider was climbing to the top of a garden wall, starting from the bottom. After climbing up 1/3 of the wall, it began to rain. During the rain the spider slipped down 30 cm and stayed there until the rain stopped. Then it climbed up the remaining 5/6 of the height of the wall to reach the top. What was the total height the spider had climbed before and after the rain, in cm?
210
Let the wall be 1 whole. The spider got to 1/3, slipped down 30 cm, then climbed 5/6 of the wall to finish at the top. So (1/3 of the wall) − 30 cm + (5/6 of the wall) = the whole wall. In sixths, 1/3 + 5/6 = 2/6 + 5/6 = 7/6, so 7/6 of the wall minus 30 cm equals 1 wall — meaning the extra 1/6 of the wall is exactly the 30 cm it slipped. So the wall is 6 × 30 = 180 cm. It climbed 1/3 of 180 = 60 cm before the rain and 5/6 of 180 = 150 cm after, a total of 60 + 150 = 210 cm.
Answering 180 — that is the height of the WALL. The question asks how far the spider CLIMBED, which is more, because it had to re-climb the 30 cm it slipped.
A rectangular park measures 54 m by 28 m. A pedestrian path 2 m wide and a cyclist path 3 m wide are built along two sides of the park — the pedestrian path wraps the top and left of the park, and the cyclist path wraps the top and left of that. It costs $18 to construct each square metre of the cyclist path. How much does it cost to construct the cyclist path?
################################ -+ 3 m cyclist
##+--------------------------+## -+ 2 m pedestrian
##| | #
##| Park | # 28 m
##| | #
##+--------------------------+-+
|<-------- 54 m -------->|
# = the two paths, along the TOP and the LEFT only4806
Build outwards from the park. The park is 54 by 28. Adding the 2 m pedestrian path along the top and the left makes that block 54 + 2 = 56 by 28 + 2 = 30. Adding the 3 m cyclist path along the top and left of THAT makes the whole thing 56 + 3 = 59 by 30 + 3 = 33. The cyclist path is the difference between those two: 59 × 33 − 56 × 30 = 1947 − 1680 = 267 m². At $18 a square metre, the cost is 267 × $18 = $4806.
Measuring the cyclist path's strips from the PARK's 54 and 28 instead of from the outside of the pedestrian path. The paths are stacked, so the outer one is longer than the inner one.
In a library, there were some story books at first. Betty, the librarian, added another 17 story books and removed 36 of them. Then, Betty received 3 times as many new story books as what were left on the shelves. She then arranged all the story books equally between 2 sections. Each section had 458 story books in the end. How many story books were in the library at first?
248
Work backwards. Two sections of 458 make 2 × 458 = 916 books at the end. Just before that, the shelves held some number and Betty received 3 times as many again, so the 916 is 1 + 3 = 4 equal shares: one share = 916 ÷ 4 = 229 books on the shelves. That 229 came after adding 17 and removing 36, a net loss of 19. So at first there were 229 + 19 = 248 books.
Reading "3 times as many new books as what were left" as making the total 3 times bigger. It ADDS 3 shares to the 1 already there, giving 4 shares in all.