Answer key and worked explanations · P4 · 45 questions · 100 marks · 90 minutes
The answers below are the official ones, printed in the source paper. The explanations are not — those were written here, and are the part worth arguing with.
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 the digit ‘6’ in the numeral 36 574?
6000
Read the places from the right: 4 ones, 7 tens, 5 hundreds, 6 thousands, 3 ten-thousands. The 6 sits in the thousands place, so its value is 6 × 1000 = 6000.
Answering 6, or naming the place ("thousands") instead of the VALUE. The question asks what the digit is worth.
A number when rounded to the nearest 100 is 8000. What is the greatest possible number?
8049
Rounding to the nearest 100 gives 8000 for anything from 7950 up to 8049. Anything from 8050 upwards rounds to 8100 instead. So the greatest is 8049.
Answering 8050 — that is the first number that rounds UP to 8100, not the last that rounds to 8000. Test the boundary by rounding it back.
A van can transport 8 passengers. How many passengers can 240 vans transport?
1920
240 × 8 = 1920 passengers.
Dividing instead of multiplying. More vans means more passengers, so the answer must be bigger than 240.
The figure below is made up of 2 squares of different sizes. AG is 8 cm and DE is 6 cm. What is the length of AD? Give your answer in cm.
|<------------- ? ------------->| A B +---------------+---C-------+ D | | | | | | | 8 cm | | 6 cm | | | | | | +---------------+-----------+ G F E
14
AG = 8 cm is the side of the big square, so AB = 8 cm too. DE = 6 cm is the side of the small square, so BD (the part of the top line beyond B) = 6 cm. AD runs along the top from A to D: 8 + 6 = 14 cm.
Trying to find AD from the slanted-looking picture instead of from the two square sides. In a square all four sides are equal — that is the only fact this question needs.
Look at the figure below — shaded squares on a 6 × 6 grid, with a dashed diagonal line running from the bottom-left corner to the top-right. Which one of the options will form a symmetric figure with the above, about that dashed line?
the dashed line is the MIRROR. The option must hold the reflection of every shaded square across it — same distance from the line, other side.
(1) Option 1
The dashed diagonal is the mirror line. Take each shaded square above the line and count its steps to the line; its partner must be the same number of steps on the other side. Only option 1 places every partner correctly, including the half-shaded triangle, which must reflect into a triangle leaning the opposite way.
Checking only the big blocks and not the half-squares. The triangles are what separate option 1 from option 3 — they must flip, not just move.
Ben was standing at Point X. He turned 225° anti-clockwise and then made another 1/4-turn in the clockwise direction. He ended up facing the Bus Bay. Which location was Ben facing at first?
Library N
| ^
General Office | Bus Bay |
\ | /
Computer Lab ------- X ------- Field
/ | \
Dental Clinic | Basketball Court
|
Hall(1) Hall
Work out the NET turn first: 225° anti-clockwise, then 90° clockwise, leaves 225 − 90 = 135° anti-clockwise. He finished facing the Bus Bay (north-east), so to find where he began, undo it — turn 135° CLOCKWISE from north-east. North-east → (90°) → south-east → (45° more) → south. South is the Hall.
Applying the turns forwards from the Bus Bay instead of backwards. The Bus Bay is where he ENDED, so the working has to run in reverse.
Which of the following is an equivalent fraction of 3/4?
(4) 6/8
Multiply the top and bottom of 3/4 by the same number: × 2 gives 6/8. Checking the others: 12/24 simplifies to 1/2, 9/16 and 5/6 do not reduce to 3/4.
Picking 9/16 because 3 × 3 = 9 and 4 × 4 = 16. Both parts must be multiplied by the SAME number, not each by itself.
Zainal had 5 m of ribbon at first. He used 1.4 m of it to wrap a present and some ribbon to decorate 2 cards. He had 2.6 m of ribbon left. Which of the following is the length of ribbon for each card?
(4) 0.5 m
Find the ribbon used on the cards: 5 − 1.4 − 2.6 = 1 m for BOTH cards. Then share it: 1 ÷ 2 = 0.5 m for each card.
Stopping at 1 m — that is the ribbon for two cards, and the question asks for each one.
The table shows the favourite colours of 108 pupils in Primary Four. Some numbers have been blotted out by ink stains. There were twice as many pupils who chose green than yellow. There is a total of 38 pupils who like 2 of the colours. Which of the following options shows the two colours?
Colour | Number of Pupils -------+----------------- Green | ### stain ### Yellow | ### stain ### Blue | 22 Red | 38 108 pupils altogether · green = 2 × yellow
(2) Yellow and Blue
Green and yellow together = 108 − 22 − 38 = 48. Green is twice yellow, so that is 3 equal shares: yellow = 48 ÷ 3 = 16 and green = 32. Now test the pairs: blue + green = 54, yellow + blue = 16 + 22 = 38 ✓, green + red = 70, red + yellow = 54.
Splitting the 48 evenly into 24 and 24. "Twice as many" makes THREE shares, not two.
Which of the following figures has only ONE line of symmetry?
(3) an isosceles trapezium
Count the folds for each. A parallelogram has none — no fold makes its halves match. A regular pentagon has five. A rectangle has two, one across and one down. The isosceles trapezium has exactly one, straight down the middle.
Choosing the rectangle. It folds down the middle AND across the middle, so it has two, and the question asks for ONLY one.
Rectangle PQRS is folded to form the figure below. ∠y is three times ∠x. Find ∠x.
P Q +-------------------------------------+ | y / : | / : x | / : | \ / : +-----------------------------\-/-----+ S R the dashed line is where the corner R USED to be; the fold crease runs from Q
18
The whole corner at Q is 90°, and it is cut into three parts: ∠y, then the two angles the fold makes. A fold reflects, so the crease from Q sits exactly halfway between the edge in its new position and the edge in its old one — those two angles are equal, and each is ∠x. So y + x + x = 90. With y = 3x: 3x + 2x = 90, so 5x = 90 and x = 18°.
Writing y + x = 90 and getting 22.5°. The fold makes TWO equal angles below y, not one — that is the whole point of a crease.
A 4-digit number is made up of three different odd digits and a zero. What is the smallest possible value of the number?
1035
The odd digits are 1, 3, 5, 7, 9 and three of them must be different. To make the number small, put the smallest digit first — but not the 0, because a 4-digit number cannot start with 0. So the thousands digit is 1, the hundreds digit is the 0, and the last two places take the next smallest odd digits: 3 then 5. That gives 1035.
Starting with 0 (0135 is a 3-digit number), or forgetting the three odd digits must be DIFFERENT and writing 1011.
Study the pattern. What is the missing number in the box? star × star = 49; moon + moon + moon + moon = 100; star × moon = ?
star × star = 49 moon + moon + moon + moon = 100 star × moon = ?
175
From the first line, a number times itself is 49, so the star is 7. From the second, four moons make 100, so the moon is 100 ÷ 4 = 25. Then star × moon = 7 × 25 = 175.
Reading the second line as 4 × moon = 100 and then halving instead of dividing by 4, or taking the star as 24.5 (half of 49) instead of the number that multiplies by ITSELF.
Alice’s present age is a 2-digit number which is also a multiple of 4. 5 years ago, her age was a factor of 54. What is Alice’s age 5 years later?
37
The factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54. Adding 5 to each gives the possible present ages: 6, 7, 8, 11, 14, 23, 32, 59. Of those, only 32 is both a 2-digit number and a multiple of 4. So Alice is 32 now, and 5 years later she is 32 + 5 = 37.
Answering 32 — that is her age NOW. The question asks for five years later.
Colin, Dylan and Eden shared some sweets. Colin took 1/6 of the sweets. Dylan took the remaining sweets and shared it equally with Eden. Eden ate 12 of his sweets and has 18 sweets left. How many sweets were there altogether?
72
Work backwards for Eden: he has 18 left after eating 12, so he had 12 + 18 = 30. Colin took 1/6, leaving 5/6, and Dylan and Eden split that equally — so Eden got half of 5/6, which is 5/12 of the whole. If 5/12 is 30 sweets, then 1/12 is 6, and the whole is 12 × 6 = 72.
Taking Eden's share as half of everything (1/2) instead of half of what was LEFT (5/12).
The product of 7 and a number is 84. What is the product of 252 and that number?
3024
Find the number first: 84 ÷ 7 = 12. Then 252 × 12 = 3024.
Multiplying 252 by 84, or by 7. Find the hidden number before using it.
A rectangular piece of paper was folded differently on both ends as shown below. What is the area of the rectangular piece of paper before it was folded? Give your answer in cm².
|<- 4 cm ->|
+----------+ -+ 1 cm
+-----------------------+ / -+
/ | /
/ 9 cm | /
+------------------------- +------+
|
3 cm |
+------------
both creases are at 45 degrees84
Each crease is at 45°, so the slanted edge runs sideways exactly as far as it drops — and that distance is the paper's WIDTH. The 4 cm across the top right is that same distance, so the paper is 4 cm wide. Now unfold each end: the left flap reaches 3 + 4 = 7 cm, and the right flap reaches 1 + 4 = 5 cm. Laid flat, the length is 7 + 9 + 5 = 21 cm. Area = 21 × 4 = 84 cm².
Adding up the visible pieces as if nothing were hidden. A fold tucks paper underneath, so the flat sheet is always LONGER than the folded shape.
A rectangular piece of paper was cut into 3 smaller rectangles with dimensions 9 cm by 12 cm, 9 cm by 2 cm, and 9 cm by 5 cm. Find the length of the paper before it was cut. Give your answer in cm.
+----------+---+------+ | | | | the three pieces, side by side | 12 cm |2cm| 5 cm | every piece is 9 cm tall +----------+---+------+ |<--------- ? -------->|
19
Every piece is 9 cm on one side, so 9 cm is the paper's breadth and the cuts ran straight across it. The other sides — 12, 2 and 5 — are the pieces of the length, so the length was 12 + 2 + 5 = 19 cm.
Adding all six numbers, or adding the 9s as well. The 9 is the side the three pieces SHARE, so it is counted once and is not part of the length.
The table shows the cost of pizza: a Regular is $12 and a Large is $22. Mrs Tan ordered 6 large pizzas and 5 regular pizzas. She gave the delivery man $200. How much change did Mrs Tan receive?
8
6 large: 6 × $22 = $132. 5 regular: 5 × $12 = $60. Total = $132 + $60 = $192. Change from $200 = 200 − 192 = $8.
Answering 192 — that is what she spent, not the change. Or pairing the wrong price to the wrong size.
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
7
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.
Answering 6 and ignoring the leftover. Always ask what the remaining centimetres are long enough to reach.
Sam spent a total of 3 h 10 min revising for his English, Science and Mathematics tests over the weekend. He spent 50 min on English and another 1 h 25 min on Science. How long did Sam spend revising Mathematics? Express your answer in minutes.
55
Change everything to minutes first. Total = 3 × 60 + 10 = 190 min. English = 50 min. Science = 60 + 25 = 85 min. Mathematics = 190 − 50 − 85 = 55 min.
Subtracting hours and minutes separately without converting, which turns 3 h 10 min − 1 h 25 min into a negative number of minutes.
The figure below is made up of 7 identical rectangles. The perimeter of the whole figure is 340 cm. Find the breadth of 1 such rectangle. Give your answer in cm.
+----------+----------+ -+ | | | | breadth +--+--+--+--+--+------+ -+ | | | | | | | | | | | | length | | | | | | +--+--+--+--+--+ top row: 2 rectangles lying down bottom row: 5 rectangles standing up
20
The two rows are the same width, so 2 lengths = 5 breadths. The figure's width is 2 lengths and its height is 1 breadth + 1 length, so the perimeter is 2 × (2 lengths + breadth + length) = 2 × (3 lengths + 1 breadth) = 340, giving 3 lengths + 1 breadth = 170. From 2 lengths = 5 breadths, one length is 2½ breadths, so 3 lengths is 7½ breadths. Then 7½ + 1 = 8½ breadths = 170, and one breadth = 170 ÷ 8.5 = 20 cm.
Adding the seven rectangles' perimeters. Only the OUTSIDE edge counts — the lines where rectangles touch are inside the figure.
Yiping had 100 fewer stickers than Joanne. Joanne had 4 times as many stickers as Wendy. Wendy had 40 stickers. How many stickers did they have altogether?
260
Start from the one you know. Wendy = 40. Joanne = 4 × 40 = 160. Yiping = 160 − 100 = 60. Altogether = 40 + 160 + 60 = 260.
Working from Yiping first, or adding the 100 to Joanne instead of taking it off. "100 fewer than Joanne" means Yiping is the smaller one.
Star Bookshop is having a promotion: notebooks are $5 each, and for every 2 notebooks bought the third notebook is free (buy 3 for the price of 2). Karen bought 25 notebooks. What is the least amount of money she has spent in total?
85
Group the notebooks in threes, because every group of 3 costs only 2 × $5 = $10. In 25 notebooks there are 25 ÷ 3 = 8 groups of three (24 notebooks) with 1 left over. The 8 groups cost 8 × $10 = $80, and the last notebook has no partners so it costs the full $5. Total = $80 + $5 = $85.
Paying full price for all 25 ($125), or counting the free ones as groups of 2. The deal gives 3 notebooks for the price of 2, so it is groups of THREE that matter.
Meena left her house to watch a movie at the cinema. Her watch showed 1630 but it was actually 5 minutes slower than the actual time. She took 35 minutes to reach the cinema. In the end, she was 10 minutes early for the movie. What time did the movie begin? Give your answer in 24-hour clock format.
1720
Her watch is 5 minutes SLOW, so the real time was 1630 + 5 = 1635. The journey took 35 minutes, so she arrived at 1635 + 35 = 1710. She was 10 minutes early, so the movie began 10 minutes after she arrived: 1710 + 10 = 1720.
Taking 5 minutes off instead of adding it. A slow watch shows a time EARLIER than the real one, so the real time is later.
Ken had some money. He used 1/2 of it to buy a shirt and 1/5 of it to buy a pair of pants. He had $180 left. How much money did he have at first?
600
He spent 1/2 + 1/5 of his money. Using tenths: 5/10 + 2/10 = 7/10. So what is left is 10/10 − 7/10 = 3/10, and that is $180. One tenth is 180 ÷ 3 = $60, so the whole is 10 × $60 = $600.
Adding 1/2 and 1/5 as 2/7 by adding tops and bottoms. Fractions need a common denominator before they can be added.
A rectangular piece of paper PQRS was folded as shown below, so that the edge RQ came down onto the line RU. Find ∠TRU.
P T Q +---+---------------------------------+ | \ \ | \ \ U + \ \ | \ \ | \ \ 52 deg + - - - - - \- - - - - - + S R the dashed line S—R is where the bottom edge still is; TR is the crease
19
The corner of the rectangle at R is 90°, and it is now split into three: ∠QRT, ∠TRU and ∠URS. The fold carried RQ onto RU, and a crease always sits exactly halfway between where an edge was and where it went — so ∠QRT = ∠TRU. That leaves 90 = ∠TRU + ∠TRU + 52, so 2 × ∠TRU = 38 and ∠TRU = 19°.
Halving 52 instead. The 52° is the LEFTOVER angle outside the fold; the two equal angles are what remains of the 90° after taking it off.
The line graph shows the earnings at Mr Lim’s chicken rice stall over a 6-month period: January $3000, February $4500, March $5000, April $3500, May $3000, June $2500. Mr Lim charged $5 for every plate of chicken rice. What is the difference in the number of plates of chicken rice sold between the highest earnings month and the lowest earnings month?
Jan Feb Mar Apr May Jun 3000 4500 5000 3500 3000 2500 highest = March, lowest = June · $5 a plate
500
The highest month is March at $5000 and the lowest is June at $2500. The difference in money is $5000 − $2500 = $2500. Each plate is $5, so that is 2500 ÷ 5 = 500 plates.
Working out each month's plates and then subtracting is fine too (1000 − 500 = 500), but dividing the DIFFERENCE once is quicker. The real slip is reading May as the lowest — June is lower.
The figure is made up of 2 identical squares overlapping each other. The unshaded rectangle where they overlap has an area of 10 cm². The total area of the shaded parts is 142 cm². Find the length of the square. Give your answer in cm.
+----------+
| shaded |
| +---+------+
| | | |
+------+---+ |
| shaded |
+----------+
the small middle box is the overlap (unshaded), 10 cm29
Each square is made of its shaded part plus its half of the overlap. Both squares contain the whole overlap, so the two squares together = shaded + overlap + overlap = 142 + 10 + 10 = 162 cm². That is two squares, so one square is 162 ÷ 2 = 81 cm². A square of area 81 has side 9 cm, since 9 × 9 = 81.
Subtracting the overlap once instead of adding it twice. The overlap belongs to BOTH squares, so it has to be counted twice to rebuild them.
A carpet is laid on a rectangular floor measuring 20 m by 13 m, leaving a border of 3 m around it. Find the area of the floor that is not covered by the carpet. Give your answer in square metres.
|<------------- 20 m ------------->| +----------------------------------+ -+ | 3 m | | | +--------------------------+ | | |3m | Carpet | | 13 m | +--------------------------+ | | | | | +----------------------------------+ -+
162
The whole floor is 20 × 13 = 260 m². The border is 3 m on EVERY side, so it takes 3 m off each end of both measurements: the carpet is (20 − 3 − 3) by (13 − 3 − 3) = 14 × 7 = 98 m². Uncovered = 260 − 98 = 162 m².
Taking off 3 m once instead of twice, giving a 17 × 10 carpet. The border runs all the way round, so each dimension loses 3 m at BOTH ends.
Max bought the same number of books and files. Each book cost $7 and each file cost $2. He paid $243 altogether. How many books and files did he buy altogether?
54
Because the numbers are equal, buy them in pairs: one book and one file cost $7 + $2 = $9. Then $243 ÷ $9 = 27 pairs. Each pair is 2 items, so altogether he bought 27 × 2 = 54 books and files.
Answering 27 — that is the number of BOOKS (and of files), but the question asks for both kinds added together.
John bought 2 crates of oranges. Crate A contained 200 more oranges than Crate B. After he transferred 50 oranges from Crate B to Crate A, there were twice as many oranges in Crate A than in Crate B. How many oranges were there in Crate A at first?
550
After the move, Crate B has 50 fewer and Crate A has 50 more, so the gap between them grows from 200 to 200 + 50 + 50 = 300. At that point A is twice B, so the gap is exactly one B: B after = 300. So B at first = 300 + 50 = 350, and A at first = 350 + 200 = 550.
Thinking the gap stays at 200. Moving oranges from one crate to the other changes BOTH, so the gap shifts by twice the number moved.
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)
91
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.
Answering 90 and forgetting the leftovers, or counting the leftover 2 shapes as 2 circles. The pattern always restarts with a TRIANGLE.
Mr Tan has less than 40 sweets for a group of pupils. If he gives his pupils 4 sweets each, he will be short of 1 sweet. If he gives each pupil 5 sweets each, he will be short of 11 sweets. How many pupils does he have in the group?
10
Compare the two plans. Going from 4 sweets each to 5 sweets each means one extra sweet per pupil, and the shortfall grows from 1 to 11 — that is 10 more sweets needed. So there are 10 pupils. (Check: 10 pupils × 4 = 40, short of 1, so he has 39, which is indeed less than 40.)
Trying to find the number of sweets first. The difference between the two plans gives the number of PUPILS straight away, because each pupil accounts for exactly one extra sweet.
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?
136
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.
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)96
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.
Forgetting to add Jean herself to each count, giving 11 × 7 = 77. She is in her own row and her own column.
Mike had 72 oranges and 45 pears. He packed the fruits such that each bag contained an equal number of oranges and an equal number of pears. He packed the fruits into as many bags as possible. How many fruits were there in each bag?
13
The number of bags must divide both 72 and 45 exactly, and we want as many bags as possible — so it is the largest common factor. 72 = 8 × 9 and 45 = 5 × 9, so the highest common factor is 9. With 9 bags: 72 ÷ 9 = 8 oranges and 45 ÷ 9 = 5 pears in each, giving 8 + 5 = 13 fruits per bag.
Answering 9 — that is the number of BAGS. Or using a smaller common factor like 3, which works but does not give as many bags as possible.
The figure shows a rectangular garden 12 m by 16 m with a footpath (shaded) running along the top, down the left side, and part-way down the right side. The width of the footpath is 2 m. The area of the garden not covered by the footpath is 124 m². Find the length of AB, the part of the right-hand edge below the footpath. Give your answer in m.
|<------- 12 m ------->| +######################+ -+ #######################| | ##+----------------+###| | ##| |###| | ##| |###| | 16 m ##| |###+ A | ##| | | | ##| | | | ##+----------------+---+ B -+ # = the 2 m footpath
6
The garden is 12 × 16 = 192 m², so the footpath covers 192 − 124 = 68 m². Now build the footpath from its parts, taking care not to count the corners twice. The left strip is 2 × 16 = 32 m². The top strip, not counting the corner already used, is (12 − 2) × 2 = 20 m². That leaves 68 − 32 − 20 = 16 m² for the right-hand strip, which is 2 m wide, so it is 16 ÷ 2 = 8 m long below the top strip — reaching 2 + 8 = 10 m down from the top. AB is the rest of that edge: 16 − 10 = 6 m.
Adding the three strips as 2×16 + 2×12 + 2×h and double-counting the two corner squares where they meet.
In a General Knowledge quiz, 5 marks were awarded for a correct answer but 2 marks were deducted for an incorrect answer. Mei Ling answered all 50 questions and was awarded a total of 166 marks. How many questions did she answer correctly?
38
Suppose every answer were correct: 50 × 5 = 250 marks. She actually scored 166, which is 84 marks less. Each wrong answer costs 5 marks not gained plus 2 marks deducted — 7 marks in all. So the number wrong is 84 ÷ 7 = 12, and the number correct is 50 − 12 = 38.
Counting a wrong answer as costing only 2 marks. It costs 7: the 5 she did not win AND the 2 taken away.
Uncle Lee had some pears in his shop. He sold 36 of them in the morning and threw away 5 bad ones. He sold 10 pears in the afternoon and packed half of the remaining pears into 10 bags. Each bag contained 7 pears. How many pears did he have at first?
191
Work backwards. The 10 bags hold 10 × 7 = 70 pears, and that was HALF of what remained, so 140 pears remained after the afternoon sale. Before that he sold 10, so there were 150. Before the morning he had also lost 36 sold and 5 thrown away: 150 + 10 is already counted, so add back 36 + 5 = 41 to the 150, giving 191.
Taking the 70 bagged pears as everything that was left. It is half of it, so the remainder is 140, not 70.
Mother bought 5 more cups than plates. Each cup cost $8 and each plate cost $3. She paid $183 altogether. How many cups did she buy?
18
Set the 5 extra cups aside first: they cost 5 × $8 = $40. That leaves $183 − $40 = $143 for equal numbers of cups and plates. One cup and one plate together cost $8 + $3 = $11, so there are $143 ÷ $11 = 13 pairs. So there are 13 plates and 13 + 5 = 18 cups.
Answering 13 — that is the number of PLATES (and of the matched cups). The five extra cups still have to be added on.
Sandy and Tracy had $600 altogether. Sandy gave 1/5 of her money to Tracy and then Tracy gave 1/4 of the money she then had to Sandy. In the end, both of them had the same amount of money. How much money did Sandy have at first?
250
Work backwards from the end: they finished equal, so each had $300. The last move was Tracy giving away 1/4 of her money, keeping 3/4 — and 3/4 of Tracy's money was $300, so Tracy had $400 just before, and gave $100 to Sandy. So before that move Sandy had $300 − $100 = $200. That $200 was Sandy's money after giving away 1/5, so it is 4/5 of what she started with: 1/5 = $50, and Sandy began with 5 × $50 = $250.
Taking 1/4 of Tracy's ORIGINAL money instead of the amount she held after Sandy's gift. Each fraction refers to the money at that moment.
The total cost of 6 skirts and 3 blouses was $144. Three blouses cost as much as 2 skirts. Find the total cost of one skirt and one blouse.
30
Swap the blouses for skirts: 3 blouses cost the same as 2 skirts, so 6 skirts + 3 blouses costs the same as 6 skirts + 2 skirts = 8 skirts. So 8 skirts = $144, and one skirt = $18. Then 3 blouses = 2 skirts = $36, so one blouse = $12. One skirt and one blouse = $18 + $12 = $30.
Assuming a blouse costs the same as a skirt, or dividing $144 by 9 items. The swap is what makes everything one kind of thing.
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 squares1326
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.
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?
10
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.
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.