Sample questions

Questions from Elite's own papers, each against the clock at the real test's pace. Choose an answer to start the clock, check it, then read Elite's worked solution.

Thinking Skills
Mathematical Reasoning

Thinking Skills

Questions Rex chose from Elite's Thinking Skills papers, each with his answer and four-step worked solution.

Question 1

Thinking Skills, Rex's sample questions

1:00

The school's average spelling score has risen from fourteen out of twenty to seventeen since the start of the year. The spelling program brought in at Easter is working, the coordinator tells the staff meeting, and it should be running in every year group by next year.
Which of the following, if true, most weakens the argument?

A The same spelling test has been used in every term of this year and last. B Two classes of older students joined in Term 2 and sit the same test. C Some students find spelling harder than others do. D The program takes ten minutes out of each morning.
Worked solution
1. Understand the question. Argument / Weaken (W2, counter-explanation)
2. Use the strategy. W2 keeps the premise true but reads it a different way. Then the same fact no longer supports the conclusion. Ask yourself: could this fact mean something else?
3. Apply the strategy. Conclusion: the spelling program works, so use it everywhere. Premises: the school average went from fourteen to seventeen after the program started. Support gap: an average depends on who is in the group. If the group changes, the average changes, even if nobody got better. Check each option: B says two classes of older students joined in Term 2. Older students usually spell better. They push the average up on their own. Every student who was already there could have scored exactly the same as before. A says the test stayed the same, which supports the coordinator. C is true of any subject. D is about time, not marks.
4. Confirm the answer. B is W2 because the average really did rise, but it rose because the group changed, not because anyone improved.
This tests Argument: strengthen and weaken, part of Thinking Skills. The answer is B.
Show the worked solution
1. Understand the question. Argument / Weaken (W2, counter-explanation)
2. Use the strategy. W2 keeps the premise true but reads it a different way. Then the same fact no longer supports the conclusion. Ask yourself: could this fact mean something else?
3. Apply the strategy. Conclusion: the spelling program works, so use it everywhere. Premises: the school average went from fourteen to seventeen after the program started. Support gap: an average depends on who is in the group. If the group changes, the average changes, even if nobody got better. Check each option: B says two classes of older students joined in Term 2. Older students usually spell better. They push the average up on their own. Every student who was already there could have scored exactly the same as before. A says the test stayed the same, which supports the coordinator. C is true of any subject. D is about time, not marks.
4. Confirm the answer. B is W2 because the average really did rise, but it rose because the group changed, not because anyone improved.
This tests Argument: strengthen and weaken. The answer is B.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 60 seconds a question.
Question 2

Thinking Skills, Rex's sample questions

1:30

Library borrowing has fallen by a third since the start of the year. Two teachers point out that the new graphic novel shelf has never been busier and that the reading corner is full every lunchtime. Ms Adeyemi is not reassured by any of that. The borrowing figures are the one hard number we have, she tells the staff meeting, and a third fewer loans means students are reading a third less. The school needs a reading push next term.
Which of the following, if true, most weakens the argument?

A The borrowing figures are counted by the same computer system as last year. B Some students read books they own at home. C At the start of the year the loan period went from one week to three. D The library's opening hours have not changed this year.
Worked solution
1. Understand the question. Argument / Weaken (W2, counter-explanation)
2. Use the strategy. W2 keeps the premise true but reads it a different way. Then the same fact no longer supports the conclusion. Ask yourself: could this fact mean something else?
3. Apply the strategy. Conclusion: students are reading a third less, so the school needs a reading push. Premises: loans have fallen by a third. Support gap: Ms Adeyemi thinks one loan means one amount of reading. That is only true if a loan is the same thing this year as last year. Check each option: C says the loan time went from one week to three. Think about one book. Last year it could go out three times, one week each. That was three loans. This year it goes out once for three weeks. That is one loan. Same reading, one third of the count. So fewer loans does not mean less reading. The busy shelf and the full reading corner fit this. A says the counting is accurate, which nobody doubted. B is about books at home. D says the hours did not change, which supports Ms Adeyemi.
4. Confirm the answer. C is W2 because the fall in loans is real, but now it shows longer borrowing, not less reading.
This tests Argument: strengthen and weaken, part of Thinking Skills. The answer is C.
Show the worked solution
1. Understand the question. Argument / Weaken (W2, counter-explanation)
2. Use the strategy. W2 keeps the premise true but reads it a different way. Then the same fact no longer supports the conclusion. Ask yourself: could this fact mean something else?
3. Apply the strategy. Conclusion: students are reading a third less, so the school needs a reading push. Premises: loans have fallen by a third. Support gap: Ms Adeyemi thinks one loan means one amount of reading. That is only true if a loan is the same thing this year as last year. Check each option: C says the loan time went from one week to three. Think about one book. Last year it could go out three times, one week each. That was three loans. This year it goes out once for three weeks. That is one loan. Same reading, one third of the count. So fewer loans does not mean less reading. The busy shelf and the full reading corner fit this. A says the counting is accurate, which nobody doubted. B is about books at home. D says the hours did not change, which supports Ms Adeyemi.
4. Confirm the answer. C is W2 because the fall in loans is real, but now it shows longer borrowing, not less reading.
This tests Argument: strengthen and weaken. The answer is C.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 90 seconds a question.
Question 3

Thinking Skills, Rex's sample questions

1:00

At the school bird-watching morning, every child who spotted an owl also spotted a kookaburra. No child who spotted a kookaburra went to the wetland hide. Some children who went to the wetland hide spotted a pelican.
If the statements above are true, which of the following cannot be true?

A A child spotted a pelican and an owl. B A child spotted a kookaburra but did not spot an owl. C A child spotted a pelican but did not go to the wetland hide. D A child spotted an owl and went to the wetland hide.
Worked solution
1. Understand the question. Venn Diagram. Three statements, and the question asks what cannot be true.
2. Use the strategy. Draw first. "Every owl also kookaburra" puts the owl circle inside the kookaburra circle. "No kookaburra went to the hide" keeps the kookaburra circle and the hide circle apart. "Some hide children spotted a pelican" makes the pelican circle reach into the hide. The rest of the pelican circle is unknown, so draw it grey. Nothing keeps pelican away from kookaburra or owl. So the grey circle is free to stretch across and overlap the kookaburra circle, and even the owl circle inside it. A child could spot all three birds. "Cannot be true" means there is no picture where it happens. Check each option by looking for a picture that allows it.
3. Apply the strategy. Option pointers: A The grey pelican circle can stretch across to the owl circle. No statement stops it, because the only rule about pelican is that some hide children spotted one. Could be true. B The kookaburra circle is bigger than the owl circle, so there is room for kookaburra children with no owl. Could be true. C Some hide children spotted a pelican, but nobody said every pelican child was at the hide. A pelican child could be anywhere, even inside kookaburra. Could be true. D An owl child is inside kookaburra, and kookaburra never touches the hide. So an owl child can never be at the hide. Cannot be true.
4. Confirm the answer. D. Follow the owl child two steps: inside kookaburra, so outside the hide. The pelican circle has no such fence, so the pelican options all stay possible.
This tests Logic reasoning: Venn diagrams, part of Thinking Skills. The answer is D.
Show the worked solution
1. Understand the question. Venn Diagram. Three statements, and the question asks what cannot be true.
2. Use the strategy. Draw first. "Every owl also kookaburra" puts the owl circle inside the kookaburra circle. "No kookaburra went to the hide" keeps the kookaburra circle and the hide circle apart. "Some hide children spotted a pelican" makes the pelican circle reach into the hide. The rest of the pelican circle is unknown, so draw it grey. Nothing keeps pelican away from kookaburra or owl. So the grey circle is free to stretch across and overlap the kookaburra circle, and even the owl circle inside it. A child could spot all three birds. "Cannot be true" means there is no picture where it happens. Check each option by looking for a picture that allows it.
3. Apply the strategy. Option pointers: A The grey pelican circle can stretch across to the owl circle. No statement stops it, because the only rule about pelican is that some hide children spotted one. Could be true. B The kookaburra circle is bigger than the owl circle, so there is room for kookaburra children with no owl. Could be true. C Some hide children spotted a pelican, but nobody said every pelican child was at the hide. A pelican child could be anywhere, even inside kookaburra. Could be true. D An owl child is inside kookaburra, and kookaburra never touches the hide. So an owl child can never be at the hide. Cannot be true.
4. Confirm the answer. D. Follow the owl child two steps: inside kookaburra, so outside the hide. The pelican circle has no such fence, so the pelican options all stay possible.
This tests Logic reasoning: Venn diagrams. The answer is D.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 60 seconds a question.
Question 4

Thinking Skills, Rex's sample questions

1:00

At lunchtime, students who forget their hat stay under the veranda. Students on a lunchtime detention also stay under the veranda. Mo: "Ari was under the veranda at lunch, so Ari must have forgotten his hat." Lia: "Bella had her hat on today, so Bella was not under the veranda."
If the information above is true, whose reasoning is correct?

A Mo only B Lia only C Both Mo and Lia D Neither Mo nor Lia
Worked solution
1. Understand the question. Conditional Statement. Two sufficient conditions that lead to the same place.
2. Use the strategy. Write both rules. Rule 1: no hat, so veranda. Rule 2: detention, so veranda. Each has a contrapositive: not under the veranda, so had a hat; not under the veranda, so no detention. Now notice that two different things send a student to the veranda. So seeing someone there tells you one of the two happened, not which. And knowing one reason is missing does not clear the other reason.
3. Apply the strategy. Look at Mo. Ari was under the veranda, so he forgot his hat. That is "B, so A" on Rule 1, the converse. Ari might have been on detention with his hat on. Mo is wrong. Now Lia. Bella had her hat, so she was not under the veranda. That is "not A, so not B" on Rule 1, the inverse. Rule 1 only covers hatless students. Rule 2 could still put Bella there if she was on detention. Lia is wrong.
4. Confirm the answer. D. When two rules lead to the same place, the converse of one rule ignores the other rule, and so does the inverse.
This tests Logic reasoning: conditional statements, part of Thinking Skills. The answer is D.
Show the worked solution
1. Understand the question. Conditional Statement. Two sufficient conditions that lead to the same place.
2. Use the strategy. Write both rules. Rule 1: no hat, so veranda. Rule 2: detention, so veranda. Each has a contrapositive: not under the veranda, so had a hat; not under the veranda, so no detention. Now notice that two different things send a student to the veranda. So seeing someone there tells you one of the two happened, not which. And knowing one reason is missing does not clear the other reason.
3. Apply the strategy. Look at Mo. Ari was under the veranda, so he forgot his hat. That is "B, so A" on Rule 1, the converse. Ari might have been on detention with his hat on. Mo is wrong. Now Lia. Bella had her hat, so she was not under the veranda. That is "not A, so not B" on Rule 1, the inverse. Rule 1 only covers hatless students. Rule 2 could still put Bella there if she was on detention. Lia is wrong.
4. Confirm the answer. D. When two rules lead to the same place, the converse of one rule ignores the other rule, and so does the inverse.
This tests Logic reasoning: conditional statements. The answer is D.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 60 seconds a question.
Question 5

Thinking Skills, Rex's sample questions

1:00

A rug lies on the floor of a room. Mia stands at one side of the room and sees the rug as shown. Ben stands at the left-hand end of the room, where the plan shows.
Which picture shows the rug as Ben sees it?

A B C D
Worked solution
1. Understand the question. Graphic Reasoning, Basic Transformation. The same rug seen from another side of the room.
2. Use the strategy. A rug on the floor stays where it is; only the person moves. Walking round to another side turns the picture but never turns it over, because you always see the same top side of the rug. So find the end of the rug that is nearest to Ben, and put that end at the bottom of the picture. Then check one shape's direction to be sure.
3. Apply the strategy. Ben stands at the left-hand end of the room, so the end of the rug on Mia's left, the end with the flag, is nearest to him. Mia's far edge, the top row, is on Ben's left. So turn Mia's picture a quarter turn so that its left end comes to the bottom: the flag ends up bottom left, the triangle goes to the right-hand column pointing left, the arrow points up on the left, and the half-square ends up top right. That is C. A is a half turn, the view from the opposite side of the room. B is the view from the other end of the rug. D is a mirror image of Mia's view, and nobody can see a rug as its mirror image without turning it over.
4. Confirm the answer. C. From Ben's end the flag end is nearest, so Mia's picture turns a quarter turn with its left end at the bottom.
This tests Graphic reasoning: transformations, part of Thinking Skills. The answer is C.
Show the worked solution
1. Understand the question. Graphic Reasoning, Basic Transformation. The same rug seen from another side of the room.
2. Use the strategy. A rug on the floor stays where it is; only the person moves. Walking round to another side turns the picture but never turns it over, because you always see the same top side of the rug. So find the end of the rug that is nearest to Ben, and put that end at the bottom of the picture. Then check one shape's direction to be sure.
3. Apply the strategy. Ben stands at the left-hand end of the room, so the end of the rug on Mia's left, the end with the flag, is nearest to him. Mia's far edge, the top row, is on Ben's left. So turn Mia's picture a quarter turn so that its left end comes to the bottom: the flag ends up bottom left, the triangle goes to the right-hand column pointing left, the arrow points up on the left, and the half-square ends up top right. That is C. A is a half turn, the view from the opposite side of the room. B is the view from the other end of the rug. D is a mirror image of Mia's view, and nobody can see a rug as its mirror image without turning it over.
4. Confirm the answer. C. From Ben's end the flag end is nearest, so Mia's picture turns a quarter turn with its left end at the bottom.
This tests Graphic reasoning: transformations. The answer is C.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 60 seconds a question.
Question 6

Thinking Skills, Rex's sample questions

1:00

Mia wants to cut this tree shape out of a square of paper. She folds the square in half first, bringing the left edge over to the right edge, and then makes one cut through both layers.
Which packet shows the cut Mia makes?

A B C D
Worked solution
1. Understand the question. Graphic Reasoning, Folding and Unfolding. Working out the cut from the finished shape.
2. Use the strategy. The tree is symmetrical: its left half is the mirror of its right half. When the paper is folded in half and cut once, whatever is cut from one layer is cut from the other layer at the mirror spot. So Mia should cut half a tree, with the tree's middle line lying exactly along the fold. Work out where the fold is first. The left edge went over to the right edge, so the fold is on the left side of the packet.
3. Apply the strategy. A has half a tree cut against the left edge of the packet, which is the fold. Opened out, the two halves join along the fold and make the whole tree, upright. B has the half tree against the right edge, where the two loose edges are, so opened out it gives two separate half trees, one at each side of the square. C cuts a whole small tree, so the square ends up with two small trees. D has the half tree against the fold but with its point at the bottom, so it opens out into a tree standing on its head.
4. Confirm the answer. A. Half a tree cut along the fold opens out into the whole tree the right way up.
This tests Graphic reasoning: folding, part of Thinking Skills. The answer is A.
Show the worked solution
1. Understand the question. Graphic Reasoning, Folding and Unfolding. Working out the cut from the finished shape.
2. Use the strategy. The tree is symmetrical: its left half is the mirror of its right half. When the paper is folded in half and cut once, whatever is cut from one layer is cut from the other layer at the mirror spot. So Mia should cut half a tree, with the tree's middle line lying exactly along the fold. Work out where the fold is first. The left edge went over to the right edge, so the fold is on the left side of the packet.
3. Apply the strategy. A has half a tree cut against the left edge of the packet, which is the fold. Opened out, the two halves join along the fold and make the whole tree, upright. B has the half tree against the right edge, where the two loose edges are, so opened out it gives two separate half trees, one at each side of the square. C cuts a whole small tree, so the square ends up with two small trees. D has the half tree against the fold but with its point at the bottom, so it opens out into a tree standing on its head.
4. Confirm the answer. A. Half a tree cut along the fold opens out into the whole tree the right way up.
This tests Graphic reasoning: folding. The answer is A.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 60 seconds a question.
Question 7

Thinking Skills, Rex's sample questions

1:00

This tray has a square bottom and four sloping sides. Its top opening is wider than its bottom.
Which net folds into the tray?

A B C D
Worked solution
1. Understand the question. Graphic Reasoning, Net of an Irregular Shape. The net of a tray with sloping sides.
2. Use the strategy. A sloping side is a trapezium, a shape with two parallel edges of different lengths. Its shorter edge is the same length as a side of the bottom, and joins the bottom. Its longer edge is at the top, where the tray is wider. So the net must be a square with four matching trapeziums joined by their short edges, all widening outwards.
3. Apply the strategy. C has a small square with four matching trapeziums that widen as they go out. Folded up, the sides lean outwards and the opening is bigger than the bottom, just like the tray. A has a big square and trapeziums that narrow as they go out, so it folds into a tray that is wider at the bottom than at the top, the wrong way up. B has rectangles, which make a box with straight sides. In D two sides widen and two narrow, so the slanting edges next to each other are different lengths and the corners cannot close.
4. Confirm the answer. C. The small square is the bottom, and the four widening trapeziums are the sloping sides.
This tests Graphic reasoning: nets, part of Thinking Skills. The answer is C.
Show the worked solution
1. Understand the question. Graphic Reasoning, Net of an Irregular Shape. The net of a tray with sloping sides.
2. Use the strategy. A sloping side is a trapezium, a shape with two parallel edges of different lengths. Its shorter edge is the same length as a side of the bottom, and joins the bottom. Its longer edge is at the top, where the tray is wider. So the net must be a square with four matching trapeziums joined by their short edges, all widening outwards.
3. Apply the strategy. C has a small square with four matching trapeziums that widen as they go out. Folded up, the sides lean outwards and the opening is bigger than the bottom, just like the tray. A has a big square and trapeziums that narrow as they go out, so it folds into a tray that is wider at the bottom than at the top, the wrong way up. B has rectangles, which make a box with straight sides. In D two sides widen and two narrow, so the slanting edges next to each other are different lengths and the corners cannot close.
4. Confirm the answer. C. The small square is the bottom, and the four widening trapeziums are the sloping sides.
This tests Graphic reasoning: nets. The answer is C.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 60 seconds a question.
Question 8

Thinking Skills, Rex's sample questions

1:30

Six children, Nora, Otto, Pip, Quinn, Rae and Sol, each sang one song at a concert, one after another. - If Nora sang before Otto, then Pip sang last. - Sol sang some time after Pip. - Nora sang straight after Quinn. - Exactly one singer came between Nora and Pip. - Otto did not sing first.
Who sang second?

A Rae B Otto C Quinn D Pip
Worked solution
1. Understand the question. Logic Sequencing, order in a line. One clue is an "if" rule, and it only tells you something when its first part happens.
2. Use the strategy. Look at what the "if" rule promises: Pip sang last. If that cannot happen, then the first part cannot have happened either. That gives you a new fact to use. Then try the places Pip could take.
3. Apply the strategy. Sol sang after Pip, so Pip did not sing last. The rule says that if Nora sang before Otto, Pip sang last. Pip did not sing last, so Nora did not sing before Otto. That means Otto sang before Nora. Now place Pip. Quinn and Nora sang one straight after the other, and exactly one singer came between Nora and Pip. That leaves two ways. Pip could sing just before Quinn. Then Quinn is the one singer between Pip and Nora: Pip, Quinn, Nora. Or Pip could sing after Nora, with one singer between them: Quinn, Nora, one singer, Pip. Try the second way. Otto sang before Nora, so he also sang before Quinn. Otto did not sing first, so someone sang before him. Sol sang after Pip. Count them: someone, Otto, Quinn, Nora, one singer, Pip, Sol. That is seven singers, but only six sang. So the second way does not work. So the order has Pip, Quinn, Nora together. Otto sang before Nora, so he sang before Pip. Otto is not first, so someone sang before him. Sol sang after Pip, so it was not Sol. The only child left is Rae. The order is Rae, Otto, Pip, Quinn, Nora, Sol. Now check each option. A) Rae sang first. She could only be second if Otto sang first, and he did not. B) Otto sang second. This is the answer. C) Quinn could only be second in an order where Nora sang before Otto. The "if" rule does not allow that. D) Pip could only be second in an order where Nora sang before Otto and Pip was not last. The "if" rule does not allow that either.
4. Confirm the answer. B. Otto sang second.
This tests Logic sequencing, part of Thinking Skills. The answer is B.
Show the worked solution
1. Understand the question. Logic Sequencing, order in a line. One clue is an "if" rule, and it only tells you something when its first part happens.
2. Use the strategy. Look at what the "if" rule promises: Pip sang last. If that cannot happen, then the first part cannot have happened either. That gives you a new fact to use. Then try the places Pip could take.
3. Apply the strategy. Sol sang after Pip, so Pip did not sing last. The rule says that if Nora sang before Otto, Pip sang last. Pip did not sing last, so Nora did not sing before Otto. That means Otto sang before Nora. Now place Pip. Quinn and Nora sang one straight after the other, and exactly one singer came between Nora and Pip. That leaves two ways. Pip could sing just before Quinn. Then Quinn is the one singer between Pip and Nora: Pip, Quinn, Nora. Or Pip could sing after Nora, with one singer between them: Quinn, Nora, one singer, Pip. Try the second way. Otto sang before Nora, so he also sang before Quinn. Otto did not sing first, so someone sang before him. Sol sang after Pip. Count them: someone, Otto, Quinn, Nora, one singer, Pip, Sol. That is seven singers, but only six sang. So the second way does not work. So the order has Pip, Quinn, Nora together. Otto sang before Nora, so he sang before Pip. Otto is not first, so someone sang before him. Sol sang after Pip, so it was not Sol. The only child left is Rae. The order is Rae, Otto, Pip, Quinn, Nora, Sol. Now check each option. A) Rae sang first. She could only be second if Otto sang first, and he did not. B) Otto sang second. This is the answer. C) Quinn could only be second in an order where Nora sang before Otto. The "if" rule does not allow that. D) Pip could only be second in an order where Nora sang before Otto and Pip was not last. The "if" rule does not allow that either.
4. Confirm the answer. B. Otto sang second.
This tests Logic sequencing. The answer is B.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 90 seconds a question.
Question 9

Thinking Skills, Rex's sample questions

1:30

Some people were waiting in one line at an ice cream van. We are not told how many. - Omar had six people in front of him. - Pete had four people behind him. - Exactly three people stood between Omar and Pete. - Rosa stood somewhere behind Omar.
How many people were in the line?

A 7 B 11 C 14 D 15
Worked solution
1. Understand the question. Logic Sequencing, order in a line where you are not told how long the line is. You have to work it out.
2. Use the strategy. Build the line out of pieces: the people in front, one person, the people between, another person, and the people behind. Add the pieces to get the total. A gap between two people can go either way round, so try both ways.
3. Apply the strategy. Omar had six people in front of him, so Omar was 7th in the line. Pete had four people behind him. Three people stood between Omar and Pete. The clue does not say who was nearer the front, so try both ways. Try Pete in front of Omar. Three people between them puts Pete 3rd. Pete's four people behind him are 4th, 5th, 6th and 7th. So the line stops at Omar, and Omar is last. But Rosa stood behind Omar, so Omar was not last. So Pete was behind Omar. Now add up the pieces from the front to the back. 6 + 1 + 3 + 1 + 4 = 15. Now check each option. A) 7 is the line with Pete in front of Omar. Then nobody could be behind Omar. B) 11 leaves out the four people behind Pete. C) 14 forgets to count Pete himself. D) 15 counts every piece. This is the answer.
4. Confirm the answer. D. There were 15 people in the line.
This tests Logic sequencing, part of Thinking Skills. The answer is D.
Show the worked solution
1. Understand the question. Logic Sequencing, order in a line where you are not told how long the line is. You have to work it out.
2. Use the strategy. Build the line out of pieces: the people in front, one person, the people between, another person, and the people behind. Add the pieces to get the total. A gap between two people can go either way round, so try both ways.
3. Apply the strategy. Omar had six people in front of him, so Omar was 7th in the line. Pete had four people behind him. Three people stood between Omar and Pete. The clue does not say who was nearer the front, so try both ways. Try Pete in front of Omar. Three people between them puts Pete 3rd. Pete's four people behind him are 4th, 5th, 6th and 7th. So the line stops at Omar, and Omar is last. But Rosa stood behind Omar, so Omar was not last. So Pete was behind Omar. Now add up the pieces from the front to the back. 6 + 1 + 3 + 1 + 4 = 15. Now check each option. A) 7 is the line with Pete in front of Omar. Then nobody could be behind Omar. B) 11 leaves out the four people behind Pete. C) 14 forgets to count Pete himself. D) 15 counts every piece. This is the answer.
4. Confirm the answer. D. There were 15 people in the line.
This tests Logic sequencing. The answer is D.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 90 seconds a question.
Question 10

Thinking Skills, Rex's sample questions

1:00

A café's lunch deal is one main, one drink and one dessert. The mains are pie, wrap and salad. The drinks are water, juice, milk and milkshake. The desserts are fruit and cake. The café will not sell the pie with a milkshake.
How many different lunch deals can the café sell?

A 24 B 23 C 22 D 9
Worked solution
1. Understand the question. Problem Solving / Counting. Count all the deals first, then take away the ones that break the rule.
2. Use the strategy. Multiply to count every deal: mains × drinks × desserts. Then count only the deals the café will not sell. Pie with a milkshake can come with either dessert. Take those deals away from the total.
3. Apply the strategy. All deals: 3 mains times 4 drinks times 2 desserts is 24. Two deals are not sold: pie and milkshake with fruit, and pie and milkshake with cake. 24 take away 2 is 22.
4. Confirm the answer. C. There are 3 × 4 × 2 = 24 deals in all. Pie with a milkshake can come with fruit or with cake, so 2 deals are not sold. 24 - 2 = 22. Careful: taking away only 1 deal gives 23, because it forgets the second dessert.
This tests Problem solving: counting and chance, part of Thinking Skills. The answer is C.
Show the worked solution
1. Understand the question. Problem Solving / Counting. Count all the deals first, then take away the ones that break the rule.
2. Use the strategy. Multiply to count every deal: mains × drinks × desserts. Then count only the deals the café will not sell. Pie with a milkshake can come with either dessert. Take those deals away from the total.
3. Apply the strategy. All deals: 3 mains times 4 drinks times 2 desserts is 24. Two deals are not sold: pie and milkshake with fruit, and pie and milkshake with cake. 24 take away 2 is 22.
4. Confirm the answer. C. There are 3 × 4 × 2 = 24 deals in all. Pie with a milkshake can come with fruit or with cake, so 2 deals are not sold. 24 - 2 = 22. Careful: taking away only 1 deal gives 23, because it forgets the second dessert.
This tests Problem solving: counting and chance. The answer is C.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 60 seconds a question.
Question 11

Thinking Skills, Rex's sample questions

1:30

An empty tank is filled by two taps. Tap A can fill the whole tank in 6 hours by itself, and Tap B can fill it in 3 hours by itself. The tank also has a small leak. The leak alone would empty a full tank in 12 hours. Both taps and the leak are open together.
How long does it take to fill the empty tank?

A 2 hours B 2 hours 24 minutes C 2 hours 40 minutes D 3 hours
Worked solution
1. Understand the question. Problem Solving / Rate. Each tap and the leak works at its own rate: a part of the tank every hour.
2. Use the strategy. Write each rate as a fraction of the tank for one hour. Add the two taps and take away the leak. Then work out how long the whole tank takes at that rate.
3. Apply the strategy. Tap A fills one sixth of the tank each hour, Tap B one third, and the leak empties one twelfth. Together five twelfths each hour, so the tank takes 12 fifths of an hour, which is 2 hours 24 minutes.
4. Confirm the answer. B. Together the taps and the leak fill 1/6 + 1/3 - 1/12 = 5/12 of the tank each hour. The whole tank takes 1 ÷ 5/12 = 12/5 hours, which is 2 hours 24 minutes. Careful: 2 hours forgets the leak, and 2 hours 40 minutes comes from reading 2.4 hours as 2 hours 40 minutes.
This tests Problem solving: rate, part of Thinking Skills. The answer is B.
Show the worked solution
1. Understand the question. Problem Solving / Rate. Each tap and the leak works at its own rate: a part of the tank every hour.
2. Use the strategy. Write each rate as a fraction of the tank for one hour. Add the two taps and take away the leak. Then work out how long the whole tank takes at that rate.
3. Apply the strategy. Tap A fills one sixth of the tank each hour, Tap B one third, and the leak empties one twelfth. Together five twelfths each hour, so the tank takes 12 fifths of an hour, which is 2 hours 24 minutes.
4. Confirm the answer. B. Together the taps and the leak fill 1/6 + 1/3 - 1/12 = 5/12 of the tank each hour. The whole tank takes 1 ÷ 5/12 = 12/5 hours, which is 2 hours 24 minutes. Careful: 2 hours forgets the leak, and 2 hours 40 minutes comes from reading 2.4 hours as 2 hours 40 minutes.
This tests Problem solving: rate. The answer is B.
Chosen by Rex for the website, 9 November 2026, with his own answer and four-step explanation. The clock runs at the real test's pace: 40 questions in 40 minutes is about 90 seconds a question.