Question 01
The greatest common factor of two numbers, Jia and Yi, is 7. Jia’s number divided by Yi’s number is 1 18. What is Yi’s number?
General Test 2
Lesson 29.2 · Grade 5 · 13 questions
Test 29.2 · Chapter 29 · General tests
General Test 2Choose a method. Check every condition.
Thirteen more questions. Turn familiar ideas into clear solutions: track what changes, use every condition, and explain why your answer works.
Comprehensive practice
13questions
10 short-answer questions
3 questions with reasoning
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Enter numbers without units; units appear beside the boxes. Decimals, fractions such as 3/2, and mixed numbers such as 1 1/2 are accepted. Use an exact fraction when a decimal repeats.
Questions 11–13 also ask for reasoning. You can write it here or work on paper. Use “Mark to revisit” for a question you want to return to.
01
Part I · 10 questions · 60 points
Fill in the answers.
Each question is worth 6 points. Keep the requested answer and unit in mind.
Question 02
Jingjing walks upstairs. It takes 36 steps to go from the first floor to the third floor. Each pair of neighboring floors has the same number of steps between them. How many steps does it take to go from the first floor to the sixth floor?
Working notes (optional)
Question 03
There are 11 consecutive natural numbers, arranged from smallest to largest. The 10th number is 1 49 times the 2nd number. What is the sum of all 11 numbers?
Working notes (optional)
Question 04
An airplane carries enough fuel for 6 hours of flight. On the outward journey, a tailwind gives it a speed of 1,500 km/h. On the return journey, a headwind gives it a speed of 1,200 km/h. What is the greatest distance it can fly outward before turning back, so that it can still return on that fuel?
Assume a constant fuel-use rate throughout both flights.
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Question 05
A point O lies inside quadrilateral ABCD. The perpendicular distance from O to each of the four sides is 2 cm. The perimeter of ABCD is 18 cm. What is the area of ABCD?
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Question 06
Multiply consecutive natural numbers starting at 1: 1 × 2 × 3 × ⋯. The product has exactly 13 zeros at its end. What is the smallest possible last natural number in the product?
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Question 07
In a test, the average score of five students A, B, C, D, and E is 90. The average of A and B is 96; the average of C and D is 92.5; and the average of A and D is 97.5. C scores 15 points fewer than D. What is B’s score?
Working notes (optional)
Question 08
Four soccer teams, A, B, C, and D, play a tournament. Every pair of teams plays one match. A finishes with 2 wins and 1 loss; B with 2 wins and 1 draw; and C with 1 win and 2 losses. What is D’s record?
Working notes (optional)
Question 09
A two-digit number leaves a remainder of 1 when divided by 5 and a remainder of 4 when divided by 8. What is the smallest possible number?
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Question 10
A natural number has exactly 8 positive divisors, including 1 and the number itself. Two of its divisors are 6 and 21. What is the number?
Working notes (optional)
02
Part II · 3 questions · 60 points
Show your reasoning.
Each question is worth 20 points. Make the steps clear enough for someone else to follow.
Question 11
Farmers in Bridge Village use a group of boats to carry 90 bags of fertilizer across a river in three trips. Every boat carries the same number of bags, and each carries at least 2 bags. How many boats could be used on each trip, and how many bags could each boat carry?
Stated convention for this page
Use the same group of at least two boats, with the same load per boat on all three trips. List every possible boat/load pair once, in any row order.
| Case | Boats per trip | Bags per boat per trip |
|---|---|---|
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 |
A matching final answer does not earn automatic method marks. Your written solution needs a person’s review.
Question 12
A shoe factory makes leather shoes in 10 quality grades. The lowest grade is grade 1, which earns a profit of 24 yuan per pair. Each increase of one grade raises the profit per pair by 6 yuan.
The factory can make 180 pairs per day at grade 1. Each increase of one grade reduces daily production by 9 pairs. Which grade gives the greatest daily profit, and what is that maximum profit?
The factory can make 180 pairs per day at grade 1. Each increase of one grade reduces daily production by 9 pairs. Which grade gives the greatest daily profit, and what is that maximum profit?
A matching final answer does not earn automatic method marks. Your written solution needs a person’s review.
Question 13
For any positive integers a and b, a new operation △ is defined by
a △ b = 6abma + 2b
where m is a fixed positive integer. Given that 1 △ 2 = 2, find 2 △ 9.A matching final answer does not earn automatic method marks. Your written solution needs a person’s review.
Revisit Chapter 19 for more practice →
56
Write the ratio in lowest terms.
Jia ÷ Yi = 1 18 = 9/8. So Jia’s number is 9k and Yi’s number is 8k for a positive whole number k.
Use the greatest common factor.
Because 9 and 8 share no factor greater than 1, the greatest common factor of 9k and 8k is k. Therefore k = 7.
Find Yi’s number and check.
Yi = 8 × 7 = 56. Jia = 9 × 7 = 63. Their greatest common factor is 7, and 63 ÷ 56 = 9/8.
Watch for this: The quotient gives Jia : Yi = 9 : 8, not 8 : 9. The question asks for Yi, the smaller number.
Revisit Chapter 22 for more practice →
90 steps
Count the first journey’s intervals.
From floor 1 to floor 3 there are 3 − 1 = 2 equal intervals.
Find the steps in one interval.
36 ÷ 2 = 18 steps between neighboring floors.
Count the new journey’s intervals.
From floor 1 to floor 6 there are 6 − 1 = 5 intervals. The total is 18 × 5 = 90 steps.
Watch for this: Six floor numbers do not make six intervals when the journey starts on the first floor.
Revisit Chapter 22 for more practice →
242
Express the two positions using one number.
Let the 2nd number be a. The 10th number is a + (10 − 2) = a + 8.
Use the multiplier.
The condition is a + 8 = 1 49 × a. Subtract a from both sides: (4/9)a = 8. Thus a = 8 × 9 ÷ 4 = 18.
Sum the whole sequence.
The first number is 17 and the last is 27. Their sum is (17 + 27) × 11 ÷ 2 = 242. As a check, the middle number is 22, and 22 × 11 = 242.
Watch for this: The 2nd number is not the first number. After finding 18, the sequence starts at 17.
Revisit Chapter 7 for more practice →
4,000 km
Use the same distance for both legs.
Let S be the greatest outward distance in kilometers. Outward time is S/1500 hours; return time is S/1200 hours.
Use the full flight-time allowance.
S/1500 + S/1200 = 6. Multiply by 6,000 to get 4S + 5S = 36,000. Hence 9S = 36,000 and S = 4,000 km.
Check the two times.
Outward: 4,000 ÷ 1,500 = 8/3 hours. Return: 4,000 ÷ 1,200 = 10/3 hours. Together they use 18/3 = 6 hours. Any greater distance would take more time.
Watch for this: Do not give each journey 3 hours. The slower return journey needs more time. The answer is the one-way distance, not the total distance flown.
Revisit Chapter 4 for more practice →
18 cm²
Split the quadrilateral into four triangles.
Join O to A, B, C, and D. The triangles OAB, OBC, OCD, and ODA together make the quadrilateral.
Add the triangle areas.
Each triangle has height 2 cm measured perpendicular to its corresponding side. Therefore the total area is ½ × 2 × (AB + BC + CD + DA).
Use the perimeter.
AB + BC + CD + DA = 18 cm. The area is ½ × 2 × 18 = 18 cm². The four side lengths do not need to be found separately.
Watch for this: The 2 cm measurements are perpendicular heights, not necessarily lengths from O to the vertices. Keep the factor ½ in every triangle area.
Revisit Chapter 18 for more practice →
55
Count factors of 5, not just multiples of 10.
In 1 × 2 × ⋯ × n, each paired factor 2 and factor 5 makes one factor 10. There are more factors 2 than factors 5, so the number of trailing zeros equals the number of factors 5.
Check just before the answer.
Up to 54, the multiples 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50 supply ten factors 5. Both 25 and 50 supply one additional factor 5. That gives 12 trailing zeros.
Add the next needed factor.
The factor 55 supplies one more factor 5, making 13. Thus 1 × 2 × ⋯ × 55 has exactly 13 trailing zeros.
Explain why it is the smallest.
All shorter products end at 54 or earlier and have at most 12 factors 5. Therefore the smallest possible last factor is 55.
Watch for this: 25 contributes two factors 5, and 50 does too. Also, “smallest possible” needs a check that no earlier stopping point works.
Revisit Chapter 3 for more practice →
97 points
Find C and D from their total and difference.
C + D = 92.5 × 2 = 185, and C = D − 15. So (D − 15) + D = 185, giving D = 100 and C = 85.
Use the A-and-D average.
A + D = 97.5 × 2 = 195. Thus A = 195 − 100 = 95.
Use the A-and-B average.
A + B = 96 × 2 = 192. Thus B = 192 − 95 = 97.
Check the remaining information.
The five scores total 90 × 5 = 450. The four known scores total 95 + 97 + 85 + 100 = 377, so E = 73. The overall average is consistent. It was not necessary to find E to answer the question.
Watch for this: 92.5 is the average of C and D, not their total. Multiply each pair average by 2 before using it.
Revisit Chapter 10 for more practice →
0 wins, 1 draw, 2 losses
Find D’s draw.
B has one draw. A and C have none, so B’s drawn match must have been against D. Thus D has one draw and two other matches.
Balance wins and losses.
A, B, and C together record 2 + 2 + 1 = 5 wins, but only 1 + 0 + 2 = 3 losses. D’s two other matches must provide two more losses and no wins.
State the full record.
D has 0 wins, 1 draw, and 2 losses. Across all four teams there are now 5 wins and 5 losses, plus two draw records for one drawn match.
Watch for this: One drawn match appears in two team records. Do not introduce a points system: only wins, draws, and losses are needed.
Revisit Chapter 16 for more practice →
36
Add the same amount for both divisions.
Adding 4 changes the remainder 1 modulo 5 into 5, a complete group of 5. It also changes the remainder 4 modulo 8 into 8, a complete group of 8.
Find a common multiple.
The number plus 4 must be a multiple of both 5 and 8. Their least common multiple is 40.
Subtract and check the size.
40 − 4 = 36, which is two-digit. Also, 36 = 5 × 7 + 1 and 36 = 8 × 4 + 4. No smaller positive common multiple gives a smaller two-digit candidate.
Watch for this: The number itself is not a common multiple of 5 and 8. The number plus 4 is.
Revisit Chapter 18 for more practice →
42
Use both known divisors.
A multiple of both 6 and 21 is a multiple of their least common multiple, 42.
List the forced divisors.
Every divisor of 42 must divide the unknown number. These are 1, 2, 3, 6, 7, 14, 21, and 42—already eight different divisors.
Use the exact divisor count.
If the unknown number were greater than 42, it would itself be a ninth divisor, different from all eight listed above. That is impossible. Therefore the number is 42.
Watch for this: Being divisible by 6 and 21 is not enough by itself. The condition “exactly 8 divisors” determines the number.
Revisit Chapter 24 for more practice →
Six distinct arrangements, listed below.
Find the load per trip.
Under the stated convention, the same total is moved on each trip. So each trip moves 90 ÷ 3 = 30 bags.
Turn the question into a product.
Let b be the boats per trip and c the bags per boat per trip. Then b × c = 30, with whole numbers b ≥ 2 and c ≥ 2.
List every permitted factor pair.
| Boats per trip | Bags per boat per trip | Bags over 3 trips |
|---|---|---|
| 2 | 15 | 3 × 2 × 15 = 90 |
| 3 | 10 | 3 × 3 × 10 = 90 |
| 5 | 6 | 3 × 5 × 6 = 90 |
| 6 | 5 | 3 × 6 × 5 = 90 |
| 10 | 3 | 3 × 10 × 3 = 90 |
| 15 | 2 | 3 × 15 × 2 = 90 |
Explain why the list is complete.
The positive divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The choice b = 1 is excluded by the stated multiple-boat convention. The choice b = 30 would give only 1 bag per boat, so it is excluded by the load rule. All six remaining choices are listed.
Watch for this: Reversing the two factors changes the arrangement: 2 boats carrying 15 bags each is different from 15 boats carrying 2 bags each. A repeated pair is not a new possibility.
Check your written method: Does your method first find 30 bags per trip, list all six distinct pairs, and explain the two exclusions? The checker accepts the pairs in any order; a person must review the explanation.
Revisit Chapter 20 for more practice →
Grade 9 · 7,776 yuan per day
Express production and unit profit for grade n.
Grade n is n − 1 upgrades above grade 1. Pairs per day = 180 − 9(n − 1) = 189 − 9n = 9(21 − n). Profit per pair = 24 + 6(n − 1) = 18 + 6n = 6(n + 3), where n is one of 1, 2, …, 10.
Write the daily profit as a product.
Daily profit = 9(21 − n) × 6(n + 3) = 54(21 − n)(n + 3) yuan. The two changing factors have a constant sum: (21 − n) + (n + 3) = 24.
Make the equal-sum factors equal.
For two positive factors with a fixed sum, their product is largest when the factors are equal. Here 21 − n = n + 3, giving n = 9. This is within the allowed grades 1–10. Both factors then equal 12.
Calculate and verify the daily profit.
Maximum daily profit = 54 × 12 × 12 = 7,776 yuan. At grade 9 the factory makes 180 − 8 × 9 = 108 pairs and earns 24 + 8 × 6 = 72 yuan per pair. The check is 108 × 72 = 7,776.
Watch for this: Grade 9 means eight upgrades, not nine. Also, 72 yuan is profit per pair; the question asks for the total profit for one day.
Check your written method: Does your explanation account for both changing quantities, identify grade 9 as the maximum among the ten permitted grades, and calculate 7,776 yuan per day? A complete comparison table is another valid method.
Check all ten grades in a comparison table
| Grade | Pairs / day | Yuan / pair | Yuan / day |
|---|---|---|---|
| 1 | 180 | 24 | 4,320 |
| 2 | 171 | 30 | 5,130 |
| 3 | 162 | 36 | 5,832 |
| 4 | 153 | 42 | 6,426 |
| 5 | 144 | 48 | 6,912 |
| 6 | 135 | 54 | 7,290 |
| 7 | 126 | 60 | 7,560 |
| 8 | 117 | 66 | 7,722 |
| 9 · maximum | 108 | 72 | 7,776 |
| 10 | 99 | 78 | 7,722 |
Revisit Chapter 25 for more practice →
4 1011 = 5411
Use the given operation to determine m.
1 △ 2 = 6 × 1 × 2m × 1 + 2 × 2 = 12m + 4 = 2. Therefore 12 = 2(m + 4), so m + 4 = 6 and m = 2.
Substitute into the same definition.
Now 2 △ 9 = 6 × 2 × 92 × 2 + 2 × 9 = 10822.
Simplify and keep the answer exact.
10822 = 5411 = 4 1011. You can use a mixed number or an equivalent fraction. The checker also accepts an equivalent fraction such as 54/11 or 108/22.
Watch for this: △ is not ordinary multiplication. The m you find is fixed, and the denominator is ma + 2b—not m(a + 2b). Do not round a repeating decimal.
Check your written method: Does your working first find m = 2, substitute the values into the numerator and denominator, and simplify the exact fraction?