Question 01
Calculate:
1.2 × 67 + 6.7 × 88
General Test 5
Lesson 29.5 · Grade 5 · 13 questions
Test 29.5 · Chapter 29 · General tests
General Test 5Connect the clues. Justify your choices.
Thirteen mixed problems, from clever calculations to answer-sheet logic and a changing river. Choose a method, explain the connections, and check every condition.
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.
Ten mixed questions · 6 points each · fill in every requested answer.
Question 02
Use 1, 2, 3, 4, 5 and +, −, ×, ÷ to form a calculation, with no parentheses. What is the greatest possible result?
Rule clarification: use each of the five numbers and each of the four operation signs exactly once. Do not join digits into a larger number. Follow ordinary operation order.
Working notes (optional)
Question 03
An item is sold at either 80% of its original price or 60% of its original price. The two selling prices differ by 4.8 yuan. What was the original price?
Working notes (optional)
Question 04
A two-digit number is divided by a nonzero one-digit number. The whole-number quotient is the smallest two-digit number. What is the greatest possible dividend?
Division convention: a remainder is allowed. The remainder must be nonnegative and smaller than the divisor.
Working notes (optional)
Question 05
A road connects A and B. Jia’s vehicle takes 60 minutes to travel from A to B; Yi’s vehicle takes 120 minutes to travel from B to A. They start at the same time from their respective ends and travel toward each other at their constant speeds. How many minutes pass before they meet?
Working notes (optional)
Question 06
For positive numbers x and y, a new operation ∗ is defined by:
x ∗ y =
Here m is a fixed natural number. Given that 1 ∗ 2 = , find m and 2 ∗ 6.
Working notes (optional)
Question 07
Start with the two numbers 515 and 53. In one operation, subtract 11 from the first number and add 11 to the second. After how many operations will the two numbers be equal?
Working notes (optional)
Question 08
In the diagram, AB = 24 cm. BDEF is a rectangle with EF = 15 cm. The shaded triangle BCE has area 60 cm². Find the area of triangle DCE.
Working notes (optional)
Question 09
A basket contains some peaches. Counting them in groups of 4 leaves 2; counting in groups of 6 leaves 4; counting in groups of 8 leaves the basket 2 short of a full group. There are at least 120 and at most 150 peaches. How many peaches are there?
Working notes (optional)
Question 10
A natural number N has ones digit 0 and exactly 8 positive divisors. What is the smallest possible value of N?
Working notes (optional)
02
Part II · 3 questions · 60 points
Show your reasoning.
Three extended problems · 20 points each · give your answers and explain the steps.
Question 11
Students A, B, and C take a test with 10 true-or-false questions. Each question is worth 10 points, for a total of 100. A statement judged true is marked ✓; a statement judged false is marked ×. Their answer sheets are shown below.
| Student / question | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| A | × | ✓ | ✓ | ✓ | × | ✓ | × | × | ✓ | × |
| B | × | × | ✓ | ✓ | ✓ | × | ✓ | ✓ | × | × |
| C | ✓ | × | ✓ | × | ✓ | ✓ | ✓ | × | ✓ | ✓ |
✓ = true · × = false. All three students scored 70 out of 100. Scroll the table sideways on a narrow screen.
All three students receive 70 points. Find the correct answer to each of the ten questions, and explain how you know.
Your reasoning is for a person to review. It is not automatically graded.
Question 12
There are square cardboard panels and rectangular cardboard panels. The ratio of the number of square panels to the number of rectangular panels is 2 : 5. All of the panels are used to make two types of open boxes without lids, as shown.
A vertical box uses 1 square panel for its bottom and 4 rectangular panels for its sides. A horizontal box uses 1 rectangular panel for its bottom, plus 2 square panels and 2 rectangular panels for its sides.
What is the ratio of the number of vertical boxes to horizontal boxes? Show your reasoning.
Keep vertical first. Equivalent positive ratios, such as a ratio scaled by the same factor on both sides, are accepted. A colon is preferred; a fraction is also accepted.
Your reasoning is for a person to review. It is not automatically graded.
Question 13
A boat travels back and forth between ports Jia and Yi. The journey from Jia to Yi is downstream; the return journey is upstream. The boat’s speed in still water is 8 km/h. Normally, the time taken upstream : the time taken downstream is 2 : 1.
One day, heavy rain makes the current twice as fast as usual. The boat takes 9 hours in total for the downstream and upstream journeys that day. How far apart are the ports? Show your reasoning.
Your reasoning is for a person to review. It is not automatically graded.
Revisit Chapter 1 for more practice →
670
Rewrite the second product.
Rewrite 6.7 × 88 as 67 × 8.8: one factor becomes ten times as large, while the other becomes one tenth as large. The product is unchanged.
Use the common factor.
67 × 1.2 + 67 × 8.8
= 67 × (1.2 + 8.8)
= 67 × 10 = 670
= 67 × (1.2 + 8.8)
= 67 × 10 = 670
Watch for this: Change both factors in opposite ways. Multiplying just one of them by 10 changes the product.
Revisit Chapter 24 for more practice →
22.5
Try this construction.
5 × 4 + 3 − 1 ÷ 2
Perform multiplication and division first.
The product is 20 and the quotient is 1/2. Thus:
20 + 3 − = 22.5
Check every rule.
Each of 1, 2, 3, 4, 5 appears once. Each operation appears once, and there are no parentheses. The optional audit below adds an exhaustive maximum check to the problem’s displayed construction.
Watch for this: Do not calculate the expression strictly from left to right, and do not leave out the subtraction or division.
Explore: is this the greatest value?
The worked example gives a maximizing expression. This check examines every ordering of the five numbers and four operation signs, using exact fractions and ordinary operation order. It is not another test question.
There are 120 number orders and 24 operation orders, so 2,880 expressions to check. Digits are not joined and parentheses are not inserted.
No audit has been run.
Exhaustive check: among the 2,880 allowed expressions, the maximum is 45/2 = 22.5.
Revisit Chapter 9 for more practice →
24 yuan
Find the difference as a share of the original.
80% − 60% = 20%
Recover the whole price.
4.8 ÷ 20% = 24 yuan
Check the two selling prices.
80% of 24 is 19.2 yuan. 60% of 24 is 14.4 yuan. Their difference is 4.8 yuan.
Watch for this: “Sold at 80% of the original price” does not mean “80% off.”
Revisit Chapter 16 for more practice →
98
Identify the quotient and the largest divisor.
The smallest two-digit number is 10. The largest one-digit divisor is 9.
Include the largest allowed remainder.
For divisor 9, the largest remainder is 8. Therefore:
9 × 10 + 8 = 98
Explain why no larger dividend works.
For any allowed divisor d, the dividend is at most 10d + (d − 1) = 11d − 1. This increases with d and reaches 98 when d = 9. Indeed, 98 ÷ 9 is 10 remainder 8.
Watch for this: Choosing 90 unnecessarily requires exact division. Choosing 99 gives quotient 11 when the divisor is 9.
Revisit Chapter 7 for more practice →
40 minutes
Turn travel times into a speed ratio.
For the same distance, speed is inversely proportional to time. Jia’s speed : Yi’s speed is 2 : 1.
Find Yi’s share of the road.
When they meet, Jia has traveled twice as far as Yi. Yi has therefore covered 1/3 of the whole road.
Find the time.
120 × = 40 minutes
In 40 minutes Jia covers 40/60 = 2/3 of the road, and Yi covers 40/120 = 1/3. Together they cover the whole road.
Watch for this: Do not average the two full-trip times. Neither vehicle completes the whole road before they meet.
Revisit Chapter 25 for more practice →
m = 1; 2 ∗ 6 = 6/7
Use the given operation value.
1 ∗ 2 = = =
Hence m + 4 = 5, so m = 1.
Use the definition again, with new inputs.
2 ∗ 6 = = =
Watch for this: ∗ is the newly defined operation, not ordinary multiplication. An approximate decimal for 6/7 is not an exact answer.
Revisit Chapter 9 for more practice →
21 operations
Find the starting difference.
515 − 53 = 462
Find how fast the gap closes.
The larger number drops by 11 while the smaller number rises by 11. The difference shrinks by 22 per operation.
Count the operations and verify.
462 ÷ 22 = 21
After 21 operations, 515 − 21 × 11 = 284 and 53 + 21 × 11 = 284.
Watch for this: Dividing the difference by 11 ignores the simultaneous increase of the second number.
Revisit Chapter 5 for more practice →
30 cm²
Find triangle ABE.
AB is a horizontal base, and EF is its perpendicular height from E.
Area ABE = 24 × 15 ÷ 2 = 180 cm²
Subtract to find BC.
Area ABC = 180 − 60 = 120 cm²
Triangle ABC is right-angled at B, so:
24 × BC ÷ 2 = 120
BC = 10 cm
BC = 10 cm
Find DC and the rectangle’s width.
The rectangle has BD = EF = 15 cm, hence DC = 15 − 10 = 5 cm. In triangle BCE, use vertical base BC and horizontal height DE:
10 × DE ÷ 2 = 60
DE = 12 cm
DE = 12 cm
Calculate the requested area.
Area DCE = DE × DC ÷ 2
= 12 × 5 ÷ 2 = 30 cm²
= 12 × 5 ÷ 2 = 30 cm²
Watch for this: The given 60 cm² belongs to the shaded triangle BCE. The question asks for the different, unshaded triangle DCE.
Revisit Chapter 16 for more practice →
142 peaches
Express all three clues in the same way.
Two more peaches would make the total divisible by 4, 6, and 8. Therefore N + 2 is a common multiple of these numbers.
Use the allowed interval.
LCM(4, 6, 8) = 24
N + 2 lies from 122 to 152. The only multiple of 24 in that interval is 144. Thus N = 144 − 2 = 142.
Check all three counts.
142 = 4 × 35 + 2
142 = 6 × 23 + 4
142 = 8 × 18 − 2
142 = 6 × 23 + 4
142 = 8 × 18 − 2
Watch for this: “Two short” of a multiple of 8 means remainder 6, not remainder 2.
Revisit Chapter 18 for more practice →
30
Start with the smallest candidates.
| Number | Positive divisors | Count |
|---|---|---|
| 10 | 1, 2, 5, 10 | 4 |
| 20 | 1, 2, 4, 5, 10, 20 | 6 |
| 30 | 1, 2, 3, 5, 6, 10, 15, 30 | 8 |
Conclude minimality.
Both smaller positive numbers ending in 0 have too few divisors, while 30 has exactly eight. Thus 30 is the smallest possible value. Zero does not have exactly eight positive divisors.
Watch for this: Count both 1 and the number itself, and do not count a divisor twice.
Revisit Chapter 11 for more practice →
× × ✓ ✓ ✓ ✓ ✓ × ✓ ×
Compare A and B’s six disagreements.
A and B differ on questions 2, 5, 6, 7, 8, 9. For each of those six questions, one student is right and one is wrong. Together, those columns account for six correct responses.
Their four agreements must all be correct.
A and B each have seven correct responses, giving 14 together. The four agreeing columns must therefore supply 14 − 6 = 8 correct responses. Each can supply at most two, so all four must be correct. This gives 1: ×, 3: ✓, 4: ✓, 10: ×.
Repeat the same comparison.
B and C also disagree in six places. Their agreements give 2: ×, 3: ✓, 5: ✓, 7: ✓. A and C’s agreements give 3: ✓, 6: ✓, 8: ×, 9: ✓. These cover every question.
| Question | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Correct answer | × | × | ✓ | ✓ | ✓ | ✓ | ✓ | × | ✓ | × |
Check all three scores.
| Student | Correct on questions | Score |
|---|---|---|
| A | 1, 3, 4, 6, 8, 9, 10 | 70 / 100 |
| B | 1, 2, 3, 4, 5, 7, 10 | 70 / 100 |
| C | 2, 3, 5, 6, 7, 8, 9 | 70 / 100 |
Every row contains exactly seven matches with the key. The pairwise deductions determine all ten entries, not merely one possible set.
Watch for this: The majority choices happen to match this key, but agreement alone is not a proof. Use the 70-point scores to show why the shared responses must be correct.
Explore: compare two answer sheets
This viewer unpacks the pairwise comparison in the worked solution. The shaded columns are the four places where the chosen pair agrees.
A and B agree on 1, 3, 4, 10; B and C agree on 2, 3, 5, 7; A and C agree on 3, 6, 8, 9. These agreements determine all ten answers.
Revisit Chapter 6 for more practice →
4 : 3
Count the panels per box.
| Type of box | Square panels | Rectangular panels |
|---|---|---|
| Vertical | 1 | 4 |
| Horizontal | 2 | 3 |
Write the problem’s ratio equation.
Let a be the number of vertical boxes and b the number of horizontal boxes. Then the total numbers of square and rectangular panels are a + 2b and 4a + 3b.
(a + 2b) : (4a + 3b) = 2 : 5
Solve for the ratio of boxes.
5(a + 2b) = 2(4a + 3b)
5a + 10b = 8a + 6b
4b = 3a
a : b = 4 : 3
5a + 10b = 8a + 6b
4b = 3a
a : b = 4 : 3
Check with the smallest whole-number mix.
Four vertical boxes and three horizontal boxes need 4 + 2 × 3 = 10 square panels, and 4 × 4 + 3 × 3 = 25 rectangular panels. Their ratio is 10 : 25 = 2 : 5.
Watch for this: The requested order is vertical : horizontal. Also, a horizontal box uses three rectangular panels, including its bottom.
Explore: check a proposed mix
Each vertical box needs 1 square and 4 rectangles. Each horizontal box needs 2 squares and 3 rectangles. This ledger checks the panel ratio only; it does not change your submitted answer.
Four vertical and three horizontal boxes use 10 square panels and 25 rectangular panels, giving the required panel ratio 2:5.
Revisit Chapter 7 for more practice →
20 km
Find the usual speed of the current.
Let the usual current speed be a km/h. Upstream speed is 8 − a and downstream speed is 8 + a. A time ratio of 2 : 1 gives a speed ratio of 1 : 2.
(8 − a) : (8 + a) = 1 : 2
8 + a = 2(8 − a)
3a = 8, so a = km/h
8 + a = 2(8 − a)
3a = 8, so a = km/h
Double only the current speed.
In the rain, the current speed becomes 2a = 16/3 km/h. The boat’s still-water speed remains 8 km/h.
| Rainy-day direction | Speed (km/h) |
|---|---|
| Downstream | 8 + 16/3 = 40/3 |
| Upstream | 8 − 16/3 = 8/3 |
Add the two travel times.
Let the distance between the ports be x km. Both journeys have this same distance.
+ = 9
+ = 9
= 9
x = 20 km
+ = 9
= 9
x = 20 km
Verify the nine-hour round trip.
Downstream: 20 ÷ (40/3) = 1.5 hours. Upstream: 20 ÷ (8/3) = 7.5 hours. The total is 9 hours.
Watch for this: The usual 2 : 1 time ratio is not the rainy-day ratio. The current doubles; the boat’s speed through still water does not.