Question 01
A two-digit number has a digit sum of 12. When its tens and ones digits are exchanged, the new two-digit number is 36 less than the original. What is the original number?
General Test 4
Lesson 29.4 · Grade 5 · 13 questions
Test 29.4 · Chapter 29 · General tests
General Test 4Look carefully. Track what changes. Explain why.
Thirteen mixed problems, from hidden shapes to changing shares. Choose a useful method, keep the conditions in view, and check the answer the question actually asks for.
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
Three consecutive natural numbers have these properties: the smallest is a multiple of 9, the middle is a multiple of 8, and the largest is a multiple of 7. What is their smallest possible sum?
Working notes (optional)
Question 03
In the figure, ABCD and CEFG are squares with equal side lengths. Points B, C, and G lie on one straight line.
How many squares and how many isosceles right triangles are there altogether?
Count shapes of every size and orientation whose sides lie completely on the drawn lines. An isosceles right triangle has two equal sides and a right angle.
Working notes (optional)
Question 04
Every small square in the grid has side length 1 cm. Find the area of the shaded part.
The shaded boundary joins the grid points shown. Use the grid lengths, not the apparent size of the drawing on your screen.
Working notes (optional)
Question 05
A factory planned to complete an order in 15 days. It actually made 300 parts per day and finished 3 days early. How many more parts did it make each day than originally planned?
Working notes (optional)
Question 06
Six numbers have an average of 53. Two of the numbers are equal. One of these equal numbers is replaced by 30, and the other by 70. The average then becomes 58. What was the original value of each of the two equal numbers?
Working notes (optional)
Question 07
A box contains 10 white and 20 black playing pieces. Except for their color, the pieces feel identical. Xiaoming takes pieces from the box with his eyes closed.
What is the smallest number he must take to guarantee at least 5 white pieces and at least 5 black pieces?
Working notes (optional)
Question 08
This year, Grandfather is twice Father’s age and six times Xiaoming’s age. In 12 years, Father will be twice Xiaoming’s age. How old is Grandfather this year?
Working notes (optional)
Question 09
Jia sells 2 chickens, 3 ducks, and 7 geese for 107.2 yuan. Yi sells 3 chickens, 4 ducks, and 9 geese for 143.2 yuan.
At the same prices for each kind of bird, how much should Bing receive for 2 chickens, 1 duck, and 1 goose?
Working notes (optional)
Question 10
To escape the sunlight, two snails move downward from the mouth of a well. In the daytime, one crawls down 20 dm per day and the other 15 dm per day. At night, both slide downward by the same distance per night.
One reaches the bottom after 5 day-and-night periods, and the other after 6 day-and-night periods. How deep is the well?
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
Passengers start queuing at a station some minutes before ticket checking begins. The same number of new passengers arrives each minute, and arrivals continue while tickets are checked.
With 5 ticket windows open at the same time, the queue disappears in 30 minutes. With 6 windows, it disappears in 20 minutes.
How many windows must be open together to make the queue disappear in 10 minutes?
Your reasoning is for a person to review. It is not automatically graded.
Question 12
Jia, Yi, and Bing have 100 extracurricular books altogether. When Jia’s number of books is divided by Yi’s, the quotient is 5 and the remainder is 1. When Bing’s number is divided by Jia’s, the quotient is again 5 and the remainder is 1.
How many books does Yi have?
Your reasoning is for a person to review. It is not automatically graded.
Question 13
A basket of eggs is shared among several people. The first person takes 1 egg and then 19 of the eggs remaining. The second takes 2 eggs and then 19 of the eggs remaining. The third takes 3 eggs and then 19 of the eggs remaining. The pattern continues in the same way.
In the end, all the eggs have been shared, and everyone receives the same total number of eggs. How many eggs were there, and how many people received them?
At each turn, the fixed number of eggs is taken first. The one-ninth share is calculated from what is left after that first part of the turn.
Your reasoning is for a person to review. It is not automatically graded.
Revisit Chapter 21 for more practice →
84
List every possible original number.
List 93, 84, 75, 66, 57, 48, 39. Each has digit sum 12.
Use “36 less,” not “a difference of 36.”
For 84, the reversed number is 48, and 84 − 48 = 36. The reverse of 48 is larger, not smaller. Checking the list leaves 84.
Verify both conditions.
8 + 4 = 12, and swapping the digits reduces 84 by 36.
Watch for this: 48 has the right digit sum, but exchanging its digits increases the number.
Revisit Chapter 16 for more practice →
1,488
Write all three numbers using the middle one.
Write them as 8k − 1, 8k, 8k + 1.
Find which k-values can work.
Since 8k − 1 = 9k − (k + 1), it is divisible by 9 exactly when k + 1 is divisible by 9. The positive candidates are 8, 17, 26, 35, 44, 53, 62, ….
Check the largest number in increasing order.
| k | 8k + 1 | Divisible by 7? |
|---|---|---|
| 8 | 65 | No |
| 17 | 137 | No |
| 26 | 209 | No |
| 35 | 281 | No |
| 44 | 353 | No |
| 53 | 425 | No |
| 62 | 497 | Yes: 7 × 71 |
The first success gives 495, 496, 497. Their divisibility checks are 495 = 9 × 55, 496 = 8 × 62, and 497 = 7 × 71.
Find the sum.
The sum of three consecutive numbers is three times the middle one: 496 × 3 = 1,488.
Watch for this: Finding any working triple is not enough. Check the smaller candidates to justify “smallest.”
Revisit Chapter 24 for more practice →
3 squares; 22 isosceles right triangles
Count the squares.
The three squares are ABCD, CEFG, and DGEB. DGEB is the tilted square formed by the added outer segments.
Name the two intersections.
Call the intersection of AC and BD O₁, and the intersection of CF and EG O₂.
Organize the triangle list.
| Size group | Triangles | Count |
|---|---|---|
| Smallest | AO₁B, BO₁C, CO₁D, DO₁A; CO₂E, EO₂F, FO₂G, GO₂C | 8 |
| Middle | ABD, BCD, ABC, ACD; DCG, BCE; CEG, EGF, CGF, CEF | 10 |
| Largest | BDG, BEG, BDE, DEG | 4 |
This gives 8 + 10 + 4 = 22 isosceles right triangles. Each triangle is counted by its three vertices, not by the number of small pieces it contains.
Explore the shapes, one at a time
Review tool added to the problem’s shape list. O₁ and O₂ name the two diagonal intersections; they are not new segments.
The highlight follows only lines already in the test figure. The full list remains in the solution above for printing.
Watch for this: A slanted square is still a square. Avoid counting the same triangle twice by listing its vertices consistently.
Revisit Chapter 4 for more practice →
3 cm²
Find the whole area.
There are 9 unit squares, each with area 1 × 1 = 1 cm², so the full area is 9 cm².
Account for all the unshaded area.
The worked example gives a total unshaded area of 6 cm². The added dashed lines below make this visible: they split the unshaded regions into four triangles, each with base 3 cm and perpendicular height 1 cm.
Each unshaded triangle: 3 × 1 ÷ 2 = 1.5 cm²
Altogether, 4 × 1.5 = 6 cm² is unshaded.
Subtract.
The shaded area is 9 − 6 = 3 cm².
Watch for this: A partially shaded cell is not a whole shaded square. Add actual areas, rather than counting every cell the shading touches.
Revisit Chapter 9 for more practice →
60 more parts per day
Find the actual duration.
Finishing 3 days early means the factory worked for 15 − 3 = 12 days.
Find the planned daily output.
The order contains 300 × 12 = 3,600 parts. Over the planned 15 days, that would be 3,600 ÷ 15 = 240 parts per day.
Compare the rates.
The increase is 300 − 240 = 60 parts per day, matching the worked example expression 300 − 300 × 12 ÷ 15.
Watch for this: The three saved days are not the actual duration. Use 12 working days to find the total order.
Revisit Chapter 3 for more practice →
35 for each number
Turn the change in average into a change in total.
The average increased by 58 − 53 = 5. Across six numbers, the total therefore increased by 5 × 6 = 30.
Recover the old pair’s sum.
The new pair sums to 30 + 70 = 100. The old pair summed to 100 − 30 = 70.
Use the fact that the two originals were equal.
Each original number was 70 ÷ 2 = 35. Replacing 35 and 35 with 30 and 70 increases the total by exactly 30.
Watch for this: The answer is each equal number, not the pair’s sum. Divide the recovered sum by two.
Revisit Chapter 23 for more practice →
25 pieces
Show why 24 might not be enough.
He could take all 20 black pieces and only 4 white pieces. That is 24 pieces, but it fails the white-piece requirement.
Show why 25 always works.
Among 25 pieces, at most 20 can be black, so at least 5 must be white. At most 10 can be white, so at least 15 must be black, which is more than enough.
State the minimum.
Since 24 can fail and 25 always works, the smallest guarantee is 25 pieces. The worked example uses the same worst-case order: 20 black, then 5 white.
Watch for this: Ten pieces might happen to contain five of each color, but “might” is not a guarantee.
Revisit Chapter 9 for more practice →
72 years
Express the present ages.
Let Xiaoming be x years old. Grandfather is 6x. Since Grandfather is twice Father’s age, Father is 3x.
Move both younger ages forward by 12 years.
3x + 12 = 2(x + 12)
Thus 3x + 12 = 2x + 24, so x = 12.
Answer for the person asked about.
Grandfather is 6 × 12 = 72 years old. Father is 36 now; in 12 years, Father and Xiaoming will be 48 and 24, respectively.
Watch for this: Add 12 years to both Father and Xiaoming before applying the future ratio. The present answer requested is Grandfather’s age.
Revisit Chapter 1 for more practice →
36.8 yuan
Scale the two known sales.
| Sale combination | Chickens | Ducks | Geese | Revenue (yuan) |
|---|---|---|---|---|
| 4 × Yi | 12 | 16 | 36 | 143.2 × 4 = 572.8 |
| 5 × Jia | 10 | 15 | 35 | 107.2 × 5 = 536 |
Subtract the quantities.
The difference is 2 chickens, 1 duck, and 1 goose—exactly Bing’s sale.
Subtract the revenues.
572.8 − 536 = 36.8 yuan
There is no need to work out the separate prices of the three kinds of birds.
Watch for this: Subtract matching multiples of the entire sales, not just convenient-looking totals.
Revisit Chapter 9 for more practice →
150 dm
Use one variable for both nighttime distances.
Let each snail slide x dm per night. The faster daytime crawler takes 5 complete periods and the other takes 6.
Equate two expressions for the same depth.
20 × 5 + 5x = 15 × 6 + 6x
This is 100 + 5x = 90 + 6x, so x = 10.
Find and check the well’s depth.
From the first snail: 20 × 5 + 10 × 5 = 150 dm. From the second: 15 × 6 + 10 × 6 = 150 dm.
Watch for this: This is not an upward-climb puzzle with an opposing nighttime slip. Both distances are added, and the worked example counts complete day-and-night periods.
Revisit Chapter 8 for more practice →
9 ticket windows
Choose a useful unit.
Call the number of passengers one window can process in one minute 1 capacity unit. This need not mean one passenger.
Find the arrival rate.
Five windows working for 30 minutes process 150 units; six working for 20 minutes process 120 units. The extra 30 units are the arrivals during the extra 10 minutes.
(5 × 30 − 6 × 20) ÷ (30 − 20) = 3 units per minute
Recover the queue present at the start.
With five windows, the queue shrinks by 5 − 3 = 2 units per minute. Thus the initial queue is 2 × 30 = 60 units. With six windows, (6 − 3) × 20 also gives 60.
Allow for the next ten minutes of arrivals.
The windows must process the initial 60 units plus 10 × 3 = 30 new units. Therefore they need a total rate of (60 + 30) ÷ 10 = 9 units per minute, which requires 9 windows.
Watch for this: The total number checked is not fixed when the clearing time changes: more passengers arrive during a longer wait.
Revisit Chapter 16 for more practice →
3 books
Build the relationships from Yi upward.
Let Yi have x books. Then Jia has 5x + 1 books. Bing has 5(5x + 1) + 1 = 25x + 6 books.
Use the combined total.
x + (5x + 1) + (25x + 6) = 100
So 31x + 7 = 100, which gives 31x = 93 and x = 3.
Check the divisions as well as the total.
Yi, Jia, and Bing have 3, 16, and 81 books. Their sum is 100. Also, 16 = 5 × 3 + 1 and 81 = 5 × 16 + 1. Each remainder is smaller than its divisor.
Watch for this: The second divisor is Jia’s number, not Yi’s. Carry the first “plus one” through the second multiplication.
Revisit Chapter 9 for more practice →
64 eggs; 8 people
Express the first person’s share.
Let there initially be x eggs, and let each total share be s. The first person takes one egg and then one-ninth of x − 1.
s = 1 + (x − 1) ÷ 9, so 9s = x + 8.
Express the second person’s share.
After the first person, x − s eggs remain. The second person first takes 2, so the fractional part is one-ninth of x − s − 2.
s = 2 + (x − s − 2) ÷ 9, so 9s = x − s + 16.
Use equal shares, as in the worked solution.
Both expressions equal 9s, so x + 8 = x − s + 16. This gives s = 8. Then 9 × 8 = x + 8, so x = 64.
Find the number of people and verify completion.
Each person receives 8 eggs. Since all 64 eggs are distributed, the number of people is 64 ÷ 8 = 8. The added ledger below checks every turn, including the last.
| Person | Before | Fixed part | One-ninth part | Total taken | After |
|---|---|---|---|---|---|
| 1 | 64 | 1 | 7 | 8 | 56 |
| 2 | 56 | 2 | 6 | 8 | 48 |
| 3 | 48 | 3 | 5 | 8 | 40 |
| 4 | 40 | 4 | 4 | 8 | 32 |
| 5 | 32 | 5 | 3 | 8 | 24 |
| 6 | 24 | 6 | 2 | 8 | 16 |
| 7 | 16 | 7 | 1 | 8 | 8 |
| 8 | 8 | 8 | 0 | 8 | 0 |
Watch for this: One-ninth is taken from a changing remainder, not from the original basket. On the last turn, one-ninth of zero is zero.