Question 01
Two prime numbers, p and q, are both less than 1,000. They satisfy p × q − 1 = x, where x is odd. What is the greatest possible value of x?
General Test 3
Lesson 29.3 · Grade 5 · 13 questions
Test 29.3 · Chapter 29 · General tests
General Test 3Read closely. Connect ideas. Justify your answer.
Thirteen mixed problems. Choose a useful method, keep track of the conditions, and show how your conclusions follow from the clues.
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
The circle has circumference 16.4 cm. Its area is exactly equal to the area of the rectangle shown. Find the perimeter of the shaded region. Use π = 3.14.
As shown: O is both the circle’s center and the rectangle’s upper-left corner. The rectangle’s bottom side is tangent to the bottom of the circle. Only the part of the rectangle outside the circle is shaded.
Working notes (optional)
Question 03
Jia, Yi, and Bing copy a manuscript. Each day, Jia copies as much as Yi and Bing together. Bing’s daily rate is 15 of the combined daily rate of Jia and Yi.
Working together, all three finish in 8 days. How many days would Yi alone need to finish the manuscript?
Working notes (optional)
Question 04
Jia’s number is 24. The least common multiple of Jia’s and Yi’s numbers is 168, and their greatest common factor is 4. What is Yi’s number?
Working notes (optional)
Question 05
A shop spends the same amount of money on each of two kinds of candy. Kind A costs 18 yuan per kg, and kind B costs 12 yuan per kg. The shop mixes all the candy together. What is the cost per kilogram of the mixture?
Working notes (optional)
Question 06
Some students calculate their average score on a Chinese-language test. If Jia’s score increases by 8 points, their average becomes 90. If Jia’s score decreases by 12 points instead, their average becomes 85.
What is the students’ actual average score?
Working notes (optional)
Question 07
In May of a certain year, there are five Fridays. Their date numbers add up to 80. On which day of the week is May 1?
Working notes (optional)
Question 08
A batch of paper is used to bind exercise books. After making 120 books, 40% of the original paper remains. After making 185 books instead, 1,350 sheets remain.
How many sheets of paper were in the original batch?
Working notes (optional)
Question 09
The older brother’s present age is three times the younger brother’s age at an earlier time. At that earlier time, the older brother was exactly as old as the younger brother is now.
Their present ages add up to 30 years. How old is the older brother now?
Working notes (optional)
Question 10
Choose n distinct numbers from 1, 2, 3, …, 15, 16. No matter which n numbers are chosen, the selection must contain two numbers such that one is exactly three times the other.
What is the smallest possible n that guarantees this?
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
The invoice below is torn. What was the unit price, in yuan?
Your reasoning is for a person to review. It is not automatically graded.
Question 12
A job’s pay for one year is 18,000 yuan plus a fully automatic washing machine. A worker works for 7 months and receives 8,400 yuan plus the same washing machine.
What is the value of the washing machine?
Your reasoning is for a person to review. It is not automatically graded.
Question 13
A king changes the ranks of four ministers, A, B, C, and D. Before the change, they make these predictions:
A: “B’s rank will fall.”
B: “C’s rank will rise, but C will not be the highest.”
C: “D’s rank will fall.”
D: “My rank will be the lowest.”
Everyone whose prediction is correct is promoted; everyone whose prediction is incorrect is demoted. A falls by two ranks, and the worked example adds that “no one rises by two ranks.”
Write the orders of the four ministers from highest to lowest, both before and after the change.
| Order | 1Highest | 2 | 3 | 4Lowest |
|---|---|---|---|---|
| Before | ||||
| After |
Your reasoning is for a person to review. It is not automatically graded.
Revisit Chapter 17 for more practice →
1,993
Use the odd-number condition.
Because x is odd, x + 1 is even. But p × q = x + 1, so the product of the two primes must be even.
Identify the even prime.
At least one of p and q must be even. The only even prime is 2; two odd primes would have an odd product.
Make the other prime as large as possible.
The greatest prime below 1,000 is 997: 998 is even and 999 is divisible by 3. To check 997, test the primes no greater than its square root: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31. None divides 997. Thus x = 2 × 997 − 1 = 1,993, which is odd.
Watch for this: Using two large odd primes would give an even value of x, so it would break the condition.
Revisit Chapter 4 for more practice →
20.5 cm
Translate the equal areas.
Write the circle’s radius as r and the rectangle’s length as a. The rectangle’s height is r, so πr² = ar. Dividing by the positive radius r gives a = πr. Therefore 2a = 2πr = 16.4 cm.
Trace just the shaded boundary.
The straight parts have lengths a along the bottom, r along the right, and a − r along the top. Together they total a + r + (a − r) = 2a. The remaining boundary is one quarter of the circle.
Add the straight and curved lengths.
The quarter-circle arc is 16.4 ÷ 4 = 4.1 cm. The shaded perimeter is 2a + 4.1 = 16.4 + 4.1 = 20.5 cm. There is no need to round the radius.
Watch for this: The rectangle’s entire top and left sides are not part of the shaded boundary. Include the quarter-circle arc, not the full circumference.
Revisit Chapter 9 for more practice →
24 days
Find Jia’s daily rate.
The whole team copies 18 of the manuscript per day. Since Jia’s rate equals Yi’s and Bing’s rates together, Jia contributes half of the total: 18 ÷ 2 = 116.
Find the combined rate of Jia and Yi.
Bing’s rate is one fifth of Jia and Yi’s combined rate. The whole team therefore has 65 of that combined rate. Jia and Yi copy 18 ÷ 65 = 548 of the manuscript per day.
Subtract Jia’s rate, then find the time.
Yi’s daily rate is 548 − 116 = 548 − 348 = 124. One complete manuscript takes 1 ÷ 124 = 24 days.
Watch for this: The one-fifth statement compares daily work rates, not the numbers of days needed alone.
Revisit Chapter 19 for more practice →
28
Factor out the shared greatest common factor.
Write the two numbers as 4a and 4b, where a and b have no common factor greater than 1. From 4a = 24, we obtain a = 6.
Use the least common multiple.
The least common multiple is 4ab = 168. Therefore ab = 42, and b = 42 ÷ 6 = 7.
Restore the factor 4 and check.
Yi’s number is 4 × 7 = 28. The numbers 24 = 4 × 6 and 28 = 4 × 7 have greatest common factor 4 and least common multiple 4 × 6 × 7 = 168.
Watch for this: 168 ÷ 24 gives b = 7, not Yi’s whole number. Restore the common factor 4.
Revisit Chapter 3 for more practice →
14.4 yuan per kg
Compare masses for equal spending.
For each 1 yuan spent on kind A, the shop gets 118 kg. For 1 yuan spent on kind B, it gets 112 kg. These equal 1-yuan shares together cost 2 yuan.
Divide the cost by the total mass.
The combined mass is 118 + 112 = 236 + 336 = 536 kg. The cost per kg is 2 ÷ 536 = 725 = 14.4 yuan.
Check with a convenient equal budget.
As an equivalent check, spend 36 yuan on each kind. That buys 2 kg and 3 kg: 5 kg for 72 yuan. The mixture again costs 72 ÷ 5 = 14.4 yuan per kg.
Watch for this: (18 + 12) ÷ 2 = 15 averages the two prices as though equal masses were mixed. The problem gives equal spending instead.
Revisit Chapter 3 for more practice →
88 points
Compare the two possible totals.
Only Jia’s score changes. The increase scenario and the decrease scenario differ by 8 + 12 = 20 points in their group totals.
Find the number of students.
The two averages differ by 90 − 85 = 5 points. The number of students is therefore 20 ÷ 5 = 4.
Return to the original average.
Adding 8 points to one score adds 8 ÷ 4 = 2 points to the average. The original average is 90 − 2 = 88. Check from the other side: 85 + 12 ÷ 4 = 88.
Watch for this: Changing one score by 8 points does not change the average by 8 points. Spread that change over all the students.
Revisit Chapter 12 for more practice →
Thursday
Find the middle Friday.
Successive Fridays are 7 days apart. The average of the five equally spaced dates is the middle one: 80 ÷ 5 = 16.
List the Friday dates.
Count back and forward by 7 from 16. The five Fridays are 2, 9, 16, 23, and 30. Their sum is 80.
Step back one day.
May 2 is a Friday, so May 1 is Thursday.
Watch for this: 16 is the middle Friday, not the first Friday. Work back by two weeks before identifying May 1.
Revisit Chapter 9 for more practice →
18,000 sheets
Find the batch’s book capacity.
The first 120 books use 100% − 40% = 60% of the paper. The whole batch can make 120 ÷ 0.6 = 200 books.
Find the paper used per book.
After 185 books, paper for 200 − 185 = 15 books remains. Those 15 books would use 1,350 sheets, so each book uses 1,350 ÷ 15 = 90 sheets.
Find the original supply.
The batch contained 200 × 90 = 18,000 sheets. Check: after 120 books, 7,200 sheets remain, which is 40% of 18,000. After 185 books, 18,000 − 16,650 = 1,350 sheets remain.
Watch for this: 40% is the fraction of paper left, not the fraction already used for the 120 books.
Revisit Chapter 9 for more practice →
18 years
Label the younger brother’s earlier age.
Let the younger brother’s age at the earlier time be x. The older brother is now 3x. Since their present ages sum to 30, the younger brother is now 30 − 3x.
Write the same age difference twice.
The older brother’s earlier age was also 30 − 3x. Thus the age difference gives 3x − (30 − 3x) = (30 − 3x) − x. Simplifying: 6x − 30 = 30 − 4x, so 10x = 60 and x = 6.
Answer and check both times.
The older brother is now 3 × 6 = 18, and the younger is 12. Six years earlier, their ages were 12 and 6. The older brother’s earlier age equals the younger brother’s present age, and 18 = 3 × 6.
Watch for this: The factor of three compares one present age with an earlier age. It does not say that their present ages are in the ratio 3 : 1.
Revisit Chapter 23 for more practice →
13
Organize the numbers into chains.
The nontrivial chains are (1, 3, 9), (2, 6), (4, 12), and (5, 15). The remaining numbers 7, 8, 10, 11, 13, 14, 16 are singletons.
Find the largest safe selection.
From (1, 3, 9), we may take 1 and 9 together: neither is exactly three times the other. At most one number can be taken from each of the other three chains. All seven singletons may be chosen. This gives at most 2 + 3 + 7 = 12 numbers without an exact three-times pair.
Show why the next number is necessary.
For example, {1, 2, 4, 5, 7, 8, 9, 10, 11, 13, 14, 16} is a safe set of 12. So 12 does not guarantee a pair, while any 13 choices exceed the maximum safe size. The answer is n = 13.
Watch for this: 1 and 9 may both be chosen safely: 9 is nine times 1, not three times 1. Also, a minimum guarantee needs both a safe smaller example and a proof that one more must work.
Revisit Chapter 15 for more practice →
1.36 yuan
Convert to an integer number of fen.
Let the two missing digits be a and b. The total is 9a.9b yuan, or 9a9b fen. Because there are 72 items, and each item’s price is a whole number of fen, 9a9b is divisible by 72.
Use divisibility by 9 and by 8.
Since 72 = 9 × 8 and 9 and 8 are relatively prime, the total must be divisible by both. The digit sum is 18 + a + b, so a + b is 0, 9, or 18. Divisibility by 8 makes b even. These facts leave the last-three-digit candidates 090, 198, 396, 594, 792, 990.
Recover the total.
Of those six endings, only 792 is divisible by 8: 792 ÷ 8 = 99. Thus a = 7 and b = 2. The total is 97.92 yuan.
Divide by the number of items.
The unit price is 97.92 ÷ 72 = 1.36 yuan. Check in fen: 72 × 136 = 9,792 fen, matching the two surviving 9s.
Watch for this: The receipt’s decimal point matters. Work in fen for divisibility, then convert back to yuan. The missing digits are not zeros unless the conditions force them.
Check your written method: Does your reasoning explain why 72 divides the total in fen, determine the two missing digits, and convert the unit price back to yuan?
Revisit Chapter 9 for more practice →
5,040 yuan
Write the monthly value of the yearly package.
Let the machine be worth x yuan. The 12-month package is worth 18,000 + x, so the monthly pay is 18,000 + x12 = 1,500 + x12 yuan.
Write the same monthly value for seven months.
The 7-month package is worth 8,400 + x. Therefore 18,000 + x12 = 8,400 + x7.
Solve for the machine.
Cross-multiply: 7(18,000 + x) = 12(8,400 + x). Then 126,000 + 7x = 100,800 + 12x, so 25,200 = 5x. Hence x = 5,040.
Check the two packages.
The yearly package is 18,000 + 5,040 = 23,040 yuan, or 1,920 yuan per month. Seven months is 7 × 1,920 = 13,440 yuan, exactly 8,400 + 5,040.
Watch for this: The same full machine appears in both packages. Comparing only 18,000 and 8,400 ignores part of the pay.
Check your written method: Does your work include the machine’s value in both packages and equate their monthly values, or give an equivalent proportional comparison?
Revisit Chapter 10 for more practice →
Before: D, A, C, B. After: B, C, D, A.
Follow the chain of true and false predictions.
A falls, so A is wrong: B does not fall. Every minister must move up or down, so B rises and B is correct. Therefore C rises, but not to first place. C is correct, so D falls. D is therefore wrong and does not finish last.
Decide where A starts and ends.
A’s two-place fall could be 1 → 3 or 2 → 4. If A starts first and finishes third, D must finish first or second: A occupies third, and D cannot finish fourth. But D cannot move down into either of the first two places because A occupied first initially. This is impossible. Hence A moves 2 → 4.
Place D without creating a two-place promotion.
D falls and does not finish fourth, so D finishes second or third. Suppose D finishes second. D must then start first. C, who rises but is not first, must move 4 → 3, leaving B to move 3 → 1. That is a forbidden rise of exactly two ranks. So D finishes third, and since A starts second, D must start first.
Complete the two orders.
After the change, B and C occupy first and second. C is not highest, so C finishes second and B first. Before the change, C cannot be fourth, since 4 → 2 would rise exactly two ranks. Thus C starts third and B fourth.
| Order, highest → lowest | Rank 1 | Rank 2 | Rank 3 | Rank 4 |
|---|---|---|---|---|
| Before | D | A | C | B |
| After | B | C | D | A |
Verify every condition—not only A’s move.
| Minister | Movement | Prediction | Consistent? |
|---|---|---|---|
| A | 2 → 4: down 2 | “B will fall”: false | Yes: false → down |
| B | 4 → 1: up 3 | “C will rise, not to first”: true | Yes: true → up |
| C | 3 → 2: up 1 | “D will fall”: true | Yes: true → up |
| D | 1 → 3: down 2 | “I will be lowest”: false | Yes: false → down |
A falls two ranks. No one rises exactly two ranks: the rises are three for B and one for C. That three-rank rise is why the wording convention must be explicit.
Watch for this: Rank 1 is highest: moving from 4 to 1 is a promotion of three ranks. Do not reverse the order, and do not replace “exactly two” with “two or more.”
Check your written method: Does your reasoning link each minister’s prediction to promotion or demotion, rule out the other placements, and verify both full orders?