Question 01
Two numbers leave remainders of 7 and 10, respectively, when divided by 13. What remainder does their sum leave when divided by 13?
General Test 1
Lesson 29.1 · Grade 5 · 13 questions
Test 29.1 · Chapter 29 begins
General Test 1Bring your strategies together.
Thirteen questions. Familiar ideas in new combinations. Read carefully, choose a method, and give your answers a final check.
Comprehensive practice
13questions
10 short-answer questions
3 questions with reasoning
Work at your own paceAnswers save automaticallyRevise and submit again
Working on this exercise
- Enter answers in the answer fields and calculations in the working fields.
- When you are ready, submit the exercise to check your answers.
- Use the help available for this submission. Edit the original answer fields, then submit again.
Your responses are saved automatically in this browser.
Your responses save automatically in this browser.
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
In the diagram, BCEF is a parallelogram and BC = 8 cm. Triangle ABC is right-angled at C, with AC = 10 cm. The total shaded area is 8 cm² greater than the area of triangle ADH. Find AH.
Working notes (optional)
Question 03
Four players, Jia, Yi, Bing, and Ding, are dealt a pack of 54 playing cards. Starting with Jia, repeat this dealing order: 3 cards to Jia, 2 to Yi, 1 to Bing, and 2 to Ding. Who receives the last card?
Working notes (optional)
Question 04
The following addition expressions follow a pattern:
3 + 4, 5 + 9, 7 + 14, 9 + 19, 11 + 24, …
Find the first and second addends of the 10th expression. Then write the 80th addition expression.Keep the first and second addends in their original order. For the 80th expression, enter both addends rather than just the sum.
Working notes (optional)
Question 05
Five boxes of apples weigh 27 kg more than three boxes of pears. Two boxes of pears weigh 17 kg more than one box of apples. How much does one box of apples weigh?
Working notes (optional)
Question 06
The numbers a, b, and c are three different prime numbers and
2a + 3b + 6c = 42.
Find a + b + c.Working notes (optional)
Question 07
Jia and Yi start at the same time from A and B, respectively, and walk toward each other. Jia walks at 60 m/min; Yi walks at 100 m/min. They first meet at C, then continue to B and A, respectively, turn around, and walk back. They meet for the second time at a point 1,000 m from C. Find the distance AB.
Working notes (optional)
Question 08
When one natural number is divided by another, the quotient is 4 and the remainder is 3. The dividend, divisor, quotient, and remainder add up to 100. Find the dividend.
Working notes (optional)
Question 09
Jia, Yi, and Bing start with the same number of pencils. Jia gives some pencils to Yi and some to Bing. Afterwards, Yi has 7 more pencils than Jia, and Bing has 2 fewer than Yi. How many pencils did Jia give to Yi and to Bing?
Working notes (optional)
Question 10
Use the eight digit cards 2, 2, 3, 3, 4, 5, 6, 7, each exactly once, to make two four-digit numbers whose sum is 6,116. What is the greatest possible value of the larger 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
A stadium ticket costs 30 yuan. After the price is reduced, attendance increases by one half (12), and total ticket revenue increases by one quarter (14). By how many yuan was the price of each ticket reduced?
A matching final answer does not earn automatic method marks. Your written solution needs a person’s review.
Question 12
Three grass fields have areas of 5 hectares, 15 hectares, and 24 hectares. They have the same initial amount of grass per hectare, and the grass grows at the same rate per hectare. The first field feeds 10 cows for 30 days. The second feeds 28 cows for 45 days. Each cow eats at the same constant daily rate, and each field is just exhausted at the end of its stated feeding period. How many cows can the third field feed for 80 days?
Each cow eats at the same constant daily rate. Each field is just exhausted at the end of its stated feeding period.
A matching final answer does not earn automatic method marks. Your written solution needs a person’s review.
Question 13
A keeper has a supply of peanuts for three groups of monkeys. If all the peanuts are shared only among the first group, each monkey receives 12 peanuts. If they are shared only among the second group, each receives 15; if only among the third group, each receives 20. Now share the same supply equally among all the monkeys in the three groups. How many peanuts does each monkey receive?
A matching final answer does not earn automatic method marks. Your written solution needs a person’s review.
Revisit Chapter 16 for more practice →
4
Keep the remainders together.
Write the two numbers as 13a + 7 and 13b + 10. Their sum is 13(a + b) + 17.
Make one more full group.
The extra 17 is 13 + 4. The full group of 13 joins the other full groups, leaving a remainder of 4.
Watch for this: 17 is not a possible remainder after division by 13: a remainder must be smaller than 13.
Revisit Chapter 5 for more practice →
4 cm
Compare the two complete shapes.
The parallelogram and triangle ABC share the unshaded region B–C–H–D. The parallelogram’s two extra pieces are shaded; the triangle’s extra piece is ADH. Subtracting their common region shows that the parallelogram is 8 cm² larger than triangle ABC.
Find the parallelogram’s area.
Triangle ABC has area ½ × 8 × 10 = 40 cm². Therefore, the parallelogram’s area is 40 + 8 = 48 cm².
Use its perpendicular height.
BC × CH = 48, so CH = 48 ÷ 8 = 6 cm. Finally, AH = AC − CH = 10 − 6 = 4 cm.
Watch for this: The 8 cm² is an area difference, not the shaded area by itself.
Revisit Chapter 12 for more practice →
Bing
Count one complete dealing cycle.
A cycle uses 3 + 2 + 1 + 2 = 8 cards. After 6 cycles, 6 × 8 = 48 cards have been dealt, leaving 6.
Follow the remaining six cards in order.
Jia receives cards 1–3 of the remaining group; Yi receives cards 4–5; Bing receives card 6. So the last card goes to Bing.
Watch for this: Six leftover cards means the sixth card within a dealing cycle—not the sixth player.
Revisit Chapter 22 for more practice →
10th: 21 and 49. 80th: 161 + 399.
Track each side of the plus sign.
The first addends increase by 2. The second addends increase by 5. In expression n, they are 3 + 2(n − 1) = 2n + 1 and 4 + 5(n − 1) = 5n − 1.
Use nine steps for the 10th expression.
3 + 9 × 2 = 21 and 4 + 9 × 5 = 49.
Use seventy-nine steps for the 80th.
3 + 79 × 2 = 161; 4 + 79 × 5 = 399. The requested expression is 161 + 399, not just its total, 560.
Watch for this: Do not multiply each step size by 10 or 80: the first expression already counts as position 1.
Revisit Chapter 9 for more practice →
15 kg
Use the second comparison first.
Let one box of pears weigh p kg. One box of apples then weighs 2p − 17 kg.
Substitute into the first comparison.
5(2p − 17) = 3p + 27. Hence 10p − 85 = 3p + 27, so 7p = 112 and p = 16.
Answer for apples, not pears.
One box of apples weighs 2 × 16 − 17 = 15 kg. Check: 5 × 15 − 3 × 16 = 27, and 2 × 16 − 15 = 17.
Watch for this: 16 kg is the weight of a box of pears. The question asks for apples.
Revisit Chapter 17 for more practice →
10
Look at multiples of 3.
Both 3b and 6c are divisible by 3, as is 42. Therefore 2a is divisible by 3. The prime a must be 3.
Use the only even prime.
Substituting a = 3 gives 3b + 6c = 36, or b + 2c = 12. Since 2c and 12 are even, b is even. The only even prime is 2.
Find the last prime and check.
2 + 2c = 12 gives c = 5. These three primes are different. Their sum is 3 + 2 + 5 = 10, and 2 × 3 + 3 × 2 + 6 × 5 = 42.
Watch for this: The answer is the sum of the three primes, not a list of the weighted terms.
Revisit Chapter 7 for more practice →
2,000 m
Locate the first meeting.
Their speed ratio is 60 : 100 = 3 : 5. So C is 3/8 of AB from A.
Count the total distance up to the second meeting.
At the first meeting, their combined walking distance is AB. By the second meeting, each has reached the opposite end and turned: their combined distance is 3AB. With unchanged speeds, the second meeting happens three times as long after the start.
Locate the second meeting.
Jia’s total distance is then 3 × (3/8 AB) = 9/8 AB: all the way from A to B, then 1/8 AB back. The second meeting is therefore 7/8 AB from A.
Compare the meeting points.
The separation is 7/8 AB − 3/8 AB = 1/2 AB. Since that is 1,000 m, AB = 2,000 m.
Check: they first meet after 12.5 minutes, 750 m from A. They meet again after 37.5 minutes, 1,750 m from A. The difference is 1,000 m.
Watch for this: The 1,000 m is the distance between the two meeting points, not the distance between the endpoints.
Revisit Chapter 16 for more practice →
75
Write the division relationship.
Let the divisor be d. Then the dividend is 4d + 3.
Include all four numbers in the sum.
(4d + 3) + d + 4 + 3 = 100. Thus 5d + 10 = 100, so d = 18.
Return to the dividend.
The dividend is 4 × 18 + 3 = 75. Check: 75 ÷ 18 is 4 remainder 3; also 75 + 18 + 4 + 3 = 100.
Watch for this: 18 is the divisor. Do not stop before calculating the dividend.
Revisit Chapter 9 for more practice →
3 to Yi; 1 to Bing.
Name the two transfers.
Suppose Jia gives x pencils to Yi and y to Bing. Since the starting amounts were equal, Yi’s final lead over Bing is x − y = 2.
Account for both giving and receiving.
Yi gains x while Jia loses x + y, so Yi’s lead over Jia is 2x + y = 7.
Solve and verify.
From x = y + 2, we get 2(y + 2) + y = 7, so y = 1 and x = 3. Jia loses 4 pencils, Yi gains 3, and Bing gains 1. Yi is then 7 ahead of Jia and 2 ahead of Bing.
Watch for this: Moving a pencil changes both the giver’s amount and the receiver’s amount.
Revisit Chapter 21 for more practice →
3,764
Find the carries from the right.
The ones digits must add to 6, with no carry: even the two largest cards cannot add to 16. The tens digits must add to 11, carrying 1. The hundreds digits must add to 10, which, with that carry, gives 11 and carries 1 again.
Fix the thousands digits.
The thousands digits must therefore add to 6 − 1 = 5. They are 2 and 3. So the larger number starts with 3.
Maximize the next place, then the next.
The largest available hundreds digit is 7; pair it with the other 3 to make 10. The remaining cards are 2, 4, 5, 6. Use 6 and 5 in the tens column, giving the larger number the 6. Then give it the 4 in the ones column.
Verify the maximum construction.
3,764 + 2,352 = 6,116, and all eight cards are used exactly once. No larger thousands digit is possible; after 3, no hundreds digit exceeds 7, no remaining tens choice exceeds 6, and no remaining ones choice exceeds 4. Thus 3,764 is the maximum.
Watch for this: A large-looking number is not enough: its partner must use exactly the leftover cards and give the required sum.
Revisit Chapter 9 for more practice →
5 yuan
Compare revenue with the number of tickets.
Let the original attendance be n. The new attendance is 3n/2 and the new revenue is (30n) × 5/4.
Calculate the new price.
New ticket price = (30n × 5/4) ÷ (3n/2) = 30 × 5/4 × 2/3 = 25 yuan.
Find the reduction and check both changes.
The reduction is 30 − 25 = 5 yuan. For example, 100 original tickets bring in 3,000 yuan. At 150 tickets and 25 yuan each, revenue is 3,750 yuan—one quarter more.
Watch for this: 25 yuan is the new price. The question asks for the reduction.
Check your written method: Does it use both increase factors, find the new price, and then subtract from 30? Other valid methods are welcome.
Revisit Chapter 8 for more practice →
42 cows
Compare equal areas.
Measure grass in cow-days: one unit is the amount one cow eats in one day. Scale the first field from 5 to 15 hectares: it would feed 30 cows for 30 days. That uses 900 units. The actual 15-hectare field feeds 28 cows for 45 days, using 1,260 units.
Separate growth from the starting supply.
For the same 15-hectare area, the extra 1,260 − 900 = 360 units grow in the extra 45 − 30 = 15 days. Daily growth is 360 ÷ 15 = 24 units.
Recover the initial grass.
The starting supply on 15 hectares is 1,260 − 24 × 45 = 180 units.
Scale the eighty-day supply to 24 hectares.
Over 80 days, 15 hectares supply 180 + 24 × 80 = 2,100 units. On 24 hectares that becomes 2,100 × 24/15 = 3,360 units. Divide by 80 days: 3,360 ÷ 80 = 42 cows.
Watch for this: A larger supply does not come only from area: grass continues to grow during the feeding period.
Check your written method: Does it normalize the field areas, find growth and initial grass separately, scale to 24 hectares, and divide the 80-day supply by 80? Cows eat at the same constant rate.
Revisit Chapter 19 for more practice →
5 peanuts per monkey
Represent a possible total without fixing its size.
The total must be a multiple of 12, 15, and 20. Their least common multiple is 60, so write the total as 60m peanuts, for a positive whole number m.
Find the group sizes.
The groups contain 60m ÷ 12 = 5m, 60m ÷ 15 = 4m, and 60m ÷ 20 = 3m monkeys. Together they contain 12m monkeys.
Share the same supply among everyone.
Each monkey receives 60m ÷ 12m = 5 peanuts. The scale m cancels, so the answer does not depend on the total number of peanuts.
Watch for this: Do not average 12, 15, and 20. These are shares of the same supply given to different-sized groups.
Check your written method: Does it keep one common peanut supply, find the group sizes, combine all the monkeys, and divide the supply by that combined number? Equivalent fraction methods are also valid.