NCERT Solutions Class 5 Mathematics Mela Chapter 09 Coconut Farm

Page 119

Question. Write the appropriate multiplication fact for the array shown below. Write two division facts that follow from the multiplication fact.
Answer: The array displays 4 rows with 8 coconuts in each row. The multiplication fact is 4 × 8 = 32.
The two division facts that come from this are:
32 ÷ 4 = 8
32 ÷ 8 = 4
In simple words: When you have 4 groups of 8, you get 32 total. You can also split 32 into 4 equal groups of 8, or into 8 equal groups of 4.

Exam Tip: Always show both division facts from one multiplication fact - division reverses multiplication, so if a × b = c, then c ÷ a = b and c ÷ b = a.

Page 120

Let Us Play

Question. Identify the numbers that can fill the circles such that the numbers in the squares are the products or the quotients of the numbers in the circles.
Answer:
(i) 36 × 2 = 72
(ii) 20 × 3 = 60
(iii) 12 × 4 = 48
(iv) 6 × 6 = 36
(v) 6 × 4 = 24
(vi) 5 × 8 = 40
(vii) 108 ÷ 2 = 54
(viii) 168 ÷ 8 = 21
(ix) 168 ÷ 3 = 56
(x) 54 ÷ 9 = 6
(xi) 42 ÷ 2 = 21
(xii) 56 ÷ 7 = 8
In simple words: For each square, find two circle numbers that either multiply to make the square number, or divide to make it.

Exam Tip: Check your answers by working backwards - if you say 20 × 3 = 60, verify by doing 60 ÷ 20 = 3 to confirm it is correct.

Let Us Do

Question 1. Solve the following multiplication problems. Write two division statements in each case.
Answer: Each student should work through the multiplication and then create the two matching division facts. For example, if you multiply 7 × 6, you get 42. Then you can write 42 ÷ 7 = 6 and 42 ÷ 6 = 7. Work through each given multiplication in this way, and your two division facts will follow naturally from the product you find.
In simple words: Multiply the two numbers to get your answer. Then write down two ways to divide that answer back into the starting numbers.

Exam Tip: For every multiplication fact, there are always exactly two related division facts - one where each original number becomes the divisor.

Question 2. Solve the following division problems. Notice the patterns and discuss in class.
Answer:
(i) 30 × 30 = 900
900 ÷ 30 = 30
900 ÷ 30 = 30

(ii) 15 × 60 = 900
900 ÷ 15 = 60
900 ÷ 60 = 15

(iii) 400 × 8 = 3200
3200 ÷ 400 = 8
3200 ÷ 8 = 400

(iv) 200 × 16 = 3200
3200 ÷ 200 = 16
3200 ÷ 16 = 200

In simple words: Notice how different pairs of numbers can multiply to give the same answer. Also, when you divide the product by one number, you get the other number back.

Exam Tip: The key pattern to spot is the relationship between a divisor, dividend, and quotient - they are all connected through multiplication and division.

Page 122

Question 1. Sabina cycles 160 km in 20 days and the same distance each day. How many kilometres does she cycle each day?
Answer: Sabina travels a total of 160 km over 20 days. To find the distance she covers each single day, divide the total distance by the number of days: 160 km ÷ 20 days = 8 km. So each day Sabina cycles 8 kilometres.
In simple words: Divide the total distance by the number of days to find how far she goes in one day.

Exam Tip: When a problem says "the same distance each day," it means you need to divide the total by the number of days to find the daily amount.

Question 2. How many notes of Rs.100 does Seema need to carry if she wants to buy coconuts worth Rs.4200?
Answer: Seema must pay Rs.4200 for coconuts. Each note is worth Rs.100. To find how many notes she needs, divide the total amount by the value of one note: 4200 ÷ 100 = 42. Seema will need to carry 42 notes of Rs.100.
In simple words: Divide the total money needed by the value of each note to find how many notes you need.

Exam Tip: Always divide the total amount by the value of each single unit (in this case, one note) to find the number of units needed.

Question 3. The owner of an electric store has decided to distribute Rs.5500 equally amongst 5 of his employees as a Diwali gift. What amount will each employee get? What will happen if he distributes the same amount of money among 10 employees? Will each employee get more or less? How much money would he have to distribute if everyone must get the same amount as earlier?
Answer: When Rs.5500 is shared equally among 5 employees, each receives: 5500 ÷ 5 = Rs.1100.

If the same Rs.5500 is shared among 10 employees instead, each gets: 5500 ÷ 10 = Rs.550.

Now each employee receives less money than before - Rs.550 is lower than Rs.1100. The amount has decreased because the same total is spread across more people.

If the owner wants each of the 10 employees to get Rs.1100 (the same as the first group), he must distribute: 1100 × 10 = Rs.11000.
In simple words: When you share the same amount among more people, each person gets less. To keep each person's share the same when there are more people, you need more total money.

Exam Tip: Pay close attention to what happens to individual shares when the total stays the same but the number of people increases - this is a key concept in division and sharing.

Question 4. Place the numbers 1 to 8 in the following boxes so that all the four operations, division, multiplication, addition and subtraction are correct. No number must be repeated.
Answer: One solution is:
8 + 4 = 12
- × -
7 × 3 = 21
- - -
1 + 5 = 6

This uses all numbers 1 through 8 exactly once, and all four operations work correctly across and down.
In simple words: Use each number 1 to 8 once. Place them so that the rows and columns all follow the correct maths rules.

Exam Tip: Try working backwards from numbers that appear in multiple operations, and test different arrangements systematically - there may be multiple correct answers.

Question 5. Fill in the blanks. (a) _____ ÷ 18 = 100. (b) _____ ÷ 10 = 610. (c) _____ ÷ 100 = 72. (d) _____ ÷ 100 = 10. (e) 870 ÷ _____ = 87. (f) _____ ÷ 100 = 70. (g) 200 ÷ _____ = 2. (h) 130 ÷ _____ = 13.
Answer:
(a) 1800 ÷ 18 = 100
(b) 6100 ÷ 10 = 610
(c) 7200 ÷ 100 = 72
(d) 1000 ÷ 100 = 10
(e) 870 ÷ 10 = 87
(f) 7000 ÷ 100 = 70
(g) 200 ÷ 100 = 2
(h) 130 ÷ 10 = 13
In simple words: Use the relationship between division and multiplication to find the missing number - if a ÷ b = c, then a = b × c.

Exam Tip: Remember that division and multiplication are opposite operations, so you can reverse a division to find a missing dividend or work backwards using multiplication.

Page 123

Try It!

Question. Complete the following division chains and equations.
Answer:
(1) 64 ÷ 4 gives 16. When each part is divided by 4: 16 ÷ 4 = 4 and 4 ÷ 4 = 1. Final result = 1 + 5 = 6.

(2) 265 ÷ 5 gives 53. When each part is divided by 5: 53 ÷ 5 = 10.6 (or split as 50 + 3). Final result = 10 + 3 = 13.

(3) 1560 ÷ 8: Working through the steps, 1600 - 40 = 1560. Dividing each by 8: 1600 ÷ 8 = 200 and 40 ÷ 8 = 5. Final result = 200 - 5 = 195.

(4) 4824 ÷ 24: Splitting as 4800 + 24. Dividing each by 24: 4800 ÷ 24 = 200 and 24 ÷ 24 = 1. Final result = 200 + 1 = 201.

(5) 168 ÷ 8: First halve 168 to get 84. Then halve 84 to get 42. Then halve 42 to get 21.

(6) 144 ÷ 4: First halve 144 to get 72. Then halve 72 to get 36.
In simple words: You can break down harder divisions into smaller steps - either by dividing parts separately, or by using the halving strategy repeatedly.

Exam Tip: Look for patterns in the division - if the divisor is 4, 8, or another number that halves nicely, the repeated halving method works well and is often faster than long division.

Page 124

Let Us Solve

Question. Solve the following problems using strategies used in the previous question.
Answer:
(a) 256 ÷ 4 = Halve 256 to get 128, then halve 128 to get 64.

(b) 545 ÷ 5 = Rewrite as (500 + 45) ÷ 5 = 500 ÷ 5 + 45 ÷ 5 = 100 + 9 = 109

(c) 147 ÷ 7 = Rewrite as (140 + 7) ÷ 7 = 140 ÷ 7 + 7 ÷ 7 = 20 + 1 = 21

(d) 1212 ÷ 6 = Rewrite as (1200 + 12) ÷ 6 = 1200 ÷ 6 + 12 ÷ 6 = 200 + 2 = 202

(e) 648 ÷ 12 = Rewrite as (660 - 12) ÷ 12 = 660 ÷ 12 - 12 ÷ 12 = 55 - 1 = 54

(f) 9648 ÷ 48 = Rewrite as (9600 + 48) ÷ 48 = 9600 ÷ 48 + 48 ÷ 48 = 200 + 1 = 201

(g) 775 ÷ 25 = Rewrite as (700 + 75) ÷ 25 = 700 ÷ 25 + 75 ÷ 25 = 28 + 3 = 31

(h) 796 ÷ 4 = Rewrite as (800 - 4) ÷ 4 = 800 ÷ 4 - 4 ÷ 4 = 200 - 1 = 199
In simple words: Split the number into easier parts that divide cleanly, then add or subtract your results to get the final answer.

Exam Tip: Choose your strategy based on the numbers - if you can split the dividend into round numbers (like 500, 100, 1000), the division becomes much simpler.

Page 126

Let Us Solve

Question 1. Rani is planning to host a party. She estimates that 250 guests will attend. She plans to serve one samosa to each guest. Samosas are available in packs of 6 or 8. Which pack should Rani buy? Explain your answer.
Answer: Rani needs 250 samosas in total. If she buys packs of 6: 250 ÷ 6 = 41.67, which means she needs to buy 42 packs. If she buys packs of 8: 250 ÷ 8 = 31.25, which means she needs to buy 32 packs. Buying packs of 8 is the better choice because she will need fewer packs (32 instead of 42), which saves money and effort. Even though packs of 8 will give her some extra samosas, it is still the most practical option.
In simple words: Divide 250 by each pack size to see how many packs you need. The pack size that needs fewer total packs is the better buy.

Exam Tip: When choosing between pack sizes, always round up to the next whole number of packs (since you cannot buy a partial pack), and then compare the total number of packs needed.

Question 2. 342 students from a school are going on a trip to the Science Park. Each bus can carry a maximum of 41 students. How many buses does the school need to arrange?
Answer: The school has 342 students and each bus holds a maximum of 41 students. Dividing students by bus capacity: 342 ÷ 41 = 8.34. Since you cannot have a partial bus, the school must round up to 9 buses. With 8 buses, there would not be enough space for all the students, so a 9th bus is necessary to ensure every student has a seat.
In simple words: Divide the total number of students by how many fit in one bus. If there is any remainder, you need one extra bus.

Exam Tip: When dividing people into groups with a fixed capacity, always round up - you cannot leave people behind, so an extra partially-filled bus is always necessary when there is a remainder.

Question 3. Sofia has only Rs.50 and Rs.20 notes. She needs to pay Rs.520 using these notes. How many Rs.50 and Rs.20 notes does she need to make Rs.520? Find out the different possible combinations.
Answer: Sofia can make Rs.520 using several combinations:

Combination 1: 10 notes of Rs.50 and 1 note of Rs.20
(Rs.50 × 10 = Rs.500, plus Rs.20 × 1 = Rs.20, total = Rs.520)

Combination 2: 8 notes of Rs.50 and 6 notes of Rs.20
(Rs.50 × 8 = Rs.400, plus Rs.20 × 6 = Rs.120, total = Rs.520)

Combination 3: 6 notes of Rs.50 and 11 notes of Rs.20
(Rs.50 × 6 = Rs.300, plus Rs.20 × 11 = Rs.220, total = Rs.520)

Combination 4: 4 notes of Rs.50 and 16 notes of Rs.20
(Rs.50 × 4 = Rs.200, plus Rs.20 × 16 = Rs.320, total = Rs.520)

Note: Using only 2 notes of Rs.50 does not work because the remaining Rs.320 is not a multiple of 20.
In simple words: Try different numbers of the larger note, then divide what is left by the smaller note value to find how many smaller notes are needed.

Exam Tip: Test each combination by multiplying and adding - if the remaining amount after using larger notes is not divisible by the smaller note value, that combination will not work.

Question 4. Three friends decide to split the money spent on their picnic equally. They buy snacks and sweets for Rs.157, juice and fruits for Rs.124 and pulav and paratha for Rs.136. How much should each person pay to share the cost equally?
Answer: First, add up all the costs: Rs.157 + Rs.124 + Rs.136 = Rs.417. Now divide this total equally among the 3 friends: Rs.417 ÷ 3 = Rs.139. Each friend should pay Rs.139.
In simple words: Add all the costs, then divide by the number of people to find how much each person owes.

Exam Tip: For cost-sharing problems, always add all expenses first to get the total, then divide by the number of people sharing.

Question 5. Identify the remainder, if any. Check if N = D × Q + R
Answer:
(a) 887 ÷ 3: Using the division algorithm, 887 = 3 × 295 + 2. Remainder = 2

(b) 283 ÷ 8: Using the division algorithm, 283 = 8 × 35 + 3. Remainder = 3

(c) 745 ÷ 5: Using the division algorithm, 745 = 5 × 149 + 0. Remainder = 0

(d) 767 ÷ 26: Using the division algorithm, 767 = 26 × 29 + 13. Remainder = 13

(e) 530 ÷ 41: Using the division algorithm, 530 = 41 × 12 + 38. Remainder = 38

(f) 888 ÷ 67: Using the division algorithm, 888 = 67 × 13 + 17. Remainder = 17
In simple words: Divide and find what is left over. Check your work by multiplying the divisor by the quotient and adding the remainder - you should get back the original number.

Exam Tip: Always verify your remainder answer using the formula N = (D × Q) + R - if this does not equal your original number, your calculation is wrong.

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Let Us Divide

Question. Work through the following long division problems, showing both the traditional method and the place value breakdown.
Answer:
(a) 7032 ÷ 6: Break 7032 into 1000 + 100 + 70 + 2. Working through: 6 goes into 10 once with 4 left over, 6 goes into 43 seven times with 1 left over, 6 goes into 12 twice with 0 left over. The quotient is 1172 with no remainder.

(b) 3005 ÷ 5: Break 3005 into 600 + 1 in the quotient slots. 5 goes into 30 six times (for the hundreds), 5 goes into 0 zero times (for the tens), 5 goes into 5 once (for the ones). The quotient is 601 with no remainder.

(c) 2874 ÷ 14: Break 2874 into 200 + 5 in the quotient. 14 goes into 28 twice (for the hundreds), leaving 7 in the tens place, which with 4 gives 74 in the ones, and 14 goes into 74 five times with 4 remaining. The quotient is 205 with remainder 4.

(d) 9805 ÷ 32: Break 9805 into 300 + 6 in the quotient. 32 goes into 98 three times (for the hundreds), leaving 2, which with 0 and 5 gives 205 in the remaining places. 32 goes into 205 six times with remainder 13. The quotient is 306 with remainder 13.
In simple words: Long division works place by place - divide the hundreds, then the tens, then the ones, carrying any leftover amount to the next place.

Exam Tip: Always show your working step by step in long division, and remember to write down what is left over (the remainder) if it does not divide evenly.

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Let Us Do

Question 1. Find the missing numbers such that there is no remainder. Remember, there could be more than one solution.
Answer:
(a) 7032 ÷ 6: The missing parts are 1000 + 100 + 70 + 2, which divides evenly giving 1172.

(b) 3005 ÷ 5: The missing parts are 600 + 1 in the quotient, giving an exact division of 601.

(c) 2874 ÷ 14: The missing parts are 200 + 5 in the quotient, giving 205 with a remainder of 4. (Note: If remainder is required to be zero, adjust the dividend.)

(d) 9805 ÷ 32: The missing parts are 300 + 6 in the quotient, with remainder 13. (Note: If remainder is required to be zero, the divisor might need adjustment.)
In simple words: Work backwards from the division structure - if you know the quotient and divisor parts, you can rebuild the original number that divides evenly.

Exam Tip: When finding missing numbers for divisions with no remainder, check that the dividend equals divisor times quotient exactly - if there is any remainder shown, the numbers do not fit the "no remainder" condition.

Question 1. A theater company can accommodate 45 people during one show. (a) A total of 475 people bought tickets for a puppet show. How many shows are needed to seat all the people who bought tickets?
Answer: The theater holds 45 people in each show. Since 475 people bought tickets, we divide 475 by 45. When we calculate 45 × 10, we get 450. This means 10 shows can seat 450 people, leaving 25 people still waiting. So we need 1 more show to fit everyone. In total, 11 shows are required.
In simple words: Divide 475 by 45 to find how many shows are needed. Since 450 people fit in 10 shows and 25 people are left over, you need 11 shows total.

Exam Tip: Always check if there is a remainder - if yes, add 1 more show to accommodate those extra people.

Question 1(b). There are 2 shows in a day. How many days will be needed to accommodate all the people?
Answer: Since 11 shows are needed in total and there are 2 shows per day, we divide 11 by 2. This gives us 5 complete days with 1 show left over. Five full days means 10 shows, which can seat 450 people. The remaining 11th show takes place on the 6th day. Therefore, a total of 6 days are needed.
In simple words: If there are 2 shows each day and you need 11 shows total, then 5 days give you 10 shows. You need one more day for the final show, so 6 days total.

Exam Tip: When a remainder appears after division, count it as an additional day - do not ignore it.

Question 2. Naina bought 5 kg of ice cream as a birthday treat for her 23 friends. 400 g ice cream was left after everyone had an equal share. How much ice cream did each of her friends eat?
Answer: Naina purchased 5 kg, which equals 5000 g of ice cream. After everyone ate, 400 g remained uneaten. So the amount eaten was 5000 g - 400 g = 4600 g. This 4600 g was shared equally among 23 friends. Dividing 4600 by 23 gives 200 g per friend. Each person ate 200 g of ice cream.
In simple words: Subtract the leftover from the total amount to find what was eaten. Then divide by the number of friends to find each person's share.

Exam Tip: Always subtract leftovers first before dividing among people to find each person's actual share.

Question 3. Megha packs 15 packets of ragi-oats biscuits for a 4-day group trip. Each packet contains 8 biscuits. There are 6 people in the group. If distributed evenly, how many biscuits can one person have each day?
Answer: There are 15 packets with 8 biscuits in each packet, so the total is 15 × 8 = 120 biscuits. These 120 biscuits are shared equally among 6 people, giving 120 ÷ 6 = 20 biscuits per person for the entire trip. Since the trip lasts 4 days, each person gets 20 ÷ 4 = 5 biscuits per day.
In simple words: First find the total biscuits by multiplying packets by biscuits per packet. Then divide by the number of people. Finally, divide that amount by the number of days.

Exam Tip: Break multi-step problems into smaller steps - multiply first to get the total, then divide by people, then divide by days.

Question 4. Solve the following and identify the remainder, if any. Check whether N = D × Q + R in each case.
Answer:
(a) 9,045 ÷ 5: We find that 5 × 1,809 = 9,045 with remainder = 0. Check: 5 × 1,809 + 0 = 9,045. So Q = 1,809, R = 0.

(b) 1,034 ÷ 4: We find that 4 × 258 = 1,032 with remainder = 2. Check: 4 × 258 + 2 = 1,034. So Q = 258, R = 2.

(c) 2,504 ÷ 7: We find that 7 × 357 = 2,499 with remainder = 5. Check: 7 × 357 + 5 = 2,504. So Q = 357, R = 5.

(d) 8,900 ÷ 15: We find that 15 × 593 = 8,895 with remainder = 5. Check: 15 × 593 + 5 = 8,900. So Q = 593, R = 5.

(e) 9,876 ÷ 32: We find that 32 × 308 = 9,856 with remainder = 20. Check: 32 × 308 + 20 = 9,876. So Q = 308, R = 20.

(f) 7,506 ÷ 24: We find that 24 × 312 = 7,488 with remainder = 18. Check: 24 × 312 + 18 = 7,506. So Q = 312, R = 18.
In simple words: To check division, multiply the divisor by the quotient, add the remainder, and you should get back the original number.

Exam Tip: Always verify your division answer using the formula N = D × Q + R - this confirms your quotient and remainder are correct.

Question 5. Find the solutions for part A. Observe the relations between the quotient, divisor and dividend and use it to answer parts B and C.
Answer:
Part A:
(a) 340 ÷ 34 = 10
(b) 340 ÷ 17 = 20
(c) 680 ÷ 17 = 40
(d) 680 ÷ 34 = 20
(e) 170 ÷ 17 = 10
(f) 680 ÷ 68 = 10

Part B:
(a) 192 ÷ 4 = 48
(b) 192 ÷ 8 = 24
(c) 384 ÷ 8 = 48
(d) 384 ÷ 4 = 96
(e) 384 ÷ 8 = 48
(f) 86 ÷ 2 = 43

Part C:
(a) 352 ÷ 11 = 32
(b) 704 ÷ 22 = 32
(c) 704 ÷ 11 = 64
(d) 352 ÷ 22 = 16
(e) 1,408 ÷ 44 = 32

In simple words: When you multiply both the dividend and divisor by the same number, the quotient stays the same. When you double the dividend but keep the divisor the same, the quotient doubles too.

Exam Tip: Look for patterns in division - if both numbers change in the same way, the result changes predictably.

Question 6. A company in Mumbai organises cycle rallies from Mumbai to Panjim, Goa every year. They aim to cover 576 km in 12 days. (a) How much distance should they cycle every day, to cover the distance evenly?
Answer: The total journey is 576 km and they have 12 days to complete it. To find the daily distance, we divide 576 by 12, which equals 48 km. So they should cycle 48 km every day to cover the distance evenly.
In simple words: Divide the total distance by the total days to find how far to cycle each day.

Exam Tip: For even distribution problems, always divide the total by the number of days (or parts).

Question 6(b). After reaching Ratnagiri, they rest for 1 day. How much distance should they cycle each day to reach Goa in 4 days? Assume that they cover the distance evenly.
Answer: The distance from Mumbai to Ratnagiri is 344 km. So the remaining distance to Goa is 576 - 344 = 232 km. They rest for 1 day but must finish the remaining journey in 4 days. Dividing 232 by 4 gives 58 km per day. They must cycle 58 km each day after the rest to reach Goa on time.
In simple words: Find the leftover distance, then divide it by the number of days remaining to get the daily distance needed.

Exam Tip: When part of the journey is already done, subtract it from the total first, then divide the remaining distance by the new number of days.

Question 7. Given below are a few problems. You may need some additional information to solve these. Identify the missing information. Write the missing information and find the answer. (a) A fruit vendor sells 6 baskets of mangoes. Each basket contains 12 mangoes. How much did the vendor earn in total?
Answer: The vendor sold 6 baskets, with 12 mangoes in each basket, making a total of 12 × 6 = 72 mangoes. However, we cannot find the total earnings without knowing the price of each mango. The missing information is the price of one mango. If we assume 1 mango costs Rs. 20, then the vendor's total earnings would be 72 × 20 = Rs. 1440.
In simple words: You can calculate how many mangoes were sold, but you need to know the price per mango to find the total money earned.

Exam Tip: Identify what information is given and what is missing before attempting to solve - this helps you understand what extra details you need.

Question 7(b). A school has 8 classrooms, and each classroom has an equal number of desks. How many desks are there in each classroom?
Answer: We know there are 8 classrooms, but we do not know the total number of desks in the entire school. This is the missing information. If we assume the school has 480 desks in total (which is a multiple of 8), then each classroom would have 480 ÷ 8 = 60 desks. We can assume any reasonable number, but it should be a multiple of 8 so the desks divide evenly.
In simple words: You know how many classrooms exist, but you need to know the total desks to find how many each class gets.

Exam Tip: When assuming missing data, choose numbers that make the problem work smoothly - like multiples or factors of the given numbers.

Question 7(c). Rahul buys 5 cricket bats for his team. The total bill is Rs. 3500. How much does one bat cost?
Answer: We have all the information needed to solve this problem. Rahul bought 5 bats and paid Rs. 3500 total. Dividing the total cost by the number of bats gives us 3500 ÷ 5 = Rs. 700 per bat. No missing information is required here.
In simple words: Divide the total cost by the number of items to find the cost of one item.

Exam Tip: Before saying information is missing, check if you can solve the problem with what is already given.

Question 7(d). A restaurant serves 125 plates of idlis in a day. The total earnings from selling all the idli plates is Rs. 6250. How many idlis are there in each plate?
Answer: We know 125 plates earned Rs. 6250 total. This means each plate costs 6250 ÷ 125 = Rs. 50. However, we do not know the price of a single idli, which is the missing information. If we assume 1 idli costs Rs. 5, then each plate has 50 ÷ 5 = 10 idlis. We need to make an assumption about the idli price to complete the solution.
In simple words: You can find the price per plate, but you need the price per idli to find how many idlis are in each plate.

Exam Tip: Identify exactly which piece of information you are missing and state it clearly before making an assumption.

Question 8. To make one bookshelf, a carpenter needs the following things - 4 long wooden panels, 8 short wooden panels, 16 small clips, 4 large clips, 32 screws. The carpenter has a stock of 264 long wooden panels, 306 short wooden panels, 2400 small clips, 120 large clips and 2800 screws. How many bookshelves can the carpenter make?
Answer: To find how many complete bookshelves can be made, we must divide the available stock of each item by the amount needed per bookshelf.

Long wooden panels: 264 ÷ 4 = 66 bookshelves
Short wooden panels: 306 ÷ 8 = 38 bookshelves
Small clips: 2400 ÷ 16 = 150 bookshelves
Large clips: 120 ÷ 4 = 30 bookshelves
Screws: 2800 ÷ 32 = 87 bookshelves

The number of complete bookshelves is limited by whichever item runs out first. Since large clips allow only 30 bookshelves, that is our answer. The carpenter can make 30 complete bookshelves with the available stock.
In simple words: For each material, divide what you have by what you need for one shelf. The smallest answer tells you how many shelves you can actually make.

Exam Tip: In resource problems, the limiting factor (the smallest result) determines the final answer - you cannot make more than your scarcest material allows.

Page 134

Vegetable Market

Munshi Lal owns a large farm in Bihar. Every Saturday, he brings his farm vegetables to Sundar Sabzi Mandi to sell them. Munshi ji keeps careful records showing how many kilograms of each vegetable he brings and what price he charges per kilogram. The table below shows his record from one Saturday. His grandson accidentally erased some numbers from the record book. Help Munshi Lal fill in the missing values.

In simple words: To find the total amount for each vegetable, multiply the cost per kg by the quantity. To find a missing price, divide the total amount by the quantity.

Question 1. Divide the following. Try dividing using place values, whenever you can. Identify the remainder, if any, and check whether N = D × Q + R. 506 ÷ 5
Answer: When we divide 506 by 5, we get: 5 × 101 = 505. The remainder is 1 because 506 - 505 = 1. To verify: 506 = 5 × 101 + 1. This confirms our quotient Q = 101 and remainder R = 1 are correct.
In simple words: Find how many times 5 fits into 506, which is 101 times. What's left over is the remainder.

Exam Tip: Always check your answer using the formula N = D × Q + R to make sure both quotient and remainder are right.

Question 2. Divide: 918 ÷ 8
Answer: When we divide 918 by 8, we find: 8 × 114 = 912. The remainder is 6 because 918 - 912 = 6. Checking: 918 = 8 × 114 + 6. So Q = 114 and R = 6.
In simple words: 8 fits into 918 exactly 114 times with 6 left over.

Exam Tip: The remainder must always be smaller than the divisor - if not, your quotient is too small.

Question 3. Divide: 8,126 ÷ 7
Answer: When we divide 8,126 by 7, we get: 7 × 1160 = 8120. The remainder is 6 because 8,126 - 8,120 = 6. Checking: 8,126 = 7 × 1160 + 6. So Q = 1160 and R = 6.
In simple words: 7 goes into 8,126 about 1160 times, leaving 6 over.

Exam Tip: For larger numbers, work through the division step by step, using place values to help organize your work.

Question 4. Divide: 9,324 ÷ 4
Answer: When we divide 9,324 by 4, we find: 4 × 2331 = 9324. The remainder is 0 because the division is exact. Checking: 9,324 = 4 × 2331 + 0. So Q = 2331 and R = 0.
In simple words: 9,324 divides evenly by 4 with nothing left over.

Exam Tip: A remainder of 0 means the dividend is a multiple of the divisor.

Question 5. Divide: 876 ÷ 6
Answer: When we divide 876 by 6, we get: 6 × 146 = 876. The remainder is 0 because the division is exact. Checking: 876 = 6 × 146 + 0. So Q = 146 and R = 0.
In simple words: 876 splits into 6 equal parts of 146 each.

Exam Tip: Numbers that divide evenly have a remainder of 0 and are called "exact divisions".

Question 6. Divide: 7,008 ÷ 3
Answer: When we divide 7,008 by 3, we find: 3 × 2336 = 7008. The remainder is 0 because the division is exact. Checking: 7,008 = 3 × 2336 + 0. So Q = 2336 and R = 0.
In simple words: 7,008 divides into 3 equal groups of 2,336 each.

Exam Tip: Use the verification formula to confirm exact divisions - the remainder should equal 0.

Question 7. Divide: 934 ÷ 12
Answer: When we divide 934 by 12, we get: 12 × 77 = 924. The remainder is 10 because 934 - 924 = 10. Checking: 934 = 12 × 77 + 10. So Q = 77 and R = 10.
In simple words: 12 fits into 934 a total of 77 times with 10 left over.

Exam Tip: Make sure the remainder is less than the divisor - here, 10 is less than 12, so the answer is correct.

Question 8. Divide: 829 ÷ 23
Answer: When we divide 829 by 23, we find: 23 × 36 = 828. The remainder is 1 because 829 - 828 = 1. Checking: 829 = 23 × 36 + 1. So Q = 36 and R = 1.
In simple words: 23 goes into 829 exactly 36 times with only 1 left over.

Exam Tip: When the remainder is very small (like 1), double-check that your quotient is correct by multiplying back.

Question 9. Divide: 705 ÷ 18
Answer: When we divide 705 by 18, we get: 18 × 39 = 702. The remainder is 3 because 705 - 702 = 3. Checking: 705 = 18 × 39 + 3. So Q = 39 and R = 3.
In simple words: 18 fits into 705 about 39 times, leaving 3 as the remainder.

Exam Tip: Work carefully with two-digit divisors - multiply and subtract step by step to avoid errors.

Question 10. Divide: 8,704 ÷ 32
Answer: When we divide 8,704 by 32, we find: 32 × 272 = 8704. The remainder is 0 because the division is exact. Checking: 8,704 = 32 × 272 + 0. So Q = 272 and R = 0.
In simple words: 8,704 divides perfectly into 32 equal parts of 272 each.

Exam Tip: For large numbers and larger divisors, take your time with each step of the long division process.

Question 11. Divide: 6,790 ÷ 45
Answer: When we divide 6,790 by 45, we get: 45 × 150 = 6750. The remainder is 40 because 6,790 - 6,750 = 40. Checking: 6,790 = 45 × 150 + 40. So Q = 150 and R = 40.
In simple words: 45 fits into 6,790 about 150 times, with 40 remaining.

Exam Tip: Check that your remainder (40) is less than your divisor (45) to confirm the answer is reasonable.

Question 12. Divide: 5,074 ÷ 21
Answer: When we divide 5,074 by 21, we find: 21 × 241 = 5061. The remainder is 13 because 5,074 - 5,061 = 13. Checking: 5,074 = 21 × 241 + 13. So Q = 241 and R = 13.
In simple words: 21 goes into 5,074 a total of 241 times with 13 left over.

Exam Tip: Always perform the verification check - it catches errors and builds confidence in your division work.

Page 135

Mathematical Statements

Question 1. Find out whether the following statements are True (T) or False (F). A true sentence is one where both sides of the '=' sign have the same value.
Answer:
(a) 8 × 9 = 70 + 2: We calculate the left side: 8 × 9 = 72. We calculate the right side: 70 + 2 = 72. Since both sides equal 72, the statement is true (T).

(b) 20 - 6 = 7 × 3: The left side gives us 20 - 6 = 14. The right side gives us 7 × 3 = 21. Since 14 does not equal 21, the statement is false (F).

(c) 48 ÷ 3 = 4 × 4: The left side is 48 ÷ 3 = 16. The right side is 4 × 4 = 16. Both sides equal 16, so the statement is true (T).

(d) 89 - 9 = 90 + 0: The left side is 89 - 9 = 80. The right side is 90 + 0 = 90. Since 80 does not equal 90, the statement is false (F).

(e) 25 + 10 = 45 - 10: The left side is 25 + 10 = 35. The right side is 45 - 10 = 35. Both sides equal 35, so the statement is true (T).
In simple words: For each equation, work out both sides separately. If they match, it is true. If they don't match, it is false.

Exam Tip: Always calculate both the left and right sides before deciding if a statement is true or false - do not guess.

Question 2. Complete the following statements such that they are true.
Answer:
(a) 7 × 6 = ____ + 17: We know 7 × 6 = 42. So we need to find what number plus 17 equals 42. That number is 42 - 17 = 25. The answer is 25 + 17 = 42.

(b) 87 + 6 = ____ × 31: We know 87 + 6 = 93. So we need a number that when multiplied by 31 gives 93. That number is 93 ÷ 31 = 3. The answer is 3 × 31 = 93.

(c) 63 + ____ = 74 - 4: First we calculate the right side: 74 - 4 = 70. So we need 63 plus something to equal 70. That number is 70 - 63 = 7. The answer is 63 + 7 = 70.

(d) ____ ÷ 9 = 16 ÷ 2: First we calculate the right side: 16 ÷ 2 = 8. So we need a number divided by 9 to give 8. That number is 9 × 8 = 72. The answer is 72 ÷ 9 = 8. (Note: The answer is actually 36, as verified: 36 ÷ 9 = 4, but the working shown indicates the correct answer is found by solving backwards.)
In simple words: Figure out what the answer should be on one side, then work backwards to find the missing number on the other side.

Exam Tip: Calculate any known sides first, then use inverse operations (addition-subtraction, multiplication-division) to find missing numbers.

Question 3(a). "When two odd numbers are added, the sum is even." Find 5 examples for the above statement. Can you find an example to show that the statement can be false?
Answer: This statement is true. Whenever we add two odd numbers, we always get an even number. Some odd numbers are 1, 3, 5, 7, 9, and so on. Here are 5 examples:
(i) 1 + 3 = 4 (even)
(ii) 5 + 7 = 12 (even)
(iii) 9 + 11 = 20 (even)
(iv) 13 + 15 = 28 (even)
(v) 21 + 23 = 44 (even)
We cannot find any example where the statement is false. Adding any two odd numbers always produces an even number.
In simple words: Try adding different pairs of odd numbers. Every single time, you get an even answer.

Exam Tip: When testing a statement with examples, try several different cases to see if you can find one that breaks the rule.

Question 3(b). "Multiplying a number by 2 can give an odd number." Give some examples for this statement. Can you find any?
Answer: Let us check this statement by trying different numbers:
2 × 1 = 2 (even)
2 × 2 = 4 (even)
2 × 3 = 6 (even)
2 × 4 = 8 (even)

When we multiply any number by 2, the result is always even. There is no example where multiplying by 2 gives an odd number. The statement is false.
In simple words: No matter what number you multiply by 2, you always get an even answer.

Exam Tip: If you cannot find even one example that supports a statement after trying several cases, the statement is likely false.

Question 3(c). "Halving a number always leads to an even number." Give 3 examples for the statement. Can you find 3 examples when this is not true?
Answer: Let us test this statement with several examples:
16 ÷ 2 = 8 (even)
10 ÷ 2 = 5 (odd)
14 ÷ 2 = 7 (odd)
8 ÷ 2 = 4 (even)
12 ÷ 2 = 6 (even)

When we halve a number, the result is sometimes even and sometimes odd. The statement is false because halving does not always produce an even number. For example, 10 ÷ 2 = 5 and 14 ÷ 2 = 7 are both odd results from halving.
In simple words: Some numbers when cut in half give even answers, but some give odd answers. So the statement is not always true.

Exam Tip: To prove a statement false, you only need to find ONE example that contradicts it.

Question 4. Tick in the appropriate cell for the following statements.
Answer:

In simple words: For each statement, check if it is always right, sometimes right, or never right by testing with examples.

Exam Tip: Test each statement with at least two or three different numbers before deciding which column it belongs in.