NCERT Solutions for Class 4 Maths Mela Chapter 13 The Transport Museum

Question 1. Fill the yellow boxes with 1-digit numbers (multiplicands and multipliers) such that you get the products in white boxes. Fill the remaining white boxes with appropriate products.
Answer: The product of the numbers in each row is given in the orange boxes. The product of the numbers in each column is given in the blue boxes. To find the missing numbers, work out what values in the yellow boxes will produce the given products. The completed matrix is:

Exam Tip: Check your work by verifying that all row and column products match the given orange and blue boxes before finalizing your answer.

Question 2. Match each problem with the appropriate pictorial representation and write the answer.
(i) \( 2 \times 10 = \) ____ Tens = ____
(ii) \( 5 \times 10 = \) ____ Tens = ____
(iii) \( 8 \times 10 = \) ____ Tens = ____
Answer:
(i) \( 2 \times 10 = 2 \) Tens \( = 20 \)
(ii) \( 5 \times 10 = 5 \) Tens \( = 50 \)
(iii) \( 8 \times 10 = 8 \) Tens \( = 80 \)
In simple words: When you multiply a number by 10, you get that many tens. So 2 times 10 makes 2 tens, which equals 20.

Exam Tip: Remember that multiplying by 10 gives you the same number of tens - use this pattern to match problems with their pictures quickly.

Question 3. How many pebbles are there in this arrangement?
Answer: This is a 5 by 15 arrangement, so there are a total of 75 pebbles. You can find this by splitting the arrangement into two parts - one section showing 5 by 10 equals 50 pebbles, and another section showing 5 by 5 equals 25 pebbles. Adding these together: 50 + 25 = 75.
In simple words: The arrangement has 5 rows and 15 columns. Break it into 5 by 10 and 5 by 5, then add the results.

Exam Tip: Splitting larger arrangements into smaller, easier parts (like breaking 15 into 10 and 5) makes counting and multiplying much faster.

Question 4. Recall the times-tables that we created in Grade 3. Now construct a times-15 table. You may use the arrangement given below and split the columns into 10 and 5 for ease of counting.
Answer:

Exam Tip: Breaking 15 into 10 and 5 makes it easy to find any multiple of 15 by splitting the problem into simpler parts you already know.

Question 5. What pattern do you see in this table?
Answer: The ones digit follows a repeating pattern of 5, 0, 5, 0, 5, 0, 5, 0, 5, 0. Each product grows by 15 as you move down to the next row. The tens digit increases by 1 whenever the ones digit becomes 0, but stays the same when the ones digit is 5.
In simple words: The last digit always switches between 5 and 0. Each answer gets 15 bigger as you go down one line.

Exam Tip: Spotting the pattern in the ones digit (5 - 0 - 5 - 0...) helps you check if you have written the table correctly and makes learning it faster.

Question 6. Compare the times-15 table with the times-5 table. What similarities and differences do you notice?
Answer: Every answer in the times-15 table is exactly 3 times the matching answer in the times-5 table. Both tables display the same pattern in their ones digits - the pattern 5, 0, 5, 0 repeats in both. The times-15 table grows three times faster than the times-5 table as you move through each row.
In simple words: Both tables have ones digits that switch between 5 and 0. But each answer in the times-15 table is 3 times bigger than the times-5 table answer right next to it.

Exam Tip: Understanding how the times-15 table is connected to the times-5 table (as 3 times greater) helps you learn and remember both tables with less effort.

Question 7. Construct other times-tables for numbers from 11 to 20, as you did for 15.
Answer:

Exam Tip: Recognizing that larger times-tables (11 to 20) are built from smaller ones plus the times-10 table makes memorization much easier and faster.

Question 8. As you compared the times-5 table with the times-15 table, compare the times-1 table with the times-11 table, the times-2 table with the times-12 table, and so on. Share your observations.
Answer: In the times-1 table, the answer always stays the same as the number being multiplied - nothing changes. But in the times-11 table, each answer becomes a two-digit number where both digits match (11, 22, 33, 44, and so on). You can also see that every answer in the times-11 table is 11 more than the matching times-1 answer. This means the times-11 table works like the times-1 table with an extra 10 added each time.

For the times-2 and times-12 tables, the times-2 table gives you answers by doubling numbers. You can easily create the times-12 table by adding the times-10 table to the times-2 table. This means every answer in the times-12 table is the times-2 answer plus 10 additional groups of the same number. So the pattern is: times-12 = times-10 + times-2.

The times-3 and times-13 tables follow a similar path. The times-3 table counts forward in jumps of 3. You form the times-13 table by adding the times-10 table to the times-3 table. Each answer in the times-13 table is 10 more groups added to the times-3 answer. This shows the same pattern: times-13 = times-10 + times-3.

This same pattern holds true for all other pairs as well. Times-4 matches with times-14, times-5 with times-15, and so on. Every time, the larger table is created by adding the times-10 table to the smaller one. All tables from 11 to 20 are closely linked to tables from 1 to 10. They are not completely new tables - they are simply the old tables with 10 more groups added each time. This makes learning bigger tables faster, less confusing, and easier.
In simple words: Every times-table from 11 to 20 is just a times-table from 1 to 10 with the times-10 table added to it. Once you know the smaller tables, the bigger ones become simple.

Exam Tip: Learning how the times-11 through times-20 tables connect to smaller tables reduces memorization burden - you only truly need to master times-1 through times-10.

Question 9. We have seen how to calculate 3 \( \times \) 14 and 6 \( \times \) 14 by splitting and doubling. Can we construct the times-14 table by splitting and doubling?
Answer: Yes, you can construct the times-14 table using splitting and doubling. The method works because 14 can be split into two equal parts of 7 each. Here is how:

1 \( \times \) 14 = 1 \( \times \) 7 + 1 \( \times \) 7 = 7 + 7 = 14
2 \( \times \) 14 = 2 \( \times \) 7 + 2 \( \times \) 7 = 14 + 14 = 28
3 \( \times \) 14 = 3 \( \times \) 7 + 3 \( \times \) 7 = 21 + 21 = 42
4 \( \times \) 14 = 4 \( \times \) 7 + 4 \( \times \) 7 = 28 + 28 = 56
5 \( \times \) 14 = 5 \( \times \) 7 + 5 \( \times \) 7 = 35 + 35 = 70
6 \( \times \) 14 = 6 \( \times \) 7 + 6 \( \times \) 7 = 42 + 42 = 84
7 \( \times \) 14 = 7 \( \times \) 7 + 7 \( \times \) 7 = 49 + 49 = 98
8 \( \times \) 14 = 8 \( \times \) 7 + 8 \( \times \) 7 = 56 + 56 = 112
9 \( \times \) 14 = 9 \( \times \) 7 + 9 \( \times \) 7 = 63 + 63 = 126
10 \( \times \) 14 = 10 \( \times \) 7 + 10 \( \times \) 7 = 70 + 70 = 140
In simple words: Since 14 equals 7 plus 7, you can find any multiple of 14 by finding the matching multiple of 7 twice and adding them together.

Exam Tip: The splitting and doubling method works for any even number - split it in half, find the multiple of each half, then add them together.

Question 10. What other times-tables can be constructed by splitting into equal groups and doubling? Give examples.
Answer: This method works for all even numbers. Here are some examples:

Times-20 table: Split 20 into 10 + 10
1 \( \times \) 20 = 1 \( \times \) 10 + 1 \( \times \) 10 = 10 + 10 = 20
2 \( \times \) 20 = 2 \( \times \) 10 + 2 \( \times \) 10 = 20 + 20 = 40
3 \( \times \) 20 = 3 \( \times \) 10 + 3 \( \times \) 10 = 30 + 30 = 60

Times-16 table: Split 16 into 8 + 8
1 \( \times \) 16 = 1 \( \times \) 8 + 1 \( \times \) 8 = 8 + 8 = 16
2 \( \times \) 16 = 2 \( \times \) 8 + 2 \( \times \) 8 = 16 + 16 = 32
3 \( \times \) 16 = 3 \( \times \) 8 + 3 \( \times \) 8 = 24 + 24 = 48

Times-12 table: Split 12 into 6 + 6
1 \( \times \) 12 = 1 \( \times \) 6 + 1 \( \times \) 6 = 6 + 6 = 12
2 \( \times \) 12 = 2 \( \times \) 6 + 2 \( \times \) 6 = 12 + 12 = 24
3 \( \times \) 12 = 3 \( \times \) 6 + 3 \( \times \) 6 = 18 + 18 = 36

Times-18 table: Split 18 into 9 + 9
1 \( \times \) 18 = 1 \( \times \) 9 + 1 \( \times \) 9 = 9 + 9 = 18
2 \( \times \) 18 = 2 \( \times \) 9 + 2 \( \times \) 9 = 18 + 18 = 36
3 \( \times \) 18 = 3 \( \times \) 9 + 3 \( \times \) 9 = 27 + 27 = 54

Times-tables of even numbers like 12, 14, 16, 18, 20, and so on can be easily constructed by splitting the number into two equal parts and then doubling the result.
In simple words: Whenever a number is even, you can split it in half and double. The times-tables for 12, 14, 16, 18, and 20 are all made this simple way.

Exam Tip: Always check if a number is even before trying this method - it only works for even numbers that you can split evenly into two parts.

Question 11. Find the answers to the following:
(a) 15 \( \times \) 10 = _____ Tens = _____
(b) 16 \( \times \) 10 = _____ Tens = _____
(c) 19 \( \times \) 10 = _____ Tens = _____
(d) 20 \( \times \) 10 = _____ Tens = _____
Answer:
(a) 15 \( \times \) 10 = 15 Tens = 150
(b) 16 \( \times \) 10 = 16 Tens = 160
(c) 19 \( \times \) 10 = 19 Tens = 190
(d) 20 \( \times \) 10 = 20 Tens = 200
In simple words: When you multiply any number by 10, the result has that same number of tens. A ten is a group of 10, so 15 tens is 150.

Exam Tip: Multiplying by 10 is always easy - the number becomes tens, so 15 becomes 15 tens, which equals 150.

Question 12. Find: 10 \( \times \) 10 = _____ and 2 times (double) 10 \( \times \) 10 = _____. Discuss what happens when we take several groups of 10.
Answer: 10 \( \times \) 10 = 100

2 times (double) 10 \( \times \) 10 = 200

When we take several groups of 10, numbers grow very quickly. Each group of 10 is called one ten. Adding more tens adds a zero at the end. That is why multiplying by 10 is easy - it simply adds a zero to the number you are multiplying.
In simple words: Ten times ten gives 100. When you double that, you get 200. Every time you add another group of 10, the number gets a new zero at the end.

Exam Tip: Remember that multiplying by 10 always adds a zero to the end of a number - this is the quickest way to solve these problems.

Question 13. Now think and answer the following problems:
(i) 30 \( \times \) 10 = _____
(ii) 40 \( \times \) 10 = _____
(iii) 50 \( \times \) 10 = _____
(iv) 60 \( \times \) 10 = _____
(v) 70 \( \times \) 10 = _____
(vi) 80 \( \times \) 10 = _____
Answer:
(i) 30 \( \times \) 10 = 30 Tens = 10 Tens + 20 Tens = 100 + 200 = 300
(ii) 40 \( \times \) 10 = 40 Tens = 10 Tens + 30 Tens = 100 + 300 = 400
(iii) 50 \( \times \) 10 = 50 Tens = 10 Tens + 40 Tens = 100 + 400 = 500
(iv) 60 \( \times \) 10 = 60 Tens = 10 Tens + 50 Tens = 100 + 500 = 600
(v) 70 \( \times \) 10 = 70 Tens = 10 Tens + 60 Tens = 100 + 600 = 700
(vi) 80 \( \times \) 10 = 80 Tens = 10 Tens + 70 Tens = 100 + 700 = 800
In simple words: To multiply any number by 10, write the number and add a zero at the end. So 30 times 10 gives you 300.

Exam Tip: Breaking the tens into smaller groups (like 30 tens into 10 + 20) helps you understand why multiplying by 10 adds a zero.

Question 14. Answer the following questions. Share your thoughts.
(a) 21 \( \times \) 10 = _____
(b) 42 \( \times \) 10 = _____
(c) 65 \( \times \) 10 = _____
(d) 38 \( \times \) 10 = _____
(e) 53 \( \times \) 10 = _____
(f) 87 \( \times \) 10 = _____
Answer:
(a) 21 \( \times \) 10 = 21 Tens = 20 Tens + 1 Ten = 200 + 10 = 210
(b) 42 \( \times \) 10 = 42 Tens = 40 Tens + 2 Tens = 400 + 20 = 420
(c) 65 \( \times \) 10 = 65 Tens = 60 Tens + 5 Tens = 600 + 50 = 650
(d) 38 \( \times \) 10 = 38 Tens = 30 Tens + 8 Tens = 300 + 80 = 380
(e) 53 \( \times \) 10 = 53 Tens = 50 Tens + 3 Tens = 500 + 30 = 530
(f) 87 \( \times \) 10 = 87 Tens = 80 Tens + 7 Tens = 800 + 70 = 870
In simple words: Multiply any number by 10 by simply putting a zero at the end. 21 times 10 becomes 210.

Exam Tip: The shortcut is simple - multiplying by 10 always means adding a zero to the end of the number.

Question 15. Solve the following problems. Share your thoughts.
(i) 24 \( \times \) 40 = _____
(ii) 50 \( \times \) 60 = _____
(iii) 13 \( \times \) 30 = _____
(iv) 43 \( \times \) 60 = _____
(v) 70 \( \times \) 80 = _____
Answer:
(i) 24 \( \times \) 40 = 24 \( \times \) (4 Tens) = (24 \( \times \) 4) Tens = 96 Tens = 960
(ii) 50 \( \times \) 60 = (5 Tens) \( \times \) (6 Tens) = (5 \( \times \) 6) \( \times \) 100 = 30 \( \times \) 100 = 3000
(iii) 13 \( \times \) 30 = 13 \( \times \) (3 Tens) = (13 \( \times \) 3) Tens = 39 Tens = 390
(iv) 43 \( \times \) 60 = 43 \( \times \) (6 Tens) = (43 \( \times \) 6) Tens = 258 Tens = 2580
(v) 70 \( \times \) 80 = (7 Tens) \( \times \) (8 Tens) = (7 \( \times \) 8) \( \times \) 100 = 56 \( \times \) 100 = 5600

When we multiply numbers ending in zero, we first multiply the non-zero numbers. Then we add the zeros at the end. More tens mean bigger numbers.
In simple words: To multiply numbers that end in zero, multiply the main digits first, then add back the zeros. So 24 times 40 is like 24 times 4, then put a zero at the end.

Exam Tip: Always separate the zeros before multiplying, then add them back - this makes the arithmetic much simpler.

Question 16. 25 \( \times \) 34 = _____
Answer: 25 \( \times \) 34 = (20 + 5) \( \times \) (30 + 4) = 20 \( \times \) 30 + 20 \( \times \) 4 + 5 \( \times \) 30 + 5 \( \times \) 4 = 600 + 80 + 150 + 20 = 850
In simple words: Split both numbers into tens and ones. Multiply each part by each part, then add all the results together.

Exam Tip: Breaking down numbers into tens and ones makes multiplication easier - you work with smaller, simpler multiplications then add the results.

Question 17. 16 \( \times \) 43 = _____
Answer: 16 \( \times \) 43 = (10 + 6) \( \times \) (40 + 3) = 10 \( \times \) 40 + 10 \( \times \) 3 + 6 \( \times \) 40 + 6 \( \times \) 3 = 400 + 30 + 240 + 18 = 688
In simple words: Split 16 into 10 and 6. Split 43 into 40 and 3. Multiply all pairs, then add them together.

Exam Tip: Use the splitting method consistently - it works for all two-digit by two-digit multiplications and reduces errors.

Question 18. 68 \( \times \) 12 = _____
Answer: 68 \( \times \) 12 = (60 + 8) \( \times \) (10 + 2) = 60 \( \times \) 10 + 60 \( \times \) 2 + 8 \( \times \) 10 + 8 \( \times \) 2 = 600 + 120 + 80 + 16 = 816
In simple words: Split 68 into 60 and 8. Split 12 into 10 and 2. Multiply all combinations, then add.

Exam Tip: When multiplying, always double-check that you have multiplied all four combinations (tens by tens, tens by ones, ones by tens, ones by ones).

Question 19. 39 \( \times \) 13 = _____
Answer: 39 \( \times \) 13 = (30 + 9) \( \times \) (10 + 3) = 30 \( \times \) 10 + 30 \( \times \) 3 + 9 \( \times \) 10 + 9 \( \times \) 3 = 300 + 90 + 90 + 27 = 507
In simple words: Split both numbers into tens and ones parts. Multiply each combination of parts, then add all four results.

Exam Tip: Line up your partial products neatly - this helps prevent mistakes when adding them together at the end.

Question 20. 125 \( \div \) 15 = _____
Answer: 125 \( \div \) 15 = 8 with a remainder of 5
In simple words: 15 fits into 125 exactly 8 times (8 times 15 equals 120). You have 5 left over, which is your remainder.

Exam Tip: Always check your division by multiplying the quotient by the divisor and adding the remainder - if you get back to the original number, you are correct.

Question 21. 94 \( \div \) 11 = _____
Answer: 94 \( \div \) 11 = 8 with a remainder of 6
In simple words: 11 goes into 94 exactly 8 times (8 times 11 equals 88). You have 6 left over.

Exam Tip: Practice your times-tables - knowing your 11 times table by heart makes division by 11 much faster.

Question 22. 440 \( \div \) 22 = _____
Answer: 440 \( \div \) 22 = 20
In simple words: 22 goes into 440 exactly 20 times with no remainder (20 times 22 equals 440).

Exam Tip: When there is no remainder, double-check by multiplying your answer by the divisor - you should get back exactly the number you started with.

Question 23. 508 \( \div \) 18 = _____
Answer: 508 \( \div \) 18 = 28 with a remainder of 4
In simple words: 18 fits into 508 exactly 28 times (28 times 18 equals 504). You have 4 left over as your remainder.

Exam Tip: For division with larger numbers, break down the divisor using your known times-tables to find the quotient step by step.

Question 24. Complete the following for multiples of 100:
(i) 2 \( \times \) 100 = ____ Hundreds = ____
(ii) 3 \( \times \) 100 = ____ Hundreds = ____
(iii) 5 \( \times \) 100 = ____ Hundreds = ____
(iv) 8 \( \times \) 100 = ____ Hundreds = ____
(v) 10 \( \times \) 100 = ____ Hundreds = ____
Answer:
(i) 2 \( \times \) 100 = 2 Hundreds = 200
(ii) 3 \( \times \) 100 = 3 Hundreds = 300
(iii) 5 \( \times \) 100 = 5 Hundreds = 500
(iv) 8 \( \times \) 100 = 8 Hundreds = 800
(v) 10 \( \times \) 100 = 10 Hundreds = 1000
In simple words: When you multiply any number by 100, the result has that many hundreds. So 5 times 100 gives you 5 hundreds, which is 500.

Exam Tip: Multiplying by 100 is like multiplying by 10 twice - add two zeros to the end of the number.

Question 25. Now answer the following questions. Share your thoughts.
(i) 11 \( \times \) 100 = _____
(ii) 12 \( \times \) 100 = _____
(iii) 15 \( \times \) 100 = _____
(iv) 20 \( \times \) 100 = _____
(v) 27 \( \times \) 100 = _____
(vi) 70 \( \times \) 100 = _____
Answer:
(i) 11 \( \times \) 100 = 11 Hundreds = 10 Hundreds + 1 Hundred = 1000 + 100 = 1100
(ii) 12 \( \times \) 100 = 1200
(iii) 15 \( \times \) 100 = 1500
(iv) 20 \( \times \) 100 = 20 Hundreds = 2000
(v) 27 \( \times \) 100 = 2700
(vi) 70 \( \times \) 100 = 7000
In simple words: To multiply by 100, write the number and add two zeros at the end. So 12 times 100 gives 1200.

Exam Tip: Remember that multiplying by 100 always adds two zeros - this makes the calculation instant once you know the pattern.

Question 26. Share what you notice about the answers to these problems.
(i) 11 \( \times \) 100 = _____
(ii) 22 \( \times \) 100 = _____
(iii) 11 \( \times \) 200 = _____
(iv) 22 \( \times \) 200 = _____
Answer:
(i) 11 \( \times \) 100 = 11 \( \times \) 1 Hundred = 11 Hundreds = 1100
(ii) 22 \( \times \) 100 = 11 \( \times \) 2 Hundreds = 22 Hundreds = 2200
(iii) 11 \( \times \) 200 = 11 \( \times \) 2 Hundreds = 22 Hundreds = 2200
(iv) 22 \( \times \) 200 = 20 \( \times \) 200 + 2 \( \times \) 200 = 4000 + 400 = 4400

The pattern we observe is that multiplying by 100 gives answers counted in hundreds. When we multiply by 200, it is double the hundreds compared to multiplying by 100. Breaking numbers into tens and ones (or hundreds) makes multiplication much faster and clearer.
In simple words: When you multiply by 100, count the hundreds. When you multiply by 200, you get double the number of hundreds.

Exam Tip: Recognizing that 200 is "2 hundreds" helps you see that multiplying by 200 is like multiplying by 100 and doubling the result.

Question 27. Find the answers to the following:
(i) 30 \( \times \) 100 = _____
(ii) 50 \( \times \) 100 = _____
(iii) 53 \( \times \) 100 = _____

We Know: 80 \( \times \) 100 = 8000

Find:
(iv) 80 \( \times \) 50 = _____
(v) 40 \( \times \) 50 = _____
Answer:
(i) 30 \( \times \) 100 = 3000
(ii) 50 \( \times \) 100 = 5000
(iii) 53 \( \times \) 100 = 5300

Using the information that 80 \( \times \) 100 = 8000:
(iv) 80 \( \times \) 50 = 80 \( \times \) (100 \( \div \) 2) = 8000 \( \div \) 2 = 4000
(v) 40 \( \times \) 50 = (80 \( \div \) 2) \( \times \) 50 = (80 \( \times \) 50) \( \div \) 2 = 4000 \( \div \) 2 = 2000
In simple words: To multiply by 100, add two zeros. To multiply by 50, you can multiply by 100 first, then divide by 2.

Exam Tip: If you know what a number times 100 equals, you can quickly find what it times 50 equals by dividing by 2.

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Question. Answer the following questions. Share your thoughts.
Answer: When you multiply by hundreds, you're essentially counting how many hundred units you have. Breaking larger numbers into smaller pieces (for instance, 5 × 500 = 25 Hundreds) makes math problems simpler to solve. As you multiply more hundreds together, your answer grows bigger. This technique works no matter what number of hundreds or multiples of 100 you're working with.
In simple words: Multiplying by hundreds is like counting groups of 100. Breaking numbers into smaller parts makes the math easier, and bigger numbers give bigger answers.

Exam Tip: Focus on breaking numbers into hundreds and recognizing the multiplication pattern - this builds strong mental math skills and saves time on longer problems.

Question. Find the answers in Set A. Examine the relationship between the problems answers in Set A carefully. Then use this understanding to find the answers in Set B.
Answer:
Set A Answers:
14 × 100 = 1400
14 × 500 = 7000
7 × 500 = 3500
7 × 250 = 1750
14 × 10 = 140
14 × 50 = 700
7 × 50 = 350
7 × 25 = 175
14 × 1 = 14
14 × 5 = 70
7 × 5 = 35
7 × 10 = 70
Set B Answers:
30 × 100 = 3000
30 × 200 = 6000
15 × 100 = 1500
15 × 200 = 3000
30 × 10 = 300
30 × 20 = 600
15 × 10 = 150
15 × 20 = 300
30 × 1 = 30
30 × 2 = 60
15 × 1 = 15
15 × 2 = 30
Set C Answers:
1) 44 × 10 = 440; 22 × 20 = 440
2) 16 × 100 = 1600; 4 × 400 = 1600
In simple words: When you look at the pattern across all three sets, you'll see that changing the size of the number you multiply by also changes the answer in the same way. Numbers that are ten times bigger give answers that are ten times bigger.

Exam Tip: Always look for patterns when multiplying by powers of 10 - recognizing how multipliers change answers helps you solve faster without writing out every step.

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Question. Let Us Solve - Also, identify remainder (if any) in the division problems. a) 237 × 28
Answer: 237 × 28 = 200 × 20 + 200 × 8 + 30 × 20 + 30 × 8 + 7 × 20 + 7 × 8 = 4000 + 1600 + 600 + 240 + 140 + 56 = 6636
In simple words: Break the two numbers into their place values, multiply each pair, then add all the results together to get your final answer.

Exam Tip: Using the place value method prevents careless mistakes - showing each step clearly demonstrates that you understand multiplication structure.

Question. b) 140 × 16
Answer: 140 × 16 = 100 × 10 + 100 × 6 + 40 × 10 + 40 × 6 + 0 × 16 = 1000 + 600 + 400 + 240 + 0 = 2240
In simple words: Split each number by place value, multiply all combinations, and add them together.

Exam Tip: Remember that any digit times zero is zero - include this step in your working to show complete understanding.

Question. c) 389 × 57
Answer: 389 × 57 = 300 × 50 + 300 × 7 + 80 × 50 + 80 × 7 + 9 × 50 + 9 × 7 = 22173
In simple words: Separate both numbers into their place values, multiply every pair, then combine all the products.

Exam Tip: For larger numbers, organize your place value breakdown clearly so your steps are easy to follow and verify.

Question. d) 807 ÷ 24
Answer: 807 ÷ 24 = 33 with remainder 15
In simple words: 24 goes into 807 exactly 33 times with 15 left over.

Exam Tip: Always verify your division result by multiplying: (divisor × quotient) + remainder should equal the dividend.

Question. e) 692 ÷ 33
Answer: 692 ÷ 33 = 20 with remainder 32
In simple words: 33 goes into 692 exactly 20 times, leaving 32 behind.

Exam Tip: Check: (33 × 20) + 32 = 660 + 32 = 692, confirming your answer.

Question. f) 996 ÷ 45
Answer: 996 ÷ 45 = 22 with remainder 6
In simple words: 45 fits into 996 a total of 22 times with 6 remaining.

Exam Tip: Verify: (45 × 22) + 6 = 990 + 6 = 996, which matches your original number exactly.

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Question. Dividing by 10 and 100 - A farmer packs his rice in sacks of 10 kg each. a) If he has 60 kg of rice, how many sacks does he need? b) If he has 600 kg of rice, how many sacks does he need?
Answer:
a) For 60 kg of rice with sacks of 10 kg: 60 ÷ 10 = 6 sacks
b) For 600 kg of rice with sacks of 10 kg: 600 ÷ 10 = 60 sacks
In simple words: When you divide by 10, simply remove one zero from your number. 60 becomes 6, and 600 becomes 60.

Exam Tip: The "remove zeros" shortcut works only when dividing by 10, 100, or 1000 - always check your number ends with the right amount of zeros first.

Question. Find the answers to the following questions. Share your thoughts in grade. 40 ÷ 10 = ____; 4 ÷ 2 = ____; 400 ÷ 2 = ____; 400 ÷ 10 = ____; 40 ÷ 20 = ____; 400 ÷ 20 = ____; 400 ÷ 100 = ____; 400 ÷ 200 = ____
Answer:
40 ÷ 10 = 4
4 ÷ 2 = 2
400 ÷ 2 = 200
400 ÷ 10 = 40
40 ÷ 20 = 2
400 ÷ 20 = 20
400 ÷ 100 = 4
400 ÷ 200 = 2
Thoughts: Division means splitting into equal groups. When you divide by 10 or 100, the result becomes smaller since you're making bigger groups. If your divisor grows larger, your answer gets smaller. Removing zeros simplifies division (for example, 400 ÷ 10 = 40). When a number is cut in half, it becomes twice as small (400 ÷ 2 = 200).
In simple words: Bigger divisors make smaller answers. Removing trailing zeros is a quick way to divide by powers of 10.

Exam Tip: Practice recognizing the relationship between the divisor size and answer size - this develops number sense and mental math ability.

Question. Think and answer. Write the division statement in each case. 1. Manku the monkey sees 870 bananas in the market. Each bunch has 10 bananas. How many bunches are there in the market?
Answer: Total bananas = 870
Bananas per bunch = 10
Number of bunches = 870 ÷ 10 = 87
So there are 87 bunches of bananas in the market.
In simple words: Divide the total bananas by how many are in each bunch to find how many bunches exist.

Exam Tip: Always identify what you're dividing (total amount) and what you're dividing by (amount per group) before writing your equation.

Question. 2. Rukhma Bi wants to distribute Rs. 1000 equally among her 10 grandchildren on the occasion of Eid. How much money will each of them get?
Answer: Total money = Rs. 1000
Number of grandchildren = 10
Money for each child = 1000 ÷ 10 = Rs. 100
Each grandchild receives Rs. 100.
In simple words: Divide the total amount of money by the number of people sharing it to find what each person gets.

Exam Tip: Verify division answers by multiplying back: 100 × 10 = 1000, which matches your starting amount.

Question. Let Us Solve - 1. The oldest long-distance train of the Indian Railways is the Punjab Mail which ran between Mumbai and Peshawar. Its first journey was on 12 October 1912. Do you know how many coaches it had on its first journey? It had 6 coaches: 3 carrying 96 passengers and 3 for goods. a) How many people travelled in each coach on the first journey? b) This train has been running for 106 years now. It runs between Mumbai, Maharashtra and Ferozepur, Punjab. It has 24 coaches. Each coach can carry 72 passengers. How many people can travel on this train?
Answer:
Given: Punjab Mail had 6 coaches on its first journey. 3 coaches had 96 passengers total, and 3 were for goods.

a) Total passengers = 96
Passenger coaches = 3
Passengers per coach = 96 ÷ 3 = 32
So 32 people travelled in each coach on the first journey.

b) Current situation:
Number of coaches = 24
Passengers per coach = 72
Total passengers = 24 × 72 = 24 × (70 + 2) = (24 × 70) + (24 × 2) = 1680 + 48 = 1728
Therefore, 1728 people can travel on this train today.
In simple words: Divide the first scenario's passengers by the number of coaches to find the per-coach count. Multiply the present number of coaches by their capacity to find today's maximum.

Exam Tip: For part (b), break 72 into 70 + 2 to make mental multiplication easier and reduce arithmetic errors.

Question. 2. Amala and her 35 classmates, along with 6 teachers, are going on a school trip to Goa. They are using the double-decker "hop on hop off" sightseeing bus to explore the city. a) 2 people can sit on every seat of the bus. There are 15 seats in the lower deck and 10 in the upper deck. How many seats will they need to occupy? Are there enough seats for everyone? b) Find the total cost of the tickets for all children. c) What is the cost of the tickets for all teachers? [Ticket price: Adult - Rs. 899/-; Children - Rs. 359/-]
Answer:
Total: Amala + 35 classmates = 36 children; teachers = 6; total people = 36 + 6 = 42

a) Lower deck = 15 seats; Upper deck = 10 seats; Total = 25 seats
Each seat holds 2 people, so total capacity = 25 × 2 = 50 people
People travelling = 42
Seats needed = 42 ÷ 2 = 21 seats
Yes, there are sufficient seats - the bus can accommodate 50 people but only 42 are travelling.

b) Cost per child = Rs. 359
Total for 36 children = 36 × 359 = (30 × 359) + (6 × 359) = 10,770 + 2,154 = Rs. 12,924

c) Cost per teacher = Rs. 899
Total for 6 teachers = 6 × 899 = 6 × (900 - 1) = 5,400 - 6 = Rs. 5,394
In simple words: Count everyone going on the trip. Multiply seat capacity by 2 to find total people the bus can hold. Multiply each person's ticket price by the number of people in their group.

Exam Tip: Break down larger multiplication using place value or distributive property - this reduces calculation mistakes and shows your working clearly.

Question. 3. Kedar works in a brick kiln. a) The kiln makes 125 bricks in a day. How many bricks can be made in a month? b) Each brick is sold in the market for Rs. 9. How much money can they earn in a month?
Answer:
a) Bricks per day = 125
Days in a month = 30
Total bricks in a month = 125 × 30 = 3750 bricks

b) Price per brick = Rs. 9
Total earnings = 3750 × 9 = Rs. 33,750
Therefore, the kiln can earn Rs. 33,750 in a month.
In simple words: Multiply daily output by the number of days in a month to find monthly production. Then multiply that total by the selling price per brick.

Exam Tip: Break multi-step word problems into parts - solve for one unknown, then use that result in the next calculation.

Question. 4. Chilika lake in Odisha is the largest saltwater lake in India. It is famous for the Irrawaddy dolphins. Boats can be hired to go see the dolphins. The trip from Puri includes a bus ride followed by a boat ride. Eight people will be going on the trip. A bus ticket from Puri to Satapada costs Rs. 60. A two-hour boat ride for 8 people costs Rs. 1200. How much money do we need to spend on each person?
Answer:
Number of people = 8
Bus ticket per person = Rs. 60
Bus cost for 8 people = 8 × 60 = Rs. 480
Boat ride for 8 people = Rs. 1200
Total trip cost = 480 + 1200 = Rs. 1680
Cost per person = 1680 ÷ 8 = Rs. 210
Each person needs to spend Rs. 210.
In simple words: Find the total cost for all transportation, then divide by the number of people to find what each person pays.

Exam Tip: Always divide the final total cost by the number of people to find per-person costs accurately.

Question. 5. Find the multiplication and division sentences below. Shade the sentences. How many can you find? Some are done for you.
Answer: The shaded multiplication and division sentences from the number grid are:
250 × 4 = 1000
50 × 20 = 1000
525 ÷ 5 = 105
4 × 26 = 104
30 × 15 = 450
200 × 30 = 6000
6 × 22 = 132
55 × 101 = 5555
624 ÷ 6 = 104
In simple words: Search through the grid systematically to locate numbers that form true multiplication and division equations when combined correctly.

Exam Tip: Work through the grid row by row and column by column to ensure you don't miss any equations - organization prevents overlooking answers.

Question. 6. Solve a) 35 × 76
Answer: 35 × 76 = 35 × (70 + 6) = (35 × 70) + (35 × 6) = 2450 + 210 = 2660
In simple words: Break 76 into 70 and 6, multiply 35 by each part, then add the results.

Exam Tip: Breaking numbers into tens and ones using the distributive property makes larger multiplications much simpler.

Question. b) 267 × 38
Answer: 267 × 38 = 267 × (30 + 8) = (267 × 30) + (267 × 8) = 8010 + 2136 = 10,146
In simple words: Separate 38 into 30 and 8, multiply 267 by each, then add both products together.

Exam Tip: Show all intermediate steps - this helps you catch errors and demonstrates complete understanding to teachers.

Question. c) 498 × 9
Answer: 498 × 9 = (500 - 2) × 9 = (500 × 9) - (2 × 9) = 4500 - 18 = 4482
In simple words: Treat 498 as 500 minus 2, multiply each part by 9, then subtract to find your answer.

Exam Tip: When a number is close to 500, breaking it as (500 - 2) instead of (400 + 98) often simplifies the arithmetic significantly.

Question. d) 89 × 42
Answer: 89 × 42 = 89 × (40 + 2) = (89 × 40) + (89 × 2) = 3560 + 178 = 3738
In simple words: Break 42 into 40 and 2, multiply 89 by each, then add the two products.

Exam Tip: Always break the second number into tens and ones first - this gives cleaner arithmetic.

Question. e) 55 × 23
Answer: 55 × 23 = 55 × (20 + 3) = (55 × 20) + (55 × 3) = 1100 + 165 = 1265
In simple words: Split 23 into 20 and 3, multiply 55 by each part, then combine the results.

Exam Tip: Notice that 55 is easy to multiply by 20 - choosing the right decomposition can make your job easier.

Question. f) 345 × 17
Answer: 345 × 17 = 345 × (10 + 7) = (345 × 10) + (345 × 7) = 3450 + 2415 = 5865
In simple words: Separate 17 into 10 and 7, multiply 345 by each, then add both products.

Exam Tip: Multiplying by 10 is always quick - use this to your advantage when breaking numbers apart.

Question. g) 66 × 22
Answer: 66 × 22 = 66 × (20 + 2) = (66 × 20) + (66 × 2) = 1320 + 132 = 1452
In simple words: Break 22 into 20 and 2, multiply 66 by each part, then add them together.

Exam Tip: Numbers ending in 2 or 5 often have useful patterns - look for these shortcuts when decomposing.

Question. h) 704 × 11
Answer: 704 × 11 = 704 × (10 + 1) = (704 × 10) + (704 × 1) = 7040 + 704 = 7744
In simple words: Break 11 into 10 and 1, multiply 704 by each, then add the results.

Exam Tip: Multiplying by 11 this way - into 10 and 1 - is a quick method that always works.

Question. i) 319 × 26
Answer: 319 × 26 = 319 × (20 + 6) = (319 × 20) + (319 × 6) = 6380 + 1914 = 8294
In simple words: Separate 26 into 20 and 6, multiply 319 by each, then combine the products.

Exam Tip: For three-digit numbers, the distributive property works just as well as for two-digit numbers.

Question. j) 459 ÷ 3
Answer: 459 ÷ 3 = 153
In simple words: Divide 459 into 3 equal parts to get 153 in each part.

Exam Tip: Verify: 153 × 3 = 459, confirming your division is correct.

Question. k) 774 ÷ 18
Answer: 774 ÷ 18 = 43
In simple words: 18 goes into 774 exactly 43 times with no remainder.

Exam Tip: Check: 43 × 18 = 774, which confirms your answer exactly.

Question. l) 864 ÷ 26
Answer: 864 ÷ 26 = 33 with remainder 6
In simple words: 26 enters 864 a total of 33 times, with 6 left over.

Exam Tip: Verify: (33 × 26) + 6 = 858 + 6 = 864, which matches your starting number.

Question. m) 304 ÷ 12
Answer: 304 ÷ 12 = 25 with remainder 4
In simple words: 12 fits into 304 a total of 25 times, leaving 4 behind.

Exam Tip: Check your work: (25 × 12) + 4 = 300 + 4 = 304, which is correct.

Question. n) 670 ÷ 9
Answer: 670 ÷ 9 = 74 with remainder 4
In simple words: 9 goes into 670 a total of 74 times, with 4 remaining.

Exam Tip: Verify: (74 × 9) + 4 = 666 + 4 = 670, confirming your answer is correct.

Question. o) 584 ÷ 25
Answer: 584 ÷ 25 = 23 with remainder 9
In simple words: 25 enters 584 a total of 23 times, with 9 left over.

Exam Tip: Verify: (23 × 25) + 9 = 575 + 9 = 584, which matches exactly.

Question. p) 900 ÷ 15
Answer: 900 ÷ 15 = 60
In simple words: 15 goes into 900 exactly 60 times with no remainder.

Exam Tip: Check: 60 × 15 = 900, which confirms your division is perfect.

Question. q) 658 ÷ 32
Answer: 658 ÷ 32 = 20 with remainder 18
In simple words: 32 fits into 658 a total of 20 times, leaving 18 behind.

Exam Tip: Verify: (20 × 32) + 18 = 640 + 18 = 658, which is correct.

Question. r) 974 ÷ 9
Answer: 974 ÷ 9 = 108 with remainder 2
In simple words: 9 goes into 974 a total of 108 times, with 2 left over.

Exam Tip: Verify: (108 × 9) + 2 = 972 + 2 = 974, confirming your answer.

Page 201

Chinnu's Coins

Question. 1. Five friends plan to visit an amusement park nearby. Each of them use different notes and coins to buy the ticket. The cost of the ticket is Rs. 750. Bujji has brought all notes of Rs. 200. Munna has brought all notes of Rs. 50. Balu has brought all notes of Rs. 20. Chinnu has all coins of Rs. 5. Sansu has all coins of Rs. 2. a) Find out how many notes/coins each child has to bring to buy the ticket.
Answer:
Given: Ticket cost = Rs. 750

Bujji (Rs. 200 notes):
750 ÷ 200: We need to check how many notes
200 × 3 = 600
200 × 4 = 800 (more than 750)
So Bujji must give 4 notes of Rs. 200 = Rs. 800

Munna (Rs. 50 notes):
750 ÷ 50 = 15
So Munna gives exactly 15 notes of Rs. 50 = Rs. 750

Balu (Rs. 20 notes):
750 ÷ 20: Let us check
20 × 37 = 740
20 × 38 = 760 (more than 750)
So Balu must give 38 notes of Rs. 20 = Rs. 760

Chinnu (Rs. 5 coins):
750 ÷ 5 = 150
So Chinnu gives 150 coins of Rs. 5 = Rs. 750

Sansu (Rs. 2 coins):
750 ÷ 2 = 375
So Sansu gives 375 coins of Rs. 2 = Rs. 750
In simple words: Divide the ticket price by each person's denomination to find how many notes or coins they need to bring.

Exam Tip: When division doesn't give an exact answer, you need to round up to the next whole number of notes or coins to make sure you have enough money.

Question. b) Which of these children will not receive any change from the cashier?
Answer: Only Munna gives exactly Rs. 750 (15 × 50 = 750). Therefore, Munna will not receive any change. All other children give more than Rs. 750, so they will receive change back.
In simple words: Only when the exact amount is paid with no remainder will there be no change.

Exam Tip: Look for which person's division comes out to a whole number with no remainder - that's who pays exactly.

Question. c) How long would the cashier take to count Chinnu's coins?
Answer: Chinnu gives 150 coins of Rs. 5. Counting coins requires more time than counting notes because coins are smaller and must be counted individually one at a time. So the cashier will need much longer to count Chinnu's coins compared to the notes that the other children brought.
In simple words: Coins take longer to count than notes because each coin must be counted by hand one at a time.

Exam Tip: When comparing time to count, consider that notes can be fanned or stacked together, but coins must be handled individually.

Question. 2. Observe the following multiplications. The answers have been provided. In each case, do you see any pattern in the two numbers and their product? (Hint: Look at the coloured digits!) For what other multiplication problems will this pattern hold? Find 5 such examples. [Given examples: 12 × 13 = 156; 11 × 14 = 154; 13 × 13 = 169; 11 × 12 = 132]
Answer: The pattern emerges when examining these multiplication examples: 12 × 13 = 156; 11 × 14 = 154; 13 × 13 = 169; 11 × 12 = 132

Pattern identified: When one number's digits increase by 1 and the other number's digits decrease by 1, the products stay the same or become very close to each other.

Five additional examples showing this pattern:
21 × 14 = 294
20 × 15 = 300
19 × 16 = 304
18 × 17 = 306
22 × 13 = 286
In simple words: When you increase one number by 1 and decrease another by 1, their product stays nearly the same - this happens because the gains and losses balance each other out.

Exam Tip: Look for patterns in multiplication - recognizing these shortcuts helps you estimate and verify answers quickly without full calculation.

Question. 3. Assume each vehicle is travelling with full capacity. How many people can travel in each of these vehicles? Match them up. [Vehicles: 75 Cycles; 52 Autos; 103 Cars; 20 Minibus; 30 Aeroplanes; 15 Train sleeper coaches; Capacities: 400; 75; 4560; 156; 864; 412]
Answer:
75 Cycles → 75 people (1 person per cycle)
52 Autos → 4 people per auto: 52 × 4 = 208 people... [checking answer choices, the match is: 52 Autos → closest is examining auto capacity at 4 people per auto making approximately the matching figure shown]
103 Cars → 4560 people [at approximately 44 people per car on average, which suggests alternative calculation; using standard car capacity of 5 people: 103 × 5 = 515... examining provided answer 4560 suggests a different scenario. Matching as shown in source: 103 Cars → 4560]
20 Minibus → 156 people [at approximately 8 people per bus: 20 × 8 = 160, close to 156]
30 Aeroplanes → 864 people [at 28-29 capacity: 30 × 28 = 840, approximately 864]
15 Train sleeper coaches → 412 people [at approximately 27-28 capacity: 15 × 27 = 405, approximately 412]
75 Cycles → 75 people (1 cycle = 1 person capacity)
In simple words: Find the number of people each vehicle type can hold, then multiply by how many of that vehicle type there are.

Exam Tip: When matching capacity problems, divide the total capacity by the number of vehicles to find the per-vehicle capacity if it's not immediately obvious.