NCERT Solutions Class 5 Mathematics Mela Chapter 06 The Dairy Farm

Question 1. The given shapes stand for numbers between 1 and 24. The same shape denotes the same number across all problems. Find the numbers hiding in all the shapes.
(a)
(b)
(c)
(d)
(e)
(f)
Answer: By solving each equation systematically, we can determine that: (a) Blue rectangle = 3, Pink circle = 4, Triangle = 12; (b) Blue rectangle = 3, so 3 × 3 = 9 (empty shape); (c) Yellow parallelogram = 2, Orange diamond = 6, Triangle = 12; (d) Orange diamond = 6, Green pentagon = 1; (e) Yellow parallelogram = 2, Pink circle = 4, Light blue curved shape = 8; (f) Light blue curved shape = 8, Blue rectangle = 3, Yellow parallelogram = 2, Triangle = 12. Each shape stays the same value throughout all problems, which lets us work out the missing numbers by using the given equations.
In simple words: Each shape is a hidden number. Once you find what one shape equals in one problem, you know it in all the other problems too.

Exam Tip: Always check that your answer for each shape works in every equation it appears in - this confirms your solution is correct.

Question 2. Place the digits 2, 5 and 3 appropriately to get a product close to 100. Share your reasoning in class.
Answer: To find a product close to 100, arrange the digits as 53 × 2 = 106. This gives us the nearest result to 100 among all possible arrangements of these three digits. Other arrangements like 52 × 3 = 156, 35 × 2 = 70, and 32 × 5 = 160 produce values farther from 100, making 106 the best choice.
In simple words: Try different ways of putting 2, 5, and 3 together. The arrangement 53 × 2 gives 106, which is closest to 100.

Exam Tip: Test all possible arrangements of the given digits systematically to find which one produces the target value.

Question 3. A dairy has packed buttermilk pouches in the following manner. Find the number of pouches kept in each arrangement. One is done for you.
Answer: The arrangements show different ways to group 60 pouches total. The first arrangement (30 × 2 = 60) is already worked out. Other arrangements of 60 pouches can be formed as: 20 × 3 = 60, 10 × 6 = 60, and 12 × 5 = 60. Each of these multiplications gives the same total of 60 pouches, just arranged in different row and column combinations. This demonstrates that multiplication is flexible - the same total can be made by grouping items in various ways.
In simple words: Sixty pouches can be arranged in many different patterns, but they all add up to the same number of pouches.

Exam Tip: Remember that multiplication shows different ways to make the same total - find all factor pairs of the given number.

Question 4. Which number am I? I am a two-digit number. Find me with the help of the following clues.
(a) I am greater than 8.
(b) I am not a multiple of 4.
(c) I am a multiple of 9.
(d) I am an odd number.
(e) I am not a multiple of 11.
(f) I am less than 50.
(g) My ones digit is even.
(h) My tens digit is odd.
Answer: Using each clue step by step: From clue (a), the number is between 10 and 99. From clue (c), it must be a multiple of 9 (9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99...). Combining with clue (b), we exclude multiples of 4, leaving 18, 27, 45, 63, 81, 90, 99. From clue (d), it must be odd, so we remove 18, 90, leaving 27, 45, 63, 81, 99. From clue (e), we exclude 99 (since 99 = 11 × 9), leaving 27, 45, 63, 81. From clue (f), we pick numbers under 50: 27 and 45. From clues (g) and (h), the ones digit must be even and the tens digit must be odd. Only 18 satisfies both: the tens digit is 1 (odd) and the ones digit is 8 (even). Therefore, the number is 18.
In simple words: Work through each clue one at a time, crossing out numbers that don't fit. Keep going until only one number is left.

Exam Tip: When solving clue-based puzzles, eliminate numbers that fail each condition systematically to narrow down the answer.

Question 5. Make your own numbers. Choose any two numbers and one operation from the grid. Try to make all the numbers between 0 and 20. For example, 2 can be formed as 4 - 2. Could you make all the numbers? Which numbers could you not make? Is it possible to make these numbers using three numbers? You can use two operations, if needed. Which numbers between 0 - 20 can you get in more than one way?
Answer: Using the numbers 100, 25, 5, 10, 2, 36, 12, 4, 3 and the operations subtraction, multiplication, addition, and division: 0 = 4 - 4; 1 = 5 - 4; 2 = 5 - 3; 3 = 36 ÷ 12; 4 = 100 ÷ 25; 5 = 25 ÷ 5; 6 = 10 - 4; 7 = 5 + 2; 8 = 5 + 3; 9 = 12 - 3; 10 = 100 ÷ 10; 11 = 36 - 25; 12 = 10 + 2; 13 = 25 - 12; 14 = 10 + 4; 15 = 25 - 10; 16 = 12 + 4; 17 = 12 + 5; 18 = 36 ÷ 2; 19 cannot be made with two numbers; 20 = 100 ÷ 5. Using three numbers and two operations, 19 can be made as (25 - 10) + 4 = 19 or other combinations. All numbers except 19 can be made with just two numbers and one operation.
In simple words: You can build most numbers 0 through 20 by picking two numbers from the grid and doing one operation. The number 19 is tricky - you need three numbers and two operations to make it.

Exam Tip: Explore combinations methodically, noting which numbers need more operations - this teaches flexibility in problem-solving.

Order of Numbers in Multiplication

Question 6. Daljeet Kaur runs a milk processing unit. She has arranged the butter packets in the following ways. Find the number of butter packets in each case. What pattern do you notice (or observe)? Discuss in class.
Answer: Looking at each arrangement: (a) 2 × 3 = 6 packets, and 3 × 2 = 6 packets; (b) 5 × 8 = 40 packets, and 8 × 5 = 40 packets; (c) 10 × 5 = 50 packets, and 5 × 10 = 50 packets; (d) 12 × 9 = 108 packets, and 9 × 12 = 108 packets. The pattern shown is that when you multiply two numbers, the order does not matter - swapping the numbers gives the same result. This property is called the commutative property of multiplication. Mathematically, a × b = b × a for any two numbers a and b. This pattern holds true for the multiplication of any two numbers, not just the ones shown in these examples.
In simple words: Multiplication works the same way no matter which number you put first. Three times two equals six, and two times three also equals six.

Exam Tip: Remember that multiplication is commutative - changing the order of the numbers doesn't change the product, which can help you check your answer.

Patterns in Multiplication by 10s and 100s

Question 7. Let us revise multiplication by 10s and 100s. Complete each multiplication.
(a) 4 × 10 = _____
(b) 20 × 10 = _____
(c) 10 × 40 = _____
(d) 10 × 10 = _____
(e) 20 × 50 = _____
(f) 80 × 10 = _____
(g) 3 × 100 = 100 × 3 = _____
(h) 8 × 100 = _____ = _____
(i) 10 × 100 = _____ = _____
Answer: (a) 40; (b) 200; (c) 400; (d) 100; (e) 1,000; (f) 800; (g) 300; (h) 800 = 100 × 8 = 800; (i) 1,000 = 100 × 10 = 1,000. Each answer follows a simple rule: when multiplying by 10, add one zero to the number. When multiplying by 100, add two zeros to the number. This pattern stays true whether the 10 or 100 comes first or second in the multiplication.
In simple words: To multiply by 10, just put a zero at the end of the number. To multiply by 100, put two zeros at the end.

Exam Tip: Use the zero-counting method to multiply quickly by 10 or 100 - this saves time on tests.

Question 8. Find answers to the following questions. Fill in the table below and describe the pattern. Discuss in class.
Answer: When multiplying numbers that end with zeros, first multiply the non-zero digits, then add the zeros at the end. For example, 30 × 20: multiply 3 × 2 = 6, then count the total zeros (two zeros), so the answer is 600. Applying this pattern: 100 × 90 = 9,000; 400 × 10 = 4,000; 60 × 50 = 3,000; 30 × 20 = 600; 700 × 4 = 2,800; 10 × 45 = 450; 80 × 90 = 7,200; 10 × 63 = 630; 40 × 12 = 480; 60 × 50 = 3,000; 220 × 20 = 4,400; 11 × 300 = 3,300. The pattern is consistent: the non-zero digits combine, and the total count of zeros from both numbers gets added to the product.
In simple words: Ignore the zeros for a moment. Multiply the main numbers first. Then add back all the zeros you saw.

Exam Tip: Practice grouping numbers by their zeros to spot the pattern quickly - this makes mental math much faster.

Question 9. For the multiplication problems shown, fill in the place-value table (Thousands, Hundreds, Tens, Ones) and observe the patterns in the results.
Answer: Filling the place-value table: 2 × 1,000 = 2,000 (2 in thousands); 5 × 1,000 = 5,000 (5 in thousands); 10 × 1,000 = 10,000 (1 in ten-thousands, 0 in thousands); 20 × 1,000 = 20,000 (2 in ten-thousands, 0 in thousands); 3 × 5,000 = 15,000 (1 in ten-thousands, 5 in thousands); 8 × 3,000 = 24,000 (2 in ten-thousands, 4 in thousands); 5 × 7,000 = 35,000 (3 in ten-thousands, 5 in thousands); 20 × 100 = 2,000 (2 in thousands); 40 × 500 = 20,000 (2 in ten-thousands); 60 × 300 = 18,000 (1 in ten-thousands, 8 in thousands); 600 × 30 = 18,000 (same as above); 80 × 900 = 72,000 (7 in ten-thousands, 2 in thousands); 70 × 600 = 42,000 (4 in ten-thousands, 2 in thousands); 5 × 7,000 = 35,000 (3 in ten-thousands, 5 in thousands). The pattern shows that multiplying by larger values (those with zeros) builds up the place values in an orderly way based on the count of zeros and the non-zero digits.
In simple words: When you multiply by thousands and hundreds, the answer gets bigger and fills up the higher place-value columns in the table.

Exam Tip: Use place-value tables to see how zeros shift the answer into different columns - this helps you understand why the answer gets bigger.

Doubling and Halving

Question 10. Butter packets are arranged in the following ways. Let us find some strategies to calculate the total number of packets. This halving and doubling strategy works well when we have to multiply with numbers like 5 and 25. Discuss why.
Answer: The halving and doubling strategy (also called the compensation method) is useful for multiplying by 5, 25, 50, and similar numbers because these have simple relationships with powers of 10 (10, 100, 1,000, etc.), which are the easiest numbers to work with in multiplication. For example, 3 × 18: instead of computing 3 × 18 directly, we can double 3 to get 6 and halve 18 to get 9, giving us 6 × 9 = 54. Similarly, 22 × 5: halving 22 gives 11 and doubling 5 gives 10, so we get 11 × 10 = 110. This works because doubling one factor and halving the other keeps the product the same (the changes cancel out), but it creates easier numbers to multiply. When working with 5, 25, or 50, halving and doubling often produces a power of 10, which is trivial to multiply by.
In simple words: Make one number twice as big and the other number half as small. The answer stays the same, but the new numbers are easier to multiply.

Exam Tip: Look for factors that are multiples of 5 or 25 - these are your signal to try halving and doubling.

Question 11. Solve the following problems using the halving and doubling strategy.
(a) 14 × 3
(b) 22 × 5
(c) 16 × 4
(d) 38 × 5
(e) 35 × 14
Answer: (a) 14 × 3: Halve 14 to get 7, double 3 to get 6. Then 7 × 6 = 42; (b) 22 × 5: Halve 22 to get 11, double 5 to get 10. Then 11 × 10 = 110; (c) 16 × 4: Halve 16 to get 8, double 4 to get 8. Then 8 × 8 = 64; (d) 38 × 5: Halve 38 to get 19, double 5 to get 10. Then 19 × 10 = 190; (e) 35 × 14: Double 35 to get 70, halve 14 to get 7. Then 70 × 7 = 490. Each calculation uses halving and doubling to convert at least one factor into a number that is simpler to work with mentally.
In simple words: Pick one number to make twice as big and the other to make half as small. Then multiply the new, easier numbers.

Exam Tip: Always check that you halved and doubled correctly - mistakes here will make your final answer wrong.

Question 12. Find the product by halving and doubling either the multiplier or the multiplicand.
(a) 5 × 18
(b) 50 × 28
(c) 15 × 22
(d) 25 × 12
(e) 12 × 45
(f) 16 × 45
Answer: (a) 5 × 18: Halve 18 to get 9, double 5 to get 10. Result: 9 × 10 = 90; (b) 50 × 28: Halve 28 to get 14, double 50 to get 100. Result: 14 × 100 = 1,400; (c) 15 × 22: Halve 22 to get 11, double 15 to get 30. Result: 11 × 30 = 330; (d) 25 × 12: Halve 12 to get 6, double 25 to get 50. Result: 6 × 50 = 300; (e) 12 × 45: Halve 12 to get 6, double 45 to get 90. Result: 6 × 90 = 540; (f) 16 × 45: Halve 16 to get 8, double 45 to get 90. Result: 8 × 90 = 720. In each case, the process transforms at least one factor into a round number, making the multiplication fast and easy.
In simple words: Use halving and doubling to turn one of your numbers into 10, 50, or 100. Multiplying by these is super quick.

Exam Tip: Recognize when a factor can be halved evenly - this signals that the halving and doubling method will work smoothly.

Question 13. Give 5 examples of multiplication problems where halving and doubling will help in finding the product easily. Find the products as well.
Answer: Five examples where halving and doubling simplifies the calculation: 1. 5 × 16 = (10 × 8) = 80; 2. 25 × 8 = (50 × 4) = 200; 3. 15 × 6 = (30 × 3) = 90; 4. 50 × 18 = (100 × 9) = 900; 5. 35 × 4 = (70 × 2) = 140. Each example converts at least one factor through halving and doubling into a power-of-10 or near-power-of-10 value, which makes the final multiplication nearly instant. This method is especially powerful when one factor is odd (or when halving an even factor results in a nice number) and the other can be doubled to reach 10, 50, 100, or 1000.
In simple words: Pick multiplication problems where one number can be split in half and the other can be doubled to make a round number like 10, 50, or 100.

Exam Tip: Choose problems where at least one factor is even (so it can be halved) and the other is 5, 25, 35, or 15 (which can be doubled to make round numbers).

Nearest Multiple

Question 14. Find the product of 4 × 19 using the nearest multiple strategy.
Answer: To multiply 4 × 19, it is easier to work with 20 (the nearest multiple of 10) rather than 19. Calculate 4 × 20 = 80. Since we used 20 instead of 19, we added one extra group of 4. Subtract that extra amount: 80 - 4 = 76. Therefore, 4 × 19 = 76. This strategy works because 20 is much simpler to multiply by than 19, and the adjustment (subtracting one group of 4) is quick and easy to do mentally.
In simple words: Round 19 up to the closer number 20. Multiply 4 × 20 to get 80. Then subtract 4 to account for the extra group. The answer is 76.

Exam Tip: Always check whether rounding up or down gives a nearest multiple that is easier to work with - this choice determines how much adjustment you need.

Question 15. Find the product of 14 × 21 using the nearest multiple strategy.
Answer: To multiply 14 × 21, round 21 down to 20 (the nearest multiple of 10). Calculate 14 × 20 = 280. Since we used 20 instead of 21, we removed one group of 14. Add that back: 280 + 14 = 294. Therefore, 14 × 21 = 294. This strategy turns the problem into a simpler multiplication by 20, with just a small addition step to correct for the rounding.
In simple words: Pretend 21 is really 20 and multiply 14 × 20 to get 280. Then add back one group of 14 since you rounded down. The answer is 294.

Exam Tip: Decide whether to round up or down by checking which multiple of 10 is truly nearest - smaller adjustments lead to fewer calculation errors.

Question 16. Give 5 examples of problems where you can use the nearest multiple to find the product easily. Find the products as well.
Answer: Five examples using the nearest multiple strategy: 1. 6 × 49 = 6 × 50 - 6 = 300 - 6 = 294; 2. 8 × 52 = 8 × 50 + 16 = 400 + 16 = 416; 3. 13 × 19 = 13 × 20 - 13 = 260 - 13 = 247; 4. 11 × 102 = 11 × 100 + 22 = 1,100 + 22 = 1,122; 5. 25 × 39 = 25 × 40 - 25 = 1,000 - 25 = 975. Each example rounds one factor to its nearest multiple of 10 or 100, multiplies, then adjusts by either adding or subtracting based on the direction of rounding. This method trades a tricky multiplication for an easy one plus a simple addition or subtraction.
In simple words: Round one number to a nearby multiple of 10 or 100, multiply, then add or subtract a small correction to get the true answer.

Exam Tip: Choose numbers that are close to (but not quite) multiples of 10 or 100 - this is when the nearest multiple strategy saves the most work.

Question 17. Find the products of the following numbers by finding the nearest multiple.
(a) 7 × 52
(b) 12 × 28
(c) 75 × 31
(d) 99 × 15
(e) 8 × 25
(f) 22 × 42
Answer: (a) 7 × 52: Round 52 to 50. Calculate 7 × 50 = 350. Add 7 × 2 = 14. Result: 350 + 14 = 364; (b) 12 × 28: Round 28 to 30. Calculate 12 × 30 = 360. Subtract 12 × 2 = 24. Result: 360 - 24 = 336; (c) 75 × 31: Round 31 to 30. Calculate 75 × 30 = 2,250. Add 75 × 1 = 75. Result: 2,250 + 75 = 2,325; (d) 99 × 15: Round 99 to 100. Calculate 100 × 15 = 1,500. Subtract 1 × 15 = 15. Result: 1,500 - 15 = 1,485; (e) 8 × 25: Round 25 to 20. Calculate 8 × 20 = 160. Add 8 × 5 = 40. Result: 160 + 40 = 200; (f) 22 × 42: Round 42 to 40. Calculate 22 × 40 = 880. Add 22 × 2 = 44. Result: 880 + 44 = 924. Each uses the nearest multiple method to convert a difficult multiplication into simpler steps.
In simple words: Round one number to make it easier to multiply. Then add or subtract a small amount to fix the rounding.

Exam Tip: After rounding, multiply the adjustment part (the difference) by the other number - don't forget this correction step, as it's what makes the answer exact.

Question 18. A school has an auditorium with 35 rows, with 42 seats in each row. How many people can sit in this auditorium?
Answer: The auditorium has 35 rows of seats. Each row holds 42 people. To find the total number of people who can sit, multiply 35 × 42. Using the nearest multiple strategy: round 42 to 40. Calculate 35 × 40 = 1,400. Since we used 40 instead of 42, we are short by two groups of 35. Add 35 × 2 = 70. Total: 1,400 + 70 = 1,470 people. Therefore, 1,470 people can sit in the auditorium.
In simple words: Multiply the number of rows by the number of seats in each row. That gives the total number of seats in the whole auditorium.

Exam Tip: Always identify what you are multiplying (rows and seats per row in this case) and make sure your answer is reasonable for that context.

Question 19. Priya jogs 4 kilometres every day. How many kilometres will she jog in 31 days?
Answer: Priya jogs 4 kilometres each day. Over 31 days, her total distance is 4 × 31. Using the nearest multiple strategy: round 31 to 30. Calculate 4 × 30 = 120. Since we used 30 instead of 31, we are short by one group of 4. Add 4 × 1 = 4. Total: 120 + 4 = 124 kilometres. Therefore, Priya will jog 124 kilometres in 31 days.
In simple words: Multiply the number of kilometres jogged each day by the number of days. That tells you the total distance.

Exam Tip: For daily or repeated amounts, always multiply the single amount by the number of repetitions to get the total.

Question 20. A school has received 36 boxes of books with 48 books in each box. How many total books did the school receive in the boxes?
Answer: The school received 36 boxes. Each box contains 48 books. To find the total number of books, multiply 36 × 48. Using the nearest multiple strategy: round 48 to 50. Calculate 36 × 50 = 1,800. Since we used 50 instead of 48, we added two extra groups of 36. Subtract 36 × 2 = 72. Total: 1,800 - 72 = 1,728 books. Therefore, the school received 1,728 books in the boxes.
In simple words: Multiply the number of boxes by the number of books in each box to get the total number of books.

Exam Tip: Check if your multiplication makes sense by estimating first - 36 × 48 should be roughly 36 × 50 = 1,800, and 1,728 is close to that.

Question 21. Priya uses 16 metres of cloth to make 4 kurtas. How much cloth would she need to make 8 kurtas?
Answer: Priya uses 16 metres of cloth to make 4 kurtas. First, find how much cloth is needed for one kurta: 16 ÷ 4 = 4 metres per kurta. To make 8 kurtas, multiply: 4 × 8 = 32 metres. Therefore, Priya would need 32 metres of cloth to make 8 kurtas. This problem requires a two-step process: first divide to find the amount per unit, then multiply by the new quantity.
In simple words: Find out how much cloth goes into one kurta. Then multiply that by 8 to see how much you need for eight kurtas.

Exam Tip: For problems involving scaling (changing quantities), always find the "per unit" rate first before multiplying for the new total.

Question 22. Gollappa has 29 cows on his farm. Each cow produces 5 litres of milk per day. How many litres of milk do the cows produce in total, each day?
Answer: Gollappa has 29 cows. Each cow produces 5 litres of milk daily. The total daily milk production is found by multiplying: 29 × 5. Using the halving and doubling strategy: halve 29 (approximately) or use nearest multiple - round 29 to 30. Calculate 30 × 5 = 150. Since we used 30 instead of 29, subtract one group of 5: 150 - 5 = 145 litres. Therefore, the cows produce 145 litres of milk per day in total.
In simple words: Multiply the number of cows by the litres each cow makes. That gives you the total milk produced in one day.

Exam Tip: For "per unit" problems, always multiply the number of units by the amount each unit produces or uses.

Question 23. Maska Cow Farm has 297 cows. Each cow requires 18 kg of fodder per day. How much total fodder is needed to feed 297 cows every day?
Answer: Maska Cow Farm has 297 cows. Each cow needs 18 kg of fodder daily. The total daily fodder required is 297 × 18. Using the nearest multiple strategy: round 297 to 300. Calculate 300 × 18 = 5,400. Since we used 300 instead of 297, we added three extra groups of 18. Subtract 3 × 18 = 54. Total: 5,400 - 54 = 5,346 kg. Therefore, 5,346 kg of fodder is needed to feed all the cows each day.
In simple words: Multiply the number of cows by the amount of food each cow eats. That shows how much food you need for everyone each day.

Exam Tip: For large numbers, use rounding and nearest multiple strategies to break the multiplication into simpler mental steps - this reduces calculation errors.

Page 79

Let Us Do

Question 1. Solve the following problems like Nida did.
(a) 78 × 4
(b) 83 × 9
(c) 67 × 28
(d) 53 × 37
Answer:
(a) 78 × 4

Exam Tip: Always set up the grid table neatly with the tens place on the left and ones place on the right to avoid missing any partial products.

Question 2. Solve the following problems like Kanti.
(a) 94 × 5
(b) 49 × 6
(c) 37 × 53
(d) 28 × 79
Answer:
(a) 94 × 5

94 (90 + 4)
× 5
20 (5 × 4)
450 (5 × 90)
470

(b) 49 × 6

49 (40 + 9)
× 6
54 (6 × 9)
240 (6 × 40)
294

(c) 37 × 53

37 (30 + 7)
× 53 (50 + 3)
21 (3 × 7)
90 (3 × 30)
350 (50 × 7)
1500 (50 × 30)
1961

(d) 28 × 79

28 (20 + 8)
× 79 (70 + 9)
72 (9 × 8)
180 (9 × 20)
560 (70 × 8)
1400 (70 × 20)
2212

In simple words: Write each number as tens plus ones. Multiply each piece by each piece. Add up all the small answers at the end.

Exam Tip: Kanti's method works well because you multiply smaller numbers instead of big ones, and it is easy to spot mistakes.

Question 3. Solve the following problems like John.
(a) 86 × 3
(b) 72 × 7
(c) 94 × 36
(d) 66 × 22
Answer:
(a) 86 × 3

86 (80 + 6)
× 3
240 + 18 = 258 (3 × 86)

(b) 72 × 7

72 (70 + 2)
× 7
490 + 14 = 504 (7 × 72)

(c) 94 × 36

94 (90 + 4)
× 36 (30 + 6)
540 + 24 = 564 (6 × 94)
2700 + 120 = 2820 (30 × 94)
3384

(d) 66 × 22

66 (60 + 6)
× 22 (20 + 2)
120 + 12 = 132 (2 × 66)
1200 + 120 = 1320 (20 × 66)
1452

In simple words: Separate the second number into tens and ones. Multiply the first number by each part. Then add both answers.

Exam Tip: John's method is fast because you only do two main multiplications instead of four, making it quicker and with fewer chances for errors.

Question 4. Solve the following problems.
(a) A movie theater has 8 rows of seats, and each row has 12 seats. If half the seats are filled, how many people are watching the movie? If 3 more rows get filled, how many total people will be there?
(b) In a test match between India and West Indies, the Indian team hit twenty-four 4s and eighteen 6s across the two innings. How many runs were scored in 4s and 6s each? 234 runs were made by running between the wickets. If 23 runs were extras, how many runs were scored by Indian team in the two innings?
(c) Anjali buys 15 bulbs and 12 tube lights from Sudha Electricals. Each bulb costs Rs.25 and each tube light costs Rs.34. How much money should Anjali give to the shopkeeper?
(d) A shopkeeper sold 28 bags of rice. Each bag costs Rs.350. How much money did he earn by selling rice bags?
(e) A school library has 86 shelves and each shelf has 162 books. Find the number of books in the library.
Answer:
(a) Total rows in theater = 8
Seats in each row = 12
Total seats in theater = 12 × 8 = 96
Number of people watching movie = 96 ÷ 2 = 48
[Since half the seats are filled]
When 3 more rows fill up, more people come in
= 3 × 12 = 36
Total people = 48 + 36 = 84 people

(b) Runs from 4s: 24 fours × 4 = 96 runs
Runs from 6s: 18 sixes × 6 = 108 runs
Runs as Extras = 23
Runs from running between wickets = 234
Total runs: 96 + 108 + 234 + 23 = 461 runs

(c) Total Bulbs Buy = 15
Cost of one Bulb = Rs.25
Cost of 15 Bulbs = Rs.15 × 25 = Rs.375
Number of Tube lights = 12
Cost of one tube light = Rs.34
Cost of 12 tube lights = Rs.12 × 34 = Rs.408
Total money given by Anjali to shopkeeper = Rs.375 + Rs.408 = Rs.783

(d) Total rice bags sold = 28
Each rice bag cost = Rs.350
Total Money earned by selling rice bags = 28 × Rs.350 = Rs.9800

(e) Total Shelves in library = 86
Number of books in each shelf = 162
Total number of books in library = 86 × 162 = 13932

In simple words: Read the problem carefully. Write down what you know. Do one step at a time. Check your answer makes sense.

Exam Tip: Always show your working step by step and label each part clearly so the teacher can follow your thinking.

Page 83

Let Us Solve

Question 1. Solve the following problems like Nida.
(a) 548 × 6
(b) 682 × 3
(c) 324 × 18
(d) 507 × 23
(e) 190 × 65
Answer:
(a) 548 × 6

Exam Tip: Keep your grid organized and line up your digits properly - this is where most mistakes happen with this method.

Question 2. Solve the following problems like John.
(a) 123 × 84
(b) 368 × 32
(c) 159 × 324
(d) 239 × 401
(e) 592 × 5
(f) 101 × 22
Answer:
(a) 123 × 84

123 (100 + 20 + 3)
× 84 (80 + 4)
400 + 80 + 12 = 492 (4 × 123)
8000 + 1600 + 240 = 9840 (80 × 123)
10332

(b) 368 × 32

638 (300 + 60 + 8)
× 32 (30 + 2)
600 + 120 + 16 = 736 (2 × 368)
9000 + 1800 + 240 = 11040 (30 × 368)
11776

(c) 159 × 324

159 (100 + 50 + 9)
× 324 (300 + 20 + 4)
400 + 200 + 36 = 636 (4 × 159)
2000 + 1000 + 180 = 3180 (20 × 159)
30000 + 15000 + 2700 = 47700 (300 × 159)
51516

(d) 239 × 401

239 (200 + 30 + 9)
× 401 (400 + 1)
200 + 30 + 9 = 239 (1 × 239)
80000 + 12000 + 3600 = 95600 (300 × 159)
95839

(e) 592 × 5

592 (500 + 90 + 2)
× 5
2500 + 450 + 10 = 2960 (5 × 592)

(f) 101 × 22

101 (100 + 0 + 1)
× 22 (20 + 2)
200 + 0 + 2 = 202 (2 × 101)
2000 + 0 + 2 = 2020 (2 × 101)
2222

In simple words: Break both numbers into their parts. Multiply the first number by each part of the second number. Add all the answers together.

Exam Tip: John's method works for any size number because you break down both numbers completely, not just one.

Question 3. Let us solve a few questions like Mili's father.
Answer:
(a) 123 × 84

123 (100 + 20 + 3)
× 84 (80 + 4)
= 400 + 80 + 12 = 492 (4 × 123)
= 8000 + 1600 + 240 = 9840 (80 × 123)
= 10332

(b) 368 × 32

638 (300 + 60 + 8)
× 32 (30 + 2)
= 600 + 120 + 16 = 736 (2 × 368)
= 9000 + 1800 + 240 = 11040 (30 × 368)
= 11776

(c) 159 × 324

159 (100 + 50 + 9)
× 324 (300 + 20 + 4)
= 400 + 200 + 36 = 636 (4 × 159)
= 2000 + 1000 + 180 = 3180 (20 × 159)
= 30000 + 15000 + 2700 = 47700 (300 × 159)
= 51516

(d) 239 × 401

239 (200 + 30 + 9)
× 401 (400 + 1)
= 200 + 30 + 9 = 239 (1 × 239)
= 80000 + 12000 + 3600 = 95600 (300 × 239)
= 95839

(e) 592 × 5

592 (500 + 90 + 2)
× 5
= 2500 + 450 + 10 = 2960 (5 × 592)

(f) 101 × 22

101 (100 + 0 + 1)
× 22 (20 + 2)
= 200 + 0 + 2 = 202 (2 × 101)
= 2000 + 0 + 2 = 2020 (2 × 101)
= 2222

In simple words: Use place value columns to keep track of tens, hundreds, and ones. This helps you see where each digit goes and makes it harder to make mistakes.

Exam Tip: Mili's father's method uses the column system to organize your work neatly and keep track of place values.

Question 4. Now use Mili's father's method to solve the following questions.
(a) 807 × 5
(b) 143 × 28
(c) 309 × 9
(d) 450 × 38
(e) 584 × 23
(f) 302 × 13
(g) 604 × 54
(h) 112 × 23
(i) 237 × 19
Answer:
(a) 807 × 5

Th H T O
8 0 7
× 5
= 40 + 35 = 4035

(b) 143 × 28

Th H T O
1 4 3
× 2 8
= 1144 + 11830 = 4004

(c) 309 × 9

Th H T O
3 0 9
× 9
= 27 + 81 = 2781

(d) 450 × 38

Th H T O
4 5 0
× 3 8
= 3600 + 13500 = 17100

(e) 584 × 23

Th H T O
5 8 4
× 2 3
= 1752 + 11680 = 13432

(f) 302 × 13

Th H T O
3 0 2
× 1 3
= 906 + 3020 = 3926

(g) 604 × 54

Th H T O
6 0 4
× 5 4
= 2416 + 30200 = 32616

(h) 112 × 23

Th H T O
1 1 2
× 2 3
= 336 + 2240 = 2576

(i) 237 × 19

Th H T O
2 3 7
× 1 9
= 2133 + 2370 = 4503

In simple words: Multiply by the ones digit first, then the tens digit. Write down each answer below in its proper place. Finally, add all the answers together.

Exam Tip: When you multiply by the tens digit, remember to shift one place to the left (or add a zero) to keep the place value correct.

Page 86

Let Us Do

Question 1. Identify the problems that have the same answer as the one given at the top of each box. Do not calculate.
(a) 807 × 5 (answer at top: 4035)
(b) 143 × 28 (answer at top: 4004)
(c) 309 × 9 (answer at top: 2781)
(d) 450 × 38 (answer at top: 17100)
(e) 584 × 23 (answer at top: 13432)
(f) 302 × 13 (answer at top: 3926)
(g) 604 × 54 (answer at top: 32616)
(h) 112 × 23 (answer at top: 2576)
(i) 237 × 19 (answer at top: 4503)
Answer:
(a) 807 × 5 matches: Same (no alternatives needed to match 4035)
(b) 143 × 28 matches: Same (no alternatives needed to match 4004)
(c) 309 × 9 matches: Same (no alternatives needed to match 2781)
(d) 450 × 38 matches: Same (no alternatives needed to match 17100)
(e) 584 × 23 matches: Same (no alternatives needed to match 13432)
(f) 302 × 13 matches: Same (no alternatives needed to match 3926)
(g) 604 × 54 matches: Same (no alternatives needed to match 32616)
(h) 112 × 23 matches: Same (no alternatives needed to match 2576)
(i) 237 × 19 matches: Same (no alternatives needed to match 4503)

In simple words: Look for ways to break apart the numbers differently but still get the same result using multiplication properties.

Exam Tip: This question tests whether you understand that different groupings and breakdowns of numbers can produce the same product.

Question 2. Find easy ways of solving these problems.
(a) 16 × 25
(b) 12 × 125
(c) 24 × 250
(d) 36 × 25
(e) 28 × 75
(f) 300 × 15
(g) 50 × 78
(h) 199 × 63
(i) 128 × 35
Answer:
(a) 16 × 25
= (10 + 6) × (20 + 5)
= 200 + 50 + 120 + 30
= 400

(b) 12 × 125
= (10 + 2) × (100 + 25)
= 1000 + 250 + 200 + 50
= 1500

(c) 24 × 250
= (20 + 4) × (200 + 50)
= 4000 + 1000 + 800 + 200
= 6000

(d) 36 × 25
= (30 + 6) × (20 + 5)
= 600 + 150 + 120 + 30
= 900

(e) 28 × 75
= (20 + 8) × (70 + 5)
= 1400 + 100 + 560 + 40
= 2100

(f) 300 × 15
= (300) × (10 + 5)
= 3000 + 1500
= 4500

(g) 50 × 78
= (50) × (70 + 8)
= 3500 + 400
= 3900

(h) 199 × 63
= (200 - 1) × (60 + 3)
= 12000 + 600 - 60 - 3
= 12537

(i) 128 × 35
= (120 + 8) × (30 + 5)
= 3600 + 600 + 240 + 40
= 4480

In simple words: Break numbers into parts that are easy to multiply. Use place value blocks to see what each part is. Then add all the smaller products.

Exam Tip: Looking for "easy" ways means finding numbers close to 25, 50, 100, 125, or other round numbers that you can work with quickly.

Question 3. Write 5 other examples for which you can find easy ways of getting products.
Answer:
1. 44 × 25
= (44 ÷ 4) × 100
= 11 × 100
= 1100

2. 8 × 375
= (10 × 375) - (2 × 375)
= 3750 - 750
= 3000

3. 99 × 28
= (100 × 28) - (1 × 28)
= 2800 - 28
= 2772

4. 125 × 16
= (125 × 8) × 2
= 1000 × 2
= 2000

5. 102 × 45
= (100 × 45) + (2 × 45)
= 4500 + 90
= 4590

In simple words: Good examples are ones where you can use numbers close to 100 or a multiple of 25 or 50 to make the work easier.

Exam Tip: When you make your own examples, pick numbers that have simple shortcuts - this shows you understand the connection between place value and multiplication.

Question 4. Find the answers to the following questions based on the given information.
(a) 17 × 23 = 391
(b) 17 × 24 = _______
(c) 17 × 22 = _______
(d) 16 × 23 = _______
(e) 8 × 9 = 72
(f) 18 × 9 = _______
(g) 28 × 9 = _______
(h) 108 × 9 = _______
(i) 18 × 23 = _______
Answer:
(a) 17 × 23 = 391 (given)

(b) 17 × 24 = 17 × (23 + 1)
= (17 × 23) + (17 × 1)
= 391 + 17 = 408

(c) 17 × 22 = 17 × (23 - 1)
= (17 × 23) - (17 × 1)
= 391 - 17 = 374

(d) 16 × 23 = (17 - 1) × 23
= (17 × 23) - (1 × 23)
= 391 - 23 = 368

(e) 8 × 9 = 72 (given)

(f) 18 × 9 = (10 + 8) × 9
= (10 × 9) + (8 × 9)
= 90 + 72 = 162

(g) 28 × 9 = (20 + 8) × 9
= (20 × 9) + (8 × 9)
= 180 + 72 = 252

(h) 108 × 9 = (100 + 8) × 9
= (100 × 9) + (8 × 9)
= 900 + 72 = 972

(i) 18 × 23 = (17 + 1) × 23
= (17 × 23) + (1 × 23)
= 391 + 23 = 414

In simple words: Start with what you already know. Rewrite the new number as the known number plus or minus a small amount. Then use the distributive property to break it apart.

Exam Tip: This strategy is called "relating to known facts" - it shows the examiner you can think flexibly about numbers instead of just calculating blindly.

Page 87

Let Us Think

Question 1. Find the possible values of the coloured boxes in each of the following problems. The same colour indicates the same number in a problem. Some problems can have more than one answer.
Answer: [This question requires visual analysis of boxes to be filled in. The specific answers depend on the place value constraints shown in the diagrams. Students should work systematically by testing different single-digit values and checking whether they satisfy the multiplication or division operation shown.]

In simple words: Think about what single digit each colour box can be. Use the answer given to work backwards and figure out what number was multiplied or divided.

Exam Tip: Test each digit from 0 to 9 one at a time. Multiply or divide to see if you get the right answer. Cross out the digits that don't work.

Question 2. Estimate the products on the left and match them to the numbers given on the right.
Answer:
25 × 31 → 25 × 30 → 750
132 × 19 → 130 × 20 → 2600
101 × 11 → 100 × 10 → 1000
248 × 49 → 250 × 50 → 12500
12 × 25 → 10 × 30 → 300

In simple words: Round each number to the nearest ten. Then multiply the rounded numbers together. This gives you a quick, close answer without exact calculation.

Exam Tip: Estimating helps you check if your exact answer is sensible - if your exact answer is very far from your estimate, you may have made a mistake.

Page 88

King's Reward

A king offers three choices to his ministers. Choice 1 involves taking 5 gold coins and doubling them every day for 7 days. Choice 2 means accepting 3 gold coins and tripling them daily for 7 days. Choice 3 requires taking just 1 gold coin and multiplying it by 5 each day for 7 days. Three ministers each picked a different option. After one week passes, all three are astonished by how much gold they have collected. Who ends up with the greatest amount of gold coins?

To find out, calculate the total gold coins each minister received.

Answer: For Choice 1 - doubling 5 coins daily:
Day 1: 5 coins
Day 2: 10 coins
Day 3: 20 coins
Day 4: 40 coins
Day 5: 80 coins
Day 6: 160 coins
Day 7: 320 coins
Total = 5 + 10 + 20 + 40 + 80 + 160 + 320 = 635 coins

For Choice 2 - tripling 3 coins daily:
Day 1: 3 coins
Day 2: 9 coins
Day 3: 27 coins
Day 4: 81 coins
Day 5: 243 coins
Day 6: 729 coins
Day 7: 2,187 coins
Total = 3 + 9 + 27 + 81 + 243 + 729 + 2,187 = 3,379 coins

For Choice 3 - multiplying 1 coin by 5 daily:
Day 1: 1 coin
Day 2: 5 coins
Day 3: 25 coins
Day 4: 125 coins
Day 5: 625 coins
Day 6: 3,125 coins
Day 7: 15,625 coins
Total = 1 + 5 + 25 + 125 + 625 + 3,125 + 15,625 = 19,531 coins

Minister 3 received the most gold coins: 19,531 coins. So, Choice 3 gives the greatest reward.
In simple words: Even though Minister 3 started with only 1 coin, multiplying by 5 each day grows much faster than doubling or tripling. Bigger multipliers make numbers grow much larger, much quicker.

Exam Tip: This problem shows exponential growth - when you multiply by a fixed number repeatedly, growth happens very rapidly. Always calculate all choices fully rather than guessing which might be best.

Page 89

Multiplication Patterns

Question 1. Notice how the multiplier, multiplicand, and products are changing in each of the following. What is the relationship of the new product with the original product? Solve a) completely, and then predict the answers for the rest. (a) 16 × 44 = 704 (1) 8 × 88 = _____ (2) 8 × 22 = _____ (3) 16 × 22 = _____ (4) 32 × 44 = _____
Answer:
Starting with 16 × 44 = 704

(1) 8 × 88 = (16 ÷ 2) × (44 × 2) = 704

(2) 8 × 22: Here we halve both numbers. So 8 × 22 = (16 ÷ 2) × (44 ÷ 2) = 704 ÷ 4 = 176

(3) 16 × 22 = (16) × (44 ÷ 2) = 704 ÷ 2 = 352

(4) 32 × 44 = (16 × 2) × (44) = 704 × 2 = 1,408

The key pattern: When you multiply one number by "a" and divide the other by "a", the product stays the same. When both numbers are cut in half, the result becomes one-fourth of the original.
In simple words: If one number gets bigger and the other gets smaller by the same amount, the answer stays put. If both get smaller, the answer gets smaller too.

Exam Tip: Use this property to simplify hard multiplications - break down one factor and multiply by the other in parts, rather than computing the full problem at once.

Question 1(b). (b) 12 × 32 = 384 (1) 6 × 16 = _____ (2) 24 × 16 = _____ (3) 24 × 64 = _____ (4) 12 × 16 = _____
Answer:
Starting with 12 × 32 = 384

(1) 6 × 16 = (12 ÷ 2) × (32 ÷ 2) = 384 ÷ 4 = 96

(2) 24 × 16 = (12 × 2) × (32 ÷ 2) = 384 × 1 = 384

(3) 24 × 64 = (12 × 2) × (32 × 2) = 384 × 4 = 1,536

(4) 12 × 16 = (12) × (32 ÷ 2) = 384 ÷ 2 = 192
In simple words: Grow one side and shrink the other equally, and the answer doesn't shift. Grow both sides, and the answer grows by the product of how much each grew.

Exam Tip: Recognise when you can swap multiplication factors for easier arithmetic - this trick saves time in longer calculations.

Question 2. Observe and complete patterns:

Pattern 1 (left box):

1 × 1 = 1
11 × 11 = 121
111 × 111 = 12,321
1111 × 1111 = _____

Answer: 1111 × 1111 = 12,34,321. The pattern shows that when you multiply repdigits (numbers made of all 1s) by themselves, the digits climb up and then back down: 1; 1, 2, 1; 1, 2, 3, 2, 1; and so on. For four 1s, the sequence goes 1, 2, 3, 4, 3, 2, 1.
In simple words: Multiply a number made of all 1s by itself. The digits go up (1, 2, 3, 4) and then come back down (3, 2, 1).

Exam Tip: Look for digit sequences and symmetry in these pattern problems - the answer often follows a simple rule about how numbers are ordered, not hard arithmetic.

Pattern 2 (right box):

5 × 5 = 25
15 × 15 = 225
25 × 25 = 625
35 × 35 = 1,225
45 × 45 = _____
55 × 55 = _____

Answer: 45 × 45 = 2,025 and 55 × 55 = 3,025. For numbers ending in 5, multiply the first part by its next whole number, then add 25 to the end. For 45, the first part is 4. So 4 × 5 = 20, and the answer is 2,025. For 55, the first part is 5. So 5 × 6 = 30, and the answer is 3,025.
In simple words: If a number ends in 5, take the first digits, multiply by the next number after it, and stick 25 on the end.

Exam Tip: This shortcut works for all numbers ending in 5 - it saves time on squaring and is great for mental maths practice.

Pattern 3 (left bottom box):

11 × 12 = 132
11 × 34 = 374
11 × 56 = 616
11 × 78 = 858
11 × 54 = _____
11 × 82 = _____

Answer: 11 × 54 = 594 and 11 × 82 = 902. When multiplying by 11, add the two digits of the other number and place the sum in the middle. For 54, add 5 + 4 = 9, so the answer is 594. For 82, add 8 + 2 = 10. Since the sum is 10 (two digits), write down 0 in the middle and carry 1 to the left: 902.
In simple words: To multiply any two-digit number by 11, add its two digits and put the sum between them. If the sum is 10 or more, carry the extra 1 to the left.

Exam Tip: This is a fast mental maths trick - master it to multiply by 11 instantly without long multiplication.

Pattern 4 (right bottom box):

1 × 9 + 1 = 10
12 × 9 + 2 = 110
123 × 9 + 3 = 1,110
1,234 × 9 + 4 = _____

Answer: 1,234 × 9 + 4 = 11,110. The pattern is: take consecutive digits starting from 1, multiply by 9, then add the last digit. For 1,234 × 9 + 4, you get 11,110. Each time, the answer is a string of 1s followed by a 0.
In simple words: Build a number from 1, 2, 3, 4... Multiply by 9 and add the last digit. You always get 1s and a 0 at the end.

Exam Tip: Pattern recognition questions ask you to spot the rule, not just fill blanks - always explain why the pattern works, not just state the answer.

Page 90

Question. Here are some numbers. Remember number pairs from Grade 4? Any two adjacent numbers in a row or a column are number pairs. Can you identify the pair whose product is the smallest and another pair whose product is the largest? Do you need to find every product or can you find this by looking at the numbers?

The grid shows:

Answer: The smallest product comes from the pair 11 and 14 (both from the bottom row, adjacent to each other): 11 × 14 = 154. The largest product comes from the pair 35 and 75 (both from the top row, adjacent to each other): 35 × 75 = 2,625. You do not need to calculate every product - simply observe which two numbers are largest and which two are smallest. The largest numbers next to each other give the biggest product, and the smallest numbers next to each other give the smallest product.
In simple words: Find the two biggest numbers touching each other - that pair gives the biggest answer. Find the two smallest numbers touching each other - that gives the smallest answer. No need to work out all of them.

Exam Tip: Train yourself to estimate and compare without computing every single step - this saves time on tests and builds number sense.

Page 91

Let Us Solve

Question 1. Mala went to a book exhibition and bought 18 books. The shop was selling 3 books for Rs.150. After buying the books, she still had Rs.20 left. How much money did Mala have at the beginning?
Answer: First, find the cost of one book. Since 3 books cost Rs.150, each book costs Rs.150 ÷ 3 = Rs.50. Mala bought 18 books, so the amount she spent is 18 × Rs.50 = Rs.900. Since she had Rs.20 left after this purchase, the amount she started with was Rs.900 + Rs.20 = Rs.920.
In simple words: Find how much one book costs by dividing the price of 3 books by 3. Then multiply that by 18 to get what she spent. Add what was left to find her starting amount.

Exam Tip: Break word problems into steps: find the unit price first, then the total spent, then work backwards to the starting amount.

Question 2. A village sports club organises a women's football tournament. The club earned money by selling match tickets and charging fees for team participation. They sold 57 tickets for Rs.115 each. They had 3 teams joining the tournament, with each team paying a participation fee of Rs.1,599. The teams paid Rs.1,750 in total to rent the football ground and Rs.1,129 for food and water. (a) How much money did the club collect in total from ticket sales and team participation fees?
Answer: Money from ticket sales: 57 × Rs.115 = (57 × 100) + (57 × 15) = Rs.5,700 + Rs.855 = Rs.6,555. Money from team participation fees: 3 teams × Rs.1,599 = Rs.4,797. Total money collected = Rs.6,555 + Rs.4,797 = Rs.11,352.
In simple words: Multiply the number of tickets by the price per ticket to get ticket money. Multiply the number of teams by the fee per team to get participation money. Add them together.

Exam Tip: When finding total income, identify all sources (tickets, fees, etc.) and calculate each one separately before adding.

Question 2(b). (b) What were the total expenses on renting the ground and food and water?
Answer: Total expenses = cost of renting the ground + cost of food and water = Rs.1,750 + Rs.1,129 = Rs.2,879.
In simple words: Add the rent and the food cost together to get the total amount spent on these two things.

Exam Tip: Always identify which items are expenses and which are income - mixing them up is a common error.

Question 3. Ananya is watching Republic Day celebrations on the city's public ground. There are 12 rows of students sitting in front of her and 17 rows behind her. There are 18 students to her right and 22 students to her left. (a) How many rows of students are there in total?
Answer: Rows in front of Ananya = 12. Rows behind Ananya = 17. Ananya herself occupies 1 row. Total rows = 12 + 17 + 1 = 30 rows.
In simple words: Count the rows ahead, count the rows behind, and count her own row. Add all three groups.

Exam Tip: In spatial reasoning problems, always remember to include the reference person (here, Ananya) as a separate unit when counting.

Question 3(b). (b) How many students are there in Ananya's row?
Answer: Students to her right = 18. Students to her left = 22. Ananya herself = 1. Total students in her row = 18 + 22 + 1 = 41 students.
In simple words: Add the students on one side, the students on the other side, and count Ananya as 1 more.

Exam Tip: Count in all directions from the reference person - left, right, and the person themselves.

Question 3(c). (c) What is the total number of students on the ground?
Answer: Total rows = 30 (from part a). Students per row = 41 (from part b). Total students = 30 × 41 = 1,230. However, we need to recount because only 29 rows have 41 students each - Ananya's row is already counted once. So the correct calculation is 29 × 41 = 1,189 students.
In simple words: Multiply the number of full rows by the number of students in each row. But remember that Ananya is already part of the count, so don't double-count her.

Exam Tip: Be careful with multi-part problems - results from earlier parts are building blocks, but reconsider whether all data applies to the final step.

Question 4. Multiply. (a) 67 × 78 (b) 34 × 56 (c) 45 × 263 (d) 86 × 542 (e) 432 × 107 (f) 310 × 120
Answer:
(a) 67 × 78 = (67 × 70) + (67 × 8) = 4,690 + 536 = 5,226

(b) 34 × 56 = (34 × 50) + (34 × 6) = 1,700 + 204 = 1,904

(c) 45 × 263 = (45 × 200) + (45 × 60) + (45 × 3) = 9,000 + 2,700 + 135 = 11,835

(d) 86 × 542 = (86 × 500) + (86 × 40) + (86 × 2) = 43,000 + 3,440 + 172 = 46,612

(e) 432 × 107 = (432 × 100) + (432 × 7) = 43,200 + 3,024 = 46,224

(f) 310 × 120 = (31 × 12) × 100 = 372 × 100 = 37,200
In simple words: Break the bigger number into tens, hundreds, and ones. Multiply the smaller number by each part. Then add all the answers together.

Exam Tip: Using the distributive property (breaking numbers into place values) is faster and less error-prone than standard long multiplication for many two- and three-digit products.

Question 5. If 67 × 67 = 4,489, without multiplication find 67 × 68.
Answer: Since 68 = 67 + 1, we can write 67 × 68 = 67 × (67 + 1) = (67 × 67) + (67 × 1) = 4,489 + 67 = 4,556.
In simple words: When two numbers are close and one result is known, use that result. Rewrite one number in terms of the other, then use the distributive property.

Exam Tip: Look for clever shortcuts using given information - adding or subtracting 1 is a common trick that saves you from recalculating the entire product.

Question 6. If 99 × 100 = 9,900, without multiplication find 99 × 99.
Answer: Since 99 = 100 - 1, we can write 99 × 99 = 99 × (100 - 1) = (99 × 100) - (99 × 1) = 9,900 - 99 = 9,801.
In simple words: If you know what 99 times 100 is, subtract 99 once to get what 99 times 99 is.

Exam Tip: This method works whenever you know the product of one number with a neighbour - use subtraction or addition to shift by 1 without multiplying afresh.