GSEB Class 6 Maths Solutions Chapter 2 Whole Numbers Exercise 2.2

Question 1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Answer:
(a) \( 837 + 208 + 363 \)
We rearrange the numbers to group those that sum to a round figure.
\( = 837 + 363 + 208 \)
\( = (837 + 363) + 208 \)
\( = 1200 + 208 \)
\( = 1408 \)
(b) \( 1962 + 453 + 1538 + 647 \)
We group the numbers to make calculations easier.
\( = (1962 + 1538) + (453 + 647) \)
\( = (3500) + (1100) \)
\( = 4600 \)
In simple words: To find the total sum quickly, we put numbers together that add up to a round number like 100 or 1000 first. This makes the addition simpler.

Exam Tip: Look for numbers ending in 0 or 5 when combined. This strategy of grouping numbers (associative property) simplifies calculations significantly.

Question 2. Find the product by suitable rearrangement:
(a) 2 x 1768 x 50
(b) 4 x 166 x 25
(c) 8 x 291 x 125
(d) 625 x 279 x 16
(e) 285 x 5 x 60
(f) 125 x 40 x 8 x 25
Answer:
(a) \( 2 \times 1768 \times 50 \)
We multiply 2 and 50 first because they give 100.
\( = (2 \times 50) \times 1768 \)
\( = 100 \times 1768 \)
\( = 176800 \)
(b) \( 4 \times 166 \times 25 \)
We multiply 4 and 25 first to get 100.
\( = (4 \times 25) \times 166 \)
\( = 100 \times 166 \)
\( = 16600 \)
(c) \( 8 \times 291 \times 125 \)
We multiply 8 and 125 first because they give 1000.
\( = (8 \times 125) \times 291 \)
\( = (1000) \times 291 \)
\( = 291000 \)
(d) \( 625 \times 279 \times 16 \)
We multiply 625 and 16 first to simplify.
\( = (625 \times 16) \times 279 \)
\( = (10000) \times 279 \)
\( = 2790000 \)
(e) \( 285 \times 5 \times 60 \)
We multiply 5 and 60 first for an easy calculation.
\( = 285 \times (5 \times 60) \)
\( = 285 \times 300 \)
\( = (285 \times 3) \times 100 \)
\( = 855 \times 100 \)
\( = 85500 \)
(f) \( 125 \times 40 \times 8 \times 25 \)
We group 125 with 8, and 40 with 25 for simple multiplication.
\( = (125 \times 8) \times (40 \times 25) \)
\( = (1000) \times (1000) \)
\( = 1000000 \)
In simple words: When multiplying several numbers, it's often simpler to group two numbers that result in a round number (like 100, 1000, etc.) first. This makes the overall multiplication much easier.

Exam Tip: Always look for pairs like \( 2 \times 50 \), \( 4 \times 25 \), \( 8 \times 125 \) as they simplify multiplication into powers of 10. This is the Associative Property of Multiplication.

Question 3. Find the value of the following:
(a) 297 x 17 + 297 x 3
(b) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218
Answer:
(a) \( 297 \times 17 + 297 \times 3 \)
We use the distributive property to factor out 297.
\( = 297 \times (17 + 3) \)
\( = 297 \times 20 \)
\( = 5940 \)
(b) \( 54279 \times 92 + 8 \times 54279 \)
Again, using the distributive property, we take 54279 as a common factor.
\( = 54279 \times (92 + 8) \)
\( = 54279 \times 100 \)
\( = 5427900 \)
(c) \( 81265 \times 169 - 81265 \times 69 \)
By applying the distributive property, we factor out 81265.
\( = 81265 \times (169 - 69) \)
\( = 81265 \times 100 \)
\( = 8126500 \)
(d) \( 3845 \times 5 \times 782 + 769 \times 25 \times 218 \)
First, notice that \( 3845 = 5 \times 769 \). So, \( 25 = 5 \times 5 \).
\( = (769 \times 5) \times 5 \times 782 + 769 \times (5 \times 5) \times 218 \)
\( = 769 \times 25 \times 782 + 769 \times 25 \times 218 \)
Now, we can take \( 769 \times 25 \) as a common factor.
\( = (769 \times 25) \times (782 + 218) \)
\( = 19225 \times 1000 \)
\( = 19225000 \)
In simple words: When you see a number that is being multiplied by two different numbers and then added or subtracted, you can pull that common number out. Then, you just add or subtract the other two numbers and multiply once. This is called the distributive property.

Exam Tip: Recognizing common factors and applying the distributive property \( a \times b + a \times c = a \times (b + c) \) is crucial for simplifying such expressions efficiently.

Question 4. Find the product using suitable properties.
(a) 738 x 103
(b) 854 x 102
(c) 258 x 1008
(d) 1005 x 168
Answer:
(a) \( 738 \times 103 \)
We can write 103 as \( 100 + 3 \).
\( = 738 \times (100 + 3) \)
Using the distributive property \( a \times (b + c) = a \times b + a \times c \).
\( = (738 \times 100) + (738 \times 3) \)
\( = 73800 + 2214 \)
\( = 76014 \)
(b) \( 854 \times 102 \)
We write 102 as \( 100 + 2 \).
\( = 854 \times (100 + 2) \)
Applying the distributive property.
\( = (854 \times 100) + (854 \times 2) \)
\( = 85400 + 1708 \)
\( = 87108 \)
(c) \( 258 \times 1008 \)
We express 1008 as \( 1000 + 8 \).
\( = 258 \times (1000 + 8) \)
Using the distributive property.
\( = (258 \times 1000) + (258 \times 8) \)
\( = 258000 + 2064 \)
\( = 260064 \)
(d) \( 1005 \times 168 \)
We write 1005 as \( 1000 + 5 \).
\( = (1000 + 5) \times 168 \)
Applying the distributive property.
\( = (1000 \times 168) + (5 \times 168) \)
\( = 168000 + 840 \)
\( = 168840 \)
In simple words: To multiply a number by a slightly larger number (like 103 or 1008), you can split the larger number into a round number (like 100 or 1000) plus a small extra number. Then, you multiply by each part separately and add the results together. This method makes big multiplications simpler.

Exam Tip: Breaking down numbers into sums or differences involving multiples of 10, 100, or 1000, and then using the distributive property, simplifies complex multiplications significantly.

Question 5. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs Rs. 44 per litre, how much did he spend in all on petrol?
Answer:
Quantity of petrol filled on Monday \( = 40 \) litres
Quantity of petrol filled on the next day \( = 50 \) litres
Per litre cost of petrol \( = \text{Rs. } 44 \)
Total quantity of petrol filled \( = 40 + 50 = 90 \) litres
Total cost of petrol for the two days \( = \text{Rs. } 44 \times (40 + 50) \)
\( = \text{Rs. } 44 \times 90 \)
\( = \text{Rs. } 3960 \)
Alternatively, using the distributive property:
Total cost \( = \text{Rs. } (44 \times 40) + (44 \times 50) \)
\( = \text{Rs. } 1760 + \text{Rs. } 2200 \)
\( = \text{Rs. } 3960 \)
The taxi driver spent Rs. 3960 in total on petrol.
In simple words: First, add up all the petrol bought on both days. Then, multiply that total amount by the price of petrol for each litre. This gives you the total money spent.

Exam Tip: For problems involving total cost, you can either sum the quantities first and then multiply by the unit price, or multiply each quantity by the unit price and then sum the individual costs. Both methods should yield the same correct answer.

Question 6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs Rs. 15 per litre, how much money is due to the vendor per day?
Answer:
Milk supplied in the morning \( = 32 \) litres
Milk supplied in the evening \( = 68 \) litres
Cost of milk per litre \( = \text{Rs. } 15 \)
Total quantity of milk supplied per day \( = 32 + 68 \) litres \( = 100 \) litres
Total cost of milk per day \( = \text{Rs. } 15 \times (32 + 68) \)
\( = \text{Rs. } 15 \times 100 \)
\( = \text{Rs. } 1500 \)
The money due to the vendor per day is Rs. 1500.
In simple words: Add the milk supplied in the morning and evening to find the total milk. Then, multiply this total milk by the price per litre to discover the full amount owed to the vendor.

Exam Tip: Combining the quantities first (using the distributive property in reverse) often simplifies calculations when the unit price is common, helping you get to a round number for easier multiplication.

Question 7. Match the following:
(i) \( 425 \times 136 = 425 \times (6 + 30 + 100) \)
(ii) \( 2 \times 49 \times 50 = 2 \times 50 \times 49 \)
(iii) \( 80 + 2005 + 20 = 80 + 20 + 2005 \)
(a) Commutativity under multiplication
(b) Commutativity under addition
(c) Distributivity of multiplication over addition
Answer:
(i) \( 425 \times 136 = 425 \times (6 + 30 + 100) \)
This shows how multiplication is spread over addition, matching "Distributivity of multiplication over addition".
(ii) \( 2 \times 49 \times 50 = 2 \times 50 \times 49 \)
This illustrates that the order of numbers in multiplication can be changed without altering the product, which is "Commutativity under multiplication".
(iii) \( 80 + 2005 + 20 = 80 + 20 + 2005 \)
This demonstrates that the order of numbers in addition can be changed without altering the sum, which is "Commutativity under addition".
So, the matches are:
(i) \( \rightarrow \) (c)
(ii) \( \rightarrow \) (a)
(iii) \( \rightarrow \) (b)
In simple words: You need to link each math example to the correct math rule. Distributivity is when you multiply a number by a sum, like \( a \times (b+c) \). Commutativity means you can change the order of numbers in adding or multiplying without changing the result, like \( a+b = b+a \) or \( a \times b = b \times a \).

Exam Tip: Understand the core meaning of each property: Commutativity (order doesn't matter for addition/multiplication), Associativity (grouping doesn't matter for addition/multiplication), and Distributivity (multiplication distributes over addition/subtraction).