GSEB Class 6 Maths Solutions Chapter 2 Whole Numbers Exercise 2.3

Question 1. Which of the following will not represent zero?
(a) \( 1 + 0 \)
(b) \( 0 \times 0 \)
(c) \( \frac { 0 }{ 2 } \)
(d) \( \frac { 10-10 }{ 2 } \)
Answer: (a) \( 1 + 0 \)
In simple words: When we add 1 and 0, the result is 1. All other options here will give you 0 as the answer.

Exam Tip: Remember that adding zero to any number keeps the number the same. Multiplying by zero, or dividing zero by any non-zero number, always results in zero.

Question 2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Answer: The product of any whole number and zero is always equal to zero. For example, \( 0 \times 0 = 0 \), \( 1 \times 0 = 0 \), \( 2 \times 0 = 0 \), \( 0 \times 3 = 0 \), and so on. Yes, if two whole numbers multiply to give zero, then at least one of them must be zero.
In simple words: Yes, if you multiply two numbers and get zero, then one of those numbers, or both of them, has to be zero.

Exam Tip: This is a fundamental property of multiplication with zero. Always provide clear examples to support your explanation in such questions.

Question 3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Answer: We understand that multiplying any whole number by 1 results in that same whole number. For instance, \( 5 \times 1 = 5 \), \( 109 \times 1 = 109 \), \( 1 \times 17 = 17 \), \( 1 \times 0 = 0 \), and \( 1 \times 1 = 1 \). The product will only equal 1 if both of the whole numbers are 1 themselves.
In simple words: Yes, if two whole numbers multiply to give 1, then both of those numbers must be 1.

Exam Tip: Illustrate with various examples, including cases where one number is 1 but the product is not 1 (e.g., \( 1 \times 5 = 5 \)), to show that both numbers must be 1 for their product to be 1.

Question 4. Find using distributive property:
(a) \( 728 \times 101 \)
(b) \( 5437 \times 1001 \)
(c) \( 824 \times 25 \)
(d) \( 4275 \times 125 \)
(e) \( 504 \times 35 \)
Answer:
(a) \( 728 \times 101 \)
\( = 728 \times [100 + 1] \) (since \( 100 + 1 = 101 \))
\( = (728 \times 100) + (728 \times 1) \)
\( = 72800 + 728 = 73528 \)

(b) \( 5437 \times 1001 \)
\( = 5437 \times [1000 + 1] \)
\( = (5437 \times 1000) + (5437 \times 1) \)
\( = 5437000 + 5437 = 5442437 \)

(c) \( 824 \times 25 \)
\( = 824 \times (20 + 5) \)
\( = (824 \times 20) + (824 \times 5) \)
\( = 16480 + 4120 = 20600 \)

(d) \( 4275 \times 125 \)
\( = 4275 \times [100 + 20 + 5] \)
\( = (4275 \times 100) + (4275 \times 20) + (4275 \times 5) \)
\( = 427500 + 85500 + 21375 = 534375 \)

(e) \( 504 \times 35 \)
\( = (500 + 4) \times 35 \)
\( = (500 \times 35) + (4 \times 35) \)
\( = 17500 + 140 = 17640 \)
In simple words: The distributive property helps to multiply large numbers easily by breaking one number into smaller, simpler parts (like tens and units) and then multiplying each part separately before adding them up.

Exam Tip: When using the distributive property, always break down one of the numbers into a sum that makes multiplication simpler (e.g., \( 101 \) becomes \( 100 + 1 \)), ensuring you multiply each component correctly.

Question 5. Study the pattern:
\( 1 \times 8 + 1 = 9 \)
\( 12 \times 8 + 2 = 98 \)
\( 123 \times 8 + 3 = 987 \)
\( 1234 \times 8 + 4 = 9876 \)
\( 12345 \times 8 + 5 = 98765 \)
Write the next two steps. Can you say how the pattern works?
Answer:
The next two steps will be:
\( 123456 \times 8 + 6 = 987654 \)
\( 1234567 \times 8 + 7 = 9876543 \)

The working of the pattern:
Since,
\( 1 + 1 = 2 \)
\( 11 + 1 = 12 \)
\( 111 + 11 + 1 = 123 \)
\( 1111 + 111 + 11 + 1 = 1234 \)
\( 11111 + 1111 + 111 + 11 + 1 = 12345 \)

We can also explain the pattern as:
\( 1 \times 8 + 1 = 9 \)
\( 12 \times 8 + 2 = 98 = (11 + 1) \times 8 + 2 \)
\( 123 \times 8 + 3 = 987 = (111 + 11 + 1) \times 8 + 3 \)
\( 1234 \times 8 + 4 = 9876 = (1111 + 111 + 11 + 1) \times 8 + 4 \)
\( 12345 \times 8 + 5 = 98765 = (11111 + 1111 + 111 + 11 + 1) \times 8 + 5 \)
For the next two steps, this pattern continues:
\( 123456 \times 8 + 6 = 987654 = (111111 + 11111 + 1111 + 111 + 11 + 1) \times 8 + 6 \)
\( 1234567 \times 8 + 7 = 9876543 = (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) \times 8 + 7 \)
In simple words: This pattern shows that as you increase the first number by adding the next digit (1, 12, 123, etc.) and also increase the number you add by one, the answer builds up with the numbers 9, 98, 987, and so on. The working reveals how each number can be seen as a sum of ones (like 11, 111) which relates to the next number in the sequence.

Exam Tip: When studying patterns, carefully observe how each part of the equation changes from one step to the next (e.g., the digits in the first multiplier, the added number, and the resulting digits). This helps in predicting subsequent steps and understanding the underlying rule.