Math worksheet featuring questions on rational numbers, covering properties, inverses, and representations.

Here are the step-by-step solutions to the questions in your homework.

Questions carrying 1 Mark each

Q.1 Is zero a rational number? If yes, give two examples.
* Reasoning: A rational number is any number that can be written as a fraction $\frac{p}{q}$ where $q$ is not zero. Zero can be written as $\frac{0}{1}$, $\frac{0}{2}$, etc.
* Answer: Yes, zero is a rational number. Two examples are $\frac{0}{1}$ and $\frac{0}{5}$.

Q.2 What are the identity elements for the addition and multiplication of rational numbers?
* Reasoning:
* For addition: Adding 0 to any number leaves it unchanged ($a + 0 = a$). So, 0 is the additive identity.
* For multiplication: Multiplying any number by 1 leaves it unchanged ($a \times 1 = a$). So, 1 is the multiplicative identity.
* Answer: The identity element for addition is 0. The identity element for multiplication is 1.

Q.3 Name the property used in the following: $\frac{-6}{6} + \frac{+6}{6} = 1$ (Note: The image likely implies the sum is 0 or asks about inverses, but let's look at the structure. Actually, looking closely at the image text: $\frac{-6}{6} + \frac{+6}{6} = 0$. Wait, the image says $= 1$? No, usually this question demonstrates the Additive Inverse Property. Let's re-read carefully. Ah, the image actually shows: $\frac{-6}{6} + \frac{6}{6} = 0$. Wait, looking at crop 1, it says `+ -6/6 + 6/6 = 1`? That is mathematically incorrect. $-1 + 1 = 0$. However, standard curriculum questions asking to "Name the property" for a number added to its negative usually refer to the Additive Inverse Property. Let's assume there is a typo in the question paper's result or I am misreading a blurry '0' as a '1'. Given the context of Q4 asking for additive inverse, Q3 is definitely about Additive Inverse.)
* *Correction*: Looking very closely at the third crop, it says $\frac{-6}{6} + \frac{6}{6} = 0$ (The digit looks like a 0, though slightly blurry). If it equals 0, the property is Additive Inverse. If the question strictly says "= 1", the question itself contains an error, but the *property* being tested by adding a number to its opposite is the Additive Inverse Property.
* Answer: Additive Inverse Property (since a number added to its negative equals zero).

Q.4 Find the additive inverse of $\frac{5}{5} / \frac{-5}{10}$
* Reasoning: First, simplify the expression.
* $\frac{5}{5} = 1$
* $\frac{-5}{10} = \frac{-1}{2}$
* The expression is $1 \div (\frac{-1}{2}) = 1 \times (\frac{-2}{1}) = -2$.
* The additive inverse of a number $x$ is $-x$. So, the additive inverse of $-2$ is $2$.
* Answer: 2

Q.5 Write the reciprocal of 0.
* Reasoning: The reciprocal of a number $x$ is $\frac{1}{x}$. Division by zero is undefined in mathematics. Therefore, 0 does not have a reciprocal.
* Answer: Undefined (or "Does not exist").

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Questions carrying 2 marks each

Q.6 Using property find the value of $\frac{2}{7} \times \frac{-2}{2} - \frac{2}{2} \times \frac{2}{2} - \frac{2}{2} \times \frac{2}{7}$
* Reasoning: Let's rearrange the terms to use the Distributive Property. Notice that $\frac{2}{7}$ and $\frac{-2}{7}$ (from the last term) share a common factor if we group them, or we can just calculate directly since the numbers are simple.
* Term 1: $\frac{2}{7} \times (-1) = \frac{-2}{7}$
* Term 2: $1 \times 1 = 1$ (Wait, $\frac{-2}{2} = -1$ and $\frac{2}{2}=1$. So $-1 \times 1 = -1$. The middle term is subtracted, so $- (-1)$? No, the expression is minus $\frac{2}{2} \times \frac{2}{2}$. $\frac{2}{2}=1$. So it is $- (1 \times 1) = -1$.
* Term 3: $\frac{2}{2} \times \frac{2}{7} = 1 \times \frac{2}{7} = \frac{2}{7}$. This term is subtracted, so $-\frac{2}{7}$.
* Let's re-read the expression carefully from the image: $\frac{2}{7} \times \frac{-2}{2} - \frac{2}{2} \times \frac{2}{2} - \frac{2}{2} \times \frac{2}{7}$
* Simplify fractions first: $\frac{-2}{2} = -1$, $\frac{2}{2} = 1$.
* Expression becomes: $\frac{2}{7}(-1) - (1)(1) - (1)(\frac{2}{7})$
* $= \frac{-2}{7} - 1 - \frac{2}{7}$
* Combine fractions: $\frac{-2}{7} - \frac{2}{7} = \frac{-4}{7}$
* Subtract 1: $\frac{-4}{7} - 1 = \frac{-4}{7} - \frac{7}{7} = \frac{-11}{7}$
* *Alternative interpretation using Distributive Property:* Maybe the question intended $\frac{2}{7} \times \frac{-2}{2} - \frac{2}{2} \times \frac{2}{7}$? If we group the first and last term: $\frac{2}{7} \times (-1) - \frac{2}{7} \times 1 = \frac{2}{7}(-1 - 1) = \frac{2}{7}(-2) = \frac{-4}{7}$. Then subtract the middle term ($1$). Result $\frac{-11}{7}$.
* Answer: $\frac{-11}{7}$

Q.7 Is 0.5 the multiplicative inverse of $2\frac{2}{5}$? Why or Why not?
* Reasoning:
* Convert $2\frac{2}{5}$ to an improper fraction: $\frac{2 \times 5 + 2}{5} = \frac{12}{5}$.
* Convert 0.5 to a fraction: $\frac{1}{2}$.
* Multiply them: $\frac{12}{5} \times \frac{1}{2} = \frac{12}{10} = \frac{6}{5} = 1.2$.
* For two numbers to be multiplicative inverses, their product must be 1.
* Since $1.2 \neq 1$, they are not inverses. (The inverse of $\frac{12}{5}$ is $\frac{5}{12}$).
* Answer: No, because their product is $\frac{6}{5}$ (or 1.2), not 1.

Q.8 Write four rational numbers which are greater than -31 and less than 4.
* Reasoning: We need numbers between -31 and 4. Integers are rational numbers. We can pick any integers in this range, or fractions/decimals.
* Answer: -30, -10, 0, 2 (Any four numbers such that $-31 < x < 4$ are correct).

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Questions carrying 3 marks each

Q.9 Represent $-\frac{6}{6}$ and $\frac{-6}{11}$ on number line.
* Reasoning:
* $-\frac{6}{6}$ simplifies to -1. On the number line, this is one unit to the left of 0.
* $\frac{-6}{11}$ is approximately -0.54. This is located between 0 and -1, slightly to the right of the midpoint (-0.5).
* Answer:
1. Draw a horizontal line with markings for integers (... -2, -1, 0, 1 ...).
2. Place a dot exactly on -1 and label it $-\frac{6}{6}$.
3. Divide the segment between 0 and -1 into 11 equal parts. Count 6 parts to the left from 0. Place a dot there and label it $\frac{-6}{11}$.

Q.10 Find ten rational numbers between 2 and 3.
* Reasoning: To find rational numbers between two integers, we can convert them to fractions with a larger denominator.
* Let's use a denominator of 11 (since we need 10 numbers, denominator $n+1$ works well for integers).
* $2 = \frac{22}{11}$
* $3 = \frac{33}{11}$
* The numbers between $\frac{22}{11}$ and $\frac{33}{11}$ are: $\frac{23}{11}, \frac{24}{11}, \frac{25}{11}, \frac{26}{11}, \frac{27}{11}, \frac{28}{11}, \frac{29}{11}, \frac{30}{11}, \frac{31}{11}, \frac{32}{11}$.
* Answer: $\frac{23}{11}, \frac{24}{11}, \frac{25}{11}, \frac{26}{11}, \frac{27}{11}, \frac{28}{11}, \frac{29}{11}, \frac{30}{11}, \frac{31}{11}, \frac{32}{11}$

Q.11 Find the value of $\frac{2}{x} + \frac{2}{y}$ given $x = \frac{-2}{5}$ and $y = \frac{2}{5}$.
* Reasoning:
* Substitute the values of $x$ and $y$ into the expression.
* Term 1: $\frac{2}{x} = \frac{2}{\frac{-2}{5}} = 2 \times \frac{5}{-2} = \frac{10}{-2} = -5$.
* Term 2: $\frac{2}{y} = \frac{2}{\frac{2}{5}} = 2 \times \frac{5}{2} = \frac{10}{2} = 5$.
* Add the results: $-5 + 5 = 0$.
* Answer: 0