RBSE Solutions Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1

Question 1. Add the following rational numbers. (Solve any two on number line)
(i) \( \frac { 5 }{ 2 } + \left( \frac { -3 }{ 4 } \right) \)
(ii) \( \frac { 2 }{ 3 } + \left( \frac { -4 }{ 5 } \right) + \frac { 5 }{ 6 } \)
(iii) \( 0 + \frac { -2 }{ 3 } \)
(iv) \( -2\frac { 1 }{ 3 } + 4\frac { 3 }{ 5 } \)
(v) \( \frac { -6 }{ 5 } + \left( \frac { -13 }{ 7 } \right) \)
(vi) \( \frac { -8 }{ 19 } + \left( \frac { -4 }{ 57 } \right) \)
Answer:
(i) First, let's add the numbers:
\( \frac { 5 }{ 2 } + \left( \frac { -3 }{ 4 } \right) \)
To add these, we find the Least Common Multiple (LCM) of the denominators 2 and 4, which is 4.
So, we convert the fractions to have a denominator of 4:
\( = \frac { 5 \times 2 }{ 2 \times 2 } + \left( \frac { -3 }{ 4 } \right) \)
\( = \frac { 10 }{ 4 } + \frac { -3 }{ 4 } \)
Now, we add the numerators:
\( = \frac { 10 + (-3) }{ 4 } \)
\( = \frac { 7 }{ 4 } \)

Next, let's show this addition on a number line.
The distance between two consecutive numbers on the number line is \( \frac { 1 }{ 4 } \).
Adding \( \frac { -3 }{ 4 } \) to \( \frac { 5 }{ 2 } \) is the same as adding \( \frac { -3 }{ 4 } \) to \( \frac { 10 }{ 4 } \).
Since \( \frac { -3 }{ 4 } \) is negative, we move 3 gaps to the left from \( \frac { 10 }{ 4 } \).
Moving 3 units to the left from \( \frac { 10 }{ 4 } \) on the number line takes us to \( \frac { 7 }{ 4 } \).
(iii) First, let's add the numbers:
\( 0 + \frac { -2 }{ 3 } \)
Adding zero to any number results in the same number.
\( = 0 - \frac { 2 }{ 3 } \)
\( = \frac { -2 }{ 3 } \)

Next, let's show this addition on a number line.
The distance between two consecutive points on the number line is \( \frac { 1 }{ 3 } \).
Adding \( \frac { -2 }{ 3 } \) to 0 means we start at 0 and move 2 gaps to the left because it's a negative number.
This movement takes us to \( \frac { -2 }{ 3 } \) on the number line.
(v) First, let's add the numbers:
\( \frac { -6 }{ 5 } + \left( \frac { -13 }{ 7 } \right) \)
To add these, we find the LCM of the denominators 5 and 7, which is 35.
So, we convert the fractions to have a denominator of 35:
\( = \frac { -6 \times 7 }{ 5 \times 7 } + \frac { -13 \times 5 }{ 7 \times 5 } \)
\( = \frac { -42 }{ 35 } + \frac { -65 }{ 35 } \)
Now, we add the numerators:
\( = \frac { -42 + (-65) }{ 35 } \)
\( = \frac { -107 }{ 35 } \)
We can also write this as a mixed fraction: \( -3\frac { 2 }{ 35 } \)

(vi) First, let's add the numbers:
\( \frac { -8 }{ 19 } + \left( \frac { -4 }{ 57 } \right) \)
To add these, we find the LCM of the denominators 19 and 57, which is 57.
So, we convert the fractions to have a denominator of 57:
\( = \frac { -8 \times 3 }{ 19 \times 3 } + \frac { -4 }{ 57 } \)
\( = \frac { -24 }{ 57 } + \frac { -4 }{ 57 } \)
Now, we add the numerators:
\( = \frac { -24 + (-4) }{ 57 } \)
\( = \frac { -28 }{ 57 } \)
In simple words: To add rational numbers, find a common bottom number (LCM), change the top numbers accordingly, and then add them. For showing on a number line, positive means moving right, and negative means moving left.

🎯 Exam Tip: Always look for the Least Common Multiple (LCM) of denominators to simplify calculations and prevent errors. Remember to move left for negative numbers on the number line and right for positive numbers.

Question 2. Find the value (Solve any two on number line)-
(i) \( \frac { 2 }{ 3 } + \frac { 5 }{ 4 } \)
(ii) \( -2\frac { 1 }{ 9 } + \frac { 7 }{ 9 } \)
(iii) \( 0 + \left( \frac { -2 }{ 3 } \right) \)
(iv) \( \frac { -7 }{ 63 } + \left( \frac { -5 }{ 21 } \right) \)
(v) \( \frac { -2 }{ 13 } + \frac { -1 }{ 7 } \)
(vi) \( 4\frac { 3 }{ 5 } - \left( -2\frac { 1 }{ 3 } \right) \)
Answer:
(i) First, let's find the sum:
\( \frac { 2 }{ 3 } + \frac { 5 }{ 4 } \)
The LCM of 3 and 4 is 12. We convert the fractions:
\( = \frac { 2 \times 4 }{ 3 \times 4 } + \frac { 5 \times 3 }{ 4 \times 3 } \)
\( = \frac { 8 }{ 12 } + \frac { 15 }{ 12 } \)
\( = \frac { 8 + 15 }{ 12 } = \frac { 23 }{ 12 } \)

Next, let's show this on a number line:
The distance between two consecutive points (gaps) on the number line is \( \frac { 1 }{ 12 } \).
Adding \( \frac { 15 }{ 12 } \) to \( \frac { 8 }{ 12 } \) means we move 15 gaps to the right from \( \frac { 8 }{ 12 } \) because it is a positive value.
This movement takes us to \( \frac { 23 }{ 12 } \).
(ii) Let's find the sum:
\( -2\frac { 1 }{ 9 } + \frac { 7 }{ 9 } \)
First, convert the mixed fraction to an improper fraction: \( -2\frac { 1 }{ 9 } = -\frac { (2 \times 9) + 1 }{ 9 } = -\frac { 18 + 1 }{ 9 } = -\frac { 19 }{ 9 } \)
Now, add the fractions:
\( = -\frac { 19 }{ 9 } + \frac { 7 }{ 9 } \)
Since the denominators are already the same, we just add the numerators:
\( = \frac { -19 + 7 }{ 9 } \)
\( = \frac { -12 }{ 9 } \)
This fraction can be simplified by dividing both the numerator and the denominator by 3:
\( = -\frac { 4 }{ 3 } \)

(iii) Let's find the sum:
\( 0 + \left( \frac { -2 }{ 3 } \right) \)
Adding zero to any number does not change the number.
\( = \frac { -2 }{ 3 } \)

(iv) Let's find the sum:
\( \frac { -7 }{ 63 } + \left( \frac { -5 }{ 21 } \right) \)
The LCM of 63 and 21 is 63.
\( = \frac { -7 }{ 63 } + \frac { -5 \times 3 }{ 21 \times 3 } \)
\( = \frac { -7 }{ 63 } + \frac { -15 }{ 63 } \)
\( = \frac { -7 + (-15) }{ 63 } \)
\( = \frac { -22 }{ 63 } \)

Next, let's show this on a number line:
The distance between two consecutive points on the number line is \( \frac { 1 }{ 63 } \).
Adding \( \frac { -15 }{ 63 } \) to \( \frac { -7 }{ 63 } \) means we start at \( \frac { -7 }{ 63 } \) and move 15 gaps to the left because it's a negative value.
This movement takes us to \( \frac { -22 }{ 63 } \).
(v) Let's find the sum:
\( \frac { -2 }{ 13 } + \frac { -1 }{ 7 } \)
The LCM of 13 and 7 is 91.
\( = \frac { -2 \times 7 }{ 13 \times 7 } + \frac { -1 \times 13 }{ 7 \times 13 } \)
\( = \frac { -14 }{ 91 } + \frac { -13 }{ 91 } \)
\( = \frac { -14 + (-13) }{ 91 } \)
\( = \frac { -27 }{ 91 } \)

(vi) Let's find the value:
\( 4\frac { 3 }{ 5 } - \left( -2\frac { 1 }{ 3 } \right) \)
First, convert the mixed fractions to improper fractions:
\( 4\frac { 3 }{ 5 } = \frac { (4 \times 5) + 3 }{ 5 } = \frac { 20 + 3 }{ 5 } = \frac { 23 }{ 5 } \)
\( -2\frac { 1 }{ 3 } = -\frac { (2 \times 3) + 1 }{ 3 } = -\frac { 6 + 1 }{ 3 } = -\frac { 7 }{ 3 } \)
Now, substitute these back into the expression:
\( = \frac { 23 }{ 5 } - \left( -\frac { 7 }{ 3 } \right) \)
Remember that subtracting a negative number is the same as adding a positive number:
\( = \frac { 23 }{ 5 } + \frac { 7 }{ 3 } \)
The LCM of 5 and 3 is 15. We convert the fractions:
\( = \frac { 23 \times 3 }{ 5 \times 3 } + \frac { 7 \times 5 }{ 3 \times 5 } \)
\( = \frac { 69 }{ 15 } + \frac { 35 }{ 15 } \)
\( = \frac { 69 + 35 }{ 15 } \)
\( = \frac { 104 }{ 15 } \)
In simple words: To add or subtract rational numbers, first make sure they are in fraction form. If they are mixed numbers, change them to improper fractions. Then, find the lowest common bottom number (denominator) to combine them.

🎯 Exam Tip: When dealing with mixed numbers, always convert them to improper fractions before performing addition or subtraction. Pay close attention to negative signs, especially when subtracting a negative number.

Question 3. Multiply the following rational numbers-
(i) \( \frac { 13 }{ 5 } \times \left( -5 \right) \)
(ii) \( \left( \frac { 4 }{ -5 } \right) \times \left( \frac { -5 }{ 4 } \right) \)
(iii) \( \frac { -2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) \)
(iv) \( \frac { 15 }{ 18 } \times \frac { 5 }{ 6 } \times \frac { 21 }{ 5 } \)
(v) \( \frac { 9 }{ 4 } \times \left( \frac { -7 }{ 5 } \right) \times \left( \frac { -6 }{ 21 } \right) \)
(vi) \( 2\frac { 1 }{ 9 } \times \left( -3\frac { 1 }{ 2 } \right) \)
Answer:
(i) Let's multiply the numbers:
\( \frac { 13 }{ 5 } \times \left( -5 \right) \)
We can write -5 as \( \frac { -5 }{ 1 } \).
\( = \frac { 13 }{ 5 } \times \frac { -5 }{ 1 } \)
Multiply the numerators and the denominators:
\( = \frac { 13 \times (-5) }{ 5 \times 1 } \)
\( = \frac { -65 }{ 5 } \)
Simplify the fraction:
\( = -13 \)

(ii) Let's multiply the numbers:
\( \left( \frac { 4 }{ -5 } \right) \times \left( \frac { -5 }{ 4 } \right) \)
We can simplify this by cancelling common terms diagonally:
\( = \frac { 4 }{ -5 } \times \frac { -5 }{ 4 } \)
The 4 in the numerator and 4 in the denominator cancel out. The -5 in the numerator and -5 in the denominator also cancel out.
\( = \frac { 1 }{ 1 } \times \frac { 1 }{ 1 } \)
\( = 1 \)
This also shows that \( \frac { 4 }{ -5 } \) and \( \frac { -5 }{ 4 } \) are multiplicative inverses of each other.

(iii) Let's multiply the numbers:
\( \frac { -2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) \)
Multiply the numerators and the denominators:
\( = \frac { (-2) \times (-3) }{ 5 \times 7 } \)
\( = \frac { 6 }{ 35 } \)

(iv) Let's multiply the numbers:
\( \frac { 15 }{ 18 } \times \frac { 5 }{ 6 } \times \frac { 21 }{ 5 } \)
First, simplify fractions before multiplying, if possible:
\( \frac { 15 }{ 18 } \) simplifies to \( \frac { 5 }{ 6 } \) (divide by 3).
So, the expression becomes:
\( = \frac { 5 }{ 6 } \times \frac { 5 }{ 6 } \times \frac { 21 }{ 5 } \)
We can cancel out one 5 from the numerator and one 5 from the denominator:
\( = \frac { \cancel{5} }{ 6 } \times \frac { 5 }{ 6 } \times \frac { 21 }{ \cancel{5} } \)
\( = \frac { 5 \times 21 }{ 6 \times 6 } \)
\( = \frac { 105 }{ 36 } \)
This fraction can be simplified by dividing both the numerator and denominator by 3:
\( = \frac { 105 \div 3 }{ 36 \div 3 } \)
\( = \frac { 35 }{ 12 } \)

(v) Let's multiply the numbers:
\( \frac { 9 }{ 4 } \times \left( \frac { -7 }{ 5 } \right) \times \left( \frac { -6 }{ 21 } \right) \)
First, simplify the fraction \( \frac { -6 }{ 21 } \) by dividing both numerator and denominator by 3:
\( \frac { -6 }{ 21 } = \frac { -2 }{ 7 } \)
Now, substitute this back into the expression:
\( = \frac { 9 }{ 4 } \times \left( \frac { -7 }{ 5 } \right) \times \left( \frac { -2 }{ 7 } \right) \)
We can cancel out the 7 in the numerator and 7 in the denominator:
\( = \frac { 9 }{ 4 } \times \left( \frac { -\cancel{7} }{ 5 } \right) \times \left( \frac { -2 }{ \cancel{7} } \right) \)
\( = \frac { 9 \times (-1) \times (-2) }{ 4 \times 5 \times 1 } \)
\( = \frac { 18 }{ 20 } \)
This fraction can be simplified by dividing both the numerator and denominator by 2:
\( = \frac { 18 \div 2 }{ 20 \div 2 } \)
\( = \frac { 9 }{ 10 } \)

(vi) Let's multiply the numbers:
\( 2\frac { 1 }{ 9 } \times \left( -3\frac { 1 }{ 2 } \right) \)
First, convert the mixed fractions to improper fractions:
\( 2\frac { 1 }{ 9 } = \frac { (2 \times 9) + 1 }{ 9 } = \frac { 18 + 1 }{ 9 } = \frac { 19 }{ 9 } \)
\( -3\frac { 1 }{ 2 } = -\frac { (3 \times 2) + 1 }{ 2 } = -\frac { 6 + 1 }{ 2 } = -\frac { 7 }{ 2 } \)
Now, multiply the improper fractions:
\( = \frac { 19 }{ 9 } \times \left( -\frac { 7 }{ 2 } \right) \)
Multiply the numerators and the denominators:
\( = \frac { 19 \times (-7) }{ 9 \times 2 } \)
\( = \frac { -133 }{ 18 } \)
In simple words: When multiplying fractions, just multiply the top numbers together and the bottom numbers together. If there are mixed numbers, change them to regular fractions first. Always try to make fractions simpler by dividing common numbers before or after multiplying.

🎯 Exam Tip: Remember that two negative numbers multiplied together give a positive result. Always simplify fractions before and after multiplication to make calculations easier and ensure your final answer is in its simplest form.

Question 4. Find the values
(i) \( (-6) + \frac { 3 }{ 5 } \)
(ii) \( \frac { -27 }{ 5 } + \left( \frac { -54 }{ 10 } \right) \)
(iii) \( \frac { 21 }{ 36 } \div \left( \frac { -7 }{ 18 } \right) \)
(iv) \( \frac { -7 }{ 12 } \div \left( \frac { -13 }{ 3 } \right) \)
(v) \( -2\frac { 1 }{ 9 } \div 6\frac { 1 }{ 9 } \)
(vi) \( \frac { 2 }{ 15 } \div \left( \frac { -8 }{ 45 } \right) \)
Answer:
(i) Let's find the sum:
\( (-6) + \frac { 3 }{ 5 } \)
To add a whole number and a fraction, we can write the whole number as a fraction with a denominator of 1.
\( = \frac { -6 }{ 1 } + \frac { 3 }{ 5 } \)
The LCM of 1 and 5 is 5.
\( = \frac { -6 \times 5 }{ 1 \times 5 } + \frac { 3 }{ 5 } \)
\( = \frac { -30 }{ 5 } + \frac { 3 }{ 5 } \)
\( = \frac { -30 + 3 }{ 5 } \)
\( = \frac { -27 }{ 5 } \)

(ii) Let's find the sum:
\( \frac { -27 }{ 5 } + \left( \frac { -54 }{ 10 } \right) \)
First, we can simplify \( \frac { -54 }{ 10 } \) by dividing both numerator and denominator by 2:
\( \frac { -54 \div 2 }{ 10 \div 2 } = \frac { -27 }{ 5 } \)
Now, substitute this back into the expression:
\( = \frac { -27 }{ 5 } + \frac { -27 }{ 5 } \)
Since the denominators are the same, we add the numerators:
\( = \frac { -27 + (-27) }{ 5 } \)
\( = \frac { -54 }{ 5 } \)

(iii) Let's find the value:
\( \frac { 21 }{ 36 } \div \left( \frac { -7 }{ 18 } \right) \)
Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of \( \frac { -7 }{ 18 } \) is \( \frac { 18 }{ -7 } \).
\( = \frac { 21 }{ 36 } \times \frac { 18 }{ -7 } \)
Simplify by cancelling common factors: 21 and -7 (21 = 3 x 7, -7 = -1 x 7) and 18 and 36 (36 = 2 x 18).
\( = \frac { 3 \times \cancel{7} }{ 2 \times \cancel{18} } \times \frac { \cancel{18} }{ -1 \times \cancel{7} } \)
\( = \frac { 3 }{ 2 } \times \frac { 1 }{ -1 } \)
\( = \frac { 3 \times 1 }{ 2 \times (-1) } \)
\( = \frac { 3 }{ -2 } \)
\( = -\frac { 3 }{ 2 } \)

(iv) Let's find the value:
\( \frac { -7 }{ 12 } \div \left( \frac { -13 }{ 3 } \right) \)
Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of \( \frac { -13 }{ 3 } \) is \( \frac { 3 }{ -13 } \).
\( = \frac { -7 }{ 12 } \times \frac { 3 }{ -13 } \)
Simplify by cancelling common factors: 3 and 12 (12 = 4 x 3).
\( = \frac { -7 }{ 4 \times \cancel{3} } \times \frac { \cancel{3} }{ -13 } \)
\( = \frac { -7 \times 1 }{ 4 \times (-13) } \)
\( = \frac { -7 }{ -52 } \)
Since both numerator and denominator are negative, the result is positive:
\( = \frac { 7 }{ 52 } \)

(v) Let's find the value:
\( -2\frac { 1 }{ 9 } \div 6\frac { 1 }{ 9 } \)
First, convert the mixed fractions to improper fractions:
\( -2\frac { 1 }{ 9 } = -\frac { (2 \times 9) + 1 }{ 9 } = -\frac { 19 }{ 9 } \)
\( 6\frac { 1 }{ 9 } = \frac { (6 \times 9) + 1 }{ 9 } = \frac { 54 + 1 }{ 9 } = \frac { 55 }{ 9 } \)
Now, divide the improper fractions:
\( = -\frac { 19 }{ 9 } \div \frac { 55 }{ 9 } \)
Multiply by the reciprocal of the second fraction:
\( = -\frac { 19 }{ 9 } \times \frac { 9 }{ 55 } \)
Cancel out the common factor 9:
\( = -\frac { 19 }{ \cancel{9} } \times \frac { \cancel{9} }{ 55 } \)
\( = -\frac { 19 }{ 55 } \)

(vi) Let's find the value:
\( \frac { 2 }{ 15 } \div \left( \frac { -8 }{ 45 } \right) \)
Multiply by the reciprocal of the second fraction:
\( = \frac { 2 }{ 15 } \times \frac { 45 }{ -8 } \)
Simplify by cancelling common factors: 2 and -8 ( -8 = -4 x 2), and 15 and 45 (45 = 3 x 15).
\( = \frac { \cancel{2} }{ \cancel{15} } \times \frac { 3 \times \cancel{15} }{ -4 \times \cancel{2} } \)
\( = \frac { 1 \times 3 }{ 1 \times (-4) } \)
\( = \frac { 3 }{ -4 } \)
\( = -\frac { 3 }{ 4 } \)
In simple words: To divide fractions, flip the second fraction upside down (find its reciprocal) and then multiply it by the first fraction. Remember to change any mixed numbers into simple fractions first.

🎯 Exam Tip: When dividing rational numbers, remember the rule "keep, change, flip": Keep the first fraction, change the division to multiplication, and flip the second fraction. Always simplify fractions before or after operations.

Question 5. Find the values
(i) \( \frac { 3 }{ 5 } + \frac { 7 }{ 10 } + \left( \frac { -8 }{ 12 } \right) + \frac { 4 }{ 3 } \)
(ii) \( \left( 2\frac { 1 }{ 2 } \right) + \left( -3\frac { 1 }{ 2 } \right) + \left( -2\frac { 1 }{ 3 } \right) + \left( 2\frac { 1 }{ 9 } \right) \)
(iii) \( \left( \frac { -7 }{ 5 } \right) \times \frac { 2 }{ 3 } \times \frac { 15 }{ 16 } \times \left( \frac { -8 }{ 9 } \right) \)
(iv) \( \frac { 2 }{ 5 } + \left[ \left( \frac { -1 }{ 3 } \right) \times \frac { 7 }{ 2 } \right] \)
Answer:
(i) Let's find the sum:
\( \frac { 3 }{ 5 } + \frac { 7 }{ 10 } + \left( \frac { -8 }{ 12 } \right) + \frac { 4 }{ 3 } \)
First, simplify \( \frac { -8 }{ 12 } \) by dividing both numerator and denominator by 4: \( \frac { -8 \div 4 }{ 12 \div 4 } = \frac { -2 }{ 3 } \)
So, the expression becomes:
\( = \frac { 3 }{ 5 } + \frac { 7 }{ 10 } + \left( \frac { -2 }{ 3 } \right) + \frac { 4 }{ 3 } \)
The LCM of the denominators 5, 10, 3 is 30. We convert all fractions to have a denominator of 30:
\( = \frac { 3 \times 6 }{ 5 \times 6 } + \frac { 7 \times 3 }{ 10 \times 3 } + \frac { -2 \times 10 }{ 3 \times 10 } + \frac { 4 \times 10 }{ 3 \times 10 } \)
\( = \frac { 18 }{ 30 } + \frac { 21 }{ 30 } + \frac { -20 }{ 30 } + \frac { 40 }{ 30 } \)
Now, add the numerators:
\( = \frac { 18 + 21 + (-20) + 40 }{ 30 } \)
\( = \frac { 39 - 20 + 40 }{ 30 } \)
\( = \frac { 19 + 40 }{ 30 } \)
\( = \frac { 59 }{ 30 } \)

(ii) Let's find the sum:
\( \left( 2\frac { 1 }{ 2 } \right) + \left( -3\frac { 1 }{ 2 } \right) + \left( -2\frac { 1 }{ 3 } \right) + \left( 2\frac { 1 }{ 9 } \right) \)
First, convert all mixed fractions to improper fractions:
\( 2\frac { 1 }{ 2 } = \frac { (2 \times 2) + 1 }{ 2 } = \frac { 5 }{ 2 } \)
\( -3\frac { 1 }{ 2 } = -\frac { (3 \times 2) + 1 }{ 2 } = -\frac { 7 }{ 2 } \)
\( -2\frac { 1 }{ 3 } = -\frac { (2 \times 3) + 1 }{ 3 } = -\frac { 7 }{ 3 } \)
\( 2\frac { 1 }{ 9 } = \frac { (2 \times 9) + 1 }{ 9 } = \frac { 19 }{ 9 } \)
Now, substitute these back into the expression:
\( = \frac { 5 }{ 2 } + \left( -\frac { 7 }{ 2 } \right) + \left( -\frac { 7 }{ 3 } \right) + \frac { 19 }{ 9 } \)
First, combine the fractions with the same denominator:
\( = \left( \frac { 5 }{ 2 } - \frac { 7 }{ 2 } \right) + \left( -\frac { 7 }{ 3 } + \frac { 19 }{ 9 } \right) \)
\( = \frac { 5 - 7 }{ 2 } + \left( -\frac { 7 \times 3 }{ 3 \times 3 } + \frac { 19 }{ 9 } \right) \)
\( = \frac { -2 }{ 2 } + \left( -\frac { 21 }{ 9 } + \frac { 19 }{ 9 } \right) \)
\( = -1 + \frac { -21 + 19 }{ 9 } \)
\( = -1 + \frac { -2 }{ 9 } \)
\( = \frac { -9 }{ 9 } + \frac { -2 }{ 9 } \)
\( = \frac { -9 - 2 }{ 9 } \)
\( = \frac { -11 }{ 9 } \)

(iii) Let's find the product:
\( \left( \frac { -7 }{ 5 } \right) \times \frac { 2 }{ 3 } \times \frac { 15 }{ 16 } \times \left( \frac { -8 }{ 9 } \right) \)
Group terms for easier multiplication and cancellation:
\( = \left( \frac { -7 }{ 5 } \times \frac { 15 }{ 16 } \right) \times \left( \frac { 2 }{ 3 } \times \frac { -8 }{ 9 } \right) \)
For the first group, cancel 5 and 15 (15 = 3 x 5):
\( = \left( \frac { -7 }{ \cancel{5} } \times \frac { 3 \times \cancel{5} }{ 16 } \right) \times \left( \frac { 2 }{ 3 } \times \frac { -8 }{ 9 } \right) \)
\( = \left( \frac { -7 \times 3 }{ 16 } \right) \times \left( \frac { 2 \times (-8) }{ 3 \times 9 } \right) \)
\( = \frac { -21 }{ 16 } \times \frac { -16 }{ 27 } \)
Now, cancel 16 from numerator and denominator:
\( = \frac { -21 }{ \cancel{16} } \times \frac { -\cancel{16} }{ 27 } \)
\( = \frac { (-21) \times (-1) }{ 27 } \)
\( = \frac { 21 }{ 27 } \)
Simplify by dividing both numerator and denominator by 3:
\( = \frac { 21 \div 3 }{ 27 \div 3 } \)
\( = \frac { 7 }{ 9 } \)

(iv) Let's find the value:
\( \frac { 2 }{ 5 } + \left[ \left( \frac { -1 }{ 3 } \right) \times \frac { 7 }{ 2 } \right] \)
First, perform the multiplication inside the brackets:
\( \left( \frac { -1 }{ 3 } \right) \times \frac { 7 }{ 2 } = \frac { -1 \times 7 }{ 3 \times 2 } = \frac { -7 }{ 6 } \)
Now, substitute this back into the expression:
\( = \frac { 2 }{ 5 } + \left( \frac { -7 }{ 6 } \right) \)
The LCM of 5 and 6 is 30. We convert the fractions:
\( = \frac { 2 \times 6 }{ 5 \times 6 } + \frac { -7 \times 5 }{ 6 \times 5 } \)
\( = \frac { 12 }{ 30 } + \frac { -35 }{ 30 } \)
\( = \frac { 12 - 35 }{ 30 } \)
\( = \frac { -23 }{ 30 } \)
In simple words: When you have mixed operations with fractions, always follow the order of operations: first things inside brackets, then multiplication/division, and finally addition/subtraction. Make sure all mixed numbers are changed to improper fractions.

🎯 Exam Tip: Remember to always follow the BODMAS/PEMDAS rule (Brackets, Orders, Division/Multiplication, Addition/Subtraction) for operations. Convert mixed fractions to improper fractions before performing any calculation to avoid mistakes.

Question 6. Find the value of the following using the appropriate properties
(i) \( \frac { 3 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) - \frac { 2 }{ 5 } \times \frac { 3 }{ 5 } + \frac { 1 }{ 15 } \times \frac { 3 }{ 5 } \)
(ii) \( \frac { 5 }{ 2 } \times \frac { 3 }{ 7 } - \frac { 3 }{ 7 } \times \frac { 3 }{ 2 } + \frac { 3 }{ 5 } \times \left( \frac { -2 }{ 3 } \right) \)
Answer:
(i) Let's find the value:
\( \frac { 3 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) - \frac { 2 }{ 5 } \times \frac { 3 }{ 5 } + \frac { 1 }{ 15 } \times \frac { 3 }{ 5 } \)
We can use the distributive property here because \( \frac { 3 }{ 5 } \) is common in two terms. Let's rewrite the terms to make it clear:
\( = \frac { 3 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) + \left( \frac { -2 }{ 5 } \right) \times \frac { 3 }{ 5 } + \frac { 1 }{ 15 } \times \frac { 3 }{ 5 } \)
However, the middle term has \( \frac{3}{5}\) but is \( \frac{-2}{5} \times \frac{3}{5} \), not \( \frac{3}{5} \times \text{something} \). Let's factor out \( \frac{3}{5} \) where it is available directly.
\( = \frac { 3 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) + \frac { 3 }{ 5 } \times \left( \frac { -2 }{ 5 } \right) + \frac { 3 }{ 5 } \times \frac { 1 }{ 15 } \)
Now, factor out \( \frac { 3 }{ 5 } \):
\( = \frac { 3 }{ 5 } \times \left[ \left( \frac { -3 }{ 7 } \right) + \left( \frac { -2 }{ 5 } \right) + \frac { 1 }{ 15 } \right] \)
First, find the sum inside the brackets. The LCM of 7, 5, and 15 is 105.
\( = \frac { 3 }{ 5 } \times \left[ \frac { -3 \times 15 }{ 7 \times 15 } + \frac { -2 \times 21 }{ 5 \times 21 } + \frac { 1 \times 7 }{ 15 \times 7 } \right] \)
\( = \frac { 3 }{ 5 } \times \left[ \frac { -45 }{ 105 } + \frac { -42 }{ 105 } + \frac { 7 }{ 105 } \right] \)
\( = \frac { 3 }{ 5 } \times \left[ \frac { -45 - 42 + 7 }{ 105 } \right] \)
\( = \frac { 3 }{ 5 } \times \left[ \frac { -87 + 7 }{ 105 } \right] \)
\( = \frac { 3 }{ 5 } \times \frac { -80 }{ 105 } \)
Now, multiply and simplify. We can simplify 3 and 105 (105 = 3 x 35), and 5 and -80 (-80 = -16 x 5).
\( = \frac { \cancel{3} }{ \cancel{5} } \times \frac { -16 \times \cancel{5} }{ \cancel{3} \times 35 } \)
\( = \frac { -16 }{ 35 } \)

(ii) Let's find the value:
\( \frac { 5 }{ 2 } \times \frac { 3 }{ 7 } - \frac { 3 }{ 7 } \times \frac { 3 }{ 2 } + \frac { 3 }{ 5 } \times \left( \frac { -2 }{ 3 } \right) \)
Rearrange the terms to group common factors using the commutative property of multiplication:
\( = \frac { 5 }{ 2 } \times \frac { 3 }{ 7 } - \frac { 3 }{ 2 } \times \frac { 3 }{ 7 } + \frac { 3 }{ 5 } \times \left( \frac { -2 }{ 3 } \right) \)
Factor out \( \frac { 3 }{ 7 } \) from the first two terms using the distributive property:
\( = \frac { 3 }{ 7 } \times \left( \frac { 5 }{ 2 } - \frac { 3 }{ 2 } \right) + \frac { 3 }{ 5 } \times \left( \frac { -2 }{ 3 } \right) \)
Perform the operations inside the parentheses:
\( = \frac { 3 }{ 7 } \times \left( \frac { 5 - 3 }{ 2 } \right) + \frac { 3 }{ 5 } \times \left( \frac { -2 }{ 3 } \right) \)
\( = \frac { 3 }{ 7 } \times \frac { 2 }{ 2 } + \frac { 3 }{ 5 } \times \left( \frac { -2 }{ 3 } \right) \)
\( = \frac { 3 }{ 7 } \times 1 + \frac { \cancel{3} }{ 5 } \times \frac { -2 }{ \cancel{3} } \)
\( = \frac { 3 }{ 7 } + \frac { -2 }{ 5 } \)
Now, add these two fractions. The LCM of 7 and 5 is 35.
\( = \frac { 3 \times 5 }{ 7 \times 5 } + \frac { -2 \times 7 }{ 5 \times 7 } \)
\( = \frac { 15 }{ 35 } + \frac { -14 }{ 35 } \)
\( = \frac { 15 - 14 }{ 35 } \)
\( = \frac { 1 }{ 35 } \)
In simple words: When solving problems with many fractions, look for numbers that appear in more than one part. You can pull them out using the "distributive property" to make the calculation easier. Also, change mixed numbers to simple fractions first.

🎯 Exam Tip: The distributive property \( a \times (b+c) = a \times b + a \times c \) is very useful for simplifying expressions with multiple fractions. Recognizing common factors can significantly reduce the number of calculations needed.

Question 7. Find the additive inverse of the following rational numbers
(i) \( \frac { 7 }{ 19 } \)
(ii) \( \frac { -9 }{ 5 } \)
(iii) \( \frac { -3 }{ -7 } \)
(iv) \( \frac { 5 }{ -9 } \)
(v) \( \frac { -13 }{ -17 } \)
(vi) \( \frac { 21 }{ -31 } \)
Answer:
(i) The additive inverse of \( \frac { 7 }{ 19 } \) is \( -\frac { 7 }{ 19 } \).
(ii) The additive inverse of \( \frac { -9 }{ 5 } \) is \( - \left( \frac { -9 }{ 5 } \right) = \frac { 9 }{ 5 } \).
(iii) First, simplify \( \frac { -3 }{ -7 } = \frac { 3 }{ 7 } \) because a negative divided by a negative is positive.
The additive inverse of \( \frac { 3 }{ 7 } \) is \( -\frac { 3 }{ 7 } \).
(iv) First, simplify \( \frac { 5 }{ -9 } = -\frac { 5 }{ 9 } \).
The additive inverse of \( -\frac { 5 }{ 9 } \) is \( - \left( -\frac { 5 }{ 9 } \right) = \frac { 5 }{ 9 } \).
(v) First, simplify \( \frac { -13 }{ -17 } = \frac { 13 }{ 17 } \) because a negative divided by a negative is positive.
The additive inverse of \( \frac { 13 }{ 17 } \) is \( -\frac { 13 }{ 17 } \).
(vi) First, simplify \( \frac { 21 }{ -31 } = -\frac { 21 }{ 31 } \).
The additive inverse of \( -\frac { 21 }{ 31 } \) is \( - \left( -\frac { 21 }{ 31 } \right) = \frac { 21 }{ 31 } \).
In simple words: The additive inverse of a number is the number you add to it to get zero. It's the same number but with the opposite sign. For fractions, if it's positive, its inverse is negative, and if it's negative, its inverse is positive.

🎯 Exam Tip: To find the additive inverse of any rational number, simply change its sign. If the number is positive, its additive inverse is negative, and vice versa. Always simplify fractions with negative signs before finding their additive inverse.

Question 8. Find multiplicative inverse of the following rational numbers
(i) \( - 17 \)
(ii) \( \frac{-11}{17} \)
(iii) \( -1 \times \frac{-3}{5} \)
(iv) \( \frac{13}{-19} \)
Answer:
(i) The multiplicative inverse of \( -17 \) is \( \frac{1}{-17} \) or \( -\frac{1}{17} \). A number's multiplicative inverse, when multiplied by the original number, always results in 1.
(ii) The multiplicative inverse of \( \frac{-11}{17} \) is \( \frac{17}{-11} \) or \( -\frac{17}{11} \).
(iii) First, calculate the product: \( -1 \times \frac{-3}{5} = \frac{3}{5} \). The multiplicative inverse of \( \frac{3}{5} \) is \( \frac{5}{3} \).
(iv) The multiplicative inverse of \( \frac{13}{-19} \) is \( \frac{-19}{13} \).
In simple words: The multiplicative inverse of a number is what you multiply it by to get 1. For a fraction, just flip the top and bottom numbers. For a whole number, put 1 over it.

🎯 Exam Tip: Remember that the sign of a number does not change when finding its multiplicative inverse. Only the reciprocal is taken.

Question 9. Multiply the rational number \( \frac{5}{7} \) to inverse of \( \frac{-7}{15} \)
Answer:
First, find the inverse (reciprocal) of \( \frac{-7}{15} \), which is \( \frac{15}{-7} \).
Now, multiply \( \frac{5}{7} \) by \( \frac{15}{-7} \):
\( \frac{5}{7} \times \frac{15}{-7} \)
\( = \frac{5 \times 15}{7 \times (-7)} \)
\( = \frac{75}{-49} \)
\( = -\frac{75}{49} \). This calculation demonstrates how to perform multiplication with rational numbers.
In simple words: First, flip the second fraction to find its inverse. Then, multiply the first fraction by this flipped fraction, multiplying the top numbers together and the bottom numbers together.

🎯 Exam Tip: Always simplify fractions before multiplying if possible to make calculations easier. Also, be careful with negative signs in your fractions.

Question 10. Fill in the blanks-
(i) Product of two rational number is always __(Rational/Integers)
(ii) Additive inverse of any negative rational number is _(positive/negative)
(iii) Inverse of zero is _(zero/In determined)
(iv) Additive identity of rational number is _(zero/one)
(v) Multiplicative identity of rational number is ___(zero/one)
(vi) Reciprocal of rational number is _of that. (inverse/same)
(vii) Negative rational number on number line is always lies on of zero (right/left)
(viii) Positive rational number on number line is always lies on of zero. (right/left)
(ix) When rational number is added with its additive inverse then result is always (zero/same)
(x) When a rational number is divided by same rational number then result is always (zero/one)
Answer:
(i) Product of two rational number is always Rational.
(ii) Additive inverse of any negative rational number is positive.
(iii) Inverse of zero is In determined.
(iv) Additive identity of rational number is zero.
(v) Multiplicative identity of rational number is one. This is because any number multiplied by 1 remains itself.
(vi) Reciprocal of rational number is inverse of that.
(vii) Negative rational number on number line is always lies on left of zero.
(viii) Positive rational number on number line is always lies on right of zero.
(ix) When rational number is added with its additive inverse then result is always zero.
(x) When a rational number is divided by same rational number then result is always one.
In simple words: Rational numbers behave in specific ways when multiplied or added. Zero acts as the additive identity, and one acts as the multiplicative identity. Knowing these rules helps us understand how numbers work.

🎯 Exam Tip: Understand the definitions of identity and inverse operations for both addition and multiplication, as these are fundamental concepts in number systems.

Question 11. By mean method-
(i) Write any five rational numbers between – 3 and 0.
(ii) Write any four rational numbers larger than 0 and smaller than \( \frac{5}{6} \)
(iii) Find any three rational numbers between \( \frac{-3}{4} \) and \( \frac{5}{6} \)
Answer:
(i) To find five rational numbers between \( -3 \) and \( 0 \) using the mean method:
1. The first rational number (mean of \( -3 \) and \( 0 \)): \( \frac{-3 + 0}{2} = \frac{-3}{2} \)
2. The second rational number (mean of \( -3 \) and \( \frac{-3}{2} \)): \( \frac{-3 + (\frac{-3}{2})}{2} = \frac{\frac{-6-3}{2}}{2} = \frac{-9}{4} \)
3. The third rational number (mean of \( \frac{-3}{2} \) and \( 0 \)): \( \frac{\frac{-3}{2} + 0}{2} = \frac{-3}{4} \)
4. The fourth rational number (mean of \( -3 \) and \( \frac{-9}{4} \)): \( \frac{-3 + (\frac{-9}{4})}{2} = \frac{\frac{-12-9}{4}}{2} = \frac{-21}{8} \)
5. The fifth rational number (mean of \( \frac{-9}{4} \) and \( \frac{-3}{2} \)): \( \frac{\frac{-9}{4} + (\frac{-3}{2})}{2} = \frac{\frac{-9-6}{4}}{2} = \frac{-15}{8} \)
So, five rational numbers between \( -3 \) and \( 0 \) are \( -\frac{21}{8}, -\frac{9}{4}, -\frac{15}{8}, -\frac{3}{2}, -\frac{3}{4} \). These numbers help visualize the density of rational numbers.
(ii) To find four rational numbers larger than \( 0 \) and smaller than \( \frac{5}{6} \) using the mean method:
1. The first rational number (mean of \( 0 \) and \( \frac{5}{6} \)): \( \frac{0 + \frac{5}{6}}{2} = \frac{5}{12} \)
2. The second rational number (mean of \( 0 \) and \( \frac{5}{12} \)): \( \frac{0 + \frac{5}{12}}{2} = \frac{5}{24} \)
3. The third rational number (mean of \( \frac{5}{12} \) and \( \frac{5}{6} \)): \( \frac{\frac{5}{12} + \frac{5}{6}}{2} = \frac{\frac{5+10}{12}}{2} = \frac{\frac{15}{12}}{2} = \frac{15}{24} = \frac{5}{8} \)
4. The fourth rational number (mean of \( \frac{5}{8} \) and \( \frac{5}{6} \)): \( \frac{\frac{5}{8} + \frac{5}{6}}{2} = \frac{\frac{15+20}{24}}{2} = \frac{\frac{35}{24}}{2} = \frac{35}{48} \)
So, four rational numbers between \( 0 \) and \( \frac{5}{6} \) are \( \frac{5}{24}, \frac{5}{12}, \frac{5}{8}, \frac{35}{48} \).
(iii) To find any three rational numbers between \( \frac{-3}{4} \) and \( \frac{5}{6} \) using the mean method:
1. The first rational number (mean of \( \frac{-3}{4} \) and \( \frac{5}{6} \)): \( \frac{\frac{-3}{4} + \frac{5}{6}}{2} = \frac{\frac{-9+10}{12}}{2} = \frac{\frac{1}{12}}{2} = \frac{1}{24} \)
2. The second rational number (mean of \( \frac{-3}{4} \) and \( \frac{1}{24} \)): \( \frac{\frac{-3}{4} + \frac{1}{24}}{2} = \frac{\frac{-18+1}{24}}{2} = \frac{\frac{-17}{24}}{2} = \frac{-17}{48} \)
3. The third rational number (mean of \( \frac{1}{24} \) and \( \frac{5}{6} \)): \( \frac{\frac{1}{24} + \frac{5}{6}}{2} = \frac{\frac{1+20}{24}}{2} = \frac{\frac{21}{24}}{2} = \frac{21}{48} = \frac{7}{16} \)
So, three rational numbers between \( \frac{-3}{4} \) and \( \frac{5}{6} \) are \( \frac{1}{24}, -\frac{17}{48}, \frac{7}{16} \). This method guarantees finding a rational number between any two given rational numbers.
In simple words: To find rational numbers between two others, you can find their average. This average will always be a new rational number that sits right in the middle of the first two. You can keep doing this over and over to find as many as you need.

🎯 Exam Tip: The mean method is one way to find rational numbers between two given numbers. Another common method is to convert both numbers to equivalent fractions with a larger common denominator, then pick integers between their numerators. Both methods are valid.