RBSE Solutions Class 8 Maths Chapter 1 Rational Numbers Exercise

Rajasthan Board RBSE Class 8 Maths Chapter 1 Rational Numbers In Text Exercise

 

Question 1. Find the value
(i) \( \frac { -11 }{ 7 } + \frac { 4 }{ 7 } \)
(ii) \( \frac { 3 }{ 5 } + ( \frac { -2 }{ 5 } ) \)
(iii) \( \frac { -3 }{ 4 } + ( \frac { -5 }{ 4 } ) \)
Answer:
(i) First, we add the two rational numbers. Since they have the same denominator, we add the numerators directly:
\( \frac { -11 }{ 7 } + \frac { 4 }{ 7 } = \frac { -11+4 }{ 7 } = \frac { -7 }{ 7 } = -1 \)
(ii) We add these rational numbers. The denominators are the same, so we combine the numerators:
\( \frac { 3 }{ 5 } + ( \frac { -2 }{ 5 } ) = \frac { 3 }{ 5 } - \frac { 2 }{ 5 } = \frac { 3-2 }{ 5 } = \frac { 1 }{ 5 } \)
(iii) For these rational numbers, the denominators are also the same. We add the numerators and keep the common denominator:
\( \frac { -3 }{ 4 } + ( \frac { -5 }{ 4 } ) = \frac { -3 }{ 4 } - \frac { 5 }{ 4 } = \frac { -3-5 }{ 4 } = \frac { -8 }{ 4 } = -2 \)
In simple words: When fractions have the same bottom number (denominator), you can just add or subtract the top numbers (numerators) and keep the bottom number the same. Always simplify the answer if you can.

🎯 Exam Tip: When dealing with fractions, always check if they have a common denominator first. If they do, the operation is straightforward; if not, find the LCM.

 

Question 2. Find the value
(i) \( \frac { 2 }{ 5 } + \frac { 1 }{ 6 } \)
(ii) \( \frac { 3 }{ 8 } + ( \frac { -5 }{ 2 } ) \)
(iii) \( \frac { -7 }{ 20 } + \frac { 7 }{ 3 } \)
(iv) \( \frac { -5 }{ 7 } + ( \frac { -2 }{ 4 } ) \)
Answer:
(i) We need to add \( \frac { 2 }{ 5 } \) and \( \frac { 1 }{ 6 } \). First, find the Least Common Multiple (LCM) of the denominators 5 and 6, which is 30. Then, convert each fraction to an equivalent fraction with a denominator of 30:
\( \frac { 2 }{ 5 } = \frac { 2 \times 6 }{ 5 \times 6 } = \frac { 12 }{ 30 } \)
\( \frac { 1 }{ 6 } = \frac { 1 \times 5 }{ 6 \times 5 } = \frac { 5 }{ 30 } \)
Now, add the equivalent fractions:
\( \frac { 12 }{ 30 } + \frac { 5 }{ 30 } = \frac { 12+5 }{ 30 } = \frac { 17 }{ 30 } \)
(ii) We need to add \( \frac { 3 }{ 8 } \) and \( \frac { -5 }{ 2 } \). The LCM of 8 and 2 is 8. Convert \( \frac { -5 }{ 2 } \) to an equivalent fraction with a denominator of 8:
\( \frac { -5 }{ 2 } = \frac { -5 \times 4 }{ 2 \times 4 } = \frac { -20 }{ 8 } \)
Now, add the fractions:
\( \frac { 3 }{ 8 } + \frac { -20 }{ 8 } = \frac { 3 + ( -20 ) }{ 8 } = \frac { -17 }{ 8 } \)
(iii) We need to add \( \frac { -7 }{ 20 } \) and \( \frac { 7 }{ 3 } \). The LCM of 20 and 3 is 60. Convert both fractions to equivalent fractions with a denominator of 60:
\( \frac { -7 }{ 20 } = \frac { -7 \times 3 }{ 20 \times 3 } = \frac { -21 }{ 60 } \)
\( \frac { 7 }{ 3 } = \frac { 7 \times 20 }{ 3 \times 20 } = \frac { 140 }{ 60 } \)
Now, add the fractions:
\( \frac { -21 }{ 60 } + \frac { 140 }{ 60 } = \frac { -21+140 }{ 60 } = \frac { 119 }{ 60 } \)
(iv) We need to add \( \frac { -5 }{ 7 } \) and \( \frac { -2 }{ 4 } \). The LCM of 7 and 4 is 28. Convert both fractions:
\( \frac { -5 }{ 7 } = \frac { -5 \times 4 }{ 7 \times 4 } = \frac { -20 }{ 28 } \)
\( \frac { -2 }{ 4 } = \frac { -2 \times 7 }{ 4 \times 7 } = \frac { -14 }{ 28 } \)
Now, add the fractions:
\( \frac { -20 }{ 28 } + \frac { -14 }{ 28 } = \frac { -20 + ( -14 ) }{ 28 } = \frac { -34 }{ 28 } \)
This fraction can be simplified by dividing both the numerator and the denominator by 2:
\( \frac { -34 }{ 28 } = \frac { -17 }{ 14 } \)
In simple words: When fractions have different bottom numbers, you must first find a common bottom number for both. You do this by finding the smallest number that both denominators can divide into (the LCM). Then, change both fractions to have this new common bottom number before adding or subtracting their top numbers.

🎯 Exam Tip: Always look for the Least Common Multiple (LCM) of denominators to make calculations simpler and to avoid dealing with larger numbers, and simplify the final answer if possible.

 

Question 3. Find the value
(i) \( \frac { 10 }{ 7 } - \frac { 4 }{ 7 } \)
(ii) \( \frac { -4 }{ 5 } - ( \frac { -2 }{ 5 } ) \)
(iii) \( \frac { 7 }{ 9 } - ( \frac { -4 }{ 9 } ) \)
Answer:
(i) To subtract \( \frac { 4 }{ 7 } \) from \( \frac { 10 }{ 7 } \), since they have the same denominator, we simply subtract the numerators:
\( \frac { 10 }{ 7 } - \frac { 4 }{ 7 } = \frac { 10-4 }{ 7 } = \frac { 6 }{ 7 } \)
(ii) To subtract \( ( \frac { -2 }{ 5 } ) \) from \( \frac { -4 }{ 5 } \), remember that subtracting a negative number is the same as adding a positive number:
\( \frac { -4 }{ 5 } - ( \frac { -2 }{ 5 } ) = \frac { -4 }{ 5 } + \frac { 2 }{ 5 } = \frac { -4+2 }{ 5 } = \frac { -2 }{ 5 } \)
(iii) To subtract \( ( \frac { -4 }{ 9 } ) \) from \( \frac { 7 }{ 9 } \), again, subtracting a negative number changes to adding a positive number:
\( \frac { 7 }{ 9 } - ( \frac { -4 }{ 9 } ) = \frac { 7 }{ 9 } + \frac { 4 }{ 9 } = \frac { 7+4 }{ 9 } = \frac { 11 }{ 9 } \)
In simple words: When fractions have the same bottom number and you need to subtract, just subtract the top numbers. If you subtract a negative number, it's like adding a positive one.

🎯 Exam Tip: Pay close attention to negative signs, especially when subtracting. "Minus a minus" always becomes a "plus".

 

Question 4. Find the value
(i) \( \frac { 4 }{ 3 } - \frac { 3 }{ 8 } \)
(ii) \( \frac { -3 }{ 7 } - \frac { 2 }{ 14 } \)
(iii) \( \frac { 5 }{ 9 } - ( \frac { -2 }{ 11 } ) \)
(iv) \( \frac { -2 }{ 9 } - \frac { 7 }{ 6 } \)
Answer:
(i) To subtract \( \frac { 3 }{ 8 } \) from \( \frac { 4 }{ 3 } \), we first find the LCM of 3 and 8, which is 24. Then, we convert the fractions:
\( \frac { 4 }{ 3 } = \frac { 4 \times 8 }{ 3 \times 8 } = \frac { 32 }{ 24 } \)
\( \frac { 3 }{ 8 } = \frac { 3 \times 3 }{ 8 \times 3 } = \frac { 9 }{ 24 } \)
Now, subtract the fractions:
\( \frac { 32 }{ 24 } - \frac { 9 }{ 24 } = \frac { 32-9 }{ 24 } = \frac { 23 }{ 24 } \)
(ii) To subtract \( \frac { 2 }{ 14 } \) from \( \frac { -3 }{ 7 } \), the LCM of 7 and 14 is 14. Convert \( \frac { -3 }{ 7 } \) to an equivalent fraction:
\( \frac { -3 }{ 7 } = \frac { -3 \times 2 }{ 7 \times 2 } = \frac { -6 }{ 14 } \)
Now, subtract the fractions:
\( \frac { -6 }{ 14 } - \frac { 2 }{ 14 } = \frac { -6-2 }{ 14 } = \frac { -8 }{ 14 } \)
This fraction can be simplified by dividing both numerator and denominator by 2:
\( \frac { -8 }{ 14 } = \frac { -4 }{ 7 } \)
(iii) To subtract \( ( \frac { -2 }{ 11 } ) \) from \( \frac { 5 }{ 9 } \), we change the subtraction of a negative to addition. The LCM of 9 and 11 is 99. Convert both fractions:
\( \frac { 5 }{ 9 } = \frac { 5 \times 11 }{ 9 \times 11 } = \frac { 55 }{ 99 } \)
\( \frac { -2 }{ 11 } = \frac { -2 \times 9 }{ 11 \times 9 } = \frac { -18 }{ 99 } \)
Now, add the fractions:
\( \frac { 55 }{ 99 } - ( \frac { -18 }{ 99 } ) = \frac { 55 }{ 99 } + \frac { 18 }{ 99 } = \frac { 55+18 }{ 99 } = \frac { 73 }{ 99 } \)
(iv) To subtract \( \frac { 7 }{ 6 } \) from \( \frac { -2 }{ 9 } \), the LCM of 9 and 6 is 18. Convert both fractions:
\( \frac { -2 }{ 9 } = \frac { -2 \times 2 }{ 9 \times 2 } = \frac { -4 }{ 18 } \)
\( \frac { 7 }{ 6 } = \frac { 7 \times 3 }{ 6 \times 3 } = \frac { 21 }{ 18 } \)
Now, subtract the fractions:
\( \frac { -4 }{ 18 } - \frac { 21 }{ 18 } = \frac { -4-21 }{ 18 } = \frac { -25 }{ 18 } \)
In simple words: When subtracting fractions with different bottom numbers, first find the smallest common bottom number (LCM). Then change both fractions to use this common number before doing the subtraction. Remember, subtracting a negative is like adding a positive.

🎯 Exam Tip: Always be careful when calculating the LCM and converting fractions. A common mistake is forgetting to multiply the numerator when changing the denominator.

 

Question 5. Find the value
(i) \( 4 \times ( \frac { -1 }{ 3 } ) \)
(ii) \( ( \frac { -3 }{ 5 } ) \times 7 \)
(iii) \( ( \frac { -4 }{ 5 } ) \times ( -3 ) \)
(iv) \( ( \frac { -3 }{ 7 } ) \times \frac { 2 }{ 5 } \)
(v) \( \frac { 2 }{ 3 } \times ( \frac { -1 }{ 4 } ) \)
(vi) \( ( \frac { -3 }{ 2 } ) \times ( \frac { -9 }{ 7 } ) \)
Answer:
(i) To multiply a whole number by a fraction, multiply the whole number by the numerator and keep the denominator:
\( 4 \times ( \frac { -1 }{ 3 } ) = \frac { 4 \times ( -1 ) }{ 3 } = \frac { -4 }{ 3 } \)
(ii) Similarly, multiply the whole number by the numerator:
\( ( \frac { -3 }{ 5 } ) \times 7 = \frac { -3 \times 7 }{ 5 } = \frac { -21 }{ 5 } \)
(iii) When multiplying two negative numbers, the result is positive. Multiply the numerators and keep the denominator:
\( ( \frac { -4 }{ 5 } ) \times ( -3 ) = \frac { ( -4 ) \times ( -3 ) }{ 5 } = \frac { 12 }{ 5 } \)
(iv) To multiply two fractions, multiply the numerators together and multiply the denominators together:
\( ( \frac { -3 }{ 7 } ) \times \frac { 2 }{ 5 } = \frac { ( -3 ) \times 2 }{ 7 \times 5 } = \frac { -6 }{ 35 } \)
(v) Multiply the numerators and denominators. Then, simplify the result if possible:
\( \frac { 2 }{ 3 } \times ( \frac { -1 }{ 4 } ) = \frac { 2 \times ( -1 ) }{ 3 \times 4 } = \frac { -2 }{ 12 } \)
This simplifies to \( \frac { -1 }{ 6 } \) by dividing both by 2.
(vi) Multiply the numerators and denominators. Since both fractions are negative, the product will be positive:
\( ( \frac { -3 }{ 2 } ) \times ( \frac { -9 }{ 7 } ) = \frac { ( -3 ) \times ( -9 ) }{ 2 \times 7 } = \frac { 27 }{ 14 } \)
In simple words: When you multiply fractions, you multiply the top numbers together and the bottom numbers together. If you multiply numbers with different signs (one positive, one negative), the answer is negative. If you multiply numbers with the same sign (both positive or both negative), the answer is positive. Always look to simplify your final fraction.

🎯 Exam Tip: Remember the rules for multiplying positive and negative numbers: same signs give positive, different signs give negative. Always simplify fractions before and after multiplying if possible.

 

Question 6. Find the value
(i) \( \frac { -7 }{ 2 } \div 4 \)
(ii) \( \frac { -12 }{ 7 } \div \frac { 3 }{ 4 } \)
(iii) \( \frac { 5 }{ 9 } \div ( \frac { -4 }{ 5 } ) \)
(iv) \( \frac { -7 }{ 11 } \div \frac { -7 }{ 11 } \)
Answer:
(i) Dividing by a whole number is the same as multiplying by its reciprocal. The reciprocal of 4 is \( \frac { 1 }{ 4 } \):
\( \frac { -7 }{ 2 } \div 4 = \frac { -7 }{ 2 } \times \frac { 1 }{ 4 } = \frac { ( -7 ) \times 1 }{ 2 \times 4 } = \frac { -7 }{ 8 } \)
(ii) Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of \( \frac { 3 }{ 4 } \) is \( \frac { 4 }{ 3 } \):
\( \frac { -12 }{ 7 } \div \frac { 3 }{ 4 } = \frac { -12 }{ 7 } \times \frac { 4 }{ 3 } \)
Now, multiply the numerators and denominators:
\( \frac { ( -12 ) \times 4 }{ 7 \times 3 } = \frac { -48 }{ 21 } \)
This fraction can be simplified by dividing both by 3:
\( \frac { -48 }{ 21 } = \frac { -16 }{ 7 } \)
(iii) Divide \( \frac { 5 }{ 9 } \) by \( ( \frac { -4 }{ 5 } ) \). First, find the reciprocal of \( ( \frac { -4 }{ 5 } ) \), which is \( ( \frac { 5 }{ -4 } ) \). Then multiply:
\( \frac { 5 }{ 9 } \div ( \frac { -4 }{ 5 } ) = \frac { 5 }{ 9 } \times ( \frac { 5 }{ -4 } ) \)
Multiply the numerators and denominators:
\( \frac { 5 \times 5 }{ 9 \times ( -4 ) } = \frac { 25 }{ -36 } = \frac { -25 }{ 36 } \)
(iv) Any non-zero number divided by itself is 1. Here, \( \frac { -7 }{ 11 } \) is divided by \( \frac { -7 }{ 11 } \):
\( \frac { -7 }{ 11 } \div \frac { -7 }{ 11 } = 1 \)
We can also think of this as multiplying by the reciprocal: \( \frac { -7 }{ 11 } \times \frac { 11 }{ -7 } = \frac { ( -7 ) \times 11 }{ 11 \times ( -7 ) } = \frac { -77 }{ -77 } = 1 \)
In simple words: To divide by a fraction, you just flip the second fraction (find its reciprocal) and then multiply it by the first fraction. Remember that dividing a number by itself always gives 1.

🎯 Exam Tip: When dividing fractions, always convert the operation to multiplication by the reciprocal of the divisor. Be careful with negative signs throughout the process.

 

Question 8. Fill in the blanks
Answer:

This table shows if a set of numbers is "closed" under different math operations. Being closed means that if you take any two numbers from the set and perform the operation, the answer will also be in that same set. For example, adding two natural numbers always gives a natural number, so natural numbers are closed under addition.
In simple words: A set of numbers is "closed" for an operation if doing that math problem with numbers from the set always gives an answer that is also in that set. For instance, if you add any two whole numbers, you always get another whole number, so whole numbers are closed for addition.

🎯 Exam Tip: Understand the concept of "closure property" for number systems. It helps classify different types of numbers and their behavior under basic arithmetic operations.

 

Question 9. Fill in the blanks
Answer:

This table shows if different sets of numbers are "commutative" under various operations. An operation is commutative if changing the order of the numbers does not change the result (e.g., a + b = b + a). This property is important for simplifying calculations.
In simple words: An operation is "commutative" if you can swap the numbers around and still get the same answer. For example, 2 + 3 is the same as 3 + 2, so addition is commutative. But 5 - 2 is not the same as 2 - 5, so subtraction is not commutative.

🎯 Exam Tip: Remember the commutative property: A + B = B + A and A × B = B × A. Subtraction and division are generally not commutative.

 

Question 10. Are the addition on both sides the same?
(i) \( \frac { -3 }{ 5 } + ( \frac { 2 }{ 3 } + \frac { 4 }{ 7 } ) = ( \frac { -3 }{ 5 } + \frac { 2 }{ 3 } ) + \frac { 4 }{ 7 } \)
(ii) \( \frac { 1 }{ 2 } + ( \frac { -3 }{ 4 } + \frac { -5 }{ 8 } ) = ( \frac { 1 }{ 2 } + \frac { -3 }{ 4 } ) + \frac { -5 }{ 8 } \)
Answer:
(i) We need to check if the associative property holds for the given expression.
First, calculate the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \frac { -3 }{ 5 } + ( \frac { 2 }{ 3 } + \frac { 4 }{ 7 } ) \)
First, calculate inside the parenthesis: \( \frac { 2 }{ 3 } + \frac { 4 }{ 7 } \). The LCM of 3 and 7 is 21.
\( \frac { 2 }{ 3 } + \frac { 4 }{ 7 } = \frac { 2 \times 7 }{ 3 \times 7 } + \frac { 4 \times 3 }{ 7 \times 3 } = \frac { 14 }{ 21 } + \frac { 12 }{ 21 } = \frac { 14+12 }{ 21 } = \frac { 26 }{ 21 } \)
Now substitute this back:
\( \text{L.H.S.} = \frac { -3 }{ 5 } + \frac { 26 }{ 21 } \)
The LCM of 5 and 21 is 105.
\( \frac { -3 }{ 5 } + \frac { 26 }{ 21 } = \frac { -3 \times 21 }{ 5 \times 21 } + \frac { 26 \times 5 }{ 21 \times 5 } = \frac { -63 }{ 105 } + \frac { 130 }{ 105 } = \frac { -63+130 }{ 105 } = \frac { 67 }{ 105 } \)
So, \( \text{L.H.S.} = \frac { 67 }{ 105 } \)
Next, calculate the Right Hand Side (R.H.S.):
\( \text{R.H.S.} = ( \frac { -3 }{ 5 } + \frac { 2 }{ 3 } ) + \frac { 4 }{ 7 } \)
First, calculate inside the parenthesis: \( ( \frac { -3 }{ 5 } + \frac { 2 }{ 3 } ) \). The LCM of 5 and 3 is 15.
\( \frac { -3 }{ 5 } + \frac { 2 }{ 3 } = \frac { -3 \times 3 }{ 5 \times 3 } + \frac { 2 \times 5 }{ 3 \times 5 } = \frac { -9 }{ 15 } + \frac { 10 }{ 15 } = \frac { -9+10 }{ 15 } = \frac { 1 }{ 15 } \)
Now substitute this back:
\( \text{R.H.S.} = \frac { 1 }{ 15 } + \frac { 4 }{ 7 } \)
The LCM of 15 and 7 is 105.
\( \frac { 1 }{ 15 } + \frac { 4 }{ 7 } = \frac { 1 \times 7 }{ 15 \times 7 } + \frac { 4 \times 15 }{ 7 \times 15 } = \frac { 7 }{ 105 } + \frac { 60 }{ 105 } = \frac { 7+60 }{ 105 } = \frac { 67 }{ 105 } \)
So, \( \text{R.H.S.} = \frac { 67 }{ 105 } \)
Since \( \text{L.H.S.} = \text{R.H.S.} \), the addition on both sides is the same.

(ii) We need to check if the associative property holds for the second expression.
First, calculate the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \frac { 1 }{ 2 } + ( \frac { -3 }{ 4 } + \frac { -5 }{ 8 } ) \)
First, calculate inside the parenthesis: \( ( \frac { -3 }{ 4 } + \frac { -5 }{ 8 } ) \). The LCM of 4 and 8 is 8.
\( \frac { -3 }{ 4 } + \frac { -5 }{ 8 } = \frac { -3 \times 2 }{ 4 \times 2 } + \frac { -5 }{ 8 } = \frac { -6 }{ 8 } + \frac { -5 }{ 8 } = \frac { -6+(-5) }{ 8 } = \frac { -11 }{ 8 } \)
Now substitute this back:
\( \text{L.H.S.} = \frac { 1 }{ 2 } + \frac { -11 }{ 8 } \)
The LCM of 2 and 8 is 8.
\( \frac { 1 }{ 2 } + \frac { -11 }{ 8 } = \frac { 1 \times 4 }{ 2 \times 4 } + \frac { -11 }{ 8 } = \frac { 4 }{ 8 } + \frac { -11 }{ 8 } = \frac { 4-11 }{ 8 } = \frac { -7 }{ 8 } \)
So, \( \text{L.H.S.} = \frac { -7 }{ 8 } \)
Next, calculate the Right Hand Side (R.H.S.):
\( \text{R.H.S.} = ( \frac { 1 }{ 2 } + \frac { -3 }{ 4 } ) + \frac { -5 }{ 8 } \)
First, calculate inside the parenthesis: \( ( \frac { 1 }{ 2 } + \frac { -3 }{ 4 } ) \). The LCM of 2 and 4 is 4.
\( \frac { 1 }{ 2 } + \frac { -3 }{ 4 } = \frac { 1 \times 2 }{ 2 \times 2 } + \frac { -3 }{ 4 } = \frac { 2 }{ 4 } + \frac { -3 }{ 4 } = \frac { 2-3 }{ 4 } = \frac { -1 }{ 4 } \)
Now substitute this back:
\( \text{R.H.S.} = \frac { -1 }{ 4 } + \frac { -5 }{ 8 } \)
The LCM of 4 and 8 is 8.
\( \frac { -1 }{ 4 } + \frac { -5 }{ 8 } = \frac { -1 \times 2 }{ 4 \times 2 } + \frac { -5 }{ 8 } = \frac { -2 }{ 8 } + \frac { -5 }{ 8 } = \frac { -2+(-5) }{ 8 } = \frac { -7 }{ 8 } \)
So, \( \text{R.H.S.} = \frac { -7 }{ 8 } \)
Since \( \text{L.H.S.} = \text{R.H.S.} \), the addition on both sides is the same.
Yes, the total of both sides is equal.
In simple words: This question checks if the way we group numbers when adding changes the answer. For rational numbers, the answer stays the same even if we group them differently for addition. This is called the associative property.

🎯 Exam Tip: The associative property (a + (b + c) = (a + b) + c) holds true for addition and multiplication of rational numbers, but not for subtraction or division. This allows flexibility in calculation order.

 

Question 11. Verify:
(i) \( \frac{-4}{3} \times \left[ \left( \frac{-2}{5} \right) \times \frac{1}{7} \right] = \left[ \frac{-4}{3} \times \left( \frac{-2}{5} \right) \right] \times \frac{1}{7} \)
(ii) \( \left( \frac{-3}{5} \right) \times \left[ \frac{4}{11} \times \left( \frac{-2}{22} \right) \right] = \left[ \left( \frac{-3}{5} \right) \times \frac{4}{11} \right] \times \left( \frac{-2}{22} \right) \)
Answer:
(i) We need to check if the associative property of multiplication holds true for these rational numbers.
LHS: \( \frac{-4}{3} \times \left[ \left( \frac{-2}{5} \right) \times \frac{1}{7} \right] \)
\( = \frac{-4}{3} \times \left[ \frac{-2 \times 1}{5 \times 7} \right] \)
\( = \frac{-4}{3} \times \left[ \frac{-2}{35} \right] \)
\( = \frac{(-4) \times (-2)}{3 \times 35} \)
\( = \frac{8}{105} \)
RHS: \( \left[ \frac{-4}{3} \times \left( \frac{-2}{5} \right) \right] \times \frac{1}{7} \)
\( = \left[ \frac{(-4) \times (-2)}{3 \times 5} \right] \times \frac{1}{7} \)
\( = \left[ \frac{8}{15} \right] \times \frac{1}{7} \)
\( = \frac{8 \times 1}{15 \times 7} \)
\( = \frac{8}{105} \)
Since LHS = RHS, the statement is verified.

(ii) We need to verify the associative property of multiplication for the given rational numbers.
LHS: \( \left( \frac{-3}{5} \right) \times \left[ \frac{4}{11} \times \left( \frac{-2}{22} \right) \right] \)
\( = \left( \frac{-3}{5} \right) \times \left[ \frac{4 \times (-2)}{11 \times 22} \right] \)
\( = \left( \frac{-3}{5} \right) \times \left[ \frac{-8}{242} \right] \)
\( = \frac{(-3) \times (-8)}{5 \times 242} \)
\( = \frac{24}{1210} \)
\( = \frac{12}{605} \)
RHS: \( \left[ \left( \frac{-3}{5} \right) \times \frac{4}{11} \right] \times \left( \frac{-2}{22} \right) \)
\( = \left[ \frac{(-3) \times 4}{5 \times 11} \right] \times \left( \frac{-2}{22} \right) \)
\( = \left[ \frac{-12}{55} \right] \times \left( \frac{-2}{22} \right) \)
\( = \frac{(-12) \times (-2)}{55 \times 22} \)
\( = \frac{24}{1210} \)
\( = \frac{12}{605} \)
Since LHS = RHS, the statement is verified. The associative property states that how numbers are grouped in multiplication does not change the product.
In simple words: For both parts, we multiply the numbers on the left side and the right side of the equals sign. We found that the answers are the same. This shows that the way we group rational numbers when multiplying does not change the final result.

🎯 Exam Tip: When verifying properties, always calculate both the Left Hand Side (LHS) and Right Hand Side (RHS) separately and simplify completely to show they are equal.

 

Question 12. Fill in the blanks

Answer: We need to complete the table by checking if each set of numbers is closed under the given operations. "Closed" means that performing the operation on any two numbers from the set will always result in a number that is also in that set. For example, adding two natural numbers always gives a natural number.In simple words: The table shows if a group of numbers (like natural numbers) stays within that group after you add, subtract, multiply, or divide them. For example, if you add two natural numbers, you always get another natural number (Yes). But if you subtract them, you might get a negative number, which is not natural (No).

🎯 Exam Tip: Remember specific counterexamples for operations that are not closed (e.g., \( 2-3 = -1 \) for natural numbers and whole numbers, \( 2 \div 3 = \frac{2}{3} \) for integers).

 

Question 13. Fill in the blanks
(i) \( 3 + \boxed{\phantom{0}} = 3 \)
(ii) \( \boxed{\phantom{0}} + 0 = -7 \)
(iii) \( \frac{-4}{9} + \boxed{\phantom{0}} = \frac{-4}{9} \)
(iv) \( 0 + \frac{9}{13} = \frac{9}{13} \)
(v) \( \frac{-5}{11} + 0 = \frac{-5}{11} \)
Answer: We need to find the numbers that, when added, result in the same original number. This involves the additive identity, which is zero.
(i) \( 3 + 0 = 3 \)
(ii) \( -7 + 0 = -7 \)
(iii) \( \frac{-4}{9} + 0 = \frac{-4}{9} \)
(iv) \( 0 + \frac{9}{13} = \frac{9}{13} \)
(v) \( \frac{-5}{11} + 0 = \frac{-5}{11} \)
This property shows that adding zero to any number does not change the number's value.
In simple words: We fill in the empty boxes with the number zero. Zero is special because when you add it to any number, the number stays the same.

🎯 Exam Tip: Zero is the additive identity; adding zero to any number, whether positive, negative, or a fraction, will always return the original number.

 

Question 14. Fill in the blanks
(i) \( \frac{8}{5} \times \boxed{\phantom{0}} = 8 \)
(ii) \( \boxed{\phantom{0}} \times (-5) = -5 \)
(iii) \( \frac{2}{3} \times \boxed{\phantom{0}} = \frac{2}{3} \)
(iv) \( \boxed{\phantom{0}} \times \frac{-4}{7} = \frac{-4}{7} \)
Answer: We need to fill the blanks with the number that, when multiplied, keeps the original number unchanged. This is the multiplicative identity, which is one.
(i) \( \frac{8}{5} \times 1 = \frac{8}{5} \) and \( 1 \times 8 = 8 \)
(ii) \( 1 \times (-5) = -5 \)
(iii) \( \frac{2}{3} \times 1 = \frac{2}{3} \)
(iv) \( 1 \times \frac{-4}{7} = \frac{-4}{7} \)
The blank spaces should all be filled with 1, as multiplying any number by 1 yields the same number.
In simple words: For each problem, we need to put the number '1' in the empty box. This is because multiplying any number by 1 always gives you the same number back.

🎯 Exam Tip: The number 1 is the multiplicative identity; multiplying any number by 1 will always give the original number.

 

Question 15. Fill in the blanks
(i) \( 2 + \boxed{\phantom{0}} = 0 \)
(ii) \( \boxed{\phantom{0}} + 2 = 0 \)
(iii) \( -3 + \boxed{\phantom{0}} = 0 \)
(iv) \( \boxed{\phantom{0}} + (-3) = 0 \)
(v) \( \frac{3}{4} + \boxed{\phantom{0}} = 0 \)
(vi) \( \boxed{\phantom{0}} + \frac{3}{4} = 0 \)
(vii) \( \frac{-5}{7} + \boxed{\phantom{0}} = 0 \)
(viii) \( \boxed{\phantom{0}} + \left( \frac{-5}{7} \right) = 0 \)
Answer: We need to find the number that, when added to the given number, results in zero. This is the additive inverse.
(i) \( 2 + (-2) = 0 \)
(ii) \( -2 + 2 = 0 \)
(iii) \( -3 + 3 = 0 \)
(iv) \( 3 + (-3) = 0 \)
(v) \( \frac{3}{4} + \left( \frac{-3}{4} \right) = 0 \)
(vi) \( \frac{-3}{4} + \frac{3}{4} = 0 \)
(vii) \( \frac{-5}{7} + \frac{5}{7} = 0 \)
(viii) \( \frac{5}{7} + \left( \frac{-5}{7} \right) = 0 \)
The additive inverse of a number is the number that has the same absolute value but the opposite sign.
In simple words: We need to put the opposite number in each blank space. If the number is positive, we use the same number but negative. If the number is negative, we use the same number but positive. This makes the total sum zero.

🎯 Exam Tip: The additive inverse of any number 'a' is '-a', because \( a + (-a) = 0 \). For fractions, simply change the sign of the fraction.

 

Question 16. Write the additive Inverse of the following rational numbers-
(i) \( 4 \)
(ii) \( \frac{-1}{5} \)
(iii) \( \frac{7}{3} \)
(iv) \( \frac{-3}{2} \)
(v) \( \frac{9}{2} \)
Answer: The additive inverse of a number is the number that, when added to the original number, gives a sum of zero. We find this by changing the sign of the given number.
(i) The additive inverse of \( 4 \) is \( -4 \).
(ii) The additive inverse of \( \frac{-1}{5} \) is \( \frac{1}{5} \).
(iii) The additive inverse of \( \frac{7}{3} \) is \( \frac{-7}{3} \).
(iv) The additive inverse of \( \frac{-3}{2} \) is \( \frac{3}{2} \).
(v) The additive inverse of \( \frac{9}{2} \) is \( \frac{-9}{2} \).
This concept is crucial for solving equations and working with rational numbers effectively.
In simple words: To find the additive inverse, just flip the sign of the number. If it's positive, make it negative. If it's negative, make it positive.

🎯 Exam Tip: Always remember that the sum of a number and its additive inverse is always zero. This is a fundamental concept in number systems.

 

Question 17. Fill in the blanks
(i) \( 5 \times \boxed{\phantom{0}} = 1 \)
(ii) \( \boxed{\phantom{0}} \times 5 = 1 \)
(iii) \( 7 \times \boxed{\phantom{0}} = 1 \)
(iv) \( \boxed{\phantom{0}} \times (-7) = 1 \)
(v) \( \frac{2}{3} \times \boxed{\phantom{0}} = 1 \)
(vi) \( \boxed{\phantom{0}} \times \frac{2}{3} = 1 \)
(vii) \( \frac{-2}{3} \times \boxed{\phantom{0}} = 1 \)
(viii) \( \boxed{\phantom{0}} \times \left( \frac{-2}{3} \right) = 1 \)
Answer: We need to fill the blanks with the number that, when multiplied by the given number, results in 1. This is the multiplicative inverse (or reciprocal).
(i) \( 5 \times \frac{1}{5} = 1 \)
(ii) \( \frac{1}{5} \times 5 = 1 \)
(iii) \( 7 \times \frac{1}{7} = 1 \)
(iv) \( \frac{-1}{7} \times (-7) = 1 \)
(v) \( \frac{2}{3} \times \frac{3}{2} = 1 \)
(vi) \( \frac{3}{2} \times \frac{2}{3} = 1 \)
(vii) \( \frac{-2}{3} \times \frac{-3}{2} = 1 \)
(viii) \( \frac{-3}{2} \times \left( \frac{-2}{3} \right) = 1 \)
The multiplicative inverse of a number 'a' is '1/a', as their product is always 1.
In simple words: To fill the blanks, we need to find the reciprocal of each number. This means flipping the fraction (top number becomes bottom, bottom becomes top) or putting 1 over the number. If the number is negative, its reciprocal should also be negative to make the product 1.

🎯 Exam Tip: The multiplicative inverse (reciprocal) of any non-zero number 'a' is \( \frac{1}{a} \), such that \( a \times \frac{1}{a} = 1 \). Be careful with the signs; if the original number is negative, its reciprocal must also be negative.

 

Question 18. Write the multiplicative Inverse of the following rational numbers: \( 3, \frac{1}{3}, \frac{-3}{7}, \frac{2}{3}, \frac{-5}{6} \)
Answer: The multiplicative inverse (or reciprocal) of a number is obtained by flipping the fraction (numerator becomes denominator and vice versa). For whole numbers, we consider them as a fraction over 1.
The multiplicative inverse of \( 3 \) is \( \frac{1}{3} \).
The multiplicative inverse of \( \frac{1}{3} \) is \( 3 \).
The multiplicative inverse of \( \frac{-3}{7} \) is \( \frac{7}{-3} \).
The multiplicative inverse of \( \frac{2}{3} \) is \( \frac{3}{2} \).
The multiplicative inverse of \( \frac{-5}{6} \) is \( \frac{6}{-5} \).
This operation is helpful in solving division problems by converting them into multiplication with the reciprocal.
In simple words: To find the multiplicative inverse, simply flip the fraction. For a whole number, put 1 over it. Make sure the sign stays the same so the product is positive 1.

🎯 Exam Tip: Remember that the multiplicative inverse of a positive number is positive, and the multiplicative inverse of a negative number is negative.

 

Question 20. Find the rational number between \( -1 \) and \( 2 \).
Answer: To find a rational number between two numbers, we can find their average. The average of two numbers \( a \) and \( b \) is \( \frac{a+b}{2} \).
Here, \( a = -1 \) and \( b = 2 \).
Rational number \( = \frac{-1+2}{2} = \frac{1}{2} \).
So, \( -1 < \frac{1}{2} < 2 \). We can always find an infinite number of rational numbers between any two distinct rational numbers.
In simple words: To find a number exactly in the middle of -1 and 2, we add them up and divide by 2. The answer is \( \frac{1}{2} \).

🎯 Exam Tip: The simplest way to find *one* rational number between two given rational numbers is to calculate their average. If more numbers are needed, repeat the process with the new numbers.

 

Question 21. Find the rational number between \( \frac{2}{3} \) and \( \frac{3}{4} \).
Answer: To find a rational number between \( \frac{2}{3} \) and \( \frac{3}{4} \), we can find their average.
Rational number \( = \frac{\frac{2}{3} + \frac{3}{4}}{2} \)
First, add the fractions in the numerator:
\( \frac{2}{3} + \frac{3}{4} = \frac{2 \times 4}{3 \times 4} + \frac{3 \times 3}{4 \times 3} = \frac{8}{12} + \frac{9}{12} = \frac{8+9}{12} = \frac{17}{12} \)
Now, divide this sum by 2:
\( \frac{17}{12} \div 2 = \frac{17}{12} \times \frac{1}{2} = \frac{17 \times 1}{12 \times 2} = \frac{17}{24} \)
So, \( \frac{2}{3} < \frac{17}{24} < \frac{3}{4} \). Converting to common denominator for comparison: \( \frac{2}{3} = \frac{16}{24} \) and \( \frac{3}{4} = \frac{18}{24} \). So, \( \frac{16}{24} < \frac{17}{24} < \frac{18}{24} \).
In simple words: To find a rational number in between \( \frac{2}{3} \) and \( \frac{3}{4} \), we add them together and then divide by two. First, we make them have the same bottom number (12), add them to get \( \frac{17}{12} \), and then cut that in half to get \( \frac{17}{24} \).

🎯 Exam Tip: When finding a rational number between two fractions, ensure you find a common denominator before adding them. Then, divide the sum by 2. This is a reliable method for finding one intermediate rational number.

 

Question 22. Find the rational number between \( 2 \) and \( 3 \).
Answer: We will find a few rational numbers between 2 and 3 by repeatedly taking averages. This process shows how to generate multiple rational numbers.
First rational number between 2 and 3:
\( \frac{2+3}{2} = \frac{5}{2} \)
So, \( 2 < \frac{5}{2} < 3 \) ...(i)
Next, find a rational number between 2 and \( \frac{5}{2} \):
\( \frac{2 + \frac{5}{2}}{2} = \frac{\frac{4}{2} + \frac{5}{2}}{2} = \frac{\frac{4+5}{2}}{2} = \frac{\frac{9}{2}}{2} = \frac{9}{2 \times 2} = \frac{9}{4} \)
So, \( 2 < \frac{9}{4} < \frac{5}{2} \) ...(ii)
Next, find a rational number between \( \frac{5}{2} \) and 3:
\( \frac{\frac{5}{2} + 3}{2} = \frac{\frac{5}{2} + \frac{6}{2}}{2} = \frac{\frac{5+6}{2}}{2} = \frac{\frac{11}{2}}{2} = \frac{11}{2 \times 2} = \frac{11}{4} \)
So, \( \frac{5}{2} < \frac{11}{4} < 3 \) ...(iii)
From (i), (ii), and (iii), three rational numbers between 2 and 3 are \( \frac{9}{4}, \frac{5}{2}, \frac{11}{4} \).
In simple words: We want to find numbers between 2 and 3. We can find the middle number by adding 2 and 3 and dividing by 2, which gives us \( \frac{5}{2} \). Then, we can find numbers between 2 and \( \frac{5}{2} \), and between \( \frac{5}{2} \) and 3 using the same method. This gives us numbers like \( \frac{9}{4} \), \( \frac{5}{2} \), and \( \frac{11}{4} \).

🎯 Exam Tip: To find multiple rational numbers between two given numbers, repeatedly use the averaging method. Always simplify fractions to their lowest terms for clarity.