CBSE Class 7 Mathematics Rational Numbers Worksheet Set 05

Question 1. The product of a number and its multiplicative inverse is _________
Answer: 1
In simple words: When you multiply any number by its flipped version, the answer is always 1.

Exam Tip: Remember that zero is the only rational number that does not have a multiplicative inverse.

Question 2. Sum of a number and its negative is _________
Answer: 0
In simple words: If you add a number to its opposite sign version, they cancel each other out to make 0.

Exam Tip: This property shows that the negative of a rational number is also its additive inverse.

Question 3. _________ is the multiplicative identity for rational numbers.
Answer: 1
In simple words: Multiplying any number by 1 does not change its value.

Exam Tip: Do not confuse this with the additive identity, which is 0.

Question 4. _________ is the additive identity for rational numbers.
Answer: 0
In simple words: Adding 0 to any number keeps the number exactly the same.

Exam Tip: This means for any rational number \( a \), \( a + 0 = a \).

Question 5. The reciprocal of a negative rational number is _________
Answer: negative
In simple words: Flipping a negative fraction upside down keeps the fraction negative.

Exam Tip: Taking the reciprocal of a rational number changes its numerator and denominator but never changes its sign.

Question 6. There are _________ number of rational numbers between any two rational numbers.
Answer: infinite (or countlessly many)
In simple words: You can find endless fractions between any two fractions.

Exam Tip: Unlike integers, there is no next rational number since you can always find another one in between.

Question 7. Additive inverse of 0 is _________
Answer: 0
In simple words: The opposite of 0 is just 0.

Exam Tip: Zero is the only rational number that is its own opposite.

Question 8. Rational Number _________ has no reciprocal.
Answer: 0
In simple words: You cannot flip 0 upside down because dividing by 0 is not allowed.

Exam Tip: Make sure to explain that \( \frac{1}{0} \) is not defined in mathematics.

Question 9. The numbers _________ and _________ are their own reciprocals.
Answer: 1 and -1
In simple words: Flipping 1 or -1 upside down gives the exact same number back.

Exam Tip: These are the only two rational numbers that are equal to their own multiplicative inverses.

Question 10. The rational number that is equal to its negative is _________
Answer: 0
In simple words: Zero is the only number that is the same as its negative.

Exam Tip: Since zero is neither positive nor negative, it is equal to its own opposite.

Question 11. Using appropriate properties find:
(a) \( -\frac{2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6} \)
(b) \( \frac{1}{2} \times \frac{1}{4} + \left(-\frac{7}{18}\right) \times \frac{15}{7} - \frac{1}{4} \times \frac{1}{3} \)
(c) \( -\frac{5}{4} \times \left( \frac{8}{5} + 16 \right) \)
(d) \( \frac{7}{12} \times \left( -\frac{3}{5} \right) + \left( -\frac{5}{3} \right) \times \frac{7}{12} \)
(e) \( \frac{2}{21} \times \frac{-3}{13} + \frac{-7}{9} - \frac{2}{21} \times \frac{10}{13} \)
(f) \( -\frac{4}{5} \times \frac{3}{7} \times \frac{15}{16} \times \left( \frac{-14}{9} \right) \)
Answer:
(a) For the standard school worksheet question (based on NCERT Exercise 1.1 Q1):
\( -\frac{2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6} \)
Using commutative property to group the terms with \( \frac{3}{5} \):
\( = -\frac{2}{3} \times \frac{3}{5} - \frac{3}{5} \times \frac{1}{6} + \frac{5}{2} \)
Using distributive property to factor out \( -\frac{3}{5} \):
\( = -\frac{3}{5} \times \left( \frac{2}{3} + \frac{1}{6} \right) + \frac{5}{2} \)
\( = -\frac{3}{5} \times \left( \frac{4 + 1}{6} \right) + \frac{5}{2} \)
\( = -\frac{3}{5} \times \frac{5}{6} + \frac{5}{2} \)
\( = -\frac{1}{2} + \frac{5}{2} \)
\( = \frac{4}{2} = 2 \)

For the literal handwritten values on the sheet:
\( -\frac{2}{3} \times \frac{2}{5} + \frac{7}{2} - \frac{2}{5} \times \frac{1}{6} \)
Using commutative property to group the terms with \( \frac{2}{5} \):
\( = -\frac{2}{3} \times \frac{2}{5} - \frac{2}{5} \times \frac{1}{6} + \frac{7}{2} \)
Using distributive property to factor out \( -\frac{2}{5} \):
\( = -\frac{2}{5} \times \left( \frac{2}{3} + \frac{1}{6} \right) + \frac{7}{2} \)
\( = -\frac{2}{5} \times \left( \frac{4 + 1}{6} \right) + \frac{7}{2} \)
\( = -\frac{2}{5} \times \frac{5}{6} + \frac{7}{2} \)
\( = -\frac{2}{6} + \frac{7}{2} \)
\( = -\frac{1}{3} + \frac{7}{2} \)
\( = \frac{-2 + 21}{6} = \frac{19}{6} \)

(b) \( \frac{1}{2} \times \frac{1}{4} + \left(-\frac{7}{18}\right) \times \frac{15}{7} - \frac{1}{4} \times \frac{1}{3} \)
Using commutative property of addition to group the terms with \( \frac{1}{4} \):
\( = \frac{1}{2} \times \frac{1}{4} - \frac{1}{4} \times \frac{1}{3} + \left(-\frac{7}{18}\right) \times \frac{15}{7} \)
Using commutative property of multiplication on the first two terms:
\( = \frac{1}{4} \times \frac{1}{2} - \frac{1}{4} \times \frac{1}{3} + \left(-\frac{7}{18} \times \frac{15}{7}\right) \)
Using distributive property to factor out \( \frac{1}{4} \):
\( = \frac{1}{4} \left( \frac{1}{2} - \frac{1}{3} \right) + \left( -\frac{15}{18} \right) \)
\( = \frac{1}{4} \left( \frac{3 - 2}{6} \right) - \frac{5}{6} \)
\( = \frac{1}{4} \times \frac{1}{6} - \frac{5}{6} \)
\( = \frac{1}{24} - \frac{5}{6} \)
Make denominators equal (LCM = 24):
\( = \frac{1}{24} - \frac{20}{24} = -\frac{19}{24} \)

(c) \( -\frac{5}{4} \times \left( \frac{8}{5} + 16 \right) \)
Using distributive property:
\( = \left( -\frac{5}{4} \times \frac{8}{5} \right) + \left( -\frac{5}{4} \times 16 \right) \)
\( = -2 - 20 = -22 \)

(d) \( \frac{7}{12} \times \left( -\frac{3}{5} \right) + \left( -\frac{5}{3} \right) \times \frac{7}{12} \)
Using commutative property of multiplication on the second term:
\( = \frac{7}{12} \times \left( -\frac{3}{5} \right) + \frac{7}{12} \times \left( -\frac{5}{3} \right) \)
Using distributive property to factor out \( \frac{7}{12} \):
\( = \frac{7}{12} \times \left( -\frac{3}{5} - \frac{5}{3} \right) \)
Find LCM of 5 and 3 inside the bracket (LCM = 15):
\( = \frac{7}{12} \times \left( \frac{-9 - 25}{15} \right) \)
\( = \frac{7}{12} \times \left( -\frac{34}{15} \right) \)
Simplify by dividing 12 and -34 by 2:
\( = \frac{7}{6} \times \left( -\frac{17}{15} \right) \)
\( = -\frac{119}{90} \)

(e) \( \frac{2}{21} \times \frac{-3}{13} + \frac{-7}{9} - \frac{2}{21} \times \frac{10}{13} \)
Using commutative property of addition to group the terms with \( \frac{2}{21} \):
\( = \frac{2}{21} \times \frac{-3}{13} - \frac{2}{21} \times \frac{10}{13} + \frac{-7}{9} \)
Using distributive property to factor out \( \frac{2}{21} \):
\( = \frac{2}{21} \times \left( -\frac{3}{13} - \frac{10}{13} \right) - \frac{7}{9} \)
\( = \frac{2}{21} \times \left( -\frac{13}{13} \right) - \frac{7}{9} \)
\( = \frac{2}{21} \times (-1) - \frac{7}{9} \)
\( = -\frac{2}{21} - \frac{7}{9} \)
Find LCM of 21 and 9 (LCM = 63):
\( = \frac{-6 - 49}{63} = -\frac{55}{63} \)

(f) \( -\frac{4}{5} \times \frac{3}{7} \times \frac{15}{16} \times \left( \frac{-14}{9} \right) \)
Rearranging terms using commutative and associative properties of multiplication:
\( = \left( -\frac{4}{5} \times \frac{15}{16} \right) \times \left( \frac{3}{7} \times -\frac{14}{9} \right) \)
Simplify each bracket:
\( = \left( -\frac{1}{1} \times \frac{3}{4} \right) \times \left( \frac{1}{1} \times -\frac{2}{3} \right) \)
\( = -\frac{3}{4} \times -\frac{2}{3} \)
\( = \frac{6}{12} = \frac{1}{2} \)
In simple words: To solve these math sentences, we do the same action to both sides until our letter is left all alone on one side.

Exam Tip: Always look for common fractions across terms so that you can factor them out using the distributive property to make calculations much easier.

Question 12. Represent \( -\frac{2}{7} \) and \( \frac{4}{7} \) on a number line.
Answer:
To show these fractions on a number line, we divide the space between 0 and 1 into 7 equal parts. We also do the same for the space between 0 and -1.
Each tick mark represents \( \frac{1}{7} \).
- \( \frac{4}{7} \) is 4 steps to the right of 0.
- \( -\frac{2}{7} \) is 2 steps to the left of 0.
In simple words: The number line is split into equal segments of 1/7. We count 4 marks to the right of zero for 4/7, and 2 marks to the left of zero for -2/7.

Exam Tip: Always make sure that the segments between 0 and 1 are exactly equal in size to maintain accuracy.

Question 13. Verify that \( -(-x) = x \) for \( x = \frac{-2}{5} \)
Answer:
Given \( x = -\frac{2}{5} \)
Let us put this value into the Left-Hand Side (L.H.S.):
L.H.S. \( = -(-x) \)
\( = -\left( -\left( -\frac{2}{5} \right) \right) \)
Since two minus signs make a plus, \( -\left( -\frac{2}{5} \right) = \frac{2}{5} \):
\( = -\left( \frac{2}{5} \right) \)
\( = -\frac{2}{5} \)

Now let us check the Right-Hand Side (R.H.S.):
R.H.S. \( = x = -\frac{2}{5} \)

Since L.H.S. = R.H.S. = \( -\frac{2}{5} \), the equation is verified.
In simple words: Adding two negative signs together keeps the sign the same as the original, so we get the exact same fraction back.

Exam Tip: Be careful with the number of negative signs. Three negative signs multiply to yield a negative result.

Question 14. Is \( -\frac{4}{17} \) the multiplicative inverse of \( -4\frac{1}{4} \)? Why or why not?
Answer:
Let us convert the mixed fraction \( -4\frac{1}{4} \) into an improper fraction:
\( -4\frac{1}{4} = -\frac{4 \times 4 + 1}{4} = -\frac{17}{4} \)

To check if \( -\frac{4}{17} \) is the multiplicative inverse, we multiply them together:
\( \left( -\frac{17}{4} \right) \times \left( -\frac{4}{17} \right) = 1 \)
Since the product is exactly 1, yes, \( -\frac{4}{17} \) is the multiplicative inverse of \( -4\frac{1}{4} \).
In simple words: Yes, because converting \( -4\frac{1}{4} \) gives \( -\frac{17}{4} \). Flipping this fraction upside down gives \( -\frac{4}{17} \).

Exam Tip: Always convert mixed fractions into improper fractions before testing for reciprocals.

Question 15. Is 0.5 the multiplicative inverse of 2? Why or why not?
Answer:
Yes, 0.5 is the multiplicative inverse of 2.
Let us convert the decimal 0.5 into a fraction:
\( 0.5 = \frac{5}{10} = \frac{1}{2} \)
To find if it is the multiplicative inverse of 2, we multiply them:
\( 2 \times \frac{1}{2} = 1 \)
Since their product is 1, they are multiplicative inverses of each other.
In simple words: Yes, because 0.5 is the same as half (1/2), and multiplying half by 2 gives exactly 1.

Exam Tip: If the product of two rational numbers is 1, they are always multiplicative inverses of each other.

Question 16. Write five rational numbers greater than -5.
Answer:
Five rational numbers greater than -5 are:
-4, -3, -2, -1, 0 (or any other fractions like \( -\frac{1}{2} \) and \( \frac{3}{4} \))
In simple words: Any numbers to the right of -5 on the number line are larger, like -4, -3, -2, -1, and 0.

Exam Tip: Positive integers or zero are always greater than any negative rational number.

Question 17. Find five rational numbers between:
(a) 0 and -2
(b) \( -\frac{3}{7} \) and \( \frac{5}{11} \)
(c) -3 and -4
Answer:
(a) Between 0 and -2:
We can write 0 as \( \frac{0}{6} \) and -2 as \( -\frac{12}{6} \).
Five rational numbers between them are:
\( -\frac{1}{6}, -\frac{2}{6}, -\frac{3}{6}, -\frac{4}{6}, -\frac{5}{6} \) which simplify to \( -\frac{1}{6}, -\frac{1}{3}, -\frac{1}{2}, -\frac{2}{3}, -\frac{5}{6} \).

(b) Between \( -\frac{3}{7} \) and \( \frac{5}{11} \):
Find the LCM of 7 and 11 (LCM = 77):
\( -\frac{3}{7} = -\frac{33}{77} \)
\( \frac{5}{11} = \frac{35}{77} \)
Five rational numbers between them are:
\( -\frac{10}{77}, -\frac{5}{77}, 0, \frac{5}{77}, \frac{10}{77} \).

(c) Between -3 and -4:
We can write -3 as \( -\frac{18}{6} \) and -4 as \( -\frac{24}{6} \).
Five rational numbers between them are:
\( -\frac{19}{6}, -\frac{20}{6}, -\frac{21}{6}, -\frac{22}{6}, -\frac{23}{6} \).
In simple words: To find fractions in between, convert the numbers to have the same large denominator so you can pick values from the middle.

Exam Tip: Convert integers into fractions with a common denominator first to make it easy to list numbers in between.

Question 18. Find the multiplicative inverse of \( -2 \times -\frac{3}{4} \)
Answer:
First, let us simplify the expression:
\( -2 \times -\frac{3}{4} = \frac{6}{4} = \frac{3}{2} \)
The multiplicative inverse is the reciprocal of \( \frac{3}{2} \):
\( = \frac{2}{3} \)
In simple words: Multiplying the terms gives 3/2. Flipping this fraction upside down gives the answer, which is 2/3.

Exam Tip: Always simplify the expression completely before finding its reciprocal.

Question 19. Find the additive inverse of \( -\frac{2}{9} + \frac{7}{9} \)
Answer:
First, simplify the expression:
\( -\frac{2}{9} + \frac{7}{9} = \frac{5}{9} \)
The additive inverse is the negative of this value:
\( = -\frac{5}{9} \)
In simple words: Adding the two parts gives 5/9. Its opposite sign version is -5/9.

Exam Tip: Additive inverse means changing the final sign of the simplified answer.

Question 20. Multiply \( \frac{8}{13} \) by the reciprocal of \( -\frac{17}{26} \)
Answer:
The reciprocal of \( -\frac{17}{26} \) is \( -\frac{26}{17} \).
Now, multiply \( \frac{8}{13} \) by \( -\frac{26}{17} \):
\( = \frac{8}{13} \times \left( -\frac{26}{17} \right) \)
\( = \frac{8 \times -2}{17} \)
\( = -\frac{16}{17} \)
In simple words: Flip the second fraction to get -26/17, then multiply it by 8/13 to get -16/17.

Exam Tip: Simplify the terms during multiplication by cancelling common factors to keep calculations simple.

Question 21. Reciprocal of \( -\frac{1}{y} \), where \( y \neq 0 \) is _________
Answer: -y
In simple words: Flipping \( -\frac{1}{y} \) upside down gives \( -y \).

Exam Tip: The reciprocal of \( \frac{1}{x} \) is \( x \), and the negative sign is retained.

Question 22. The product of two negative rational nos is always a _________ rational number.
Answer: positive
In simple words: Multiplying two negative numbers always gives a positive answer.

Exam Tip: Remember the basic rule: \( \text{Negative} \times \text{Negative} = \text{Positive} \).

Question 23. \( -\frac{25}{23} \div 0 = \) _________
Answer: Not defined (or undefined)
In simple words: Dividing any number by zero is not possible in math.

Exam Tip: Division by zero is always written as "Not defined" or "Undefined".

Question 24. \( 0 \div \left( -\frac{25}{23} \right) = \) _________
Answer: 0
In simple words: Dividing zero by any non-zero number always equals zero.

Exam Tip: Note that \( 0 \div a = 0 \) for any non-zero number \( a \).

Question 25. State the property used in the following:
(a) \( -\frac{25}{29} \times \frac{29}{-25} = 1 \)
(b) \( \frac{8}{5} + 0 = 0 + \frac{8}{5} = \frac{8}{5} \)
(c) \( -\frac{2}{9} \left( \frac{3}{5} + \frac{2}{9} \right) = -\frac{2}{9} \times \frac{3}{5} + \left( -\frac{2}{9} \right) \times \frac{2}{9} \)
Answer:
(a) Multiplicative Inverse Property (or existence of reciprocal)
(b) Additive Identity Property (or the identity property of zero)
(c) Distributive Property of Multiplication over Addition
In simple words: Part (a) shows multiplying by a reciprocal gives 1. Part (b) shows adding 0 does nothing. Part (c) shows how to distribute a number into a bracket.

Exam Tip: Memorize the standard names of these properties to ensure you get full marks on terminology questions.