CBSE Class 8 Mathematics Rational Numbers Worksheet Set 03

RATIONAL NUMBER (NUMBER LINE)

Question. Draw the number line and represent following rational number on it:
(i) (2/3)
(ii) (3/4)
(iii) (3/8)
(iv) (-5/8)
Solution:
(i) We know that (2/3) is greater than 2 and less than 3.
∴ it lies between 2 and 3. It can be represented on number line as,

(ii) We know that (3/4) is greater than 0 and less than 1.
∴ it lies between 0 and 1. It can be represented on number line as,

(iii) We know that (3/8) is greater than 0 and less than 1.
∴ it lies between 0 and 1. It can be represented on number line as,

(iv) We know that (-5/8) is greater than -1 and less than 0.
∴ it lies between 0 and -1. It can be represented on number line as,

Question. Which of the two rational numbers in each of the following pairs of rational number is greater? (i) (-3/8), 0
(ii) (5/2), 0
(iii) (– 4/11), (3/11)
Solution:
(i) Given (-3/8), 0
We know that every positive rational number is greater than zero and every negative
rational number is smaller than zero. Thus, – (3/8) > 0

(ii) Given (5/2), 0
We know that every positive rational number is greater than zero and every negative rational number is smaller than zero. Thus, (5/2) > 0

(iii) Given (– 4/11), (3/11)
We know that every positive rational number is greater than zero and every negative rational number is smaller than zero, also the denominator is same in given question now we have to compare the numerator, thus – 4/11 < 3/11.

Question. Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
(i) (-6/-13), (7/13)
(ii) (16/-5), 3
Solution:
(i) Given (-6/-13), (7/13)
Here denominator is same therefore compare the numerator,
Thus (-6/-13) < (7/13)

(ii) Given (16/-5), 3
We know that 3 is a whole number with positive sign
Therefore (16/-5) < 3

Question. Fill in the blanks by the correct symbol out of >, =, or <:
(i) (-6/7) …. (7/13)
(ii) (-3/5) …. (-5/6)
(iii) (-2/3) …. (5/-8)
(iv) 0 …. (-2/5)
Solution:
(i) (- 6/7) < (7/13)
Explanation:
Because every positive number is greater than a negative number.

(ii) (-3/5) > (-5/6)
Explanation:
Consider (-3/5)
Multiply both numerator and denominator by 6 then we get
(-3/5) × (6/6) = (-18/30)…… (1)
Now consider (-5/6)
Multiply both numerator and denominator by 5 we get
(-5/6) × (5/5) = (-25/30)…… (2)
The denominator is same in equation (1) and (2) now we have to compare the numerator, thus (-3/5) > (-5/6)

(iii) (-2/3) < (5/-8)
Explanation:
Consider (-2/3)
Multiply both numerator and denominator by 8 then we get
(-2/3) × (8/8) = (-16/24)…… (1)
Now consider (5/-8)
Multiply both numerator and denominator by 3 we get
(5/-8) × (3/3) = (15/-24)…… (2)
The denominator is same in equation (1) and (2) now we have to compare the numerator, thus (-2/3) < (5/-8)

(iv) 0 > (-2/5)
Explanation:
Because every positive number is greater than a negative number

Question. Arrange the following rational numbers in ascending order:
(i) (3/5), (-17/-30), (8/-15), (-7/10)
(ii) (-4/9), (5/-12), (7/-18), (2/-3)
Solution:
(i) Given (3/5), (-17/-30), (8/-15), (-7/10)
The LCM of 5, 30, 15 and 10 is 30
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 30
Consider (3/5)
Multiply both numerator and denominator by 6, then we get
(3/5) × (6/6) = (18/30) ….. (1)
Consider (8/-15)
Multiply both numerator and denominator by 2, then we get
(8/-15) × (2/2) = (16/-30) ….. (2)
Consider (-7/10)
Multiply both numerator and denominator by 3, then we get
(-7/10) × (3/3) = (-21/30) ….. (3)
In the above equation, denominators are same
Now on comparing the ascending order is:
(-7/10) < (8/-15) < (-17/30) < (3/5)

(ii) Given (-4/9), (5/-12), (7/-18), (2/-3)
The LCM of 9, 12, 18 and 3 is 36
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 36
Consider (-4/9)
Multiply both numerator and denominator by 4, then we get
(-4/9) × (4/4) = (-16/36) ….. (1)
Consider (5/-12)
Multiply both numerator and denominator by 3, then we get
(5/-12) × (3/3) = (15/-36) ….. (2)
Consider (7/-18)
Multiply both numerator and denominator by 2, then we get
(7/-18) × (2/2) = (14/-36) ….. (3)
Consider (2/-3)
Multiply both numerator and denominator by 12, then we get
(2/-3) × (12/12) = (24/-36) ….. (4)
In the above equation, denominators are same
Now on comparing the ascending order is:
(2/-3) < ((-4/9) < (5/-12) < (7/-18)

Question. Arrange the following rational numbers in descending order:
(i) (7/8), (64/16), (39/-12), (5/-4), (140/28)
(ii) (-3/10), (17/-30), (7/-15), (-11/20)
Solution:
(i) Given (7/8), (64/16), (39/-12), (5/-4), (140/28)
The LCM of 8, 16, 12, 4 and 28 is 336
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 336
Consider (7/8)
Multiply both numerator and denominator by 42, then we get
(7/8) × (42/42) = (294/336) ….. (1)
Consider (64/16)
Multiply both numerator and denominator by 21, then we get
(64/16) × (21/21) = (1344/336) ….. (2)
Consider (39/-12)
Multiply both numerator and denominator by 28, then we get
(39/-12) × (28/28) = (-1008/336) ….. (3)
Consider (5/-4)
Multiply both numerator and denominator by 84, then we get
(5/-4) × (84/84) = (-420/336) ….. (4)
In the above equation, denominators are same
Now on comparing the descending order is:
(140/28) > (64/16) > (7/8) > (5/-4) > (36/-12)

(ii) Given (-3/10), (17/-30), (7/-15), (-11/20)
The LCM of 10, 30, 15 and 20 is 60
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 60
Consider (-3/10)
Multiply both numerator and denominator by 6, then we get
(-3/10) × (6/6) = (-18/60) ….. (1)
Consider (17/-30)
Multiply both numerator and denominator by 2, then we get
(17/-30) × (2/2) = (34/-60) ….. (2)
Consider (7/-15)
Multiply both numerator and denominator by 4, then we get
(7/-15) × (4/4) = (28/-60) ….. (3)
In the above equation, denominators are same
Now on comparing the descending order is:
(-3/10) > (7/-15) > (-11/20) > (17/20)

Question. Which of the following statements are true:
(i) The rational number (29/23) lies to the left of zero on the number line.
(ii) The rational number (-12/-17) lies to the left of zero on the number line.
(iii) The rational number (3/4) lies to the right of zero on the number line.
(iv) The rational number (-12/-5) and (- 7/17) are on the opposite side of zero on the number line.
(v) The rational number (-2/15) and (7/-31) are on the opposite side of zero on the number line.
Solution:
(i) False
Explanation:
It lies to the right of zero because it is a positive number.

(ii) False
Explanation:
It lies to the right of zero because it is a positive number.

(iii) True
Explanation:
Always positive number lie on the right of zero

(iv) True
Explanation:
Because they are of opposite sign

(v) False
Explanation:
Because they both are of same sign

SECTION A

Question 1. \(\left| \frac{-3}{4} \right|\) is
(a) \(\frac{3}{4}\)
(b) \(\frac{-3}{4}\)
(c) \(\frac{4}{3}\)
(d) \(\frac{-4}{3}\)
Answer: (a) \(\frac{3}{4}\)
The absolute value of any rational number is always non-negative, so \(\left| \frac{-3}{4} \right| = \frac{3}{4}\).
In simple words: The absolute value bars make any negative number positive.

Exam Tip: Remember that modulus or absolute value always strips away the negative sign.

Question 2. Commutative property of rational numbers states that \((a + b) =\)
(a) \((a \cdot b)\)
(b) \((b + a)\)
(c) a
(d) b
Answer: (b) \((b + a)\)
The commutative property of addition states that changing the order of the addends does not change the sum, meaning \(a + b = b + a\).
In simple words: Order does not matter in addition; \(2 + 3\) is the same as \(3 + 2\).

Exam Tip: Do not confuse commutative property (changing order) with associative property (changing grouping).

Question 3. The additive identity of rational numbers is
(a) 0
(b) 1
(c) -1
(d) 2
Answer: (a) 0
Zero is the additive identity because adding 0 to any rational number leaves the number unchanged, i.e., \(a + 0 = a\).
In simple words: Adding zero to a number gives the same number back.

Exam Tip: Additive identity is 0, whereas multiplicative identity is 1.

Question 4. If we add additive inverse of any rational number to the number the result is
(a) 1
(b) -1
(c) 0
(d) none
Answer: (c) 0
The additive inverse of a number \(a\) is \(-a\). Their sum is \(a + (-a) = 0\).
In simple words: A number plus its opposite always equals zero.

Exam Tip: Additive inverse means changing the sign of the number.

Question 5. The multiplicative inverse of \(\left(\frac{-5}{9}\right)\) is
(a) \(\frac{-5}{9}\)
(b) \(\frac{-9}{5}\)
(c) \(\frac{9}{5}\)
(d) \(\frac{5}{9}\)
Answer: (b) \(\frac{-9}{5}\)
The multiplicative inverse (reciprocal) of \(\frac{a}{b}\) is \(\frac{b}{a}\). For \(\frac{-5}{9}\), it is \(\frac{9}{-5}\) or \(\frac{-9}{5}\).
In simple words: Flip the fraction upside down while keeping the negative sign in place.

Exam Tip: The product of a number and its multiplicative inverse is always 1.

Question 6. \((9^2 \div 9^{-2})^2 \times 9^{-4} =\)
(a) 1
(b) 729
(c) 6561
(d) \(1 / 6561\)
Answer: (b) 729
Using laws of exponents:
\(9^2 \div 9^{-2} = 9^{2 - (-2)} = 9^4\)
\((9^4)^2 = 9^8\)
\(9^8 \times 9^{-4} = 9^{8 + (-4)} = 9^4 = 6561\) - wait, let us check calculation:
\(9^4 = 9 \times 9 \times 9 \times 9 = 81 \times 81 = 6561\). Let us re-read question options or check if option (c) 6561 is correct.
Let us re-verify: \(9^2 \div 9^{-2} = 9^{2 - (-2)} = 9^4\).
Then \((9^4)^2 = 9^8\).
Then \(9^8 \times 9^{-4} = 9^{8-4} = 9^4 = 6561\).
Therefore, correct option is (c) 6561.
In simple words: Simplify the exponents step by step using quotient and product rules.

Exam Tip: Subtract powers during division and add powers during multiplication with the same base.

Question 7. \((7^0 + 8^0 + 9^0)^{-2} =\)
(a) \(1 / 625\)
(b) 9
(c) \(1 / 9\)
(d) 625
Answer: (c) \(1 / 9\)
Any non-zero number raised to power 0 is 1:
\(7^0 = 1, 8^0 = 1, 9^0 = 1\)
\((1 + 1 + 1)^{-2} = (3)^{-2} = \frac{1}{3^2} = \frac{1}{9}\)
In simple words: Any number to the power of zero is one, so \(1 + 1 + 1 = 3\), and \(3^{-2} = 1/9\).

Exam Tip: Remember \(a^0 = 1\) for any \(a \neq 0\).

Question 8. \((3^{-5})^{-1} =\)
(a) \(1 / 243\)
(b) 81
(c) \(1 / 81\)
(d) 243
Answer: (d) 243
Using \((a^m)^n = a^{m \times n}\):
\((3^{-5})^{-1} = 3^{(-5) \times (-1)} = 3^5 = 243\)
In simple words: Multiply the two exponent powers together, minus times minus gives plus.

Exam Tip: Power of a power rule multiplies the indices.

Question 9. The reciprocal of \(8^{-7}\) is
(a) \(8^7\)
(b) \(8^{-7}\)
(c) \(8^7\)
(d) \(1 / 8^7\)
Answer: (a) \(8^7\)
The reciprocal of \(8^{-7}\) is \(\frac{1}{8^{-7}} = 8^7\).
In simple words: Reciprocal inverts the fraction and changes the negative exponent to positive.

Exam Tip: Reciprocal of \(a^{-n}\) is \(a^n\).

Question 10. The standard form of \(0.0000009\) is
(a) \(9 \times 10^7\)
(b) \(9 \times 10^5\)
(c) \(9 \times 10^4\)
(d) \(9 \times 10^6\)
Note: Examining page 2 continuation, options are listed as (a) \(9 \times 10^7\) (b) \(9 \times 10^5\) (c) \(9 \times 10^4\) (d) \(9 \times 10^6\). Wait, let us check actual standard form: moving decimal 7 places to the right gives \(9 \times 10^{-7}\). Since \(9 \times 10^{-7}\) is not listed, let us check if the option was meant to be \(9 \times 10^{-7}\) or if there is a typo in the worksheet. Assuming standard form, \(0.0000009 = 9 \times 10^{-7}\). If restricted to options printed, let us provide the mathematical correct value and note it.
Answer: \(9 \times 10^{-7}\) (None of the options match exactly; printed options have positive exponents)
Moving the decimal point 7 places to the right gives \(9 \times 10^{-7}\).
In simple words: For small decimal numbers, the exponent of 10 is negative.

Exam Tip: Standard form is expressed as \(k \times 10^n\) where \(1 \le k < 10\).

SECTION - B

Question 11. Represent \(\frac{-9}{11}\) on number line.
Answer:
1. Draw a straight horizontal line with arrows on both sides.
2. Mark integer points like 0, -1, and 1 at equal intervals.
3. Since the number is negative and lies between 0 and -1, divide the segment between 0 and -1 into 11 equal parts.
4. Count 9 parts to the left of 0 and mark the point as \(\frac{-9}{11}\).
In simple words: Locate the position between 0 and -1 by splitting the unit segment into 11 equal slices.

Exam Tip: The denominator tells you how many equal parts to divide each unit interval into.

Question 12. What should be subtracted from \((-2/3)\) to get \((-5/6)\)?
Answer:
Let the required number be \(x\).
According to the question:
\(\left(-\frac{2}{3}\right) - x = -\frac{5}{6}\)
\(-x = -\frac{5}{6} - \left(-\frac{2}{3}\right)\)
\(-x = -\frac{5}{6} + \frac{2}{3}\)
Find the LCM of 6 and 3, which is 6:
\(\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}\)
\(-x = -\frac{5}{6} + \frac{4}{6} = \frac{-5 + 4}{6} = -\frac{1}{6}\)
\(x = \frac{1}{6}\)
In simple words: Set up a subtraction equation and solve for the unknown variable by finding a common denominator.

Exam Tip: Be very careful with signs when subtracting negative fractions.

Question 13. Write \(235.367\) in expanded form.
Answer:
\(235.367 = (2 \times 10^2) + (3 \times 10^1) + (5 \times 10^0) + \left(3 \times \frac{1}{10^1}\right) + \left(6 \times \frac{1}{10^2}\right) + \left(7 \times \frac{1}{10^3}\right)\)
Or in decimal form:
\(200 + 30 + 5 + 0.3 + 0.06 + 0.007\)
In simple words: Break down each digit according to its place value position.

Exam Tip: Fractional place values use negative powers of 10 or denominators of 10, 100, 1000.

Question 14. The thickness of a sheet of paper is \(1.5 \times 10^{-3}\) cm & the thickness of a Human hair is \(6 \times 10^{-3}\) cm. Compare the two.
Answer:
Thickness of paper = \(1.5 \times 10^{-3}\) cm
Thickness of human hair = \(6 \times 10^{-3}\) cm
To compare, find the ratio of human hair thickness to paper thickness:
\(\text{Ratio} = \frac{6 \times 10^{-3}}{1.5 \times 10^{-3}} = \frac{6}{1.5} = 4\)
Thus, human hair is 4 times as thick as a sheet of paper.
In simple words: Divide the thickness of human hair by the thickness of paper to find how many times larger it is.

Exam Tip: When comparing numbers in scientific notation with the same power of 10, simply divide their coefficients.

SECTION - C

Question 15. By what number should \(\left(\frac{4}{3}\right)^{-3}\) be divided so that the quotient is \(\left(\frac{16}{9}\right)^2\)?
Answer:
Let the required number be \(x\).
\(\left(\frac{4}{3}\right)^{-3} \div x = \left(\frac{16}{9}\right)^2\)
\(x = \frac{\left(\frac{4}{3}\right)^{-3}}{\left(\frac{16}{9}\right)^2}\)
Note that \(\frac{16}{9} = \left(\frac{4}{3}\right)^2\), so:
\(\left(\frac{16}{9}\right)^2 = \left(\left(\frac{4}{3}\right)^2\right)^2 = \left(\frac{4}{3}\right)^4\)
Therefore:
\(x = \frac{\left(\frac{4}{3}\right)^{-3}}{\left(\frac{4}{3}\right)^4} = \left(\frac{4}{3}\right)^{-3 - 4} = \left(\frac{4}{3}\right)^{-7} = \left(\frac{3}{4}\right)^7\)
In simple words: Express all terms with the same base fraction and apply exponent laws.

Exam Tip: Converting bases to a common number simplifies complex fractional exponent problems instantly.

Question 16. Evaluate : \(\left\{\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^3\right\} \div \left(\frac{1}{4}\right)^{-2}\)
Answer:
Calculate each part inside the brackets:
\(\left(\frac{1}{3}\right)^{-2} = 3^2 = 9\)
\(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)
First bracket value = \(9 - \frac{1}{8} = \frac{72 - 1}{8} = \frac{71}{8}\)
Calculate the divisor:
\(\left(\frac{1}{4}\right)^{-2} = 4^2 = 16\)
Now divide:
\(\frac{71}{8} \div 16 = \frac{71}{8 \times 16} = \frac{71}{128}\)
In simple words: Evaluate powers first, perform subtraction inside curly braces, and finally divide by the last term.

Exam Tip: Follow BODMAS order of operations strictly.

Question 17. Tamanna purchased \(10\frac{1}{2}\) kg of potatoes, \(14\frac{3}{4}\) kg of onions and \(16\frac{1}{4}\) kg of tomatoes. How many kilograms of tomatoes does she buy?
Answer:
The question directly states the quantity of tomatoes purchased:
Tamanna purchased \(16\frac{1}{4}\) kg of tomatoes.
(If the question intended total weight: \(10\frac{1}{2} + 14\frac{3}{4} + 16\frac{1}{4} = \frac{21}{2} + \frac{59}{4} + \frac{65}{4} = \frac{42 + 59 + 65}{4} = \frac{166}{4} = 41.5\) kg total).
In simple words: Read carefully - the amount of tomatoes is already given in the problem text.

Question 18. If \(x = \frac{2}{3}\), \(y = \frac{-1}{4}\), & \(z = \frac{-5}{8}\) verify that \(x(y + z) = x \times y + x \times z\).
Answer:
This verifies the distributive property of multiplication over addition.
LHS: \(x(y + z)\)
\(y + z = -\frac{1}{4} + \left(-\frac{5}{8}\right) = -\frac{2}{8} - \frac{5}{8} = -\frac{7}{8}\)
\(\text{LHS} = \frac{2}{3} \times \left(-\frac{7}{8}\right) = \frac{-14}{24} = -\frac{7}{12}\)
RHS: \((x \times y) + (x \times z)\)
\(x \times y = \frac{2}{3} \times \left(-\frac{1}{4}\right) = -\frac{2}{12} = -\frac{1}{6}\)
\(x \times z = \frac{2}{3} \times \left(-\frac{5}{8}\right) = -\frac{10}{24} = -\frac{5}{12}\)
\(\text{RHS} = -\frac{1}{6} + \left(-\frac{5}{12}\right) = -\frac{2}{12} - \frac{5}{12} = -\frac{7}{12}\)
Since LHS = RHS = \(-\frac{7}{12}\), the property is verified.
In simple words: Substitute the given fractions into both sides and show that both evaluate to the exact same value.

Exam Tip: Always show LHS and RHS calculations separately when verifying identities.

SECTION - D

Question 19. Insert a rational number between \((x + y)^{-1}\) & \(x^{-1} + y^{-1}\), where \(x = \frac{-7}{9}\), \(y = \frac{-3}{5}\).
Answer:
First, compute \(x + y\):
\(x + y = -\frac{7}{9} + \left(-\frac{3}{5}\right) = \frac{-35 - 27}{45} = -\frac{62}{45}\)
Find \((x + y)^{-1}\):
\((x + y)^{-1} = -\frac{45}{62}\)
Next, compute \(x^{-1} + y^{-1}\):
\(x^{-1} = -\frac{9}{7}\)
\(y^{-1} = -\frac{5}{3}\)
\(x^{-1} + y^{-1} = -\frac{9}{7} - \frac{5}{3} = \frac{-27 - 35}{21} = -\frac{62}{21}\)
Now, find a rational number between \(-\frac{45}{62}\) and \(-\frac{62}{21}\).
We can take their average or choose a convenient number in between:
Let us take the arithmetic mean: \(\frac{1}{2} \left(-\frac{45}{62} - \frac{62}{21}\right)\)
Alternatively, a simple rational number between them is \(-1\).
In simple words: Calculate each expression by finding reciprocals and sums, then find a value lying between them.

Exam Tip: The average of two numbers always lies strictly between them.

Question 20. Find four rational numbers between \(\frac{6}{7}\) & \(\frac{15}{2}\).
Answer:
Convert the fractions to have a common denominator or decimal equivalents:
\(\frac{6}{7} \approx 0.857\)
\(\frac{15}{2} = 7.5\)
Four rational numbers between \(0.857\) and \(7.5\) can be chosen as integers or simple fractions:
1, 2, 3, 4 (or \(1, \frac{3}{2}, 2, \frac{5}{2}\))
In simple words: Choose any four numbers that lie between 0.857 and 7.5.

Exam Tip: Converting to decimals makes it very easy to spot numbers lying in between.

Question 21. If \(5^{4x} \times 5^{-2x} = 125 \times 5^x\), find x.
Answer:
Simplify both sides using exponent rules (\(a^m \times a^n = a^{m+n}\)):
LHS: \(5^{4x + (-2x)} = 5^{2x}\)
RHS: \(125 \times 5^x = 5^3 \times 5^x = 5^{3+x}\)
Equating the exponents since the bases are equal (base 5):
\(2x = 3 + x\)
\(2x - x = 3\)
\(x = 3\)
In simple words: Combine exponents on both sides of the equation and equate the powers when bases are identical.

Exam Tip: Whenever exponential equations have the same base on both sides, set their exponents equal to each other.

Question 22. If \(\frac{x}{y} = \left(\frac{2}{3}\right)^2 \times \left(\frac{3}{5}\right)^{-2}\), find \(\left(\frac{y}{x}\right)^{-1}\).
Answer:
First, simplify \(\frac{x}{y}\):
\(\frac{x}{y} = \left(\frac{2}{3}\right)^2 \times \left(\frac{5}{3}\right)^2\)
\(\frac{x}{y} = \frac{4}{9} \times \frac{25}{9} = \frac{100}{81}\)
We need to find \(\left(\frac{y}{x}\right)^{-1}\):
Since \(\left(\frac{y}{x}\right)^{-1} = \frac{x}{y}\),
\(\left(\frac{y}{x}\right)^{-1} = \frac{100}{81}\)
In simple words: Calculate \(x/y\) first, and use the property that a negative exponent on a reciprocal inverts it back to the original fraction.

Exam Tip: \((a/b)^{-1} = b/a\), so \((y/x)^{-1} = x/y\).