CBSE Class 8 Mathematics Rational Numbers Practice Paper Worksheet Set 14 with Answers (Old NCERT Syllabus)

This Class 8 Maths Rational Numbers practice paper carries 40 marks and has 17 questions in five sections (A to E), each with a step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising this chapter. The questions test closure, commutativity and distributivity of rational numbers, reciprocals and multiplicative inverses, simplifying sums and products of fractions with negative signs, ordering rational numbers, and a word problem and a wingspan case study using fractions.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Find using distributivity: \( \left\{\frac{7}{5} \times \left(\frac{-3}{12}\right)\right\} + \left\{\frac{7}{5} \times \frac{5}{12}\right\} \)
Answer:
Both terms have the common factor \( \frac{7}{5} \).
By distributivity, a × b + a × c = a × (b + c):
\( \frac{7}{5} \times \left(\frac{-3}{12} + \frac{5}{12}\right) = \frac{7}{5} \times \frac{2}{12} \)
\( = \frac{14}{60} = \frac{7}{30} \)
So the value is \( \frac{7}{30} \).
In simple words: Take 7/5 out as a common factor, add −3/12 and 5/12 to get 2/12, then multiply.

Teacher's Note:
1. Spot the common factor first; it saves working out two separate products.
2. Always reduce the final fraction to lowest terms: 14/60 = 7/30.

Question 8. What number should be subtracted from \( \frac{-7}{8} \) so as to get \( \frac{5}{12} \)?
Answer:
Let the required number be x.
Then \( \frac{-7}{8} \) − x = \( \frac{5}{12} \).
So x = \( \frac{-7}{8} \) − \( \frac{5}{12} \).
LCM of 8 and 12 is 24: x = \( \frac{-21}{24} - \frac{10}{24} = \frac{-31}{24} \).
So the number to be subtracted is \( \frac{-31}{24} \).
Check: \( \frac{-21}{24} - \left(\frac{-31}{24}\right) = \frac{10}{24} = \frac{5}{12} \).
In simple words: Required number = given number − result = −7/8 − 5/12 = −31/24.

Teacher's Note:
1. Write the statement as an equation before computing.
2. Subtracting a negative number is the same as adding its positive.

Question 9. Using suitable rearrangement find the sum \( \frac{4}{7} + \left(-\frac{4}{9}\right) + \frac{3}{7} + \left(-\frac{13}{9}\right) \):
Answer:
Group the fractions with the same denominator together (commutative and associative properties):
\( \left(\frac{4}{7} + \frac{3}{7}\right) + \left[\left(-\frac{4}{9}\right) + \left(-\frac{13}{9}\right)\right] \)
\( = \frac{7}{7} + \left(\frac{-17}{9}\right) = 1 - \frac{17}{9} \)
\( = \frac{9 - 17}{9} = \frac{-8}{9} \)
So the sum is \( \frac{-8}{9} \).
In simple words: Add the sevenths together (they make 1) and the ninths together (−17/9); then 1 − 17/9 = −8/9.

Teacher's Note:
1. Rearranging is allowed because addition is commutative and associative.
2. Grouping like denominators avoids finding an LCM of 7 and 9 at the start.

Section C: Short Answer Questions (3 Marks Each)

Question 10. Simplify: \( \frac{8}{-15} + \frac{7}{20} - \frac{-11}{35} + \frac{1}{5} \)
Answer:
Write each fraction with a positive denominator: \( -\frac{8}{15} + \frac{7}{20} + \frac{11}{35} + \frac{1}{5} \).
LCM of 15, 20, 35 and 5 is 420.
\( = \frac{-8 \times 28 + 7 \times 21 + 11 \times 12 + 1 \times 84}{420} \)
\( = \frac{-224 + 147 + 132 + 84}{420} = \frac{139}{420} \)
So the value is \( \frac{139}{420} \).
In simple words: Make all denominators 420, then add the numerators: −224 + 147 + 132 + 84 = 139.

Teacher's Note:
1. Subtracting −11/35 turns into adding 11/35.
2. 139 is a prime number, so 139/420 is already in lowest terms.

Question 11. Verify the property x + y = y + x of rational numbers by taking \( x = \frac{-2}{5},\ y = \frac{-9}{10} \).
Answer:
LHS: \( x + y = \frac{-2}{5} + \frac{-9}{10} = \frac{-4}{10} + \frac{-9}{10} = \frac{-13}{10} \).
RHS: \( y + x = \frac{-9}{10} + \frac{-2}{5} = \frac{-9}{10} + \frac{-4}{10} = \frac{-13}{10} \).
LHS = RHS = \( \frac{-13}{10} \).
Hence x + y = y + x is verified (addition of rational numbers is commutative).
In simple words: Both orders give −13/10, so the order of adding does not change the sum.

Teacher's Note:
1. Work out each side separately, then compare.
2. Change −2/5 to −4/10 so both fractions have denominator 10.

Question 12. Verify the property x × (y + z) = x × y + x × z of rational numbers by taking \( x = \frac{-2}{3},\ y = \frac{-4}{6},\ z = \frac{-7}{9} \).
Answer:
LHS: \( y + z = \frac{-4}{6} + \frac{-7}{9} = \frac{-12}{18} + \frac{-14}{18} = \frac{-26}{18} \).
\( x \times (y + z) = \frac{-2}{3} \times \frac{-26}{18} = \frac{52}{54} = \frac{26}{27} \).
RHS: \( x \times y = \frac{-2}{3} \times \frac{-4}{6} = \frac{8}{18} \) and \( x \times z = \frac{-2}{3} \times \frac{-7}{9} = \frac{14}{27} \).
\( x \times y + x \times z = \frac{8}{18} + \frac{14}{27} = \frac{24}{54} + \frac{28}{54} = \frac{52}{54} = \frac{26}{27} \).
LHS = RHS = \( \frac{26}{27} \), so x × (y + z) = x × y + x × z is verified (distributivity of multiplication over addition).
In simple words: Both sides work out to 26/27, so multiplication distributes over addition here.

Teacher's Note:
1. LCM of 6 and 9 is 18; LCM of 18 and 27 is 54.
2. Negative × negative gives positive, so both products on the right are positive.

Question 13. Find: (i) \( \frac{6}{25} \div \frac{3}{10} \) (ii) \( \frac{-9}{44} \div \frac{3}{11} \)
Answer:
To divide by a fraction, multiply by its reciprocal.
(i) \( \frac{6}{25} \div \frac{3}{10} = \frac{6}{25} \times \frac{10}{3} = \frac{60}{75} = \frac{4}{5} \). Answer: \( \frac{4}{5} \).
(ii) \( \frac{-9}{44} \div \frac{3}{11} = \frac{-9}{44} \times \frac{11}{3} = \frac{-99}{132} = \frac{-3}{4} \). Answer: \( \frac{-3}{4} \).
In simple words: Flip the second fraction and multiply, then cancel common factors.

Teacher's Note:
1. Cancel before multiplying: in (ii), 9 and 3 share 3, and 44 and 11 share 11.
2. Keep the negative sign with the numerator.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Simplify: \( \left(\frac{-5}{9} \times \frac{72}{-125}\right) - \left(\frac{11}{17} \times \frac{34}{55}\right) + \left(\frac{28}{-13} \times \frac{-52}{21}\right) \)
Answer:
First bracket: \( \frac{-5}{9} \times \frac{72}{-125} = \frac{5 \times 72}{9 \times 125} = \frac{1 \times 8}{1 \times 25} = \frac{8}{25} \).
Second bracket: \( \frac{11}{17} \times \frac{34}{55} = \frac{1 \times 2}{1 \times 5} = \frac{2}{5} \).
Third bracket: \( \frac{28}{-13} \times \frac{-52}{21} = \frac{28 \times 52}{13 \times 21} = \frac{4 \times 4}{1 \times 3} = \frac{16}{3} \).
Now \( \frac{8}{25} - \frac{2}{5} + \frac{16}{3} \); LCM of 25, 5 and 3 is 75.
\( = \frac{24 - 30 + 400}{75} = \frac{394}{75} \)
So the value is \( \frac{394}{75} \) (that is, \( 5\frac{19}{75} \)).
In simple words: Each bracket simplifies to 8/25, 2/5 and 16/3; then 8/25 − 2/5 + 16/3 = 394/75.

Teacher's Note:
1. In each bracket the two negative signs cancel, giving a positive product.
2. Cancel common factors (5 with 125, 9 with 72, 13 with 52, 7 with 28) before multiplying.

Question 15. Arrange the rational numbers \( \frac{-3}{7},\ \frac{5}{-14},\ -\frac{7}{12} \) in ascending order.
Answer:
Write each with a positive denominator: \( \frac{-3}{7},\ \frac{-5}{14},\ \frac{-7}{12} \).
LCM of 7, 14 and 12 = 84.
\( \frac{-3}{7} = \frac{-36}{84},\quad \frac{-5}{14} = \frac{-30}{84},\quad \frac{-7}{12} = \frac{-49}{84} \).
Since −49 < −36 < −30: \( \frac{-49}{84} < \frac{-36}{84} < \frac{-30}{84} \).
Ascending order: \( -\frac{7}{12} < \frac{-3}{7} < \frac{5}{-14} \).
In simple words: Change all three to a denominator of 84 and compare the numerators −49, −36 and −30.

Teacher's Note:
1. Among negative numbers, the one with the larger absolute value is smaller.
2. Move the minus sign to the numerator before finding the LCM.

Question 16. 5/8 of total number of teachers come by bus, while 1/8 of teachers come by two wheeler. All the other teachers walk to school of which 1/4 walk on their own. If 78 teachers come to school walking on their own, how many teachers taught in that school?
Answer:
Let the total number of teachers be x.
Teachers coming by bus or two wheeler = \( \frac{5x}{8} + \frac{x}{8} = \frac{6x}{8} \).
Teachers who walk = \( x - \frac{6x}{8} = \frac{2x}{8} = \frac{x}{4} \).
Teachers who walk on their own = \( \frac{1}{4} \times \frac{x}{4} = \frac{x}{16} \).
So \( \frac{x}{16} = 78 \), which gives x = 78 × 16 = 1248.
So 1248 teachers taught in the school.
Check: walkers = 1248 ÷ 4 = 312, and 312 ÷ 4 = 78.
In simple words: Walkers are 1/4 of all teachers, and those walking alone are 1/4 of that, i.e. 1/16 of all; 78 × 16 = 1248.

Teacher's Note:
1. Find the fraction of the whole for each group step by step.
2. "1/4 of the walkers" means 1/4 of x/4, not 1/4 of x.

Section E: Case Study Based Question (4 Marks)

Question 17. The diagram shows the wingspans of different species of birds. Use the diagram to answer the questions given below:
(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull?
(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay?
Answer:
(a) Albatross: \( 3\frac{3}{5} = \frac{18}{5} \) m. Sea gull: \( 1\frac{7}{10} = \frac{17}{10} \) m.
Difference = \( \frac{18}{5} - \frac{17}{10} = \frac{36}{10} - \frac{17}{10} = \frac{19}{10} \) m.
The Albatross's wingspan is longer by \( \frac{19}{10} \) m = \( 1\frac{9}{10} \) m (1.9 m).
(b) Golden eagle: \( 2\frac{1}{2} = \frac{5}{2} \) m. Blue jay: \( \frac{41}{100} \) m.
Difference = \( \frac{5}{2} - \frac{41}{100} = \frac{250}{100} - \frac{41}{100} = \frac{209}{100} \) m.
The Golden eagle's wingspan is longer by \( \frac{209}{100} \) m = \( 2\frac{9}{100} \) m (2.09 m).
In simple words: Change the mixed numbers to improper fractions, use a common denominator and subtract.

Teacher's Note:
1. Convert mixed numbers to improper fractions before subtracting.
2. Check with decimals: 3.6 − 1.7 = 1.9 and 2.5 − 0.41 = 2.09.