CBSE Class 8 Mathematics Algebraic Expressions and Identities Practice Paper Assignment Set 11 with Answers (Old NCERT Syllabus)

This Class 8 Maths Algebraic Expressions practice paper contains 17 questions for 40 marks, arranged in Sections A to E, with a step-by-step solution for each question. It follows the earlier (old) NCERT Class 8 Maths syllabus (Algebraic Expressions and Identities), so it helps state-board students and anyone revising the chapter. It tests identifying terms, adding and subtracting expressions, multiplying monomials, binomials and polynomials, finding areas and volumes as products of monomials, and forming simple equations from a case study.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Simplify 3x (4x − 5) + 3 and find its values for (i) x = 3 (ii) x = \( \frac{1}{2} \).
Answer:
3x(4x − 5) + 3 = 12x² − 15x + 3
(i) x = 3: 12 × 9 − 15 × 3 + 3 = 108 − 45 + 3 = 66
(ii) x = \( \frac{1}{2} \): 12 × \( \frac{1}{4} \) − 15 × \( \frac{1}{2} \) + 3 = 3 − \( \frac{15}{2} \) + 3 = 6 − \( \frac{15}{2} \) = \( -\frac{3}{2} \)
In simple words: Open the bracket to get 12x² − 15x + 3, then put in each value of x.

Teacher's Note:
1. Simplify first, then substitute; it saves work when you need several values.
2. Square the value of x before multiplying by 12: 12 × (1/2)² = 12 × 1/4 = 3.

Question 8. (a) Add: p(p − q), q(q − r) and r(r − p)
(b) Add: 2x(z − x − y) and 2y(z − y − x)
Answer:
(a) (p² − pq) + (q² − qr) + (r² − rp)
= p² + q² + r² − pq − qr − rp
(b) (2xz − 2x² − 2xy) + (2yz − 2y² − 2xy)
= −2x² − 2y² − 4xy + 2yz + 2xz
In simple words: Open each bracket, then add only the like terms (in (b), −2xy and −2xy give −4xy).

Teacher's Note:
1. xy and yx are like terms, so they combine.
2. In (a) no two terms are alike, so the answer is just all six terms together.

Question 9. Subtract: 3l(l − 4m + 5n) from 4l(10n − 3m + 2l)
Answer:
4l(10n − 3m + 2l) − 3l(l − 4m + 5n)
= (40ln − 12lm + 8l²) − (3l² − 12lm + 15ln)
= 40ln − 12lm + 8l² − 3l² + 12lm − 15ln
= 5l² + 25ln
In simple words: Expand both, change every sign of the second one, and combine: the lm terms cancel, leaving 5l² + 25ln.

Teacher's Note:
1. "Subtract A from B" means B − A.
2. When you remove a bracket with a minus sign before it, change the sign of every term inside.

Section C: Short Answer Questions (3 Marks Each)

Question 10. What should be subtracted from x⁴ + 2x² − 3x + 7 to get x³ + x² + x − 1?
Answer:
Required expression = (x⁴ + 2x² − 3x + 7) − (x³ + x² + x − 1)
= x⁴ + 2x² − 3x + 7 − x³ − x² − x + 1
= x⁴ − x³ + (2x² − x²) + (−3x − x) + (7 + 1)
= x⁴ − x³ + x² − 4x + 8
Check: (x⁴ + 2x² − 3x + 7) − (x⁴ − x³ + x² − 4x + 8) = x³ + x² + x − 1.
In simple words: Take the second expression away from the first; whatever is left is what must be subtracted.

Teacher's Note:
1. If A − ? = B, then ? = A − B.
2. The constant term is 7 − (−1) = 7 + 1 = 8; it is easy to slip here.

Question 11. Subtract 4p²q − 3pq + 5pq² − 8p + 7q − 10 from 18 − 3p − 11q + 5pq − 2pq² + 5p²q
Answer:
(18 − 3p − 11q + 5pq − 2pq² + 5p²q) − (4p²q − 3pq + 5pq² − 8p + 7q − 10)
= 18 − 3p − 11q + 5pq − 2pq² + 5p²q − 4p²q + 3pq − 5pq² + 8p − 7q + 10
= (18 + 10) + (−3p + 8p) + (−11q − 7q) + (5pq + 3pq) + (−2pq² − 5pq²) + (5p²q − 4p²q)
= 28 + 5p − 18q + 8pq − 7pq² + p²q
In simple words: Change every sign of the expression being subtracted, then add like terms.

Teacher's Note:
1. p²q and pq² are unlike terms; check the powers carefully before combining.
2. Writing the like terms in brackets keeps the working tidy.

Question 12. Multiply (i) (x − 4) and (2x + 3) (ii) (x − y) and (3x + 5y)
Answer:
(i) (x − 4)(2x + 3) = x(2x + 3) − 4(2x + 3)
= 2x² + 3x − 8x − 12 = 2x² − 5x − 12
(ii) (x − y)(3x + 5y) = x(3x + 5y) − y(3x + 5y)
= 3x² + 5xy − 3xy − 5y² = 3x² + 2xy − 5y²
In simple words: Multiply each term of the first bracket by each term of the second, then combine the middle terms.

Teacher's Note:
1. A binomial times a binomial gives four products before simplifying.
2. Check (i) at x = 1: (−3)(5) = −15 and 2 − 5 − 12 = −15.

Question 13. Simplify (a + b)(2a − 3b + c) − (2a − 3b)c.
Answer:
(a + b)(2a − 3b + c) = 2a² − 3ab + ac + 2ab − 3b² + bc = 2a² − ab − 3b² + ac + bc
(2a − 3b)c = 2ac − 3bc
Difference = 2a² − ab − 3b² + ac + bc − 2ac + 3bc
= 2a² − ab − 3b² − ac + 4bc
In simple words: Expand both products, subtract the second, and combine the ac and bc terms.

Teacher's Note:
1. Put the second product in brackets before subtracting so every sign changes.
2. ac − 2ac = −ac and bc + 3bc = 4bc.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Simplify: (i) (a + b)(c − d) + (a − b)(c + d) + 2(ac + bd)
(ii) (x + y)(2x + y) + (x + 2y)(x − y)
Answer:
(i) (a + b)(c − d) = ac − ad + bc − bd
(a − b)(c + d) = ac + ad − bc − bd
2(ac + bd) = 2ac + 2bd
Sum = (ac + ac + 2ac) + (−ad + ad) + (bc − bc) + (−bd − bd + 2bd) = 4ac
(ii) (x + y)(2x + y) = 2x² + xy + 2xy + y² = 2x² + 3xy + y²
(x + 2y)(x − y) = x² − xy + 2xy − 2y² = x² + xy − 2y²
Sum = 3x² + 4xy − y²
In simple words: Expand every product and add; in (i) everything cancels except 4ac.

Teacher's Note:
1. Expand each product separately before adding, so no term is missed.
2. Check (ii) at x = y = 1: 2 × 3 + 3 × 0 = 6, and 3 + 4 − 1 = 6.

Question 15. Obtain the product of (i) xy, yz, zx (ii) a, −a², a³ (iii) 2, 4y, 8y², 16y³ (iv) a, 2b, 3c, 6abc
Answer:
(i) xy × yz × zx = x²y²z²
(ii) a × (−a²) × a³ = −a⁶
(iii) 2 × 4y × 8y² × 16y³ = (2 × 4 × 8 × 16) × y¹⁺²⁺³ = 1024y⁶
(iv) a × 2b × 3c × 6abc = (1 × 2 × 3 × 6) × a²b²c² = 36a²b²c²
In simple words: Multiply the numbers together and add the powers of each letter.

Teacher's Note:
1. For the same base, multiply by adding powers: a × a² × a³ = a⁶.
2. One negative factor makes the whole product negative.

Question 16. Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.
(i) 5a, 3a², 7a⁴ (ii) 2p, 4q, 8r (iii) xy, 2x²y, 2xy² (iv) a, 2b, 3c
Answer:
Volume = length × breadth × height.
(i) 5a × 3a² × 7a⁴ = (5 × 3 × 7)a¹⁺²⁺⁴ = 105a⁷
(ii) 2p × 4q × 8r = 64pqr
(iii) xy × 2x²y × 2xy² = (1 × 2 × 2) × x¹⁺²⁺¹ × y¹⁺¹⁺² = 4x⁴y⁴
(iv) a × 2b × 3c = 6abc
In simple words: Multiply the three dimensions: numbers with numbers, letters with letters.

Teacher's Note:
1. Volume always has three length factors multiplied together.
2. Count the powers of x and y separately in (iii).

Section E: Case Study Based Question (4 Marks)

Question 17. A playground is in the shape of a square. The area of the square PQRS is 256 m² with each side (x + 2) m. One day Suraj along with his two friends Ajay and Aman went to play there with their bicycles. Someone stole Suraj's bicycle, but Ajay and Aman helped him by contributing ₹ (4a + 60) and ₹ (6a + 10) respectively, to buy a new bicycle. The cost of the bicycle was ₹ 4200.
On the basis of the information given in the passage, answer the following questions.
(a) Find the value of x. (1)
(b) What is the value of a? (1)
(c) What was the amount given by Ajay and Aman to Suraj? (1)
(d) What is the perimeter of the playground? (1)
Answer:
(a) (x + 2)² = 256 = 16², so x + 2 = 16 and x = 14.
(b) (4a + 60) + (6a + 10) = 4200, so 10a + 70 = 4200, 10a = 4130 and a = 413.
(c) Ajay gave 4 × 413 + 60 = ₹ 1,712; Aman gave 6 × 413 + 10 = ₹ 2,488 (total ₹ 4,200).
(d) Side = 14 + 2 = 16 m, so perimeter = 4 × 16 = 64 m.
In simple words: The side is √256 = 16 m, so x = 14; the two gifts add to ₹ 4,200, giving a = 413.

Teacher's Note:
1. Area of a square = side², so the side is the square root of the area.
2. Check (c): 1712 + 2488 = 4200, the cost of the bicycle.