CBSE Class 10

CBSE Extra Questions for Class 10 English Free PDF Download: Here we are providing NCERT Extra Questions for Class 10 English First Flight, Footprints Without Feet, Literature Reader. Students can get Class 10 English NCERT Solutions, Chapter Wise CBSE Class 10 English Important Questions and Answers were designed by subject expert teachers.

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Extra Questions for Class 10 English Beehive, Moments, Literature Reader Important Questions

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Extra Questions for Class 10 English Literature Reader

1. Two Gentlemen of Verona Extra Questions
2. Mrs. Packletide’s Tiger Extra Questions
3. The Letter Extra Questions
4. A Shady Plot Extra Questions
5. Patol Babu, Film Star Extra Questions
6. Virtually True Extra Questions
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9. Not Marble, Nor the Gilded Monuments Extra Questions
10. Ozymandias Extra Questions
11. The Rime of the Ancient Mariner Extra Questions
12. Snake Extra Questions
13. The Dear Departed Extra Questions
14. Julius Caesar Extra Questions
15. From the Diary of Anne Frank Extra Questions
16. The Story of My Life Extra Questions

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3. A Tiger in the Zoo Extra Questions
4. How to Tell Wild Animals Extra Questions
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6. Amanda Extra Questions
7. Animals Extra Questions
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10. The Tale of Custard the Dragon Extra Questions
11. For Anne Gregory Extra Questions

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1. A Triumph of Surgery Extra Questions
2. The Thief’s Story Extra Questions
3. The Midnight Visitor Extra Questions
4. A Question of Trust Extra Questions
5. Footprints without Feet Extra Questions
6. The Making of a Scientist Extra Questions
7. The Necklace Extra Questions
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Here we are providing Pair of Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.

Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers

Quadratic Equations Class 10 Extra Questions Very Short Answer Type

Question 1.
What will be the nature of roots of quadratic equation 2x² + 4x – n = 0?
Solution:
D = b² – 4ac
⇒ 4² – 4 x 2 (-7)
⇒ 16 + 56 = 72 > 0
Hence, roots of quadratic equation are real and unequal.

Question 2.
If \(\frac{1}{2}\) is a root of the equation x² + kx – \(\frac{5}{4}\) = 0, then find the value of k.
Solution:
∴ \(\frac{1}{2}\) is a root of quadratic equation.
∴ It must satisfy the quadratic equation.

Question 3.
If ax² + bx + c = 0 has equal roots, find the value of c.
Solution:
For equal roots D = 0
i.e., b² – 4ac = 0
⇒ b² = 4 ac
⇒ c = \(\frac{b^{2}}{4 a}\)

Question 4.
If a and b are the roots of the equation x² + ax – b = 0, then find a and b.
Solution:
Sum of the roots = a + b = – \(\frac{B}{A}\) = – a
Product of the roots = ab = \(\frac{B}{A}\) = – b
= a + b = – a and ab = -b
⇒ 2a = -b and a = -1
⇒ b = 2 and a = -1

Question 5.
Show that x = – 2 is a solution of 3x² + 13x + 14 = 0.
Solution:
Put the value of x in the quadratic equation,
⇒ LHS = 3x² + 13x + 14
⇒ 3(-2)² + 13(-2) + 14
⇒ 12 – 26 + 14 = 0
⇒ RHS Hence, x = -2 is a solution.

Question 6.
Find the discriminant of the quadratic equation 4√2x² + 8x + 2√2 = 0).
Solution:
D = 62 – 4ac = (8)² – 4(4√2)(2√2)
⇒ 64 – 64 = 0

Quadratic Equations Class 10 Extra Questions Short Answer Type 1

Question 1.
State whether the equation (x + 1)(x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution:
(x + 1)(x – 2) + x = 0
⇒ x² – x – 2 + x = 0
⇒ x² – 2 = 0
D = b² – 4ac
⇒ (-4(1)(-2) = 8 > 0
∴ Given equation has two distinct real roots.

Question 2.
Is 0.3 a root of the equation x² – 0.9 = 0? Justify.
Solution:
∵ 0.3 is a root of the equation x² – 0.9 = 0
∴ x² – 0.9 = (0.3)² – 0.9 = 0.09 – 0.9 ≠ 0
Hence, 0.3 is not a root of given equation.

Question 3.
For what value of k, is 3 a root of the equation 2x² + x + k = 0?
Solution:
3 is a root of 2x² + x + k = 0, when
⇒ 2(3)² + 3 + k = 0
⇒ 18+ 3 + k = 0
⇒ k = – 21

Question 4.
Find the values of k for which the quadratic equation 9x² – 3kx + k = 0 has equal roots.
Solution:
For equal roots:
D = 0
⇒ b² – 4ac = 0
⇒ (- 3k)² – 4 × 9 × k = 0
⇒ 9k² = 36k
⇒ k = 4

Question 5.
Find the value of k for which the equation x² + k(2x + k – 1)+ 2 = 0 has real and equal roots.
Solution:
Given quadratic equation: x² + k(2x + k-1) + 2 = 0)
= x² + 2kx + (k² – k + 2) = 0
For equal roots, b² – 4ac = 0
⇒ 4k² – 4k² + 4k – 8 = 0
⇒ 4k = 8
⇒ k = 2

Question 6.
If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then find the value of k.
Solution:
Since – 5 is a root of the equation 2x² + px – 15 = 0
∴ 2(-5)² + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p(x² + x) + k = 0 or 7x² + 7x + k = 0 has equal roots
∴ D = 0
i.e., b² – 4ac = 0 or 49- 4 × 7k = 0
⇒ k = \(\frac{49}{28}\) = \(\frac{7}{4}\)

Question 7.
Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Solution:
Yes, x² – 4x + 1 = 0 is a quadratic equation with rational co-efficients.

Question 8.
Write the set of values of k for which the quadratic equation 2x² + kx + 8 = 0 has real roots.
Solution:
For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ k² – 4(2)(8) ≥ 0
⇒ k² – 64 ≥ 0
⇒ k² ≥ 64
⇒ k ≤ -8 and k ≥ 8

Question 9.
Solve the quadratic equation 2x² + ax – a² = 0 for x.
Solution:
2x² + ax – a² = 0
Here, a = 2, b = a and c = -a².
Using the formula,

Question 10.
Find the roots of the quadratic equation
Solution:
The given quadratic equation is
√2x² + 7x + 5√2 = 0
By applying mid term splitting, we get
√2x² + 2x + 5x + 5√2 = 0
⇒ √2x(x + √2) + 5(x + √2)
⇒ (√2x + 5) + 5(x + √2) = 0
⇒ x = \(\frac{-5}{\sqrt{2}}/latex], -√2 or [latex]\frac{-5 \sqrt{2}}{2}\), -√2

Question 11.
Find the values of p for which the quadratic equation 4x² + px + 3 = 0 has equal roots.
Solution:
For equal roots;
D = 0
⇒ b² – 4ac = 0
⇒ p² – 4 × 4 × 3 = 0
⇒ p² – 48 = 0
⇒ p² = 48
⇒ p = ± √48
⇒ p = 4√3 or -4√3

Question 12.
Solve for x: √13x? – 2√3x – 2√3 = 0
Solution:
√3x² – 2√3x – 2√3 = 0
⇒ √3x² – 3√2x + √2x – 2√3 = 0
⇒ √3x(x – √6) + √2(x – √6) = 0
⇒ (√3x + 2)(x – √6) = 0
⇒ √3x + √2 = 0 or x – √6 = 0
⇒ x = \(\frac{-√2}{√3}\) or x = √6

Question 13.
If x = \(\frac{2}{3}\) and x = -3 are roots of the quadratic equation ax² + 7x + b = 0, find the values of a and b.
Solution:
Let us assume the quadratic equation be Ax² + Bx + C = 0.
Sum of the roots = –\(\frac{B}{A}\)

Question 14.
A two-digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
Solution:
Let the ten’s digit be x and unit’s digit = y
Number 10x + y

Question 15.
Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.
Solution:
Let the roots of the given equation be a and 6α.
Thus the quadratic equation is (x – a) (x – 6α) = 0
⇒ x² – 7αx + 6α² = 0 …(i)
Given equation can be written as

Question 16.
If ad ≠ bc, then prove that the equation
(a² + b²) x² + 2(ac + bd)x + (c²+ d²) = 0 has no real roots.
Solution:
The given quadratic equation is (a² + b²⁾x² + 2(ac + bd)x +(c²+ d²) = 0
D = b² – 4ac
= 4(ac + bd)² – 4(a² + b²) (c²+ d²)
= -4(a²d² + b²c²– 2abcd) = – 4(ad – bc)²
Since ad ≠ bc
Therefore D < 0
Hence, the equation has no real roots.

Question 17.
Solve for x: √13x² – 2x – 8√3 = 0
Solution:
√3x² – 2x – 8√3 = 0
By mid term splitting
⇒ √3x² – 6x + 4x – 8√3 = 0
⇒ √3x(x – 2/3) + 4 (x – 2/3) = 0
⇒ (x – 2√3)(√3x + 4) = 0
⇒ Either (x – 2√3) = 0 or (√3x + 4) = 0
⇒ x = \(\frac{-4}{\sqrt{3}}\), 2√3

Quadratic Equations Class 10 Extra Questions Short Answer Type 2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) √2x² + 7x + 5√2 = 0 (ii) 2x² – x + \(\frac{1}{8}\) = 0
Solution:
(i) We have, √2x² + 7x + 5√2 = 0
= √2x² + 5x + 2x + 5√2 = 0
x(√2x + 5) + √2 (√2x + 5) = 0
= (√2x + 5)(x + √2) = 0
∴ Either √2x + 5 = 0 or x + √2 = 0
∴ x = – \(\frac{5}{\sqrt{2}}\) or x = -√2
Hence, the roots are – \(\frac{5}{\sqrt{2}}\) and -√2.
(ii) We have, 2x² – x + \(\frac{1}{8}\) = 0

Question 2.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² + x – 4 = 0
(ii) 4x² + 4√3x + 3 = 0
Solution:
(i) We have, 2x² + x – 4 = 0
On dividing both sides by 2, we have

(ii) We have, 4x² + 4√3x + 3 = 0

Question 3.
Find the roots of the following quadratic equations by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0
(ii) 4x² + 4√3x + 3 = 0
Solution:
(i) We have, 2x² – 7x + 3 = 0
Here, a = 2, b = -7 and c = 3
Therefore, D = b² – 4ac
⇒ D = (-7)² – 4 × 2 × 3 = 49 – 24 = 25
∵ D > 0, ∴ roots exist.


So, the roots of given equation are 3 and \(\frac{1}{2}\)

(ii) We have, 4x² + 4√3x + 3 = 0
Here, a = 4, b = 4√3 and c = 3
Therefore, D = b² – 4ac = (4√3)² – 4 × 4 × 3 = 48 – 48 = 0
∴ D = 0, roots exist and are equal.

Question 4.
Using quadratic formula solve the following quadratic equation:
p²x² + (p² – q²) x – q² = 0
Solution:
We have, p²x² + (p² – q²) x – q² = 0
Comparing this equation with ax² + bx + c = 0, we have
a = p², b = p² – q² and c = – q²
∴ D = b² – 4ac
⇒ (p² – q)² – 4 × p² × (-q²)
⇒ (p² – q²)² + 4p²q²
⇒ (p² + q³)² > 0
So, the given equation has real roots given by

Question 5.
Find the roots of the following equation:

Solution:

⇒ (x + 3) (x – 6)
⇒ -20 or x² – 3x + 2 = 0
⇒ x² – 2x -x + 2 = 0
⇒ x(x – 2) -1(x – 2) = 0)
⇒ (x – 1) (x – 2) = 0
⇒ x = 1 or x = 2
Both x = 1 and x = 2 are satisfying the given equation. Hence, x = 1, 2 are the solutions of the equation.

Question 6.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 3x² – 4√3x + 4 = 0) (ii) 2x² – 6x + 3 = 0
Solution:
(i) We have, 3x² – 4√3x + 4 = 1
Here, a = 3, b = – 4√3 and c = 4
Therefore,
D = b² – 4ac
⇒ (- 4√3)² – 4 × 3 × 4
⇒ 48 – 48 = 0
Hence, the given quadratic equation has real and equal roots.

(ii) Wehave, 2x² – 6x + 3 = 0
Here, a = 2, b = -6, c = 3
Therefore, D = b² – 4ac
= (-6)² 4 × 2 × 3 = 36 – 24 = 12 > 0
Hence, given quadratic equation has real and distinct roots.

Question 7.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Solution:
(i) We have, 2x² + kx + 3 = 0
Here, a = 2, b = k, c = 3
D = b² – 4ac = k² – 4 × 2 × 3 = k² – 24 For equal roots
D = 0
i.e., k² – 24 = 0
⇒ ķ² = 24
⇒ k = ± √24
⇒ k = + 2√6

(ii) We have, kx(x – 2) + 6 = 0
⇒ kx² – 2kx + 6 = 0
Here, a = k, b = – 2k, c = 6
For equal roots, we have
D = 0
i.e., b² – 4ac = 0
⇒ (-2k)² – 4 × k × 6 = 0
⇒ 4k² – 24k = 0
⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k = 0 6)ecause if k = 0 then given equation will not be a quadratic equation).
So, k = 6.

Question 8.
If the roots of the quadratic equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.
Solution:
Since the equation (a – b)x² + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)² – 4(a – b) (c – a) = 0
⇒ b² + c² – 2bc – 4(ac – a² – bc + ab)
⇒ b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0
⇒ 4a² + b² + c² – 4ab + 2bc – 4ac = 0
⇒ (2a)² + (- b)² + (-c)² + 2(2a) (-b) + 2(-b) (-c) + 2(-c) 2a = 0
⇒ (2a – b – c)² = 0
⇒ 2a – b – c = 0
⇒ 2a = b + c. Hence Proved

Question 9.
If the equation (1 + m²)x² + 2mcx + c²– a² = 0 has equal roots, show that c² = a² (1 + m²).
Solution:
The given equation is (1 + m²) x² + (2mc) x + (c²– a²) = 0
Here, A = 1 + m², B = 2mc and C = c² – a²
Since the given equation has equal roots, therefore D = 0 = B² – 4AC = 0.
⇒ (2mc)² – 4(1 + m²) (c² – a²) = 0
⇒ 4m²c² – 4(c² – a² + m²c² – m²a²) = 0
⇒ m²c² – c² + a² – m²c² + m²a² = 0. [Dividing throughout by 4]
⇒ – c² + a² (1 + m²) = 0
⇒ c² = a(1 + m²) Hence Proved

Question 10.
If sin θ and cos θ are roots of the equation ax² + bx + c = 0, prove that a² – b² + 2ac = 0.
Solution:

Question 11.
Determine the condition for one root of the quadratic equation ax² + bx + c = 0 to be thrice the other.
Solution:
Let the roots of the given equation be a and 3α.

Question 12.
Salve for

Solution:

⇒ (4x – 2)(2x – 1) – (3x + 9)(x + 3) = 5(x + 3)(2x – 1)
⇒ (8x² – 4x – 4x + 2) – (3x² + 9x + 9x + 27) = 5(2x² – x + 6x – 3)
⇒ 8x² – 8x + 2 – 3x² – 18x – 27 = 10x² + 25x – 15
⇒ 5x² – 26x – 25 = 10x² + 25x – 15
⇒ 5x² + 51x + 10 = 0
⇒ 5x² + 50x + x + 10 = 0
⇒ 5x (x + 10) + 1 (x + 10) = 0
⇒ (5x + 1) (x + 10) = 0
⇒ 5x + 1 = 0 or x + 10 = 0
⇒ x = \(\frac{-1}{5}\) or x = -10

Question 13.
Solve the equation

Solution:

⇒ (4 – 3x) (2x + 3) = 5x ⇒ 8x – 6x² + 12 – 9x = 5x
⇒ 6x² + 6x – 12 = 0
⇒ x² + x – 2 = 0
⇒ x² + 2x + x – 2 = 0
⇒ x(x + 2)-1(x + 2) = 0
⇒ (x – 1)(x + 2) = 0
⇒ x – 1 = 0 or x + 2 = 0
⇒ x = 1 or x = -2

Question 14.
Solve for

Solution:

⇒ (16 – x) (x + 1) = 15x
⇒ 16x – x² + 16 – x = 15x
⇒ x² + 15x – 15x – 16 = 0
⇒ x² = 16
⇒ x = ± 4

Question 15.
Solve for x: + 5x – (a² + a – 6) = 0
Solution:
⇒ x² + 5x – (a² + a – 6) = 0
⇒ x² + 5x – (a? + 3a – 2a – 6) = 0
⇒ x² + 5x – [a(a + 3) -2 (a + 3)] = 0
⇒ x² + 5x – (a – 2) (a + 3) = 0
∴ x² + (a + 3)x – (a – 2) x – (a – 2) (a + 3) = 0
⇒ x[x + (a + 3)]-(a – 2) [X + (a + 3)] = 0
⇒ [{x + (a + 3)} {x – (a – 2)}] = 0
∴ x = -(a + 3) or x = (a -2)
⇒ -(a + 3), (a – 2)
Alternative method
x² + 5x -(a² + a – 6) = 0

Question 16.
Solve for

Solution:

⇒ 2x(2x + 3) + (x – 3) + (3x + 9) = 0
⇒ 4x² + 10x + 6 = 0
⇒ 2x² + 5x + 3 = 0
⇒ (x + 1) (2x + 3) = 0
⇒ x = -1, x = – \(\frac{3}{2}\)
But x ≠ – \(\frac{3}{2}\)
∴ x = -1

Question 17.
Solve for

Solution:

⇒ 3(x – 3 + x – 1) = 2(x – 1)(x – 2)(x – 3)
⇒ 3(2x – 4) = 2(x – 1)(x – 2)(x – 3)
⇒ 3 × 2(x – 2) = 2(x – 1)(x − 2)(x – 3)
⇒ 3 = (x – 1)(x – 3) i.e., x² – 4x = 0
⇒ x(x – 4) = 0 ∴ x = 0, x = 4

Question 18.
If the roots of the equation (c- ab)x² – 2(a²– bc)x + b² – ac = 0 in x are equal, then show that either a = 0 or a³ + b³ + c³ = 3abc.
Solution:
For equal roots D = 0
Therefore 4(a² – bc)² – 4(c² – ab) (b² – ac) = 0
⇒ 4[a⁴ + b²c² – 2a²bc – b²c² + ac³ + ab³ – a²bc] = 0.
⇒ a(a³ + b³ + c³ – 3abc) = 0
Either a = 0 or a³ + b³ + c³ = 3abc

Question 19.
If the roots of the quadratic equation
(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
are equal, then show that a = b = c.
Solution:
Given (x – a) (x – b) + (x – b) (x – c) + (x – 6) (x – a) = 0
⇒ x² – ax – bx + ab + x² – bx – cx + bc + x² – cx – ax + ac = 0
⇒ 3x² – 2(a + b + c)x + ab + bc + ca = 0
Now, for equal roots D = 0
⇒ B² – 4AC = 0
⇒ 4(a + b + c)² – 12(ab + bc + ca) = 0
4a² + 4b² + 4c² + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0
⇒ 2[2a² + 2b² + 2co – 2ab – 2bc – 2ca] = 0
⇒ 2[(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca)] = 0
⇒ [(a – b)² + (b – c)² + (c – a)²] = 0
⇒ a – b = 0, b – c = 0, c – a = 0
⇒ a = b, b = c, c = a
⇒ a = b = c

Quadratic Equations Class 10 Extra Questions Long Answers

Question 1.
Using quadratic formula, solve the following equation for x:
abx² + (b² – ac) x – bc = 0
Solution:
We have, abx² + (b² – ac) x – bc = 0
Here, A = ab, B = b² – ac, C = – bc

Question 2.
Find the value of p for which the quadratic equation
(2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.
Solution:
Since the quadratic equation has equal roots, D = 0
i.e., b² – 4ac = 0
In (2p + 1 )x² – (7p + 2)x + (7p – 3) = 0
Here, a = (2p + 1), b = -(7p + 2), c = (7p – 3)

Question 3.
Solve for

Solution:

Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is Find his present age.
Solution:
Let the present age of Rehman be x years.
So, 3 years ago, Rehman’s age = (x – 3) years
And 5 years from now, Rehman’s age = (x + 5) years
Now, according to question, we have

But x ≠ -3 (age cannot be negative)
Therefore, present age of Rehman = 7 years.

Question 5.
The difference of two natural numbers is 5 and the difference of their reciprocals is \(\frac{1}{10}\). Find the numbers.
Solution:
Let the two natural numbers be x and y such that x > y.
According to the question,
Difference of numbers, x – y = 5 ⇒ x = 5 + y …..(i)
Difference of the reciprocals,

∴ y is a natural number.
∵ y = 5
Putting the value of y in (i), we have
⇒ x = 5 + 5
⇒ x = 10
The required numbers are 10 and 5.

Question 6.
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Solution:
Let the two consecutive odd numbers be x and x + 2.

Hence, the numbers are 13 and 15 or -15 and -13.

Question 7.
The sum of two numbers is 15 and the sum of their reciprocals is 3. Find the numbers.
Solution:
Let the numbers be x and 15 – x.
According to given condition,

⇒ 150 = 3x(15 – x)
⇒ 50 = 15x – x²
⇒ x² – 15x + 50 = 0
⇒ x² – 5x – 10x + 50 = 0
⇒ x(x – 5) -10(x – 5) = 0
⇒ (x – 5)(x – 10) = 0
⇒ x = 5 or 10.
When x = 5, then 15 – x = 15 – 5 = 10
When x = 10, then 15 – x = 15 – 10 = 5
Hence, the two numbers are 5 and 10.

Question 8.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics be x.
Therefore, Shefali’s marks in English is (30 – x).
Now, according to question,
⇒ (x + 2) (30 – x – 3) = 210
⇒ (x + 2) (27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ 25x – x² + 54 – 210 = 0
⇒ 25x – x² – 156 = 0
⇒ -(x² – 25x + 156) = 0
⇒ x² – 25x + 156 = 0
= x² – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13) (x – 12) = 0
Either x – 13 or x – 12 = 0
∴ x = 13 or x = 12
Therefore, Shefali’s marks in Mathematics = 13
Marks in English = 30 – 13 = 17
or Shefali’s marks in Mathematics = 12
marks in English = 30 – 12 = 18.

Question 9.
A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let the uniform speed of the train be x km/h.

But x cannot be negative, so x ≠ – 45
therefore, x = 40
Hence, the uniform speed of train is 40 km/h.

Question 10.
The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let x be the length of the side of first square and y be the length of side of the second square.
Then, x² + y² = 468 …(i)
Let x be the length of the side of the bigger square.
4x – 4y = 24
⇒ x – y = 6 or x = y + 6 …(ii)
Putting the value of x in terms of y from equation (ii), in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468 or 232 + 12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
Either y + 18 = 0 or y – 12 = 0
⇒ y = -18 or y = 12
But, sides cannot be negative, so y = 12
Therefore, x = 12 + 6 = 18
Hence, sides of two squares are 18 m and 12 m.

Question 11.
Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence, Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Solution:
Seven years ago, let Swati’s age be x years. Then, seven years ago Varun’s age was 5x² years.
∴ Swati’s present age = (x + 7) years
Varun’s present age = (5x² + 7) years
Three years hence,
Swati’s age = (x + 7 + 3) years = (x + 10) years
Varun’s age (5x² + 7 + 3) years = (5x² + 10) years
According to the question,

Hence, Swati’s present age = (2 + 7) years = 9 years
and Varun’s present age = (5 × 2² + 7) years = 27 years

Question 12.
A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.
Solution:
Let the original speed of the train = x km/h.


Therefore, the usual speed of the train = 25 km/h.

Question 13.
A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Let the digit at tens place be x.

Question 14.
If twice the area of a smaller square is subtracted from the area of a larger square; the result is 14 cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of the two squares.
Solution:
Let the sides of the larger and smaller squares be x and y respectively. Then
x² – 2y² = 14 …..(i)
and 2x² + 3y² = 203 ……(ii)
Operating (ii) -2 × (i), we get
⇒ 2x² + 3y² – (2x² – 4y²) = 203 – 2 × 14
⇒ 2x² + 3y² – 2x² + 4y² = 203 – 28
⇒ 7y² = 175
⇒ y² = 25
⇒ y ± 15
⇒ y = 5 [∵ Side cannot be negative]
By putting the value of y in equation (i), we get
x² – 2 × 5 = 14
⇒ x² – 50 = 14 or x² = 64
∴ x = 8 or x = 8
∴ Sides of the two squares are 8 cm and 5 cm.

Question 15.
If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Solution:
Let the present age of Zeba be x years
Age before 5 years = (x – 5) years According to given condition,
⇒ (x – 5)² = 5x + 11
⇒ x² + 25 – 10x = 5x + 11
⇒ x² – 10x – 5x + 25 – 11 = 0
⇒ x² – 15x + 14 = 0
⇒ x² – 14x – x + 14 = 0
⇒ x(x – 14) -1(x – 14) = 0
⇒ (x – 1)(x – 14) = 0
⇒ x – 1 = 0 or x – 14 = 0
x = 1 or x = 14
But present age cannot be 1 year.
∴ Present age of Zeba is 14 years.

Question 16.
A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/h.
Speed of motorboat in still water = 18 km/h
Speed of motorboat in upstream = (18 – x) km/h
Speed of motorboat in downstream = (18 + x) km/h
Distance travelled = 24 km.

Speed of motorboat = 6 km/h. (∵Speed cannot be negative)

Question 17.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Solution:
Let the natural number be x
According to the question,
⇒  x + 12 = \(\frac{60}{x}\)
= x² + 12x – 160 = 0
= x² + 20x – 8x – 160 = 0
= x(x + 20) -8(x + 20) = 0
= (x + 20) (x – 8) = 0
x = -20 (Not possible) or x = 8
Hence, the required natural number is 8.

Question 18.
The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Solution:
Let the two consecutive multiples of 7 be x and x +7.
x² + (x + 7)² = 637
= x² + x² + 49 + 14x = 637
= 2x² + 14x – 588 = 0
= x² + 7x – 294 = 0
= x² + 21x – 14x – 294 = 0
= x(x + 21) – 14(x + 21) = 0
= x(x + 14) (x + 21) = 0
= x = 14 or x = -21
The multiples are 14 and 21.

Question 19.
Solve for

Solution:

Question 20.
Find the positive value(s) of k for which both quadratic equations x² + kx + 64 = 0 and x² – 8x + k = 0 will have real roots.
Solution:
(i) For x² + kx + 64 = 0 to have real roots
⇒ k² – 4(1)(64) ≥ 0 i.e., k² – 256 ≥ 0
⇒ k ≥ ± 16

(ii) For x² – 8x + k = 0 to have real roots
⇒ (-8)² – 4(k) ≥ 0 i.e., 64 – 4k ≥ 0
⇒ k ≤ ± 16
For (i) and (ii) to hold simultaneously k = 16

Question 21.
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Solution:
Let the speed of stream be x km/h.
∴ Speed of boat upstream = (15 – x) km/h.
Speed of boat downstream = (15 + x) km/h.
According to question,

⇒ 30 × 2 × 30 = 9(225 – x²)
⇒ 100 × 2 = 225 – x²
⇒ 200 = 225 – x²
⇒ x² = 25
⇒ x = ±5
⇒ x = 5 (Rejecting – 5)
∴ Speed of stream = 5 km/h

Question 22.
Ram takes 6 days less than Bhagat to finish a piece of work. If both of them together can finish the work in 4 days, in how many days Bhagat alone can finish the work.
Solution:
Let Bhagat alone can do the work in x number of days
∴ Ram takes (x – 6) number of days
Work done by Bhagat in 1 day = \(\frac{1}{x}\)
Work done by Ram in 1 day = \(\frac{1}{x-6}\)
According to the question,

Quadratic Equations Class 10 Extra Questions HOTS

Question 1.
One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Solution:
Let x be the total number of camels.
Then, number of camels in the forest = \(\frac{x}{4}\)
Number of camels on mountains = 2√x
and number of camels on the bank of river = 15

But, the number of camels cannot be a fraction.
∴ y = 6
⇒ x = x² = 36
Hence, the number of camels = 36

Question 2.
Solve the following quadratic equation:
9x² – 9 (a + b) x + [2a² + 5ab + 2b²] = 0.
Solution:
Consider the equation 9x² – 9 (a + b) x + [2a² + 5ab + 2b²] = 0)
Now comparing with Ax² + Bx + C = 0, we get
A = 9, B = -9 (a + b) and C = [2a² + 5ab + 2b²]
Now discriminant, .
D = B² – 4AC

Question 3.
Two taps running together can fill a tank in 3 hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?
Solution:
Let, time taken by faster tap to fill the tank be x hours
Therefore, time taken by slower tap to fill the tank = (x + 3) hours
Since the faster tap takes x hours to fill the tank.

Hence, time taken by faster tap to fill the tank = x = 5 hours
and time taken by slower tap = x + 3 = 5 + 3 = 8 hours.

Question 4.
In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed so that the area of grass strip of uniform width surrounding the pond would be 1184 m². Find the length and breadth of the pond.
Solution:

Let ABCD be rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass area, which is same around the pond.
Given, Length of lawn = 50 m
Width of lawn = 40 m
Length of pond = (50 – 2x)m
Breadth of pond = (40 – 2x)m
Also given,
Area of grass surrounding the pond = 1184 m²
⇒ Area of rectangular lawn – Area of pond = 1184 m²
⇒ 50 × 40 – {(50 – 2x) × (40 – 2x)} = 1184
⇒ 2000 – (2000 – 80x – 100x + 4x²) = 1184
⇒ 2000 – 2000 + 180x – 4x² = 1184
⇒ 4x² – 180x + 1184 = 0
⇒ x² – 45x + 296 = 0
⇒ x² – 37x – 8x + 296 = 0
⇒ x(x – 37) – 8(x – 37) = 0
⇒ (x – 37) (x – 8) = 0
⇒ x- 37 = 0 or x – 8 = 0)
⇒ x = 37 or x = 8
x = 37 is not possible as in this case length of pond becomes 50 – 2 × 37 = -24 (not possible) Hence, x = 8 is acceptable
∴ Length of pond = 50 – 2 × 8 = 34 m
Breadth of pond = 40 – 2 × 8 = 24 m

Question 5.
A car covers a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.
Solution:
Let speed of the car be x km/h
According to question
Time taken = \(\frac{x}{2}\) h.

⇒ x = 72 km/h [Taking square root both sides]
∴ Time taken = \(\frac{x}{2}\) = 36 hours.

Here we are providing Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.

Extra Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers Solutions

Extra Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Solutions Answers

Pair of Linear Equations in Two Variables Class 10 Extra Questions Very Short Answer Type

Question 1.
If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then find value of k.
Solution:
Since the given lines are parallel

Question 2.
Find the value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions.
Solution:
The given system of equations will have infinitely many solutions if \(\frac{c}{6}=\frac{-1}{-2}=\frac{2}{3}\) which is not possible
∴ For no value of c, the given system of equations have infinitely many solutions.

Question 3.
Do the equations 4x + 3y – 1 = 5 and 12x + 9y = 15 represent a pair of coincident lines?
Solution:

Given equations do not represent a pair of coincident lines.

Question 4.
Find the co-ordinate where the line x – y = 8 will intersect y-axis.
Solution:
The given line will intersect y-axis when x = 0.
∴0 – y = 8 ⇒ y = -8
Required coordinate is (0, -8).

Question 5.
Write the number of solutions of the following pair of linear equations:
x + 2y – 8 = 0, 2x + 4y = 16
Solution:


∴The given pair of linear equations has infinitely many solutions.

Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type 1

Question 1.
Is the following pair of linear equations consistent? Justify your answer.
2ax + by = a, 4ax + 2by – 2a = 0; a, b≠ 0
Solution:
Yes,

∴ The given system of equations is consistent.

Question 2.
For all real values of c, the pair of equations
x – 2y = 8, 5x + 10y = c
have a unique solution. Justify whether it is true or false.
Solution:

So, for all real values of c, the given pair of equations have a unique solution.
∴ The given statement is true.

Question 3.
Does the following pair of equations represent a pair of coincident lines? Justify your answer.
\(\frac{x}{2}\) + y + \(\frac{2}{5}\) = 0, 4x+ 8y + \(\frac{5}{16}\) = 0.
Solution:

∴ The given system does not represent a pair of coincident lines.

Question 4.
If x = a, y = b is the solution of the pair of equation x – y = 2 and x + y = 4, then find the value of a and b.
Solution:
x – y = 2 … (i)
x + y = 4 … (ii)
On adding (i) and (ii), we get 2x = 6 or x = 3
From (i), 3 – y ⇒ 2 = y = 1
a = 3, b = 1.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) , and, \(\frac{c_{1}}{c_{2}}\) find out whether the following pair of linear equations consistent or inconsistent. is consistent or inconsistent. (Q. 5 to 6)

Question 5.
\(\frac{3}{2}\) x + \(\frac{5}{3}\) y = 7
9x – 10y = 14
Solution:

Question 6.
\(\frac{4}{3}\) x + 2y = 8;
2x + 3y = 12
Solution:

Hence, the pair of linear equations is consistent with infinitely many solutions.

On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\), and \(\frac{c_{1}}{c_{2}}\), find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: (Q. 7 to 9).

Question 7.
5x – 4y + 8 = 0
7x + 6y – 9 = 0
Solution:

Question 8.
9x + 3y + 12 = 0
18x + 6y + 24 = 0
Solution:
We have,
9x + 3y + 12 = 0
18x + 6y + 24 = 0

Question 9.
6x – 3y + 10 = 0 .
2x – y + 9 = 0
Solution:

Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type 2

Question 1.
Solve: ax + by = a – b and bx – ay = a + b
Solution:
The given system of equations may be written as
ax + by – (a – b) = 0
bx – ay – (a + b) = 0
By cross-multiplication, we have

Hence, the solution of the given system of equations is x = 1, y = -1

Question 2.
Solve the following linear equations:
152x – 378y = -74 and -378x + 152y = -604
Solution:
We have, 152x – 378y = -74 …(i)
-378x + 152y = -604 ……(ii)

Putting the value of x in (iii), we get
2 + y = 3 ⇒ y = 1
Hence, the solution of given system of equations is x = 2, y = 1.

Question 3.
Solve for x and y

Solution:

Putting the value of y in (ii), we get
x + ab = 2ab ⇒ x = 2ab – ab ⇒ x = ab
∴ x = ab, y = ab

Question 4.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
(ii) for which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Solution:
(i) We have, 2x + 3y = 7
(a – b) x + (a + b) y = 3a + b – 2 … (ii)
Here, a1 = 2, b1 = 3, c1 = 7 and
a2 = a – b, b2 = a + b, c2 = 3a + b – 2
For infinite number of solutions, we have

⇒ 9a – 7a + 3b – 75 -6 = 0 ⇒ 2a – 45 – 6 = 0 => 2a – 4b = 6
⇒ a – 2b = 3 …(iv)
Putting a = 5b in equation (iv), we get
56 – 2b = 3 or 3b = 3 i.e., b = \(\frac{3}{3}\) =1
Putting the value of b in equation (ii), we get a = 5(1) = 5
Hence, the given system of equations will have an infinite number of solutions for a = 5 and b = 1.

(ii) We have, 3x + y = 1, 3x + y − 1 = 0 …(i)
(2k – 1) x + (k – 1) y = 2k + 1
⇒ (2k – 1) x + (k – 1) y – (2k + 1) = 0 ……(ii)
Here, a1 = 3, b1 = 1, C1 = -1
a2 = 2k – 1, b2 = k – 1, C2 = -(2k + 1)
For no solution, we must have

⇒ 3k – 2k = 3 – 1 ⇒ k = 2
Hence, the given system of equations will have no solutions for k = 2.

Question 5.
Find whether the following pair of linear equations has a unique solution. If yes, find the
7x – 4y = 49 and 5x – y = 57
Solution:
We have, 7x – 4y = 49 ……..(i)
and 5x – 6y = 57 ……..(ii)

So, system has a unique solution.
Multiply equation (i) by 5 and equation (ii) by 7 and subtract

Put y = -7 in equation (ii)
5x – 6(-7)57 ⇒ 5x = 57 – 42 ⇒ x = 3
hence, x = 3 and y = -7.

Question 6.
Solve for x and y.

Solution:

Question 7.
Solve the following pair of equations for x and y.

Solution:

Question 8.
In ∆ABC, LA = x, ∠B = 3x, and ∠C = y if 3y – 5x = 30°, show that triangle is right angled.
Solution:
∠A + 2B + ∠C = 180°
(Sum of interior angles of A ABC) x + 3x + y = 180°
4x + y = 180° …(i)
3y – 5x = 30° (Given) …(ii) Multiply equation (i) by 3 and subtracting from eq. (ii), we get
-17x = – 510 = x = 910 = 30°
17 then _A = x = 30° and 2B = 3x = 3 X 30o = 90°
∠C = y = 180° – (∠A + ∠B) = 180° – 120° = 60°
∠A = 30°, ∠B = 90°, ∠C = 60° Hence ∆ABC is right triangle right angled at B.

Question 9.
In Fig. 3.1, ABCDE is a pentagon with BE|CD and BC||DE. BC is perpendicular to CD. If the perimeter of ABCDE is 21 cm. Find the value of x and y.

Solution:
Since BC||DE and BE||CD with BC||CD.
BCDE is a rectangle.
Opposite sides are equal BE = CD
∴ x + y = 5 …… (i)
DE = BC = x – y
Since perimeter of ABCDE is 21 cm.
AB + BC + CD + DE + EA = 21
3 + x – y + x + y + x – y + 3 = 21 ⇒ 6 + 3x – y = 21
3x – y = 15 ….. (iii)
Adding (i) and (ii), we get
4x = 20 ⇒ x = 5
On putting the value of x in (i), we get y = 0
Hence, x = 5 and y = 0.

Question 10.
Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?
Solution:
Let the present ages of B and A be x years and y years respectively. Then
B’s age 5 years ago = (x – 5) years
and A’s age 5 years ago = (- 5) years
(-5) = 3 (x – 5) = 3x – y = 10 …….(i)
B’s age 10 years hence = (x + 10) years
A’s age 10 years hence = (y + 10) years
y + 10 = 2 (x + 10) = 2x – y = -10 …….. (ii)
On subtracting (ii) from (i) we get x = 20
Putting x = 20 in (i) we get
(3 × 20) – y = 10 ⇒ y = 50
∴ x = 20 and y = 50
Hence, B’s present age = 20 years and A’s present age = 50 years.

Question 11.
A fraction becomes when \(\frac{1}{3}\) is subtracted from the numerator and it becomes \(\frac{1}{4}\) when 8 is added to its denominator. Find the fraction.
Solution:
Let the numerator be x and denominator be y.

Putting the value of x in equation (i), we have
3 × 5 – y = 3 ⇒ 5 – y = 3 ⇒ 15 – 3 = y
∴ y = 12
Hence, the required fraction is \(\frac{5}{12}\)

Question 12.
Solve the following pairs of equations by reducing them to a pair of linear equations:

Solution:

Putting the value of x in equation (iii), we have
3 x 1 + y = 4
⇒ y = 4 – 3 = 1
Hence, the solution of given system of equations is x = 1, y = 1.

Pair of Linear Equations in Two Variables Class 10 Extra Questions Long Answers

Question 1.
Form the pair of linear equations in this problem, and find its solution graphically: 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Solution:
Let x be the number of girls and y be the number of boys.
According to question, we have
x = y + 4
⇒ x – y = 4 ……(i)
Again, total number of students = 10
Therefore, x + y = 10 …(ii)
Hence, we have following system of equations
x – y = 4
and x + y = 10
From equation (i), we have the following table:

From equation (ii), we have the following table:

Plotting this, we have

Here, the two lines intersect at point (7,3) i.e., x = 7, y = 3.
So, the number of girls = 7
and number of boys = 3.

Question 2.
Show graphically the given system of equations
2x + 4y = 10 and 3x + 6y = 12 has no solution.
Solution:
We have, 2x + 4y = 10
⇒ 4y = 10 – 2x ⇒ y = \(\frac{5-x}{2}\)
Thus, we have the following table:

Plot the points A (1, 2), B (3, 1) and C (5,0) on the graph paper. Join A, B and C and extend it on both sides to obtain the graph of the equation 2x + 4y = 10.
We have, 3x + 6y = 12
⇒ 6y = 12 – 3x ⇒ y = \(\frac{4-x}{2}\)
Thus, we have the following table :

Plot the points D (2, 1), E (0, 2) and F (4,0) on the same graph paper. Join D, E and F and extend it on both sides to obtain the graph of the equation 3x + 6y = 12.

We find that the lines represented by equations 2x + 4y = 10 and 3x + y = 12 are parallel. So, the two lines have no common point. Hence, the given system.of equations has no solution.

Question 3.
Solve the following pairs of linear equations by the elimination method and the substitution method:
(i) 3x – 5y – 4 = 0 and 9x = 2y + 7
(ii) \(\frac{x}{2}+\frac{2 y}{3}\) = -1 and x – \(\frac{y}{3}\) = 3
Solution:
(i) We have, 3x – 5y – 4 = 0
⇒ 3x – 5y = 4 …….(i)
Again, 9x = 2y + 7
9x – 2y = 7 …(ii)

By Elimination Method:
Multiplying equation (i) by 3, we get
9x – 15y = 12 … (iii)
Subtracting (ii) from (iii), we get

By Substitution Method:
Expressing x in terms of y from equation (i), we have

By Elimination Method:
Subtracting (ii) from (i), we have
5y = – 15 or y = – \(\frac{5}{5}\) = -3
Putting the value of y in equation (i), we have
3x + 4 × (-3) = -6 ⇒ 3x = – 6 + 12
∴ 3x – 12 = -6 ⇒ 3x = 6
∴ x = \(\frac{6}{3}\) = 2
Hence, solution is x = 2, y = -3.

By Substitution Method:
Expressing x in terms of y from equation (i), we have
3 × \(\left(\frac{-6-4 y}{3}\right)\) – y = 9 ⇒ -6 – 4y – y = 9 ⇒ -6 – 5y = 9
Substituting the value of x in equation (ii), we have
∴ -5y = 9 + 6 = 15
y = – \(\frac{15}{5}\) = – 3
Putting the value of y in equation (i), we have
3x + 4 × (-3) = -6 ⇒ 3x – 12 = -6
∴ 3x = 12 – 6 = 6
∴ x = \(\frac{6}{3}\) = 2
Hence, the required solution is x = 2, y = – 3.

Question 4.
Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
We have,’ x – y + 1 = 0 and 3x + 2y – 12 = 0
Thus, x – y = -1 => x = y – 1 …(i)
3x + 2y = 12 => x = \(\frac{12-2 y}{3}\) … (ii)
From equation (i), we have

From equation (ii), we have

Plotting this, we have

ABC is the required (shaded) region and point of intersection is (2, 3).
∴ The vertices of the triangle are (-1, 0), (4, 0), (2, 3).

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method (Q. 5 to 8):

Question 5.
A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay 31000 as hostel charges whereas a student B, who takes food for 26 days, pays 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solution:
Let the fixed charge be *x and the cost of food per day be by.
Therefore, according to question,
x + 20y = 1000 …(i)
x + 26y = 1180 …(ii)
Now, subtracting equation (ii) from (i), we have

Putting the value of y in equation (i), we have
x + 20 x 30 = 1000 ⇒ x + 600 = 1000 ⇒ x = 1000 – 600 = 400
Hence, fixed charge is ₹400 and cost of food per day is ₹30.

Question 6.
Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deduced for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Solution:
Let x be the number of questions of right answer and y be the number of questions of wrong answer.
According to question,
3x – y = 40 … (i)
and 4x – 2y = 50
or 2x – y = 25 …(ii)
Subtracting (ii) from (i), we have

Putting the value of x in equation (i), we have
3 x 15 – y = 40 ⇒45 – y = 40
∴ y = 45 – 40 = 5
Hence, total number of questions is x + yi.e., 5 + 15 = 20.

Question 7.
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution:
Let the speed of two cars be x km/h and y km/h respectively.
Case I: When two cars move in the same direction, they will meet each other at P after 5 hours.

The distance covered by car from A = 5x (Distance = Speed × Time)
and distance covered by the car from B = 5y
∴ 5x – 5y = AB = 100 ⇒ x – y = \(\frac{100}{5}\)
∴ x – y = 20 ….(i)

Case II: When two cars move in opposite direction, they will meet each other at Q after one hour.

The distance covered by the car from A = x
The distance covered by the car from B = y
∴ x + y = AB = 100 ⇒ x + y = 100 …..(ii)
Now, adding equations (i) and (ii), we have
2x = 120 ⇒ x = \(\frac{120}{2}\) = 60
Putting the value of x in equation (i), we get
60 – y = 20 ⇒ – y = -40
∴ y = 40
Hence, the speeds of two cars are 60 km/h and 40 km/h respectively.

Question 8.
The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
Let the length and breadth of a rectangle be x and y respectively.
Then area of the rectangle = xy
According to question, we have
(x – 5) 6 + 3) = xy – 9 ⇒ xy + 3x – 5y – 15 = xy – 9
⇒ 3x – 5y = 15 – 9 = 6 ⇒ 3x – 5y = 6 …(i)
Again, we have
(x + 3) 6 + 2) = xy + 67 ⇒ xy + 2x + 3y + 6 = xy + 67
⇒ 2x + 3y = 67 – 6 = 61 ⇒ 2x + 3y = 61…. (ii)
Now, from equation (i), we express the value of x in terms of y as

Putting the value of y in equation (i), we have
3x – 5 x 9 = 6 ⇒ 3x = 6 + 45 = 51
∴ x = \(\frac{51}{3}\) = 17
Hence, the length of rectangle = 17 units and breadth of rectangle = 9 units.

Question 9.
Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) Roobi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by bus and the remaining by train. If she travels 100 km by bus and the remaining by train, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
(i) Let her speed of rowing in still water be x km/h and the speed of the current be y km/h.
Case I: When Ritu rows downstream
Her speed (downstream) = (x + y) km/h

Putting the value of x in equation (i), we have
6 + y = 10 ⇒ y = 10 – 6 = 4
Hence, speed of Ritu in still water = 6 km/h.
and speed of current = 4 km/h.

(ii) Let the speed of the bus be x km/h and speed of the train be y km/h.
According to question, we have

Question 10.
The sum of a two digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits in the first number. Find the first number.
Solution:
Let the digits at unit and tens places be x and y respectively.
Then, number = 10y + x …(i)
Number formed by interchanging the digits = 10x + y
According to the given condition, we have
(10y + x) + (10x + y) = 110
⇒ 11x + 11y = 110
⇒ x + y – 10 = 0
Again, according to question, we have
(10y + x) – 10 = 5 (x + y) + 4
⇒ 10y + x – 10 = 5x + 5y + 4
⇒ 10y + x – 5x – 5y = 4 + 10
5y – 4x = 14 or 4x – 5y + 14 = 0
By using cross-multiplication, we have .

Putting the values of x and y in equation (i), we get
Number 10 × 6 + 4 = 64.

Question 11.
Jamila sold a table and a chair for 1050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair she would have got 1065. Find cost price of each.
Solution:
Let cost price of table be ₹x and the cost price of the chair be ₹y.
”

From equation (i) and (ii) we get
110x + 125y = 105000
and 125x + 110y = 106500
On adding and subtracting these equations, we get
235x + 235y = 211500
and 15x – 15y = 1500
i.e., x + y = 900 …(iii)
x – y = 100 …… (iv)
Solving equation (iii) and (iv) we get
x = 500, y = 400
So, the cost price of the table is ₹500 and the cost price of the chair is ₹400.

Pair of Linear Equations in Two Variables Class 10 Extra Questions HOTS

Question 1.
8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.
Solution:
Let one man alone can finish the work in x days and one boy alone can finish the work in y days


Hence, one man alone can finish the work in 140 days and one boy alone can finish the work in 280 days.

Question 2.
A boat covers 25 km upstream and 44 km downstream in 9 hours. Also, it covers 15 km upstream and 22 km downstream in 5 hours. Find the speed of the boat in still water and that of the stream.
Solution:
Let the speed of the boat in still water be x km/h and that of the stream be y km/h. Then,
Speed upstream (x – y) km/h
Speed downstream (x + y) km/h

25u + 44v = 9 ⇒ 25u + 44v – 9 = 0 …(iii)
15u + 22v = 5 ⇒ 15u + 22v – 5 = 0 …(iv)
By cross-multiplication, we have


Solving equations (v) and (vi), we get x = 8 and y = 3.
Hence, speed of the boat in still water is 8 km/h and speed of the stream is 3 km/h.

Question 3.
Students of a class are made to stand in rows. If one student is extra in each row, there would be 2 rows less. If one student is less in each row, there would be 3 rows more. Find the number of students in the class.
Solution:
Let total number of rows be y
and total number of students in each row be x – Total number of students = xy
Case I: If one student is extra in each row, there would be two rows less. Now, number of rows = (y-2) Number of students in each row = (x + 1)
Total number of students = Number of rows x Number of students in each row
xy = 6 – 2)(x + 1) ⇒ xy = xy + y – 2x – 2
xy – xy – y + 2x = -2 ⇒ 2x – y = -2 …(i)

Case II: If one student is less in each row, there would be 3 rows more.
Now, number of rows = (y + 3)
and number of students in each row = (x – 1)
Total number of students = Number of rows x Number of students in each row
∴ xy = 6 + 3)(x – 1) ⇒ xy = xy – y + 3x – 3
xy – xy + y – 3x = -3 ⇒ – 3x + y = -3 … (ii)
On adding equations (i) and (ii), we have

or
Putting the value of x in equation (i), we get
2(5) – y = -2
⇒ 10- y = -2
– y = -2 – 10
⇒ – y = -12
or y = 12
∴ Total number of students in the class = 5 × 12 = 60.

Question 4.
Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and x-axis. Find the area of the shaded region.
Solution:
We have, 2x + y = 6, ⇒ y = 6 – 2x
When x = 0, we have y = 6 – 2 × 0 = 6
When x = 3, we have y = 6 – 2 × 3 = 0
When x = 2, we have y = 6 – 2 × 2 = 2
Thus, we get the following table:

Now, we plot the points A(0,6), B(3,0) and C(2, 2) on the graph paper. We join A, B and C and extend it on both sides to obtain the graph of the equation 2x + y = 6.

We have, 2x – y + 2 = 1 = y = 2x + 2
When x = 0, we have y = 2 × 0 + 2 = 2
When x = -1, we have y = 2 × (-1) + 2 = 0
When x = 1, we have y = 2 × 1 + 2 = 4
Thus, we have the following table:

Now, we plot the points D(0, 2), E(-1,0) and F(1,4) on the same graph paper. We join D, E and F and extend it on both sides to obtain the graph of the equation 2x – y + 2 = 0. It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. 3.7.
Thus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from Fon x-axis.
Clearly, we have
FM = y-coordinate of point F(1, 4) = 4 and BE = 4
∴ Area of the shaded region = Area of AFBE

Question 5.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the ages of Ani and Biju be x and y years respectively. Then x – y = 13
Age of Dharam = 2x years
Age of Cathy = \(\frac{y}{2}\) years
Clearly, Dharam is older than Cathy
IMMM
Thus, we have following two systems of linear equations
x – y = 3 ….. (i)
4x – y = 60 … (ii)
and x – y = -3 … (iii)
4x – y = 60 … (iv)
Subtracting equation (i) from (ii), we get

Putting x = 19 in equation (i), we get
19 – y = 3 ⇒ y = 16
Now, subtracting equation (iii) from (iv)

Putting x = 21 in equation (iii), we get
21 – y = -3 ⇒ y = 24
Hence, age of Ani = 19 years and age of Biju = 16 years
or age of Ani = 21 years and age of Biju = 24 years

Question 6.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h it would have taken 3 hours more than the scheduled time. Find distance covered by the train.
Solution:
Let actual speed of the train be x km/h and actual time taken be y hours.
Then, distance covered = speed × time = xy km … (i)
Case I: When speed is (x + 10) km/h, then time taken is (y – 2) hours
∴ Distance covered = (x + 10)(y – 2)
⇒ xy = (x + 10)(y – 2) [From (i)]
= xy = xy – 2x + 10y – 20 ⇒ 2x – 10y = -20 ….(ii)
⇒ x – 5y = -10
Case II: When speed is (x – 10) km/h, then time taken is (y + 3) hours.
∴Distance covered = (x – 10)(y + 3)
xy = (x – 10)(y + 3) [From (i)]
xy = xy + 3x – 10y – 30 …..(iii)
= 3x – 10y = 30
Multiplying equation (ii) by 2 and subtracting it from (iii), we get

Putting x = 50 in equation (ii), we get
50 – 5y = -10
⇒ 50 + 10 = 5y ⇒ y = 12
∴ Distance covered by the train = xy km = 50 × 12 km = 600 km

Here we are providing Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.

Extra Questions for Class 10 Maths Polynomials with Answers Solutions

Extra Questions for Class 10 Maths Chapter 2 Polynomials with Solutions Answers

Polynomials Class 10 Extra Questions Very Short Answer Type

The graphs of y = p(x) for some polynomials (for questions 1 to 4) are given below. Find the number of zeros in each case.

Question 1.

Answer:
There is no zero as the graph does not intersect the X-axis.

Question 2.

Answer:
The number of zeros is four as the graph intersects the X-axis at four points.

Question 3.

Answer:
The number of zeros is three as the graph intersects the X-axis at three points.

Question 4.

Answer:
The number of zeros is three as the graph intersects the X-axis at three points.

Question 5.
What will the quotient and remainder be on division of ax² + bx + c by px² + qx² + rx + 5, p ≠ 0?
Answer:
0, ax² + bx + C.

Question 6.
If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p(x) and g(x)?
Answer:
Since the quotient is zero, therefore
deg p(x) < deg g(x)

Question 7.
Can x – 2 be the remainder on division of a polynomial p(x) by x + 3?
Answer:
No, as degree (x – 2) = degree (x + 3)

Question 8.
Find the quadratic polynomial whose zeros are -3 and 4.
[NCERT Exemplar]
Answer:
Sum of zeros = -3 + 4 = 1,
Product of zeros = – 3 x 4 = -12
∴ Required polynomial = x² – x – 12

Question 9.
If one zero of the quadratic polynomial x² – 5x – 6 is 6 then find the other zero.
Answer:
Let α,6 be the zeros of given polynomial.
Then α + 6 = 5 3 ⇒ α = -1

Question 10.
If both the zeros of the quadratic polynomial ax² + bx + c are equal and opposite in sign, then find the value of b.
Answer:
Let α and -α be the roots of given polynomial.
Then α + (-α) = 0 ⇒ \(-\frac{b}{a}=0\) ⇒ b = 0.

Question 11.
What number should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the polynomial?
Answer:
Let f(x) = x² – 5x + 4
Then f(3) = 3² – 5 x 3 + 4 = -2
For f(3) = 0, 2 must be added to f(x).

Question 12.
Can a quadratic polynomial x² + kx + k have equal zeros for some odd integer k > 1?
Answer:
No, for equal zeros, k = 0,4 ⇒ k should be even.

Question 13.
If the zeros of a quadratic polynomial ax² + bx + c are both negative, then can we say a, b and c all have the same sign? Justify your answer.
Answer:
Yes, because \(-\frac{b}{a}\) = sum of zeros < 0, so that \(\frac{b}{a}=0\) > 0. Also the product of the zeros = \(\frac{c}{a}=0\) > 0.

Question 14.
If the graph of a polynomial intersects the x-axis at only one point, can it be a quadratic polynomial?
Answer:
Yes, because every quadratic polynomial has at the most two zeros.

Question 15.
If the graph of a polynomial intersects the x-axis at exactly two points, is it necessarily a quadratic polynomial?
Answer:
No, x⁴ – 1 is a polynomial intersecting the x-axis at exactly two points.

Polynomials Class 10 Extra Questions Short Answer Type 1

Question 1.
If one of the zeros of the quadratic polynomial f(x) = 4x² – 8kx – 9 is equal in magnitude but opposite in sign of the other, find the value of k.
Answer:
Let one root of the given polynomial be α.
Then the other root = -α
Sum of the roots = (-α) + α = 0
⇒ \(-\frac{b}{a}\) = 0 or \(-\frac{8k}{4}\) = 0 or k = 0

Question 2.
If one of the zeros of the quadratic polynomial (k – 1)x² + kx + 1 is -3 then find the value of k.
Answer:
Since – 3 is a zero of the given polynomial
∴ (k – 1)(-3)² + k(-3) + 1 = 0 :
⇒ 9k – 9 – 3k + 1 = 0 ⇒ k = 4/3.

Question 3.
If 1 is a zero of the polynomial p(x) = ax² – 3(a – 1)x -1, then find the value of a.
Answer:
Put x = 1 in p(x)
∴ p(1) = a(1)² – 3(a – 1) x 1 – 1 = 0
⇒ a – 3a + 3 – 1 = 0 ⇒ 2a = -2 ⇒ a = 1

Question 4.
If α and β are zeros of polynomial p(x) = x² – 5x + 6, then find the value of α + B – 3aß.
Answer:
Here, α + β = 5, αβ = 6
= α + β – 3αβ = 5 – 3 x 6 = -13

Question 5.
Find the zeros of the polynomial p(x) = 4x² – 12x + 9.
Answer:
p(x) = 4x² – 12x + 9 = (2x – 3)²
For zeros, p(x) = 0
⇒ (2x – 3)(2x – 3) = 0 ⇒ x = \(\frac{3}{2}, \frac{3}{2}\)

Question 6.
If one root of the polynomial p(y) = 5y² + 13y + m is reciprocal of other, then find the value of m.
Answer:

Question 7.
If α and β are zeros of p(x) = x² + x – 1, then find \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
Here, α + β = -1, αβ = -1,
So \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-1}=1\)

Question 8.
Given that one of the zeros of the cubic polynomial ax³ + bx² + cx + d is zero, find the product of the other two zeros.
Answer:
Let α, β, γ be the roots of the given polynomial and α = 0.
Then αβ + βγ + γα = c/a ⇒ βγ = c/a

Question 9.
If the product of two zeros of the polynomial p(x) = 2x³ + 6x² – 4x + 9 is 3, then find its third zero.
Answer:
Let α, β, γ be the roots of the given polynomial and αβ = 3
Then αβγ = \(-\frac{9}{2}\)
⇒ 3 x γ = \(\frac{-9}{2}\) or γ = \(\frac{-3}{2}\)

Question 10.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.
(i) \(-\frac{1}{4}, \frac{1}{4}\)
(ii) \(\sqrt{2}, \frac{1}{3}\)
Answer:
Let α, β be the zeros of polynomial.
(i) We have, α + β = \(-\frac{1}{4}\) and αβ = \(\frac{1}{4}\)
Thus, polynomial is
p(x) = x2 – (a + B) x + aß

Quadratic polynomial 4x² + x + 1

(ii) We have, α + β = √2 and  αβ = \(\frac{1}{3}\)
Thus, polynomial is p(x) = x² – (α + β) x + αβ
= x² – √2x + \(\frac{1}{3}\) = \(\frac{1}{3}\) (3x² – 3√2x + 1)
Quadratic polynomial = 3x² – 3√2x + 1

Polynomials Class 10 Extra Questions Short Answer Type 2

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients (Q. 1 – 2).

Question 1.
6x² – 3 – 7x
Answer:
We have, p(x) = 6x² – 3 – 7x
p(x) = 6x² – 7x – 3
(In general form)
= 6x² – 9x + 2x – 3 = 3x (2x – 3) + 1 (2x – 3)
= (2x – 3) (3x + 1)
The zeros of polynomial p(x) is given by
p(x) = 0) = (2x – 3) (3x + 1) = 0 ⇒ \(x=\frac{3}{2},-\frac{1}{3}\)
Thus, the zeros of 6x² – 7x – 3 are α = \(-\frac{3}{2}\) and β = \(-\frac{1}{3}\)
Now, sum of the zeros = α + β = \(\frac{3}{2}-\frac{1}{3}\) = \(\frac{9-2}{6}=\frac{7}{6}\)

Question 2.
4u² + 8u
Answer:
We have, p(u) = 4u² + 8u = p(u) = 4u (u + 2)
The zeros of polynomial p(u) is given by
p(u) = 0 ⇒ 4u (u + 2) = 0 .
∴ u = 0, -2
Thus, the zeros of 4u² + 8u are α = 0 and β = -2
Now, sum of the zeros = α + β = 0 – 2 = -2

Question 3.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2 (ii) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
Answer:
(i) We have,

Clearly, remainder is zero, so x2 + 3x + 1 is a factor of polynomial 3x⁴ + 5x³ – 7x² + 2x + 2

(ii) We have,

Clearly, remainder is zero, so t’ – 3 is a factor of polynomial 2t⁴ + 3t³ – 2t² – 9t – 12.

Question 4.
If α and β are the zeros of the quadratic polynomial f(x) = 2x² – 5x + 7, find a polynomial whose zeros are 2α + 3β and 3α + 2β.
Answer:
Since α and β are the zeros of the quadratic polynomial f(x) = 2x² – 5x + 7

Let S and P denote respectively the sum and product of the zeros of the required polynomial.

Question 5.
What must be subtracted from p(x) = 8x⁴ + 14x³ – 2x² + 7x – 8 so that the resulting polynomial
is exactly divisible by g(x) = 4x² + 3x – 2?
Answer:
Let y be subtracted from polynomial p(x)
: 8x⁴ + 14x³ – 2x² + 7x – 8 – y is exactly divisible by g(x)
Now,

∵ Remainder should be 0.
∴ 14x – 10 – y = 0 or 14x – 10 = y or y = 14x – 10
∴ (14x – 10) should be subtracted from p(x) so that it will be exactly divisible by g(x)

Question 6.
What must be added to f(x) = 4x⁴ + 2x³ – 2x² + x – 1 so that the resulting polynomial is divisible
by g(x) = x² + 2x – 3?
Answer:
By division algorithm, we have
f(x) = g(x) × q(x) + r(x)
= f(x) – r(x) = g(x) × q(x) ⇒ f(x) + {-r(x)} = g(x) × q(x)
Clearly, RHS is divisible by g(x). Therefore, LHS is also divisible by g(x). Thus, if we add –r(x) to f(x), then the resulting polynomial is divisible by g(x). Let us now find the remainder when f(x) is divided by g(x).

∴ r(x) = -61x + 65 or -r(x) = 61x – 65
Hence, we should add –r(x) = 61x – 65 to f(x) so that the resulting polynomial is divisible by g(x).

Question 7.
Obtain the zeros of quadratic polynomial 3x² – 8x + 4√3 and verify the relation between its zeros and coefficients.
Answer:
We have,

Question 8.
If α and β are the zeros of the polynomial 6y² – 7y + 2, find a quadratic polynomial whose zeros are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\)
Answer:
Let p(y) = 6y² – 7y + 2

Question 9.
If one zero of the polynomial 3x² – 8x + 2k + 1 is seven times the other, find the value of k.
Answer:
Let α and β be the zeros of the polynomial. Then as per question β = 7α

Question 10.
If one zero of the polynomial 2x² + 3x + λ is 1/2 find the value of and other zero.
Answer:
Let P(x) = 2x² + 3x + λ

Question 11.
If one zero of polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other, find the value of a.
Answer:
Let one zero of the given polynomial be α.
Then, the other zero is 1/α
∴ Product of zeros = α × \(\frac{1}{\alpha}\) = 1
But, as per the given polynomial product of zeros = \(\frac{6 a}{a^{2}+9}\)
∴ \(\frac{6 a}{a^{2}+9}\) = 1 ⇒ a² + 9 = 6a
⇒ a² – 6a + 9 = 0) ⇒ (a – 3)² = 0
⇒ a – 3 = 0 ⇒ a = 3
Hence, a = 3.

Question 12.
If the polynomial (x⁴ + 2x³ + 8x² + 12x + 18) is divided by another polynomial (x² + 5), the remainder comes out to be (px +q). Find values of p and q.
Answer:
Let f(x) = (x⁴ + 2x³ + 8x² + 12x + 18) and g(x) = (x² + 5)
On dividing f(x) by g(x), we get

Now, px + 9 = 2x + 3 ⇒ p = 2,q = 3 (By comparing the coefficient of x and constant term).

Polynomials Class 10 Extra Questions Long Answer Type 1

Question 1.
Verify that the numbers given alongside the cubic polynomial below are their zeros. Also verify the relationship between the zeros and the coefficients.
x³ – 4x² + 5x – 2; 2,1,1
Solution:
Let p(x) = x³ – 4 x² + 5x – 2
On comparing with general polynomial px) ax³ + bx² + cx + d, we get a = 1, b = -4, c = 5 and d = -2
Given zeros 2, 1, 1.
∴ p(2) = (2)³ – 4(2)² + 5(2) – 2 = 8 – 16 + 10 – 2 = 0
and p(1) = (1)³ – 4(1)² + 5(1) – 2 = 1 – 4 + 5 – 2 = 0
Hence, 2, 1 and I are the zeros of the given cubic polynomial.
Again, consider α = 2, β = 1, γ = 1
∴ α + 13 + y = 2 + 1 + 1 = 4

Question 2.
Find a cubic polynomial with the sum of the zeros, sum of the products of its zeros taken two at a time, and the product of its zeros as 2, -7, -14 respectively.
Solution:
Let the cubic polynomial be p(x) = ax³ + bx² + cx + d. Then

p(x) = a[x³ + (-2)x² + (-7)x + 14] ⇒ p(x) = a[x3 – 2x² – 7x + 14]
For real value of a = 1, p(x) = x³ – 2x² – 7x + 14

Question 3.
Find the zeros of the polynomial f(x) = x3 – 5x² – 2x + 24, if it is given that the product of its two zeros is 12.
Solution:
Let α, β and γ be the zeros of polynomial (fx) such that αβ = 12.

Now, α + β + γ = 5 α + β – 2 = 5
= α + β = 7 a = 7 – β
= (7 – β) β =12 ⇒ 7β – β² – 12
= β² + 7β + 12 = 0 ⇒ β² – 3β – 4β + 12 = O
= β = 4 or β = 3
β = 4 or β = 3
∴ α = 3 or α = 4

Question 4.
If the remainder on division of x3 – kx² + 13x – 21 by 2x – 1 is -21, find the quotient and the value of k. Hence, find the zeros of the cubic poIyncmia1 x³ – kx² + 13x.
Solution:
Let f(x) = x³ – kx² + 13x – 21

Question 5.
Obtain all other zeros of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeros are \(\sqrt{\frac{5}{3}}\) and \(\sqrt{\frac{5}{3}}\).
Solution:

Question 6.
Given that √2 is a zero of the cubic polynomial 6x³ + √2x² – 10x – 4√2, find its other zeros.
Solution:
The given polynomial is f(x) = (6x³ +√2x² – 10x – 4√2). Since √2 is the zero of f(x), it follows that (x – √2) is a factor of f(x).
On dividing f(x) by (x – √2), we get

Polynomials Class 10 Extra Questions HOTS

Question 1.
If α, β, γ bezerosofpo1ynomia1 6x³ + 3x² – 5x + 1, then find die value of α^-1 + β^-1 + γ^-1.
Solution:
∵ p(x) = 6x³ + 3x² – 5x + 1 so a = 6, b = 3, c = -5, d = 1
∴ α, β and γ are zeros of the polynomial p(x).

Question 2.
Find the zeros of the polynomial f(x) = – 12x² + 39x – 28, if it is given that the zeros are in AP.
Solution:
If α, β, γ are in AP., then,
β – α = γ + β ⇒ 2β = α + γ
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-(-12)}{1}\) = 12 ⇒ α + γ = 12 – β …….. (i)
From (i) and (ii)
2β = 12 – β or 3β = 12 or β = 4
Putting the value of β in (i), we have
8 = a + γ
αβγ = – \(\frac{d}{a}\) = \(\frac{-(-28)}{1}\) = 28 …….. (iii)
(αγ) 4 = 28 or αγ = 7 or γ = \(\frac{7}{α}\) ….. (iv)
Putting the value of γ = \(\frac{7}{α}\) in (iii), we get
⇒ 8 = α + \(\frac{7}{α}\) ⇒ 8α = α2 + 7
⇒ α² – 8α + 7 = 0 ⇒ α² – 7α – 1α + 7 = 0
⇒ α(α – 7)-1 (α – 7) = 0 ⇒ (α – 1)(α – 7) = 0
⇒ α = 1 or α = 7

Question 3.
If the polynomial f(x) = x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a. Find k and a.
Solution:
By division algorithm, we have Dividend = Divisor × Quotient + Remainder
⇒ Dividend – Remainder = Divisor × Quotient
⇒ Dividend – Remainder is always divisible by the divisor.
When f(x) = – 6x³ + 16x² – 25x + 10 is divided by x² – 2x + k the remainder comes out to be x + a.
∴ f(x) – (x + a) = x⁴ – 6x³+ 16x² – 25x + 10 – (x + a)
= x⁴ – 6x³ + 16x² – 25x + 10 – x – a x⁴ – 6x³ + 16x² – 26x + 10 – a
is exactly divisible by x² – 2x + k
Let us now divide x⁴ – 6x³ + 16x² – 26x + 10 – a by x² – 2x + k.

For f(x) – (x + a) = x⁴ – 6x³ + 16x² – 26x + 10 – a to be exactly divisible by x² – 2x + k, we must
have (-10 + 2k)x + (10 – a – 8k + k²) = 0 for all x
= – 10 + 2k = 0 and 10 – a – 8k + k² = 0
⇒ k = 5 and 10 – a – 40 + 25 = 0
⇒ k = 5 and a – 5

Chapter 1 Chemical Reactions and Equations Class 10 Science Important Questions and Answers PDF will help you in scoring more marks. This consists of 1 mark Questions, 3 Mark Numericals Questions, 5 Marks Numerical Questions and previous year questions from Chemical Reactions and Equations Chapter.

Chemical Reactions and Equations Class 10 Important Questions and Answers Science Chapter 1

Very Short Answer Questions

Question 1.
Why does not a wall immediately acquire a white colour when a coating of slaked lime is applied on it ?
Answer:
Slaked lime as such is not very white. When applied on the wall, CO₂ gas present in air reacts with calcium hydroxide to form calcium carbonate. It is quite white and therefore, imparts white look to the wall.
Ca(OH)₂ + CO₂(g) ———> CaCO₃(s) + H₂O(l).

More Resources

• Previous Year Question Papers for CBSE Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science

Question 2.
What is rust ?
Answer:
It is a brown mass known as hydrated ferric oxide. Its formula is Fe₂O₃.xH₂O.

Question 3.
Identify the most reactive and least reactive metal : Al, K, Ca, Au.
Answer:
Most reactive metal : K (potassium) ; Least reactive metal : Au (gold).

Question 4.
Which of the following is a combination reaction and which is a displacement reaction ?
(a) Cl₂ + 2KI ——–> 2KCl + I₂
(b) 2K + Cl₂ ——–> 2KCl.
Answer:
(a) It is a displacement reaction,
(b) It is a combination reaction.

Question 5.
What is the difference between the following two reactions ?
(a) Mg + 2HCl ——–> MgCl₂ + H₂
(b) NaOH + HCl ——–> NaCl + H₂O.
Answer:
(a) It is a single displacement reaction,
(b) It is a double displacement reaction also called neutralisation reaction.

Question 6.
Identify the compound which is oxidised in the following reaction
H₂S + Br₂ ———–> 2HBr + S.
Answer:
H₂S is oxidised to S because H₂S has lost hydrogen.

Question 7.
Suggest two ways to check the rancidity of food articles.
Answer:
(a) Keep the articles in airtight containers,
(b) Keep the articles in refrigerator.

Question 8.
Name two metals which donot get corroded.
Answer:
Gold (Au) and platinum (Pt) do not get corroded.

Question 9.
Identify the substance oxidised and reduced in the reaction :
CuO(s) + Zn(s) ———-> ZnO(s) + Cu(s).
Answer:
Zinc is oxidised to zinc oxide and copper oxide is reduced to copper.

Question 10.
How will you know whether a sample of cheese has become rancid or not ?
Answer:
If the cheese starts giving foul smell, it means that it has become rancid.

Question 11.
Why are eatables preferably packed in aluminium foils ?
Answer:
Aluminium foils donot corrode in atmosphere even if kept for a long time. Actually, a protective coating of aluminium oxide (Al₂O₃) is formed on the surface of the metal. It stops any further reaction of the metal with air (oxygen) and water. The eatables do not get spoiled.

Question 12.
What happens chemically when quick lime is added to water ?
Answer:
Calcium hydroxide (or slaked lime) is formed accompanied by a hissing sound. So much heat is evolved during the reaction that the reaction mixture starts boiling. The chemical equation for the reaction is :

Question 13.
Give an example of exothermic reaction.
Answer:
CH₄(g) + 2CO₂(g) ———-> CO₂(g) + 2H₂O(g) + heat, (evolved)

Question 14.
Give an example of endothermic reaction.
Answer:
N₂(g) + O₂(g) ———> 2NO(g) – heat, (absorbed).

Question 15.
Name the gas that can be used for the storage of fresh sample of chips for a long time.
Answer:
The gas is nitrogen (N₂). It checks rancidity of food articles.

Question 16.
Name the type of reaction
N₂(g) + 3H₂(g) ———-> 2NH₃(g)
Answer:
It is an example of combination reaction.

Question 17.
Give an example of a double displacement reaction (only reaction with complete balanced equation).
Answer:
HCl(aq) + NaOH(g) ——–> NaCl(aq) + H₂O(l)

Question 18.
Why are decomposition reactions called the opposite of combination reactions ? Write equations for these reactions.
Answer:
A decomposition reaction may be defined as the reaction in which a single substance decomposes or splits into two or more substances under suitable conditions.
For example,

It may be concluded that a certain substance is formed or synthesised in combination reaction and it breaks or splits in decomposition reaction. Therefore, the two reactions oppose each other.

Question 19.
In the reaction MnO₂ + 4HCl —————> MnCl₂ + 2H₂O + Cl₂; identify which one is reduced and which one is oxidized ?
Answer:
In this reaction HCl is oxidised to Cl₂ and MnO₂ is reduced to MnCl₂.

Question 20.
Take a small amount of calcium oxide or quick lime in a beaker and slowly add water to this. Is there any change in temperature ?
Answer:
Yes, the temperature increases since the process of dissolution of calcium oxide (CaO) in water is highly exothermic in nature.

Question 21.
Name two salts that are used in black and white photography.
Answer:
Both silver chloride and silver bromide are used in black and white photography.

Question 22.
State the chemical change that takes place when lime stone is heated
Answer:
Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide.
CaCO₃(s) heat , CaO(s) + CO₂(g)

Short Answer Questions

Question 23.
Identify the substance oxidised and substance reduced in the following reactions
(i) ZnO(s) + C(s) ———> Zn(s) + CO(g)
(ii) 2Na(s) + O₂(g) ———> 2Na₂O(s)
(iii) CuO(s) + H 2(g) ———> Cu(s) + H₂O(l).
Answer:
(i) C is oxidised to CO and ZnO is reduced to Zn.
(ii) Na is oxidised to Na₂O and O₂ is reduced.
(iii) H₂ is oxidised to H₂O and CuO is reduced to Cu.

Question 24.
Which types of reactions are represented by the following equations ?
(a) CaO + CO₂ ——-> CaCO₃
(b) Mg + CuSO₄ ——–> MgSO₄ + Cu
(c) CH4 + 2O₂ ———–> CO₂ + 2H₂O
(d) NH₄NO₂ ———-> N₂ + 2H₂O.
Answer:
(a) Combination reaction
(b) Displacement reaction
(c) Combustion reaction
(d) Decomposition reaction

Question 25.
What happens when :
CO₂(g) is bubbled through lime water (i) in small amount (ii) in excess
Answer:
(i) Solution becomes milky due to the formation of calcium carbonate

(ii) Milkiness disappears because calcium carbonate changes to calcium hydrogen carbonate which is colourless in nature.

Question 26.
Aluminium is a reactive metal but is still used for packing food articles. Why ?
Answer:
From the position of the aluminium (Al) metal in the activity series, it seems to be quite reactive. However, it is not so reactive. Actually, when the metal is kept in air or oxygen for sometime, it is converted into its oxide called aluminum oxide (Al₂O₃). This gets deposited as the surface of the metal as a thin coating. It is rather passive which means that it is not reactive. Therefore, the metal is used for packing food articles which do not get spoiled under the foil.

Question 27.
Give one example each of :
(i) Thermal decomposition reaction
(ii) Electrolytic decomposition reaction
(iii) Photo decomposition reaction. (CBSE 2014)
Answer:

Question 28.
What are neutralisation reactions ? Why are they so named ? Give one example.
Answer:
A neutralisation reaction is a chemical reaction between an acid and base dissolved in water. For example,
KOH(aq) + HNO₃(aq) ———-> KNO₃(aq) + H₂O (aq)
It is called neutralisation as both KN03 (salt) and H20 that are formed as the products, are of neutral nature.

Question 29.
(a) Why is combustion reaction an oxidation reaction ?
(b) How will you test whether the gas evolved in a reaction is hydrogen ?
(c) Why does not silver evolve hydrogen on reacting with dillute sulphuric acid ?
Answer:
(a) Combustion reaction is an oxidation reaction because it is always carried in the presence of air or oxygen. For example,
CH₄(s) + 2O₂(g) ——–> CO₂(g) + 2H₂O (l)
(b) Bring a burning match stick close to the mouth of the tube from which hydrogen gas escapes. The gas will immediately catch fire and this will be accompanied by pop sound.
(c) Silver is a less reactive metal in the sense that it occupies a place below hydrogen in the reactivity series. Therefore it does not evolve hydrogen gas on reacting with either dilute sulphuric acid or dilute hydrochloric acid.

Question 30.
What is an oxidation reaction ? Identify in the following reactions :
(i) the substance oxidised
(ii) the substance reduced.
Answer:
ZnO + C ———> Zn + CO
Oxidation involves the addition of oxygen or the removal of hydrogen in a chemical reaction. In the given reaction, carbon is oxidised to carbon monoxide while zinc oxide is reduced to zinc.

Question 31.
Identify the type of reaction in the following examples :
(i) Na₂SO₄(aq) + BaCl₂(aq) ———-> BaSO₄(s) + 2NaCl(aq)
(ii) Fe(s) + CuSO₄(aq) ———-> FeSO₄(aq) + Cu(s)
(iii) 2H₂(g) + O₂(g) ———> 2H₂O(l)

Answer:
(i) It is an example of double displacement reaction.
(ii) It is an example of displacement reaction.
(iii) It is an example of combination reaction.

Question 32.
Solid calcium oxide was taken in a container and water was added slowly to
(i) State two observations made in the experiment.
(ii) Write the name of the chemical formula of the product.
Answer:
(i) Water will start boiling and hissing noise will be produced.
(ii) Calcium hydroxide (slaked lime) will be formed.

Question 33.
A house wife wanted her house to be white washed. She bought 10 kg of quick lime from the market and dissolved in 30 litres of water.

Answer:
She noticed that water started boiling even when it was not being heated. Give reason for her observation. Write the corresponding equation and name the product formed.
A suspension of slaked lime also called calcium hydroxide is formed when water is added to quick lime.

Since the reaction is highly exothermic, the solution started boiling although it was not being heated. The suspension of slaked lime is allowed to cool for sometime, preferably overnight. It is then decanted and the liquid obtained is used for white washing.

Question 34.
(i) What is observed when a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube ?
(ii) What type of reaction is this ?
(iii)Write a balanced equation to represent the above reaction.
Answer:
(i) A yellow precipitate of lead iodide appears at the bottom of the test tube.
(ii) It is an example of double displacement reaction.
(iii) The balanced equation for the reaction is :

Question 35.
What change in colour is observed when white silver chloride is left exposed to sun light ? State the type of chemical reaction in this change.
Answer:
White colour of silver chloride changes to grey due to formation of metallic silver. The reaction is known as photochemical reaction. It is also a decomposition reaction.

Question 36.
What happens when an aqueous solution of sodium sulphate reacts with an aqueous solution of barium chloride ? State the physical conditions of reactants in which the reaction between them will not take place. Write the balanced chemical equation for the reaction and name the type of reaction. (CBSE 2016)
Answer:
A white precipitate of barium sulphate is immediately formed when the two aqueous solutions are mixed in a test tube. No reaction will be possible if the two reactants are in the solid state. The balanced chemical equation for the double displacement reaction is :

Question 37.
What is a redox reaction ? When a magnesium ribon burns in air with a dazzling flame and forms a white ash; is magnesium oxidised or reduced ? Why ?
Answer:
A redox reaction is a chemical reaction in which one of the reactants gets oxidised while the other is reduced simultaneously. In the reaction under study, magnesium is oxidised to magnesium oxide since the metal has gained oxygen.

Question 38.
(a) What happens chemically when quick lime is added to water ? (CBSE 2012)
(b) Write the chemical equation in balanced form.
MnO₂ + HCl ———-> MnCl₂ + Cl₂ + H₂O
(c) What is decomposition reaction ? Explain it with suitable example.
Answer:
(a) When quick lime (CaO) is added to water, slaked lime Ca(OH)₂ is formed. The reaction is highly exothermic in nature.
(b) The balanced chemical equation is :
MnO₂ + 4HCl ———-> MnCl₂ + 2H₂O + 2Cl₂.
(c) Decomposition reaction is a chemical reaction in which a single substance splits or breaks into two or more substances under suitable conditions. For example,

Long Answer Questions

Question 39.
(a) Why cannot a chemical change be normally reversed ?
(b) Why is it always essential to balance a chemical equation ?
(c) Why do diamond and graphite, the two allotropie forms of carbon evolve different amounts of heats on combustion ?
(d) Can rusting of iron take place in distilled water ?
Answer:
(a) In a chemical change, the products are quite different from the reactants. Therefore, it cannot be normally reversed.
(b) A chemical equation has to be balanced to meet the requirement of the law of conservation of mass. According to the law, the total mass of the reacting species taking part in the reaction is the same as that of the products formed. Since there is a direct relationship between the mass of the different species and their number, it is always essential to balance a chemical equation.
(c) Because they differ in the arrangement of carbon atoms present and have different shapes. The attractive forces among the atoms in the two cases are not same. That is why they evolve different amount of heat.
C(diamond) + O₂(g) ———–> CO₂(g) + 393.5 kj
C(graphite) + O₂(g) ———–> CO₂(g) + 395.4 kj
Please note that diamond and graphite are the two allotropie forms of carbon.
(d) No, rusting of iron cannot take place in distilled water because it neither contains dissolved oxygen nor carbon dioxide. Both are essential for the rusting of iron.

Question 40.
You are given the following materials :
(i) Iron nails
(ii) Copper sulphate solution
(iii) Barium chloride solution
(iv) Copper powder
(v) Ferrous sulphate crystals
(vi) Quick lime.
Identify the type of chemical reaction taking place when :
(a) Barium chloride solution is mixed with copper sulphate solution and a white precipitate is observed.
(b) On heating, copper powder in air in a china dish, the surface of copper powder becomes black.
(c) On heating green ferrous sulphate crystals, reddish brown solid is left and a gas having smell of burning sulphur is noticed.
(d) Iron nails when left dipped in blue copper sulphate solution become brownish in colour and blue colour of copper sulphate solution fades away.
(e) Quick lime reacts vigorously with water releasing a large amount of heat.
Answer:
(a) The reaction is double displacement in nature. ‘

(b) It is an example of combination reaction.

(c) The green crystals of ferrous sulphate have the chemical formula FeS04.7H20. Upon heating, they lose molecules of water of crystallisation.

Upon further heating, ferrous sulphate undergoes decomposition reaction as follows :

Both the gases evolved have the smell of burning sulphur.
(d) This happens because of displacement reaction. Iron displaces copper form copper sulphate solution. Brownish coating of copper gets deposited on the iron nails. As the concentration of copper sulphate in the solution decreases, the blue colour of the solution slowly fades.

(e) Calcium hydroxide is formed as a result of combination reaction. It is highly exothermic. A large amount of heat is evolved accompanied by hissing sound.

Question 41.
A silvery white metal X is in the form of ribbons. Upon ignition, it burns with a dazzling white flame to form white powder Y. When water is added to the powder Y, it partially dissolves to form a substance Z which is used as an antacid.
(a) What is metal X ?
(b) Name the white powder Y.
(c) What is the substance Z ?
(d) Write the chemical reactions that are taking place.
Answer:
(a) The metal is X is Mg.
(b) The white powder Y is MgO.
(c) White powder Y dissolves partially in water to form substance Z. It is Mg(OH), and is used as an antacid.
(d) The chemical reactions that are taking place are :

Question 42.
(i) Account for the following :
(a) White silver chloride turns grey in sunlight.
(b) Brown coloured copper powder on heating in air turns into black coloured substance.
(ii) What do you mean by
(a) Displacement reaction
(b) Reduction reaction
(c) Combination reaction ? Write balanced chemical equation.
Answer:
(i) (a) White coloured silver chloride undergoes decomposition in the presence of sunlight and forms silver (grey in colour) and chlorine.

(b) Brown coloured copper powder on heating in air gets oxidised to copper oxide which is black in colour.

(ii) For the different types of reactions,
(a) In a displacement reaction, one element takes the place of another in a compound dissolved in a solution. For example,
Fe(s) + CuSO₄ (aq) ———> FeSO₄ (aq) + Cu(s)
(b) Combination reaction may be defined as the reaction in which two or more substances combine under suitable conditions to form a new substance. For example,

(c) A decomposition reaction may be defined as the reaction in which a single substance decomposes or splits into two or more substances under suitable conditions.
For example,

It may be concluded that a certain substance is formed or synthesised in combination reaction and it breaks or splits in decomposition reaction. Therefore, the two reactions oppose each other.

Question 43.
(a) Write the chemical equation in the balanced form.
Fe(s) + H₂O(g) ———–> Fe₃O₄(s) + H₂(g)
(b) Identify the type of reaction from the equation given below :
Na₂SO₄ (aq) + BaCl₂ (aq) ———–> BaSO₄(s) + NaCl(aq)
(c) You could have noted that when copper powder is heated in a china dish, the surface of copper powder gets coated with black coloured substance.
(i) Why is this black coloured substance formed ?
(ii) What is this black substance ?
(iii) Write the chemical equation of the reaction taking place.
Answer:
(a) 3Fe(s) + 4H₂O(g) ————> Fe₃O₄(s) + 4H₂(g)
(b) It is a double displacement reaction also called precipitation reaction. A white precipitate of BaSO₄ is formed in the reaction.
(c)
(i) The black substance is formed due to the oxidation of copper.
(ii) The black substance is cupric oxide or copper (II) oxide with formula CuO.

Question 44.
Observe the given figure and answer the following questions.

(a) Write the complete balanced reaction for the reaction that takes place.
(b) Type of reaction involved.
(c) Is there any precipitate formed.
(d) If any precipitate formed, write the colour of the precipitate.
Answer:

(b) It is a double displacement reaction
(c) Yes, a precipitate of barium sulphate is formed.
(d) The precipitate is white in colour.

Question 45.
Select (i) combination reactions (ii) decomposition reactions and displacement reactions from the following

Answer:
(i) Decomposition reaction
(ii) Displacement reaction
(iii) Combination reaction
(iv) Displacement reaction
(v) Combination reaction
(vi) Decomposition reaction.

Hope given Previous Year Question Papers for CBSE Class 10 Science Chapter 1 Chemical Reactions and Equations are helpful to complete your science homework.

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