CBSE Class 10 Mathematics Quadratic Equations Assignment Set 01

Question. If the product of roots of the equation x³ – 3x + k = 10 is –2, then the value of k is
(a) –2
(b) –8
(c) 8
(d) 12

Answer: C

Question. If one root of 5x² + 13x + k = 0 be the reciprocal of the other root, then the value of k is
(a) 0
(b) 1
(c) 2
(d) 5

Answer: D

Question. If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is
(a) x² – 6x + 6 = 0
(b) x² + 6x – 6 = 0
(c) x² – 6x – 6 = 0
(d) x² + 6x + 6 = 0

Answer: A

Question. Find the product of the roots of x² + 8x – 16 = 0
(a) 8
(b) –8
(c) 16
(d) –16

Answer: D

Question. If the roots of the equation ax² + bx + c = 0 are α and β, then the quadratic equation whose roots are –α and –β is _____ .
(a) ax² – bx – c = 0
(b) ax² – bx + c = 0
(c) ax² + bx – c = 0
(d) ax² – bx + 2c = 0

Answer: B

Question. If the equation (1 + m²) x² + (2mc) x + (c² – a²) = 0 has equal roots, then
(a) c² – a² = 1 +m²
(b) c² = a² (1 + m²)
(c) c²a² = (1 + m²)
(d) c² + a² = 1 + m²

Answer: B

Question. Which of the following satisfy the equation a²b²x² + b²x – a²x – 1 = 0
(a) 1/a²
(b) 1/b²
(c) -1/b²
(d) None of these

Answer: B

Question. The roots of the quadratic equation x² – 0.04 = 0 are
(a) ± 0.2
(b) ± 0.02
(c) 0.4
(d) 2

Answer: A

Question. One of the two students, while solving a quadratic equation in x, copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term and coefficient of x² correctly as –6 and 1 respectively. The correct roots are
(a) 3, –2
(b) –3, 2
(c) –6, –1
(d) 6, –1

Answer: D

Question. If the equation x² + 2(k + 2)x + 9k = 0 has equal roots, then k = ?
(a) 1 or 4
(b) –1 or 4
(c) 1 or – 4
(d) –1 or – 4

Answer: A

Question. If the roots of 5x² – kx + 1 = 0 are real and distinct, then
(a) −2√5 < k < 2√5
(b) k > 2√5 only
(c) k < −2√5 only
(d) either k > 2√5 or k < −2√5

Answer: D

Question. If a – b, b – c are the roots of ax² + bx + c = 0, then find the value of (a - b)(b - c)/c - a
(a) b/c
(b) c/b
(c) ab/c
(d) bc/a

Answer: B

Question. The condition for one root of the quadratic equation ax² + bx + c = 0 to be twice the other, is
(a) b² = 4ac
(b) 2b² = 9ac
(c) c² = 4a + b²
(d) c² = 9a – b²

Answer: B

Question. If the ratio of the roots of the equation x² + bx + c = 0 is the same as that of x² + qx + r = 0, then
(a) r²b = qc²
(b) r²c = qb²
(c) c2r = q²b
(d) b²r = q²c

Answer: D

Question. The real roots of the equation x²/³ + x^1/3 − 2 = 0 are
(a) 1, 8
(b) –1, –8
(c) –1, 8
(d) 1, –8

Answer: D

Question. Which of the following is not a quadratic equation?
(a) x² – 2x + 2 (3 – x) = 0
(b) x (x + 1) + 1 = (x – 2) (x – 5)
(c) (2x – 1) (x – 3) = (x + 5) (x – 1)
(d) x³ – 4x² – x + 1 = (x –2)³

Answer: B

Question. If one root of the quadratic equation ax² + bx + c = 0 is the reciprocal of the other, then
(a) b = c
(b) a = b
(c) ac = 1
(d) a = c

Answer: D

Question. The roots of the equation x + 1/x  = 3(1/3), x ≠ 0, are
(a) 3, 1
(b) 3, 1/3
(c) 3, − (1/3)
(d) – 3, 1/3

Answer: B

Question. If the equation (m² + n²) x² –2 (mp + nq) x + p² + q² = 0 has equal roots, then
(a) mp = nq
(b) mq = np
(c) mn = pq
(d) mq = √np

Answer: B

Question. If x² + y² = 25, xy = 12, then x =
(a) {3, 4}
(b) {3, –3}
(c) {3, 4, –3, –4}
(d) {–3, –3}

Answer: C

Question. If x = √ 7 + 4√3 , then x + 1/x =
(a) 4
(b) 6
(c) 3
(d) 2

Answer: A

Question. If the roots of the equation px² + 2qx + r = 0 and qx² − 2√prx + q = 0 be real, then
(a) p = q
(b) q² = pr
(c) p² = qr
(d) r² = pq

Answer: B

Question. The equation 2x² + 2(p + 1) x + p = 0, where p is real, always has roots that are
(a) Equal
(b) Equal in magnitude but opposite in sign
(c) Irrational
(d) Real

Answer: D

Question. Each root of x² – bx + c = 0 is decreased by 2. The resulting equation is x² – 2x + 1 = 0, then
(a) b = 6, c = 9
(b) b = 3, c = 5
(c) b = 2, c = –1
(d) b = – 4, c = 3

Answer: A

Question. Two distinct polynomials f (x) and g(x) are defined as follows:
f (x) = x² + ax + 2; g (x) = x² + 2x + a.
If the equations f (x) = 0 and g(x) = 0 have a common root, then the sum of the roots of the equation f (x) + g(x) = 0 is
(a) - (1/2)
(b) 0
(c) 1/2
(d) l

Answer: C

Question. If a and b are the roots of the quadratic equation x² – 6x – 2 = 0 and if a_n = α_n – β_n, then the value of a₁₀ - 2a₈/2a₉ is
(a) 6.0
(b) 5.2
(c) 5.0
(d) 3.0

Answer: D

Question. If x = 3 + √5 / 2 and y = x³, then y satisfies the quadratic equation
(a) y² – 18y + 1 = 0
(b) y² + 18y + 1 = 0
(c) y² – 18y – 1 = 0
(d) y² + 18y – 1 = 0

Answer: A

Question. Let b be a non-zero real number. Suppose the quadratic equation 2x² + bx + 1/b = 0 has two distinct real roots. Then
(a) b + 1/b > 5/2
(b) b + 1/b < 5/2
(c) b² – 3b > –2
(d) b² + 1/b² < 4

Answer: C

Question. If the quadratic equations 2x² + 4x + (a + 5) = 0 have equal roots and (a + 4)x² + ax – 3b = 0 have distinct real roots then which of the following is true:
(a) a = –3, b < 3/4
(b) a = 3, b > 3/4
(c) a = –3, b > – 3/4
(d) a = 3, b < 3/4

Answer: C

Question. The value of λ such that sum of the squares of the roots of the quadratic equation, x² + (3 – l)x + 2 = λ has the least value is:
(a) 15/8
(b) 1
(c) -
(d) 2

Answer: D

Question. Consider the quadratic equation nx² + 7√nx + n = 0, where n is a positive integer. Which of the following statements are necessarily correct?
I. For any n, the roots are distinct.
II. There are infinitely many values of n for which both roots are real.
III. The product of the roots is necessarily an integer.
(a) III only
(b) I and III
(c) II and III
(d) I, II and III

Answer: B

Question. Two quadratic equations x² – bx + 6 = 0 and x² – 6x + c – 0 have a common root. If the remaining roots of the first and second equations are positive integers and are in the ration 3 : 4 respectively, then the common root is
(a) 1
(b) 2
(c) 3
(d) 4

Answer: B

Question. The values of k, so that the equations 2x² + kx – 5 = 0 and x² – 3x – 4 = 0 have one root in common, are
(a) 3, 27/2
(b) 9, 27/4
(c) – 3, −27/4
(d) 3, 4/27

Answer: C

Question. If α and β be two roots of the equation x² – 64x + 256 = 0.
Then the value of (α³/β⁵)^1/8 + (β³/α⁵)^1/8 is:
(a) 2
(b) 3
(c) 1
(d) 4

Answer: A

Question. Which one of the following is not a quadratic equation?
(a) (x + 2)² = 2(x + 3)
(b) x² + 3x = (–1) (1 – 3x)²
(c) (x + 2) (x – 1) = x² – 2x – 3
(d) x³ – x² + 2x + 1 = (x + 1)³

Answer: C

Question. If equation x² – (2 + m) x + 1 (m2 – 4m + 4) = 0 has equal roots, then:
(a) m = 0
(b) m = 6
(c) m = 2
(d) m = 3

Answer: B

Question. Which of the following equations have no real roots?
(a) x² − 2√3x + 5 = 0
(b) 2x² + 6√2 +8 = 0
(c) x² − 2√3x − 5 = 0
(d) 2x² − 6√2x − 9 = 0

Answer: A

Question. Two numbers whose sum is 8 and the absolute value of whose difference is 10 are roots of the equation
(a) x² – 8x + 9 = 0
(b) x² – 8x – 9 = 0
(c) x² + 8x – 9 = 0
(d) –x² + 8x + 9 = 0

Answer: B

Question. If α, β are roots of x² + 5x + a = 0 and 2α + 5β = –1, then
(a) α = 8
(b) β = –3
(c) α = 9
(d) a = – 24

Answer: D

Question. Which constant should be added and subtracted to solve the quadratic equation 4x² – √3x – 5 = 0 by the method of completing the square?
(a) 9/16
(b) 3/64
(c) 3/4
(d) √3/4

Answer: B

Question. Which of the following equations has 2 as a root?
(a) x² – 4x + 5 = 0
(b) x² + 3x – 12 = 0
(c) 2x² – 7x + 6 = 0
(d) 3x² – 6x – 2 = 0

Answer: C

Question. Values of k for which the quadratic equation 2x²– kx + k = 0 has equal roots is
(a) 0 only
(b) 4 only
(c) 8 only
(d) 0, 8

Answer: D

Question. If α, β are roots of the equation x² – 5x + 6 = 0, then the equation whose roots are α + 3 and β + 3 is
(a) 2x² – 11x + 30 = 0
(b) –x² + 11x = 0
(c) x² – 11x + 30 = 0
(d) 2x² – 22x + 40 = 0

Answer: C

Question. The value of p for which the difference between the roots of the equation x² + px + 8 = 0 is 2, are
(a) 4
(b) 8
(c) 6
(d) – 4

Answer: C

Question. If the roots of x² + px + 12 = 0 are in the ratio 1 : 3, then value(s) of p are
(a) 3
(b) 8
(c) 6
(d) – 3

Answer: B

Question. Roots of quadratic equation x² – 3x + 2 = 0 are
(a) 3
(b) –1
(c) 2
(d) 4

Answer: C

Question. If x = 2 and x = 3 are roots of the equation 3x² – 2px + 2q = 0, then
(a) P = 2/15
(b) p = 15
(c) q = 9
(d) 6p + 2q = 27

Answer: C

Fill in the Blanks

DIRECTIONS : Complete the following statements with an appropriate word/ term to be filled in the blank space(s).

Question. A quadratic equation in the variable x is of the form ax² + bx + c = 0, where a, b, c are real numbers and a ......
Answer: ≠ 0

Question. A quadratic equation ax² + bx + c = 0 has two distinct real roots, if b² – 4ac ..............
Answer: > 0

Question. The values of k for which the equation 2x² + kx + x + 8 = 0 will have real and equal roots are .............
Answer: 7 and –9

Question. The quadratic equation whose roots are the sum and difference of the squares of roots of the equation x² – 3x + 2 = 0 is....
Answer: x² – 8x + 15 = 0

Question. If a, b are the roots of x² + x + 1 = 0, then a2 + b2 = ...........
Answer: –1

Question. If α, β are the roots of x² + bx + c = 0 and α + h, β + h are the roots of x² + qx + r = 0, then h = ..........
Answer: 1/2 (b - q)

Question. If α, β are roots of the equation ax² + bx + c = 0, then the quadratic equation whose roots are aα + b and aβ + b is .............
Answer: x² – bx + ca = 0

Question. If r, s are roots of ax² + bx + c = 0, then is  1/r² + 1/r² is ..........
Answer: b² - 2ac/c²

Question. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides are ...............
Answer: 5 cm, 12 cm.

Question. The equation ax² + bx + c = 0, a ≠ 0 has no real roots, if .........
Answer: b² < 4ac

Question. A quadratic equation cannot have more than ........ roots.
Answer: two

Question. Let ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0, be a quadratic equation, then this equation has no real roots if and only if ........... 
Answer: b² < 4ac

LEVEL 1 (1 Mark each)

Question 1. What will be the nature of roots of the quadratic equation: \(2x^2 + 4x - 7 = 0\)?
Answer: Comparing the given equation with the standard quadratic equation \(ax^2 + bx + c = 0\), we get:
\(a = 2, b = 4, c = -7\)
Discriminant \(D = b^2 - 4ac\)
\(D = (4)^2 - 4 \times 2 \times (-7)\)
\(D = 16 + 56 = 72\)
Since \(D > 0\), the roots of the quadratic equation are real and unequal.
In simple words: Since the discriminant is positive, the equation has two distinct real solutions.

Exam Tip: Always calculate the discriminant \(D = b^2 - 4ac\) carefully, keeping track of negative signs.

Question 2. Show that \(x = -2\) is a solution of \(3x^2 + 13x + 14 = 0\).
Answer: Substitute \(x = -2\) into the left-hand side (L.H.S.) of the equation:
\(\text{L.H.S.} = 3(-2)^2 + 13(-2) + 14\)
\(\text{L.H.S.} = 3(4) - 26 + 14\)
\(\text{L.H.S.} = 12 - 26 + 14 = 0\)
Since \(\text{L.H.S.} = \text{R.H.S.}\), \(x = -2\) is a solution of the given quadratic equation.
In simple words: Substituting the value makes both sides equal, proving it is a valid solution.

Exam Tip: Use parentheses around negative numbers when substituting to avoid sign errors with exponents.

Question 3. Check if \((x + 1)^2 = 2(x - 3)\) is a quadratic or not?
Exam Tip: Always expand brackets and simplify completely before concluding the degree of the equation.

Question 4. Find the roots of the quadratic \(2x^2 - 7x + 3 = 0\).
Answer: Using the middle-term splitting method:
\(2x^2 - 6x - x + 3 = 0\)
Group the terms:
\(2x(x - 3) - 1(x - 3) = 0\)
\((2x - 1)(x - 3) = 0\)
Set each factor equal to zero:
\(2x - 1 = 0 \implies x = \frac{1}{2}\)
\(x - 3 = 0 \implies x = 3\)
Hence, the roots are \(x = \frac{1}{2}\) and \(x = 3\).
In simple words: Factor the quadratic expression into two linear factors and solve each to find the roots.

Exam Tip: Check your factored form by expanding it back to ensure it matches the original quadratic expression.

LEVEL 2 (2 Marks each)

Question 1. Find the value of \(p\) for which the quadratic equation \(4x^2 + px + 3 = 0\) has equal roots.
Answer: For a quadratic equation to have equal roots, the discriminant must be zero (\(D = 0\)).
Here, \(a = 4, b = p, c = 3\)
\(D = b^2 - 4ac = 0\)
\(p^2 - 4(4)(3) = 0\)
\(p^2 - 48 = 0\)
\(p^2 = 48\)
\(p = \pm\sqrt{48} = \pm 4\sqrt{3}\)
In simple words: Set the discriminant to zero for equal roots and solve for the unknown parameter.

Exam Tip: Do not forget to include both positive and positive-negative square root values when solving \(p^2 = k\).

Question 2. Find two consecutive positive integers, sum of whose squares is 925.
Answer: Let the two consecutive positive integers be \(x\) and \(x + 1\).
According to the question:
\(x^2 + (x + 1)^2 = 925\)
\(x^2 + x^2 + 2x + 1 = 925\)
\(2x^2 + 2x + 1 - 925 = 0\)
\(2x^2 + 2x - 924 = 0\)
Divide the entire equation by 2:
\(x^2 + x - 462 = 0\)
Factor the quadratic equation by splitting the middle term:
\(x^2 + 22x - 21x - 462 = 0\)
\(x(x + 22) - 21(x + 22) = 0\)
\((x - 21)(x + 22) = 0\)
\(x = 21 \text{ or } x = -22\)
Since the integers must be positive, we reject \(x = -22\).
Therefore, \(x = 21\) and \(x + 1 = 22\).
In simple words: Formulate the equation using consecutive numbers, expand, simplify, solve, and discard negative values.

Exam Tip: Always state why negative values are rejected based on the word problem conditions.

Question 3. Solve the quadratic \(2x^2 - 7x + 3 = 0\) by using quadratic formula.
Answer: Comparing with \(ax^2 + bx + c = 0\), we get \(a = 2, b = -7, c = 3\).
Discriminant \(D = b^2 - 4ac\)
\(D = (-7)^2 - 4(2)(3)\)
\(D = 49 - 24 = 25\)
Since \(D > 0\), real roots exist.
Using the quadratic formula \(x = \frac{-b \pm \sqrt{D}}{2a}\):
\(x = \frac{-(-7) \pm \sqrt{25}}{2(2)}\)
\(x = \frac{7 \pm 5}{4}\)
Calculating the two cases:
\(x = \frac{7 + 5}{4} = \frac{12}{4} = 3\)
\(x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}\)
Hence, the roots are \(3\) and \(\frac{1}{2}\).
In simple words: Substitute coefficients directly into the quadratic formula to find the roots.

Question 4. For what value of \(k\), is 3 a root of the equation \(2x^2 + x + k = 0\)?
Answer: Since \(3\) is a root of the equation, substituting \(x = 3\) must satisfy the equation:
\(2(3)^2 + (3) + k = 0\)
\(2(9) + 3 + k = 0\)
\(18 + 3 + k = 0\)
\(21 + k = 0\)
\(k = -21\)
In simple words: Substitute the given root into the equation to evaluate the unknown constant.

Exam Tip: Evaluate exponents before multiplying by coefficients (i.e., \(3^2 = 9\), then \(2 \times 9 = 18\)).

LEVEL 3 (3 - Marks each)

Question 1. Solve \(4x^2 + 4\sqrt{3}x + 3 = 0\) by the method of completing squares.
Answer: Given equation: \(4x^2 + 4\sqrt{3}x + 3 = 0\)
Divide both sides by 4 to make the coefficient of \(x^2\) equal to 1:
\(x^2 + \sqrt{3}x + \frac{3}{4} = 0\)
\(x^2 + \sqrt{3}x = -\frac{3}{4}\)
Add the square of half the coefficient of \(x\), which is \(\left(\frac{\sqrt{3}}{2}\right)^2\), to both sides:
\(x^2 + \sqrt{3}x + \left(\frac{\sqrt{3}}{2}\right)^2 = -\frac{3}{4} + \left(\frac{\sqrt{3}}{2}\right)^2\)
\(\left(x + \frac{\sqrt{3}}{2}\right)^2 = -\frac{3}{4} + \frac{3}{4}\)
\(\left(x + \frac{\sqrt{3}}{2}\right)^2 = 0\)
\(x + \frac{\sqrt{3}}{2} = 0 \text{ or } x + \frac{\sqrt{3}}{2} = 0\)
\(x = -\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}\)
In simple words: Make the leading coefficient 1, complete the square on the left, and solve for the equal roots.

Exam Tip: When completing the square, always halve the coefficient of the linear term carefully.

Question 2. Had Anita scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?
Answer: Let her actual marks be \(x\).
According to the question:
\(9(x + 10) = x^2\)
\(9x + 90 = x^2\)
Rearranging into standard quadratic form:
\(x^2 - 9x - 90 = 0\)
Factor by splitting the middle term:
\(x^2 - 15x + 6x - 90 = 0\)
\(x(x - 15) + 6(x - 15) = 0\)
\((x + 6)(x - 15) = 0\)
\(x = -6 \text{ or } x = 15\)
Since marks cannot be negative, \(x = -6\) is rejected.
Therefore, \(x = 15\).
Anita got 15 marks in her mathematics test.
In simple words: Translate the word statement into an equation, solve the quadratic, and discard negative marks.

Exam Tip: Verify the result: 9 times (15 + 10) = 9 × 25 = 225, and 15² = 225. Both sides match.

Question 3. Solve for \(x\): \(\frac{1}{2x - 3} + \frac{1}{x - 5} = 1\), where \(x \neq \frac{3}{2}, 5\).
Answer: Take the LCM on the left-hand side:
\(\frac{(x - 5) + (2x - 3)}{(2x - 3)(x - 5)} = 1\)
\(\frac{3x - 8}{2x^2 - 10x - 3x + 15} = 1\)
\(\frac{3x - 8}{2x^2 - 13x + 15} = 1\)
Cross-multiply:
\(3x - 8 = 2x^2 - 13x + 15\)
Rearrange all terms to one side:
\(2x^2 - 13x - 3x + 15 + 8 = 0\)
\(2x^2 - 16x + 23 = 0\)
Using the quadratic formula with \(a = 2, b = -16, c = 23\):
\(x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(2)(23)}}{2(2)}\)
\(x = \frac{16 \pm \sqrt{256 - 184}}{4}\)
\(x = \frac{16 \pm \sqrt{72}}{4}\)
\(x = \frac{16 \pm 6\sqrt{2}}{4} = \frac{8 \pm 3\sqrt{2}}{2}\)
In simple words: Combine fractions using LCM, cross-multiply to form a standard quadratic, and apply the quadratic formula.

Exam Tip: Simplify square roots completely (e.g., \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\)) before reducing fractions.

Question 4. A cottage industry produces a certain number of articles in a day. It was observed on a particular day that the cost of production of each article was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
Answer: Let the number of articles produced be \(x\).
Cost of production of each article = \(2x + 3\)
Total cost of production = \(x(2x + 3) = 90\)
\(2x^2 + 3x - 90 = 0\)
Factor by splitting the middle term:
\(2x^2 + 15x - 12x - 90 = 0\)
\(x(2x + 15) - 6(2x + 15) = 0\)
\((2x + 15)(x - 6) = 0\)
\(x = -\frac{15}{2} \text{ or } x = 6\)
Since the number of articles cannot be negative or fractional in this context, \(x = 6\).
Number of articles produced = 6
Cost of each article = \(2(6) + 3 = \text{Rs } 15\)
In simple words: Set up an equation for total cost as number of articles multiplied by unit cost, then solve.

Exam Tip: Clearly state the final values for both the number of articles and the unit cost with units.

LEVEL 4 (4 - Marks each)

Question 1. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer: Let the speed of the stream be \(x\) km/h.
Speed of the boat upstream = \((18 - x)\) km/h
Speed of the boat downstream = \((18 + x)\) km/h
Time taken upstream = \(\frac{24}{18 - x}\) hours
Time taken downstream = \(\frac{24}{18 + x}\) hours
According to the question, upstream time takes 1 hour more than downstream time:
\(\frac{24}{18 - x} - \frac{24}{18 + x} = 1\)
\(24 \left[\frac{(18 + x) - (18 - x)}{(18 - x)(18 + x)}\right] = 1\)
\(24 \left[\frac{2x}{324 - x^2}\right] = 1\)
\(48x = 324 - x^2\)
\(x^2 + 48x - 324 = 0\)
Using the quadratic formula:
\(x = \frac{-48 \pm \sqrt{48^2 - 4(1)(-324)}}{2(1)}\)
\(x = \frac{-48 \pm \sqrt{2304 + 1296}}{2}\)
\(x = \frac{-48 \pm \sqrt{3600}}{2}\)
\(x = \frac{-48 \pm 60}{2}\)
\(x = \frac{12}{2} = 6 \text{ or } x = \frac{-108}{2} = -54\)
Since speed cannot be negative, \(x = 6\).
Speed of the stream = 6 km/h.
In simple words: Equate the difference between upstream and downstream times to 1 and solve the resulting quadratic equation.

Exam Tip: Remember that upstream speed is boat speed minus stream speed, while downstream speed is boat speed plus stream speed.

Question 2. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed?
Answer: Let the original average speed be \(x\) km/h.
According to the question:
\(\frac{63}{x} + \frac{72}{x + 6} = 3\)
Take LCM and combine the fractions:
\(\frac{63(x + 6) + 72x}{x(x + 6)} = 3\)
\(\frac{63x + 378 + 72x}{x^2 + 6x} = 3\)
\(\frac{135x + 378}{x^2 + 6x} = 3\)
Cross-multiply:
\(135x + 378 = 3(x^2 + 6x)\)
\(135x + 378 = 3x^2 + 18x\)
Rearrange terms to one side:
\(3x^2 + 18x - 135x - 378 = 0\)
\(3x^2 - 117x - 378 = 0\)
Divide by 3:
\(x^2 - 39x - 126 = 0\)
Factor by splitting the middle term:
\(x^2 - 42x + 3x - 126 = 0\)
\(x(x - 42) + 3(x - 42) = 0\)
\((x + 3)(x - 42) = 0\)
\(x = -3 \text{ or } x = 42\)
Since speed cannot be negative, \(x = 42\).
Original average speed of the train = 42 km/h.
In simple words: Set up total time as the sum of times taken for both parts of the journey and solve for speed.

Exam Tip: Simplifying the equation by dividing by common factors (like 3) makes factorization much easier.

Question 3. 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.
Answer: Let the digit at the tens place be \(x\).
Since the product of the digits is 18, the digit at the units place is \(\frac{18}{x}\).
The original two-digit number is \(10x + \frac{18}{x}\).
When digits are interchanged, the new number is \(10\left(\frac{18}{x}\right) + x = \frac{180}{x} + x\).
According to the question, subtracting 63 from the original number gives the interchanged number:
\(\left(10x + \frac{18}{x}\right) - 63 = \frac{180}{x} + x\)
Rearrange terms:
\(10x - x + \frac{18}{x} - \frac{180}{x} = 63\)
\(9x - \frac{162}{x} = 63\)
Multiply through by \(x\):
\(9x^2 - 162 = 63x\)
\(9x^2 - 63x - 162 = 0\)
Divide by 9:
\(x^2 - 7x - 18 = 0\)
Factor by splitting the middle term:
\(x^2 - 9x + 2x - 18 = 0\)
\(x(x - 9) + 2(x - 9) = 0\)
\((x + 2)(x - 9) = 0\)
\(x = -2 \text{ or } x = 9\)
Since a digit cannot be negative, \(x = 9\).
Tens digit = 9, Units digit = \(\frac{18}{9} = 2\).
Therefore, the required number is \(10(9) + 2 = 92\).
In simple words: Express digits algebraically, set up the digit-interchange condition, solve for the tens digit, and find the number.

Exam Tip: Verify: 92 - 63 = 29, which has digits interchanged from 92. The product 9 × 2 = 18.

Question 4. A train takes 2 hours less for a journey of 300 km, if its speed is increased by 5 km/h from its usual speed. Find the usual speed of train.
Answer: Let the usual speed of the train be \(x\) km/h.
Usual time taken to cover 300 km = \(\frac{300}{x}\) hours.
Time taken at increased speed \((x + 5)\) km/h = \(\frac{300}{x + 5}\) hours.
According to the question, the new time is 2 hours less:
\(\frac{300}{x} - \frac{300}{x + 5} = 2\)
Take LCM and simplify:
\(\frac{300(x + 5) - 300x}{x(x + 5)} = 2\)
\(\frac{300x + 1500 - 300x}{x^2 + 5x} = 2\)
\(\frac{1500}{x^2 + 5x} = 2\)
Cross-multiply and simplify:
\(2(x^2 + 5x) = 1500\)
\(x^2 + 5x = 750\)
\(x^2 + 5x - 750 = 0\)
Factor by splitting the middle term:
\(x^2 + 30x - 25x - 750 = 0\)
\(x(x + 30) - 25(x + 30) = 0\)
\((x - 25)(x + 30) = 0\)
\(x = 25 \text{ or } x = -30\)
Since speed cannot be negative, \(x = 25\).
Usual speed of the train = 25 km/h.
In simple words: Set up time difference equation, simplify into a quadratic form, and solve for speed.

Exam Tip: Check your answer by substituting \(x = 25\): Time at 25 km/h is 300/25 = 12 hours. Time at 30 km/h is 300/30 = 10 hours. Difference is exactly 2 hours.