CBSE Class 10 Mathematics Quadratic Equations VBQs Set F

Question. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?
Answer: Let the original average speed of train be \( x \) km/hr. Therefore, \( \frac{63}{x} + \frac{72}{x + 6} = 3 \)
\( 63(x + 6) + 72x = 3x(x + 6) \)
\( 63x + 378 + 72x = 3x^2 + 18x \)
\( 135x + 378 = 3x^2 + 18x \)
\( 3x^2 - 117x - 378 = 0 \)
\( x^2 - 39x - 126 = 0 \)
\( (x - 42)(x + 3) = 0 \)
Since speed cannot be negative, \( x \neq -3 \).
\( \therefore x = 42 \)
Thus, original speed of train is 42 km/h.

Question. 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.
Answer: Let the speed of stream be \( x \) km/h. Then, the speed of boat upstream = \( (18 - x) \) km/h and speed of boat downstream = \( (18 + x) \) km/h.
According to the question, \( \frac{24}{18 - x} - \frac{24}{18 + x} = 1 \)
\( \Rightarrow \frac{24(18 + x) - 24(18 - x)}{(18 - x)(18 + x)} = 1 \)
\( \Rightarrow \frac{432 + 24x - 432 + 24x}{324 - x^2} = 1 \)
\( \Rightarrow 48x = 324 - x^2 \)
\( \Rightarrow x^2 + 48x - 324 = 0 \)
\( \Rightarrow x^2 + 54x - 6x - 324 = 0 \)
\( \Rightarrow x(x + 54) - 6(x + 54) = 0 \)
\( \Rightarrow (x + 54)(x - 6) = 0 \)
\( \Rightarrow x = -54 \) or \( x = 6 \)
Since, speed cannot be negative. Hence, the speed of stream is 6 km/h.

Question. Solve \( \frac{1}{(a + b + x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \), where \( a + b \neq 0 \).
Answer: Given, \( \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \)
\( \Rightarrow \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \)
\( \Rightarrow \frac{x - (a + b + x)}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \Rightarrow \frac{x - a - b - x}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \Rightarrow \frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \Rightarrow x(a + b + x) = -ab \)
\( \Rightarrow x^2 + (a + b)x + ab = 0 \)
\( \Rightarrow (x + a)(x + b) = 0 \)
\( \Rightarrow x = -a \) or \( x = -b \)

Question. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?
Answer: Let B complete a work in \( x \) days. Then A takes \( (x - 6) \) days to complete it. Together they complete it in 4 days.
According to work done per day,
\( \frac{1}{x - 6} + \frac{1}{x} = \frac{1}{4} \)
\( \Rightarrow \frac{x + x - 6}{x(x - 6)} = \frac{1}{4} \)
\( \Rightarrow 4(2x - 6) = x(x - 6) \)
\( \Rightarrow 8x - 24 = x^2 - 6x \)
\( \Rightarrow x^2 - 14x + 24 = 0 \)
\( \Rightarrow x^2 - 12x - 2x + 24 = 0 \)
\( \Rightarrow (x - 12)(x - 2) = 0 \)
\( \Rightarrow x = 2 \) or \( x = 12 \).
\( x = 2 \) is not possible because then \( x - 6 \) is negative.
So, B takes 12 days to finish the work.

Question. In a rectangular part of dimensions 50 m × 40 m a rectangular pond is constructed so that the area of grass strip of uniform breadth surrounding the pond would be 1184 m\(^2\). Find the length and breadth of the pond.
Answer: Let width of grass strip be \( x \) m. \( \therefore \) Length of pond = \( (50 - 2x) \) m and breadth of pond = \( (40 - 2x) \) m.
Area of park – area of pond = area of grass strip
\( \Rightarrow (50 \times 40) - (50 - 2x)(40 - 2x) = 1184 \)
\( \Rightarrow 2000 - (2000 - 100x - 80x + 4x^2) = 1184 \)
\( \Rightarrow 180x - 4x^2 = 1184 \)
\( \Rightarrow x^2 - 45x + 296 = 0 \)
\( \Rightarrow x^2 - 37x - 8x + 296 = 0 \)
\( \Rightarrow (x - 37)(x - 8) = 0 \)
\( \Rightarrow x = 8 \) or \( 37 \).
\( 37 \) is rejected, as it gives negative values for length and breadth.
Thus, the length of pond = \( 50 - 2 \times 8 = 34 \) m and breadth of pond = \( 40 - 2 \times 8 = 24 \) m.

Question. Solve for \( x \): \( \left( \frac{2x}{x - 5} \right)^2 + 5\left( \frac{2x}{x - 5} \right) - 24 = 0, x \neq 5 \)
Answer: Let \( \frac{2x}{x - 5} = y \)
\( \therefore y^2 + 5y - 24 = 0 \)
\( \Rightarrow (y + 8)(y - 3) = 0 \)
\( y = 3 \) or \( -8 \)
Putting \( y = 3 \), we get \( \frac{2x}{x - 5} = 3 \Rightarrow 2x = 3x - 15 \Rightarrow x = 15 \).
Again, for \( y = -8 \), \( \frac{2x}{x - 5} = -8 \Rightarrow 2x = -8x + 40 \Rightarrow 10x = 40 \Rightarrow x = 4 \).
Hence, \( x = 15 \) or \( 4 \).

Question. Find \( x \) in terms of \( a, b \) and \( c \): \( \frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c} \), where \( x \neq a, b, c \).
Answer: \( a(x - b)(x - c) + b(x - a)(x - c) = 2c(x - a)(x - b) \)
\( \Rightarrow a(x^2 - (b + c)x + bc) + b(x^2 - (a + c)x + ac) = 2c(x^2 - (a + b)x + ab) \)
\( \Rightarrow (a + b - 2c)x^2 + (-ab - ac - ab - bc + 2ac + 2bc)x + (abc + abc - 2abc) = 0 \)
\( \Rightarrow (a + b - 2c)x^2 + (ac + bc - 2ab)x = 0 \)
\( \Rightarrow x[(a + b - 2c)x + (ac + bc - 2ab)] = 0 \)
\( \Rightarrow x = 0 \) or \( x = \frac{2ab - ac - bc}{a + b - 2c} \).

Question. The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is \( 2\frac{16}{21} \), find the fraction.
Answer: Let numerator be \( x \). Then, the fraction = \( \frac{x}{2x + 1} \).
Again, \( \frac{x}{2x + 1} + \frac{2x + 1}{x} = \frac{58}{21} \)
\( \Rightarrow 21[x^2 + (2x + 1)^2] = 58(2x^2 + x) \)
\( \Rightarrow 21(5x^2 + 4x + 1) = 116x^2 + 58x \)
\( \Rightarrow 105x^2 + 84x + 21 = 116x^2 + 58x \)
\( \Rightarrow 11x^2 - 26x - 21 = 0 \)
\( \Rightarrow 11x^2 - 33x + 7x - 21 = 0 \)
\( \Rightarrow (x - 3)(11x + 7) = 0 \)
\( \Rightarrow x = 3 \) or \( x = -\frac{7}{11} \). (rejected negative value)
Hence, fraction = \( \frac{3}{2(3) + 1} = \frac{3}{7} \).

Question. The time taken by a person to cover 150 km was \( 2\frac{1}{2} \) hours more than the time taken in the return journey. If he returned at a speed of 10 km/h more than the speed while going, find the speed in km/h in each direction.
Answer: Let the speed while going be \( x \) km/h.
\( \therefore \) Speed while returning = \( (x + 10) \) km/h.
According to question, \( \frac{150}{x} - \frac{150}{x + 10} = \frac{5}{2} \)
\( \frac{150(x + 10 - x)}{x(x + 10)} = \frac{5}{2} \)
\( \frac{1500}{x^2 + 10x} = \frac{5}{2} \)
\( x^2 + 10x - 600 = 0 \)
\( (x + 30)(x - 20) = 0 \)
\( x = 20 \) or \( -30 \). (Rejecting negative value)
\( \therefore \) Speed while going = 20 km/h and speed while returning = 20 + 10 = 30 km/h.

Question. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is \( \frac{29}{20} \). Find the original fraction.
Answer: Let the denominator be \( x \), then numerator = \( x - 3 \).
So, the fraction = \( \frac{x - 3}{x} \).
By the given condition, new fraction = \( \frac{x - 3 + 2}{x + 2} = \frac{x - 1}{x + 2} \).
Then, \( \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20} \)
\( \Rightarrow 20[(x - 3)(x + 2) + x(x - 1)] = 29(x^2 + 2x) \)
\( \Rightarrow 20(x^2 - x - 6 + x^2 - x) = 29x^2 + 58x \)
\( \Rightarrow 11x^2 - 98x - 120 = 0 \)
\( \Rightarrow 11x^2 - 110x + 12x - 120 = 0 \)
\( \Rightarrow (11x + 12)(x - 10) = 0 \)
\( \Rightarrow x = 10 \).
\( \therefore \) The fraction is \( \frac{7}{10} \).

Question. Check whether the equation \( 5x^2 - 6x - 2 = 0 \) has real roots and if it has, find them by the method of completing the square. Also, verify that roots obtained satisfy the given equation.
Answer: Discriminant \( D = b^2 - 4ac \). Here, \( a = 5, b = -6 \) and \( c = -2 \).
Then, \( b^2 - 4ac = (-6)^2 - 4 \times 5 \times -2 = 36 + 40 = 76 > 0 \).
So the equation has real and two distinct roots.
Again, \( 5x^2 - 6x = 2 \Rightarrow x^2 - \frac{6}{5}x = \frac{2}{5} \).
On adding square of the half of coefficient of \( x \),
\( x^2 - \frac{6}{5}x + \frac{9}{25} = \frac{2}{5} + \frac{9}{25} \Rightarrow \left( x - \frac{3}{5} \right)^2 = \frac{19}{25} \)
\( \Rightarrow x - \frac{3}{5} = \pm \frac{\sqrt{19}}{5} \Rightarrow x = \frac{3 \pm \sqrt{19}}{5} \).
Verification: \( 5\left( \frac{3 + \sqrt{19}}{5} \right)^2 - 6\left( \frac{3 + \sqrt{19}}{5} \right) - 2 = \frac{9 + 6\sqrt{19} + 19}{5} - \frac{18 + 6\sqrt{19}}{5} - 2 = 0 \). Hence Verified.

Question. If the roots of the quadratic equation \( (c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0 \) in \( x \) are equal, then show that either \( a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \).
Answer: Here, \( A = (c^2 - ab), B = -2(a^2 - bc), C = (b^2 - ac) \). For real equal roots \( D \Rightarrow B^2 - 4AC = 0 \).
\( [-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0 \)
\( 4(a^4 + b^2c^2 - 2a^2bc) - 4(b^2c^2 - ac^3 - ab^3 + a^2bc) = 0 \)
\( a^4 - 3a^2bc + ac^3 + ab^3 = 0 \)
\( a(a^3 + b^3 + c^3 - 3abc) = 0 \)
\( \Rightarrow a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \). Hence Proved.

Question. The difference between the radii of the smaller circle and the larger circle is 7 cm and the difference between the areas of the two circles is 1078 sq. cm. Find the radius of the smaller circle.
Answer: We have \( r_2 - r_1 = 7 \) cm, \( r_2 > r_1 \) and \( \pi(r_2^2 - r_1^2) = 1078 \text{ cm}^2 \).
\( \pi(r_2 - r_1)(r_2 + r_1) = 1078 \Rightarrow \frac{22}{7} \times 7 \times (r_2 + r_1) = 1078 \)
\( \Rightarrow r_2 + r_1 = \frac{1078}{22} = 49 \).
On adding \( r_2 - r_1 = 7 \) and \( r_2 + r_1 = 49 \), we get \( 2r_2 = 56 \Rightarrow r_2 = 28 \) cm.
Then \( r_1 = 49 - 28 = 21 \) cm.
Hence, radii of smaller circle is 21 cm.

Question. If roots of the quadratic equation \( x^2 + 2px + mn = 0 \) are real and equal, show that the roots of the quadratic equation \( x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0 \) are also equal.
Answer: For equal roots of \( x^2 + 2px + mn = 0, 4p^2 - 4mn = 0 \Rightarrow p^2 = mn \).
For equal roots of \( x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0 \),
Discriminant = \( 4(m + n)^2 - 4(m^2 + n^2 + 2p^2) \)
= \( 4[m^2 + n^2 + 2mn - m^2 - n^2 - 2mn] = 0 \) (since \( p^2 = mn \)).
\( \therefore \) roots of the second equation are also equal.

Question. Find the positive values of \( k \) for which quadratic equations \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) both will have the real roots.
Answer: (i) For \( x^2 + kx + 64 = 0 \) to have real roots \( k^2 - 256 \ge 0 \Rightarrow k \ge 16 \).
(ii) For \( x^2 - 8x + k = 0 \) to have real roots \( 64 - 4k \ge 0 \Rightarrow k \le 16 \).
For both to hold simultaneously, \( k = 16 \).

Question. If (– 5) is a root of the quadratic equation \( 2x^2 + px + 15 = 0 \) and the quadratic equation \( p(x^2 + x) + k = 0 \) has equal roots, then find the values of \( p \) and \( k \).
Answer: Since, (–5) is a root of given quadratic equation \( 2x^2 + px - 15 = 0 \),
\( 2(-5)^2 + p(-5) - 15 = 0 \Rightarrow 50 - 5p - 15 = 0 \Rightarrow 5p = 35 \Rightarrow p = 7 \).
And \( p(x^2 + x) + k = 0 \) has equal roots \( \Rightarrow px^2 + px + k = 0 \).
So, \( b^2 - 4ac = 0 \Rightarrow p^2 - 4pk = 0 \Rightarrow (7)^2 - 4(7)k = 0 \Rightarrow 28k = 49 \Rightarrow k = \frac{49}{28} = \frac{7}{4} \).
Hence, \( p = 7 \) and \( k = \frac{7}{4} \).

Question. 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 \( a = b = c \).
Answer: Given, \( (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 \)
\( \Rightarrow 3x^2 - 2(a + b + c)x + (ab + bc + ca) = 0 \).
For equal roots, \( B^2 - 4AC = 0 \)
\( \Rightarrow \{-2(a + b + c)\}^2 = 4 \times 3(ab + bc + ca) \)
\( \Rightarrow 4(a + b + c)^2 - 12(ab + bc + ca) = 0 \)
\( \Rightarrow a^2 + b^2 + c^2 - ab - bc - ca = 0 \)
\( \Rightarrow \frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2] = 0 \)
\( \Rightarrow (a - b)^2 = 0, (b - c)^2 = 0, (c - a)^2 = 0 \)
\( \Rightarrow a = b, b = c, c = a \). Hence, \( a = b = c \). Proved.