CBSE Class 10 Mathematics Quadratic Equations VBQs Set 06

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.

Question. Write the discriminant of the quadratic equation \( (x + 5)^2 = 2(5x - 3) \).
Answer: \( D = -124 \)

Question. Find the discriminant of the quadratic equation: \( 4x^2 - \frac{2}{3}x - \frac{1}{16} = 0 \).
Answer: 3328

Question. Find the roots of the equation \( ax^2 + a = a^2x + x \).
Answer: \( a, \frac{1}{a} \)

Question. Solve the quadratic equation for \( x \): \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \).
Answer: \( \frac{a^2 + b^2}{2}, \frac{a^2 - b^2}{2} \)

Question. Solve the quadratic equation for \( x \): \( 9x^2 - 6b^2x - (a^4 - b^4) = 0 \).
Answer: \( \frac{b^2 + a^2}{3}, \frac{b^2 - a^2}{3} \)

Question. Solve the quadratic equation for \( x \): \( 4x^2 + 4bx - (a^2 - b^2) = 0 \).
Answer: \( \frac{-b + a}{2}, \frac{-b - a}{2} \)

Question. Solve the quadratic equation for \( x \): \( x^2 - 2ax - (4b^2 - a^2) = 0 \).
Answer: \( a + 2b, a - 2b \)

Question. Solve the following using quadratic formula: \( 2\sqrt{3}x^2 - 5x + \sqrt{3} = 0 \).
Answer: \( \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{3}} \)

Question. Solve the following using quadratic formula: \( 3x^2 + 2\sqrt{5}x - 5 = 0 \).
Answer: \( \frac{\sqrt{5}}{3}, -\sqrt{5} \)

Question. Find the roots of quadratic equation: \( x^2 - 3\sqrt{5}x + 10 = 0 \).
Answer: \( 2\sqrt{5}, \sqrt{5} \)

Question. Find the roots of quadratic equation: \( 5\sqrt{5}x^2 + 30x + 8\sqrt{5} = 0 \).
Answer: \( \frac{-4\sqrt{5}}{5}, \frac{-2\sqrt{5}}{5} \)

Question. Solve for \( x \): \( 4x^2 - 4ax + (a^2 - b^2) = 0 \).
Answer: \( \frac{a \pm b}{2} \)

Question. Two water taps together can fill a tank in 9 hours 36 minutes. The tap of large diameter takes 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: 16 hrs and 24 hrs

Question. A rectangular field is 20 m long and 14 m wide. There is a path of equal width all around it, having an area of 111 sq m. Find the width of the path.
Answer: 1.5 m

Question. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Answer: 12

Question. Solve for \( x \): \( \frac{x - 3}{x - 4} + \frac{x - 5}{x - 6} = \frac{10}{6} \); \( x \neq 4, 6 \).
Answer: \( 2 \pm \sqrt{10} \)

Question. Solve for \( x \): \( \frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3} \); \( x \neq 3, 5 \).
Answer: \( \frac{7}{2}, 6 \)

Question. The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.
Answer: 9 and 13

Question. Sum of the areas of two squares is 544 m\(^2\). If the difference of their perimeters is 32 m, find the sides of the two squares.
Answer: 20 m and 12 m

Water Distribution System: Delhi Jal Board (DJB) is the main body of the Delhi Government which supplies drinking water in the National Capital Territory of Delhi. Distribution system is well knit and properly planned. Maintenance of underground pipe and hose system is also performed at regular interval of time. Many rivers and canals are inter-connected in order to ensure un-interrupted water supply. It has been meeting the needs of potable water for more than 16 million people. It ensures availability of 50 gallons per capita per day of pure and filtered water with the help of efficient network of water treatment plants and pumping stations. In our locality, DJB constructed two big reservoir labelled as Reservoir–A and Reservoir–B. Reservoir–A: In order to fill it, department uses two pipes of different diameter. Reservoir–B: Department uses two taps to store water in this reservoir.

Question. Two pipes running together can fill the reservoir in \( 11\frac{1}{9} \) minutes. If one pipe takes 5 minutes more than the other to fill the reservoir, the time in which each pipe alone would fill the reservoir is
(a) 10 min, 12 min
(b) 25 min, 20 min
(c) 15 min, 18 min
(d) 22 min, 28 min
Answer: (b) 25 min, 20 min

Question. Two pipes running together can fill a reservoir in 6 minutes. If one pipe takes 5 minutes more than the other to fill the reservoir, the time in which each pipe would fill the reservoir separately is
(a) 8 min, 6 min
(b) 10 min, 15 min
(c) 12 min, 16 min
(d) 16 min, 18 min
Answer: (b) 10 min, 15 min

Question. Two water taps together can fill a reservoir in \( 9\frac{3}{8} \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the reservoir separately. The time in which each tap can separately fill the reservoir will be
(a) 15 hrs, 25 hrs
(b) 20 hrs, 22 hrs
(c) 14 hrs, 18 hrs
(d) 18 hrs, 16 hrs
Answer: (a) 15 hrs, 25 hrs

Question. Two taps running together can fill the reservoir in \( 3\frac{1}{13} \) minutes. If one tap takes 3 minutes more than the other to fill it, how many minutes each tap would take to fill the reservoir?
(a) 12 min, 15 min
(b) 6 min, 9 min
(c) 18 min, 14 min
(d) 5 min, 8 min
Answer: (d) 5 min, 8 min

Question. If two tapes function simultaneously, reservoir will be filled in 12 hours. One tap fills the reservoir 10 hours faster than the other. The time that the second tap takes to fill the reservoir is given by
(a) 25 hrs
(b) 28 hrs
(c) 30 hrs
(d) 32 hrs
Answer: (c) 30 hrs

A Hill Station: In the last summer, I enjoyed a tour to a hill station at Shimla. I was accompanied by my five friends and enjoyed the natural beauties of mountains, rivers, streams, forests etc. The beginning of the tour was the most adventurous itself! How amazingly my group win the bet! Actually, the story is that my two friends along with me prefered train to go to Shimla, but other three were forcing for a car or a bus. At last the consensus was reached and we were divided ourselves in two groups of 3 each and started for Shimla at the same time. It was decided that the group who reach the destination first, would be declared as the winner, and runner up the group have to bear the expanses of the tour. I named my group, ‘Group A’ while the second group was named as ‘Group B’. Luckily we reached Shimla 1 hour before the Group-B and enjoyed the trip for absolutely FREE!! How thrilling it was the tour!

Question. An express train takes 1 hour less than a passenger train to travel 132 km between Delhi and Shimla (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, the average speeds of the two trains will be
(a) 33 km/h, 44 km/hr
(b) 40 km/h, 45 km/h
(c) 30 km/h, 38 km/h
(d) 42 km/h, 62 km/h
Answer: (a) 33 km/h, 44 km/hr

Question. An express train makes a run of 240 km at a certain speed. Another train whose speed is 12 km/hr less takes an hour longer to make the same trip. The speed of the express train will be
(a) 60 km/h
(b) 50 km/h
(c) 65 km/h
(d) 48 km/h
Answer: (a) 60 km/h

Question. A journey of 192 km from Delhi to Shimla takes 2 hours less by a super fast train than that by an ordinary passenger train. If the average speed of the slower train is 16 km/hr less than that of the faster train, average speed of super fast train is
(a) 50 km/h
(b) 48 km/h
(c) 55 km/h
(d) 60 km/h
Answer: (b) 48 km/h

Question. A deluxe bus takes 3 hours less than a ordinary bus for a journey of 600 km. If the speed of the ordinary bus is 10 km/hr less than that of the deluxe bus, the speeds of the two buses will be
(a) 35 km/h, 42 km/h
(b) 42 km/h, 52 km/h
(c) 40 km/h, 50 km/h
(d) 30 km/h, 58 km/h
Answer: (c) 40 km/h, 50 km/h

Question. A bus travels a distance of 300 km at a uniform speed. If the speed of the bus is increased by 5 km an hour, the journey would have taken two hours less. The original speed of the bus will be
(a) 20 km/h
(b) 15 km/h
(c) 22 km/h
(d) 25 km/h
Answer: (d) 25 km/h