16. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer:
Let the speed of the train be x km/hr.
Time taken to cover 360 km = \(\frac{360}{x}\) hr
According to the given question,
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h.
17. Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer:
Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x − 10) hr
Part of tank filled by smaller pipe in 1 hour = \(\frac{1}{x}\)
Part of tank filled by larger pipe in 1 hour = \(\frac{1}{x-10}\)
It is given that the tank can be filled in 9\(\frac{3}{8}\) = \(\frac{75}{8}\) hours by both the pipes together.
Therefore,
Time taken by the smaller pipe cannot be \(\frac{30}{8}\) = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively.
18. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Answer:
Let the average speed of passenger train be x km/h.
Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.
Speed cannot be negative.
Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.
19. 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.
Answer:
Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y respectively and their areas will be x2 and y2 respectively. It is given that 4x − 4y = 24 x − y = 6 x = y + 6
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m
20. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2x2 −3x + 5 = 0
(ii) 3x2 - 4\(\sqrt{3x}\) + 4 = 0
(iii) 2x2 − 6x + 3 = 0
Answer:
We know that for a quadratic equation ax2 + bx + c = 0, discriminant is b2 − 4ac.
- If b2 − 4ac > 0 → two distinct real roots
- If b2 − 4ac = 0 → two equal real roots
- If b2 − 4ac < 0 → no real roots
(i) 2x2 −3x + 5 = 0
Comparing this equation with ax2 + bx + c = 0,
we obtain a = 2, b = −3, c = 5
Discriminant = b2 − 4ac = (− 3)2 − 4 (2) (5) = 9 – 40 = −31
As b2 − 4ac < 0,
Therefore, no real root is possible for the given equation.
(ii) 3x2 - 4\(\sqrt{3x}\) + 4 = 0
Comparing this equation with ax2 + bx + c = 0,
we obtain a = 3, b = \(-4\sqrt{3}\), c = 4
Discriminant = b2 − 4ac = \((-4\sqrt{3})^2 - 4(3)(4)\) = 48 − 48 = 0
As b2 − 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be \(\frac{-b}{2a}\) and \(\frac{-b}{2a}\) .
Therefore, the roots are \(\frac{2}{\sqrt{3}}\) and \(\frac{2}{\sqrt{3}}\) .
(iii) 2x2 − 6x + 3 = 0
Comparing this equation with ax2 + bx + c = 0,
we obtain a = 2, b = −6, c = 3
Discriminant = b2 − 4ac = (− 6)2 − 4 (2) (3)
= 36 − 24 = 12
As b2 − 4ac > 0,
Therefore, distinct real roots exist for this equation as follows.
Therefore, the roots are \(\frac{3+\sqrt{3}}{2}\) or \(\frac{3+\sqrt{3}}{2}\) .
21. 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
Answer:
We know that if an equation ax2 + bx + c = 0 has two equal roots, its discriminant (b2 − 4ac) will be 0.
(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0,
we obtain a = 2, b = k, c = 3 Discriminant = b2 − 4ac = (k)2− 4(2) (3) = k2 − 24
For equal roots,
Discriminant = 0
k2 − 24 = 0
k2 = 24
\(k = \pm\sqrt{24}=\pm 2\sqrt{6}\)
(ii) kx (x − 2) + 6 = 0
or kx2 − 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0,
we obtain a = k, b = −2k, c = 6
Discriminant = b2 − 4ac = (− 2k)2 − 4 (k) (6) = 4k2 − 24k
For equal roots, b2 − 4ac = 0
4k2 − 24k = 0
4k (k − 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2’ and ‘x’.
Therefore, if this equation has two equal roots, k should be 6 only.
