Selina Concise Class 10 Math Chapter 6 Solving (simple) problems (Based on Quadratic Equations) Exercise 6C Solutions
Solving (simple) problems
(Based on Quadratic Equations)
Exercise 6C
(Q1) The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr.
(i) Find the time taken by each train to cover 300 km
(ii) If the ordinary train takes 2hrs more than the express train;
Calculate speed of the express train
Solution:
Given that the speed of an ordinary train is x km per hr and the speed of an express train is (x + 25) km per hr.
(i) The time taken by each train to cover 300 km
Means distance = 300 km
We know that,
Speed = distance/time
or time = distance/speed
∴ Ordinary train – express train
= 2
The time taken by ordinary train to cover 300 km = 300/x hrs
and the time taken by express train to cover 300km = 300/x+25 hrs
If the ordinary train takes 2 hrs more than the express train to cover the distance of 300 kms
∴ ordinary train + express train = 2
300/x – 300/x + 25 = 2
300 (x + 25) – 300x/x (x + 25) = 2
300x + 7500 – 300x/x2 + 25x = 2
7500/x2 + 25x = 2
7500 = 2 (x2 + 25x)
= 2x2 + 50x
7500 = 2x2 + 50x
2x2 + 50x – 7500 = 0
Divide by ‘2’
2x2/2 + 50x/2 – 7500/2 = 0
x2 + 25x – 3750 = 0
x2 + 75x – 50x – 3750 = 0
x (x + 75) – 50 (x + 75) = 0
(x + 75) (x- 50) = 0
x + 75 = 0 or x – 50 = 0
x = -75 or x = 50
∴ We know that the speed cannot be negative.
∴ x = -75 is not solution.
∴ x = 50 is the only solution
∴ The speed of the express train = (x + 25) km/hr
= 50 + 25
= 75 km/hr
∴ The speed of the ordinary train x = 50 km/hr
(Q2) If the speed of a car is increased by 10km/hr, it takes 18 minutes less to cover a distance of 36km, Find the speed of the car.
Solution:
Let us consider the speed of the car is x km/hr.
Given that the distance is 36km
We know that,
Time = distance/speed
∴ The time taken to cover a distance of 36km = 36/x hrs
If the speed of a car is increased by 10 km per hr.
∴ (x + 10) km/hr
∴ The time taken to cover a distance of 36km = 36/(x+10) hrs
From question we have write,
36/x – 36/(x + 10) = 18/60
36 (x + 10) – 36x/x (x + 10) = 18/60
36x + 360 – 36x/x2 + 10x = 18/60
360/x2 + 10x = 18/60
60 (360) = 18 (x2 + 10x)
21, 600 = 18x2 + 180x
Dividing both sides by 18,
21,600/18 = 18x2/18 + 180x/18
1200 = x2 + 10x
x2 + 10x – 1200 = 0
x2 + 40x – 30x – 1200 = 0
x (x + 40) – 30 (x + 40) = 0
(x + 40) (x – 30) = 0
x + 40 = 0 or x – 30 = 0
x = -40 or x = 30
But the speed of the car cannot be negative
∴ x = 30 is the only solution
∴ The speed of the car is 30 km/hr.
(Q3) If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more t cover 1200 km. Find the speed of the aeroplane.
Solution:
Let us assume that the initial speed is x km/hr
∴ If the speed of an aroplane is reduced by 40 km/hr.
∴ (x – 40) km/hr
It takes 20 minutes more to cover 1200 km.
1200/x – 40 – 1200/x = 20/60 (Cover minute in hour)
1200 [1/x – 40 – 1/x] = 1/3
1200 [x – (x – 40)/x (x – 40)] = 1/3
1200 [x – x + 40/x2 – 40x] = 1/3
1200 [40/x2 – 40x] = 1/3
1200×40×3 = x2 – 40x
1200×120 = x2 – 40x
144000 = x2 – 40x
x2 – 40x – 144000 = 0
x2 – 400x + 360x – 14400 = 0 {144000}
x (x – 400) + 360 (x – 400) = 0
(x – 400) (x + 360) = 0
x – 400 = 0 or x + 360 = 0
x = 400 or x = -360
But the speed cannot be negative
∴ x = 400 is the solution
∴ The speed of the aeroplane is 400 km/hr.
(Q4) A car covers a distance of 400 km at a certain speed, had the speed been 12km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Solution:
Let us assume that the original speed of the car is x km/hr
A car covers a distance of 400 km at a certain speed.
(x + 12) km/h
∴ Total distance = 400 km.
∴ The time taken for the journey would have been 1 hour 40 minutes less.
∴ Tim = 1 hour 40 min
= 1 + 40/60 (Convert into hour)
= 1 + 2/3
= 3+2/3
= 5/3 hr.
∴ The original speed of the car is x km/hr.
and the distance = 400 km
∴ 400/x
∴ A car covers a distance of 400 km at certain speed.
400/x + 2
∴ 400/x – 400/x + 12 = 5/3
400 [1/x – 1/x+12] = 5/3
400 [x + 12 – x/x (x + 12)] = 5/3
400 [12/x2 + 12x] = 5/3
400×12×3 = 5 (x2 + 12x)
400×36 = 5x2 + 60x
14,400 = 5x2 + 60x
Divide by 5, on both sides, 2,880 = x2 + 12x
x2 + 12x – 2880 = 0
x2 + 60x – 48x – 2880 = 0
x (x + 60) – 48 (x + 60) = 0
(x + 60) (x – 48) = 0
x + 60 = 0 or x – 48 = 0
x = -60 or x = 48
∴ But the speed is not negative
∴ x = 48 is the only solution.
∴ The original speed of the car is 48 km/hr.
(Q5) A girl goes to her friend’s house, which is at a distance of 12km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find ‘x’
Solution:
Total distance 12km
∴ She covers half of the distance at a speed of x km/hr.
6/x
The remaining distance at a speed of (x + 2) km/hr
6/x + 2
∴ 6/x + 6/x + 2 = 2 ½
= 4+1/2
6 (x + 2) + 6x/x (x+ 2) = 5/2
6x + 12+6x/x2 + 2x = 5/2
12x + 12/x2 + 2x = 5/2
12x + 12/x2 + 2x = 5/2
2 (12x + 12) = 5 (x2 + 2x)
24x + 24 = 5x2 + 10x
5x2 + 10x – 24x – 24 = 0
5x2 – 14x – 24 = 0
5x2 – 20x + 6x – 24 = 0
5x (x – 4) + 6 (x – 4) = 0
(x – 4) (5x + 6) = 0
x – 4 = 0 or 5x + 6 = 0
5x = -6
x = 4
x = -6/5
∴ The speed cannot be negative
∴ x = 4 is the only solution.
Here is your solution of Selina Concise Class 10 Math Chapter 6 Solving (simple) problems (Based on Quadratic Equations)
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