# ICSE Solutions for Chapter 6 Solving Problems On Quadratic Equations Class 10 Mathematics

Question 1: 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 2 hrs more than the express train, calculate speed of the express train

Solution 1:
(i) Speed of ordinary train = x km/hr
Speed of express train = (x + 25) km/hr
Distance = 300 km
We know:
Time = Distance/Speed
∴ Time taken by ordinary train to cover 300 km = 300/x hrs
Time taken by express train to cover 300 km = (300/x + 25) hrs

(ii) Given that the ordinary train takes 2 hours more than the express train to cover the distance, Therefore,
300/x – 300/(x + 25) = 2
(300x + 7500 - 300x)/x(x + 25) = 2
7500 = 2x2 + 5x
2x2 + 50x - 7500 = 0
x2 + 25x - 3750 = 0
x2 + 75x – 50x - 3750 = 0
x(x + 75) - 50(x + 75) = 0
(x + 75)(x - 50) = 0
x = -75, 50
But, speed cannot be negative. So, x = 50.
Speed of the express train = (x + 25) km/hr = 75 km/hr

Question 2: A trader buys x articles for a total cost of Rs 600.
(i) Write down the cost of one article in terms of x.
If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it for x.

Solution 2: Number of articles = x
Total cost of articles = Rs 600
(i) Cost of one article = Rs 600/x

(ii) From the given information, we have:
600/(x – 4) – 600/x = 5
(600x - 600x + 2400)/{x(x – 4)} = 5
480/ x(x-4) = 1
x2 – 4x - 480 = 0
x2 - 24x – 20x - 480 = 0
x(x - 24) + 20(x - 24) = 0
(x - 24)(x + 20) = 0
x = 24, -20
Since, number of articles cannot be negative. So, x = 24.

Question 3: The product of the digits of a two digit number is 24. If its unit's digit exceeds twice its ten's digit by 2; find the number

Solution 3: Let the ten's and unit's digit of the required number be x and y respectively.
From the given information,
x + y = 24
y = 24/x  ...(1)
Also, y = 2x + 2
24/x  =  2x + 2 [Using (1)]
24 = 2x2 + 2x
2x2 + 2x – 24 = 0
x2 + x – 12 = 0
(x + 4)(x - 3) = 0
x = - 4, 3
The digit of a number cannot be negative, so, x = 3
∴ y = 24/3  = 8
Thus, the required number is 38.

Question 4:  A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

Solution 4: Let x km/h be the original speed of the car.
We know that
Time = Distance/Speed
It is given that the car covers a distance of 400 km with the speed of x km/h.
Thus, the time taken by the car to complete 400 km is
t = 400/x
Now, the speed is increased by 12 km.
∴ Increased speed = (x + 12) km/hr.
Also given that, increasing the speed of the car will decrease the time taken by 1 hour 40 minutes.
Hence,
400/x – 400/(x + 12) = 1 hour 40 minutes
⇒ 400/x – 400/(x + 12) = 1.40/60
⇒ {400(x + 12) – 400x}/{x(x + 12)}  = 1.2/3
⇒ (400x + 4800 - 400x)/{x(x + 12)} = 5/3
⇒  4800/{x(x + 12)} = 5/3
⇒ 3 × 4800 = 5 × x × (x +12)
⇒ 14400 = 5x2 + 60x
⇒ 5x2 + 60x - 14400 = 0
⇒ 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
Since speed cannot be negative, the original
Speed of the car is 48 km/h

Question 5: The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of the rectangle.

Solution 5:

Let the shorter side be x m.
Length of the other side = (x + 30) m
Length of hypotenuse = (x + 60) m
Using Pythagoras theorem,
(x + 60)2 = x2 + (x + 30)2
x2 + 3600 + 120x = x2 + x2 + 900 + 60x
x2 - 60x - 2700 = 0
x2 - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x - 90) (x + 30) = 0
x = 90, -30
But, x cannot be negative. So, x = 90.
Thus, the sides of the rectangle are 90 m and (90 + 30) m = 120 m.

Question 6: The perimeter of a rectangle is 104 m and its area is 640 m2. Find its length and breadth.

Solution 6: Let the length and the breadth of the rectangle be x m and y m.
Perimeter = 2(x + y) m
∴  104 = 2(x + y)
x + y = 52
y = 52 - x
Area = 640 m2
∴  xy = 640
x(52 - x) = 640
x2 - 52x + 640 = 0
x2 - 32x - 20x + 640 = 0
x(x - 32) - 20(x - 32) = 0
(x - 32) (x - 20) = 0
x = 32, 20
When x = 32, y = 52 - 32 = 20
When x = 20, y = 52 - 20 = 32
Thus, the length and breadth of the rectangle are 32 cm and 20 cm.

Question 7: A goods train leaves a station at 6 p.m., followed by an express train which leaved at 8 pm and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the train remain constant between the two stations, calculate their speeds.

Solution 7: Let the speed of goods train be x km/hr.
So, the speed of express train will be (x + 20) km/hr.
Distance = 1040 km
We know:
Time = Distance/Speed
Time taken by goods train to cover a distance of 1040 km  = 1040/x  hrs
Time taken by express train to cover a distance of 1040 km = 1040/(x + 20) hrs
It is given that the express train arrives at a station 36 minutes before the goods train. Also, the express train leaves the station 2 hours after the goods train. This means that the express train arrives at the station
(36/60 + 2)hrs = 13/2 hrs before the goods train
Therefore, we have:
1040/x – 1040/(x + 20) = 13/5
(1040x + 20800 - 1040x)/{x(x + 20)} = 13/5
20800/(x2 + 20x) = 13/5
1600/(x2 + 20x)  = 1/5
x2 + 20x - 8000 = 0
(x - 80)(x + 100) = 0
x = 80, -100
Since the speed cannot be negative. So, x = 80.
Thus, the speed of goods train is 80 km/hr and the speed of express train is 100 km/hr.

Question 8: Mr. Mehra sends his servant to the market to buy oranges worth Rs 15. The servant having eaten three oranges on the way. Mr. Mehra pays Rs 25 paise per orange more than the market price.
Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x Hence, find the value of x.

Solution 8: Number of oranges = y
Cost of one orange  = Rs 15/y
The servant ate 3 oranges, so Mr. Mehra received (y - 3) oranges.
So, x = y – 3 ⇒   y = x + 3 ...(1)
Cost of one orange paid by Mr. Mehra : Rs15/y + 0.25
Rs 15/(x + 3) + ¼ [using (1)]
Now, Mr. Mehra pays a total of Rs 15.
∴  {15/(x + 3) + ¼} × x = 15
(60 + x + 3)/{4(x + 3)} × x = 15
63x + x2 = 60x + 180
x2 + 3x – 180 = 0
(x + 15)(x – 12) = 0
x = -15, 12
But, the number of oranges cannot be negative. So, x = 12

Question 9: Rs 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.

Solution 9: Let the number of children be x.
It is given that Rs 250 is divided amongst x students.
So, money received by each child = Rs 250/x
If there were 25 children more, then
Money received by each child = Rs 250/(x + 25)
From the given information,
250/x -  250/(x + 25) = 50/100
(250x + 6250 – 250x)/{x(x + 25)} = 1/2
6250/(x2 + 25x) = 1/2
x2 + 25x – 12500 = 0
(x + 125)(x - 100) = 0
x = -125, 100
Since the number of students cannot be negative. So, x = 100
Hence, the number of students is 100

Question 10: An employer finds that if he increased the weekly wages of each worker by Rs 5 and employs five workers less, he increases his weekly wage bill from Rs 3,150 to Rs 3,250. Taking the original weekly wage of each worker as Rs x; obtain an equation in x and then solve it to find the weekly wages of each worker

Solution 10: Original weekly wage of each worker = Rs x
Original weekly wage bill of employer = Rs 3150
Number of workers = 3150/x
New weekly wage of each worker = Rs (x + 5)
New weekly wage bill of employer = Rs 3250
Number of workers = 3250/(x + 5)
From the given condition,
(3150/x – 5) = 3250/(x + 5)
(3150 - 5x)/x =  3250/(x + 5)
(3150 - 5x)/x = 3250/(x + 5)
3150 - 5x2 + 15750 - 25x = 3250x
-5x2 + 15750 - 125x = 0
x2 + 25x - 3150 = 0
x2 + 70x - 45x - 3150 = 0
x(x + 70) - 45(x + 70) = 0
(x + 70)(x - 45) = 0
x = -70, 45
Since, wage cannot be negative, x = 45
Thus, the original weekly wage of each worker is Rs 45.