# ICSE Solutions for Selina Concise Chapter 6 Problems based on Quadratic Equations Class 10 Maths

**Exercise 6(A) **

**1. The product of two consecutive integers is 56. Find the integers.**

**Solution**

Let us consider the two consecutive integers to be x and x + 1.

So from the question,

x(x + 1) = 56

x^{2} + x – 56 = 0

(x + 8) (x – 7) = 0

x = -8 or 7

∴ the required integers are (- 8, -7) or (7, 8).

**2. The sum of the squares of two consecutive natural numbers is 41. Find the numbers.**

**Solution**

Let us take the two consecutive natural numbers as x and x + 1.

So from the question,

x^{2} + (x + 1)^{2} = 41

2x^{2} + 2x + 1 – 41 = 0

x^{2} + x – 20 = 0

(x + 5) (x – 4) = 0

x = -5, 4

As -5 is not a natural number.

x = 4 is the only solution.

∴ the two consecutive natural numbers are 4 and 5.

**3. Find the two natural numbers which differ by 5 and the sum of whose squares is 97.**

**Solution**

Let’s assume the two natural numbers to be x and x + 5. (As given they differ by 5)

So from the question,

x^{2} + (x + 5)^{2} = 97

2x^{2} + 10x + 25 – 97 = 0

2x^{2} + 10x – 72 = 0

x^{2} + 5x – 36 = 0

(x + 9) (x – 4) = 0

x = -9 or 4

As -9 is not a natural number. x = 4 is the only valid solution

∴ the two natural numbers are 4 and 9.

**4. The sum of a number and its reciprocal is 4.25. Find the number.**

**Solution**

Let the number be x. So, its reciprocal is 1/x

Then according to the question,

⇒ 4x^{2} – 17x + 4 = 0

⇒ 4x^{2} – 16x – x + 4 = 0

⇒ 4x(x – 4) – 1(x – 4) = 0

⇒ (4x – 1) (x – 1) = 0

So, 4x -1 = 0 or x – 1 = 0

Thus, x = ¼ or x = 1

∴ the numbers are 4 and ¼.

**5. Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10.**

**Solution**

Let’s consider the two natural numbers to be x and x + 3. (As they differ by 3)

Then, from the question we have

⇒ 20x + 30 = 7x^{2} + 21x

⇒ 7x^{2} + x – 30 = 0

⇒ 7x^{2} – 14x + 15x – 30 = 0

⇒ 7x(x – 2) + 15(x – 2) = 0

⇒ (7x + 15) (x – 2) = 0

So, 7x + 15 = 0 or x – 2 = 0

x = -15/7 or x = 2

As, x is a natural number. Only x = 2 is a valid solution.

∴ the two natural numbers are 2 and 5.

**6. Divide 15 into two parts such that the sum of their reciprocals is 3/10**

**Solution**

Let’s assume the two parts to be x and 15 – x.

So, according to the question

⇒ 150 = 45x – 3x^{2}

⇒ 3x^{2} – 45x + 150 = 0

Dividing by 3, we get

x^{2} – 15x + 50 = 0

⇒ x^{2} – 10x – 5x + 50 = 0

⇒ x(x – 10) – 5(x – 10) = 0

⇒ (x – 5)(x – 10) = 0

So, x – 5 = 0 or x – 10 = 0

x = 5 or 10

Thus, if one part is 5 the other part is 10 and vice versa..

**7. The sum of the squares of two positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.**

**Solution**

Let’s assume the two numbers to be x and y, y being the larger of the two numbers.

Then, from the question

x^{2 }+ y^{2 }= 208 **... (i)** and

y^{2} = 18x **... (ii)**

From (i), we get y^{2 }= 208 – x^{2}.

Now, putting this in (ii), we have

208 – x^{2 }= 18x

⇒ x^{2} + 18x – 208 = 0

⇒ x^{2} + 26x – 8x – 208 = 0

⇒ x(x + 26) – 8(x + 26) = 0

⇒ (x – 8)(x + 26) = 0

As x can’t be a negative integer, so x = 8 is valid solution.

Using x = 8 in (ii), we get y^{2} = 18 x 8 = 144

Thus, y = 12 only as y is also a positive integer

∴ the two numbers are 8 and 12.

**8. The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.**

**Solution**

Let the two consecutive positive even numbers be taken as x and x + 2.

From the question, we have

x^{2} + (x + 2)^{2} = 52

⇒ 2x^{2} + 4x + 4 = 52

⇒ 2x^{2} + 4x – 48 = 0

⇒ x^{2} + 2x – 24 = 0

⇒ (x + 6) (x – 4) = 0

⇒ x = -6, 4

As, the numbers are positive only x = 4 is a valid solution.

∴ the numbers are 4 and 6.

**Exercise 6(B)**

**1. The sides of a right-angled triangle containing the right angle are 4x cm and (2x – 1) cm. If the area of the triangle is 30 cm ^{2}; calculate the lengths of its sides.**

**Solution**

Given, the area of triangle = 30 cm^{2}

As, x cannot be negative, only x = 3 is valid.

Hence, we have

AB = 4 × 3 cm = 12 cm

BC = (2 × 3 – 1) cm = 5 cm

CA = √(12^{2} + 5^{2}) = √169 = 13cm **(Using Pythagoras theorem)**

**2. The hypotenuse of a right-angled triangle is 26 cm and the sum of other two sides is 34 cm. Find the lengths of its sides.**

**Solution**

Given, a right triangle

Hypotenuse = 26 cm and the sum of other two sides is 34 cm.

Now, let consider the other two sides to be x cm and (34 – x) cm.

By using Pythagoras theorem,

(26)^{2} = x^{2} + (34 – x)^{2}

⇒ 676 = x^{2} + x^{2} + 1156 – 68x

⇒ 2x^{2} – 68x + 480 = 0

⇒ x^{2} – 34x + 240 = 0

⇒ x^{2} – 10x – 24x + 240 = 0

⇒ x(x – 10) – 24(x – 10) = 0

⇒ (x – 10) (x – 24) = 0

So, x = 10, 24

If x = 10; (34 – x) = 24

Or if x = 24; (34 – x) = 10

∴ the lengths of the three sides of the right-angled triangle are 10 cm, 24 cm and 26 cm.

**3. The sides of a right-angled triangle are (x – 1) cm, 3x cm and (3x + 1) cm. Find:**

**(i) the value of x,**

**(ii) the lengths of its sides,**

**(iii) its area.**

**Solution**

Given,

The longer side = Hypotenuse = (3x + 1) cm

And the lengths of other two sides are (x – 1) cm and 3x cm.

By using Pythagoras theorem, we have

(3x + 1)^{2} = (x – 1)^{2} + (3x)^{2}

⇒ 9x^{2} + 1 + 6x = x^{2} + 1 – 2x + 9x^{2}

⇒ x^{2} – 8x = 0

⇒ x(x – 8) = 0

⇒ x = 0, 8

Now, if x = 0, then one side = 3x = 0, which is not possible.

Hence, we take x = 8

∴ the lengths of sides of the triangle are (x – 1) cm = 7 cm, 3x cm = 24 cm and (3x + 1) cm = 25 cm.

And, Area of the triangle = ½ x 7 x 24 = 84 cm^{2}

^{}

**4. The hypotenuse of a right-angled triangle exceeds one side by 1 cm and the other side by 18 cm; find the lengths of the sides of the triangle.**

**Solution**

Let the hypotenuse of the right triangle be x cm.

From the question, we have

Length of one side = (x – 1) cm

Length of other side = (x – 18) cm

By using Pythagoras theorem,

x^{2} = (x – 1)^{2} + (x – 18)^{2}

⇒ x^{2} = x^{2} + 1 – 2x + x^{2} + 324 – 36x

⇒ x^{2} – 38x + 325 = 0

⇒ x^{2} – 13x – 25x + 325 = 0

⇒ x(x – 13) – 25(x – 13) = 0

⇒ (x – 13) (x – 25) = 0

⇒ x = 13, 25

But when x = 13, x – 18 = 13 – 18 = -5, which is negative and is not possible.

Hence, we take x = 25

∴ the lengths of the sides of the triangle are x = 25 cm, (x – 1) = 24 cm and (x – 18) = 7 cm.

**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**

Let’s consider the shorter side of the rectangle to be x m.

Then, the length of the other side = (x + 30) m

Length of the diagonal = (x + 60) m

By using Pythagoras theorem,

(x + 60)^{2} = x^{2} + (x + 30)^{2}

⇒ x^{2} + 3600 + 120x = x^{2} + x^{2} + 900 + 60x

⇒ x^{2} – 60x – 2700 = 0

⇒ x^{2} – 90x + 30x – 2700 = 0

⇒ x(x – 90) + 30(x – 90) = 0

⇒ (x – 90) (x + 30) = 0

⇒ x = 90, -30

As, x cannot be negative. Hence, x = 90 is only valid.

∴ the sides of the rectangle are 90 m and (90 + 30) m = 120 m.

**6. The perimeter of a rectangle is 104 m and its area is 640 m ^{2}. Find its length and breadth.**

**Solution**

Let’s take the length and the breadth of the rectangle be x m and y m.

So, the perimeter = 2(x + y) m

104 = 2(x + y)

⇒ x + y = 52

⇒ y = 52 – x

And, given area = 640 m^{2}

So, xy = 640

x(52 – x) = 640

⇒ x^{2} – 52x + 640 = 0

⇒ x^{2} – 32x – 20x + 640 = 0

⇒ x(x – 32) – 20 (x – 32) = 0

⇒ (x – 32) (x – 20) = 0

⇒ x = 32, 20

If x = 32 then, y = 52 – 32 = 20

Or if x = 20, y = 52 – 20 = 32

∴ the length and breadth of the rectangle are 32 m and 20 m.

**Exercise 6(C)**

**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**

(i) Given,

Speed of the ordinary train = x km/hr

Speed of the express train = (x + 25) km/hr

Distance = 300 km

We know that,

Time = Distance/ Speed

So, the time taken by the ordinary train to cover 300 km = 300/x hrs

And the time taken by the express train to cover 300 km = 300/ (x + 25) hrs

(ii) From the question, it’s given that the ordinary train takes 2 hours more than the express train to cover the distance of 300kms.

Hence, we can write

⇒ 7500 = 2x^{2} + 50x

⇒ 2x^{2} + 50x – 7500 = 0

⇒ x^{2} + 25x – 3750 = 0

⇒ x^{2} + 75x – 50x – 3750 = 0

⇒ x(x + 75) – 50(x + 75) = 0

⇒ (x – 50) (x + 75) = 0

Thus, x = 50 or -75

As speed cannot be negative we shall ignore x = -75

∴ The speed of the express train = (x + 25) km/hr = 75 km/hr

**2. If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.**

**Solution**

Let’s assume the speed of the car to be x km/hr.

Given, distance = 36 km

So, the time taken to cover a distance of 36 km = 36/x hrs [Since, Time = Distance/ Speed]

And, the new speed of the car = (x + 10) km/hr

So, the new time taken by the car to cover a distance of 36 km = 36/(x + 10) hrs

Then according to the question, we can write

⇒ x^{2} + 10x – 1200 = 0

⇒ x^{2} + 40x – 30x – 1200 = 0

⇒ x(x + 40) – 30(x + 40) = 0

⇒ (x + 40) (x – 30) = 0

Thus, x = -40 or 30

But, as speed cannot be negative. x = 30 is only considered.

∴ the original speed of the car is 30 km/hr.

**3. If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.**

**Solution**

Let’s consider the original speed of the aeroplane to be x km/hr.

Now, the time taken to cover a distance of 1200 km = 1200/x hrs [Since, Time = Distance/ Speed]

Let the new speed of the aeroplane be (x – 40) km/hr.

So, the new time taken to cover a distance of 1200 km = 1200/ (x – 40) hrs

According to the question, we have

⇒ x(x – 40) = 48000 x 3

⇒ x^{2} – 40x – 144000 = 0

⇒ x^{2} – 400x + 360x – 144000 = 0

⇒ x(x – 400) + 360(x – 400) = 0

⇒ (x – 400) (x + 360) = 0

As, speed cannot be negative. So we only take, x = 400.

∴ the original speed of the aeroplane is 400 km/hr.

**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**

Let’s assume x km/h to be the original speed of the car.

We know that,

Time = Distance/ Speed

From the question,

The time taken by the car to complete 400 km = 400/x hrs

Now, when the speed is increased by 12 km.

Increased speed = (x + 12) km/h

And, the new time taken by the car to complete 400 km = 400/(x + 12) hrs

Thus, according to the question we can write

⇒ 4800 x 3 = 5x(x + 12)

⇒ 5x^{2} + 60x – 14400 = 0

Dividing by 5 we get,

x^{2} + 12x – 2880 = 0

⇒ x^{2} + 60x – 48x – 2880 = 0

⇒ x(x + 60) – 48(x + 60) = 0

⇒ (x + 60) (x – 48) = 0

So, x + 60 or x – 48

⇒ x = -60 or 48

As, speed cannot be negative.

x = 48 is only valid

∴ the speed of the car is 48 km/h.

**5. A girl goes to her friend’s house, which is at a distance of 12 km. 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**

Given,

The girl covers a distance of 6 km at a speed x km/ hr.

So, the time taken to cover first 6 km = 6/x hr [Since, Time = Distance/ Speed]

Also given, the girl covers the remaining 6 km distance at a speed (x + 2) km/ hr.

So, the time taken to cover next 6 km = 6/ (x + 2)

And, the total time taken to cover the whole distance = 2 hrs 30 mins = (120 + 30)/60 = 5/2 hrs

Then the below equation can be formed,

⇒ 24 + 24x = 5x^{2} + 10x

⇒ 5x^{2} – 14x – 24 = 0

⇒ 5x^{2} – 20x + 6x – 24 = 0

⇒ 5x(x – 4) + 6(x – 4) = 0

⇒ (5x + 6) (x – 4) = 0

So, x = -6/5 or 4

As speed cannot be negative. x = 4 is only valid

∴ the value of x is 4.

**Exercise 6(D) **

**1. The sum S of n successive odd numbers starting from 3 is given by the relation: S = n(n + 2). Determine n, if the sum is 168.**

**Solution**

From the question, we have

n(n + 2) = 168

⇒ n^{2} + 2n – 168 = 0

⇒ n^{2} + 14n – 12n – 168 = 0

⇒ n(n + 14) – 12(n + 14) = 0

⇒ (n + 14) (n – 12) = 0

⇒ n = -14, 12

Since, n cannot be negative.

Thus, n = 12

**2. A stone is thrown vertically downwards and the formula d = 16t ^{2} + 4t gives the distance, d metres, that it falls in t seconds. How long does it take to fall 420 metres?**

**Solution**

According to the question,

16t^{2} + 4t = 420

⇒ 4t^{2} + t – 105 = 0

⇒ 4t^{2} – 20t + 21t – 105 = 0

⇒ 4t(t – 5) + 21(t – 5) = 0

⇒ (4t + 21)(t – 5) = 0

⇒ t = -21/4 or 5

As, time cannot be negative.

∴ the required time taken by the stone to fall 420 meter is 5 seconds.

**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**

Let’s assume the ten’s and unit’s digit of the required number to be x and y respectively.

Then, from the question we have

x × y = 24

y = 24/x **…. (1)**

Also,

y = 2x + 2

Using (1) in the above equation,

24/x = 2x + 2

⇒ 24 = 2x^{2} + 2x

⇒ 2x^{2} + 2x – 24 = 0

⇒ x^{2} + x – 12 = 0** [Dividing by 2]**

⇒ (x + 4) (x – 3) = 0

⇒ x = -4, 3

As the digit of a number cannot be negative, x = -4 is neglected

Thus, x = 3

So, y = 24/ 3 = 8

∴ the required number is 38.

**4. The ages of two sisters are 11 years and 14 years. In how many years’ time will the product of their ages be 304?**

**Solution**

Given, the ages of two sisters are 11 years and 14 years.

Let x be the number of years later when their product of their ages become 304.

So, (11 + x) (14 + x) = 304

⇒ 154 + 11x + 14x + x^{2} = 304

⇒ x^{2} + 25x – 150 = 0

⇒ (x + 30) (x – 5) = 0

⇒ x = -30, 5

As, the number of years cannot be negative. We only consider, x = 5.

∴ the required number of years is 5 years.

**5. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.**

**Solution**

Let’s consider the present age of the son to be x years.

So, the present age of the man = x^{2} years

One year ago,

Son’s age = (x – 1) years

Man’s age = (x^{2} – 1) years

From the question, it’s given that one year ago; the man was 8 times as old as his son.

(x^{2} – 1) = 8(x – 1)

⇒ x^{2} – 8x – 1 + 8 = 0

⇒ x^{2} – 8x + 7 = 0

⇒ (x – 7) (x – 1) = 0

⇒ x = 7, 1

When x = 1, then x^{2} = 1, which is not possible as father’s age cannot be equal to son’s age.

Hence, x = 7 is taken

∴ The present age of son = x years = 7 years

The present age of man = x^{2} years = 49 years

**6. The age of the father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.**

**Solution**

Let’s assume the present age of the son to be x years.

So, the present age of the father = 2x^{2} years

Eight years hence,

Son’s age = (x + 8) years

Father’s age = (2x^{2} + 8) years

From the question, it’s given that eight years hence, the age of the father will be 4 years more than three times the age of the son.

2x^{2} + 8 = 3(x + 8) +4

⇒ 2x^{2} + 8 = 3x + 24 +4

⇒ 2x^{2} – 3x – 20 = 0

⇒ 2x^{2} – 8x + 5x – 20 = 0

⇒ 2x(x – 4) + 5(x – 4) = 0

⇒ (x – 4) (2x + 5) = 0

⇒ x = 4, -5/2

As, age cannot be negative. x = 4 is considered.

∴ The present age of the son = 4 years

The present age of the father = 2(4)^{2} years = 32 years

**Exercise 6(E) **

**1. The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate:**

**(i) the time taken by the car to reach town B from A, in terms of x;**

**(ii) the time taken by the train to reach town B from A, in terms of x.**

**(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.**

**(iv) Hence, find the speed of the train.**

**Solution**

Given,

Speed of car = x km/hr

Speed of train = (x + 16) km/hr

And, we know that

Time = Distance/ Speed

(i) Time taken by the car to reach town B from town A = 216/x hrs

(ii) Time taken by the train to reach town B from A = 208/(x + 16) hrs

(iii) According to the question, we have

⇒ 4x + 1728 = x^{2} + 16x

⇒ x^{2} + 12x – 1728 = 0

⇒ x^{2} + 48x – 36x – 1728 = 0

⇒ x(x + 48) – 36(x + 48) = 0

⇒ (x + 48) (x – 36) = 0

⇒ x = -48, 36

As speed cannot be negative,

x = 36

(iv) ∴ the speed of the train is (x + 16) = (36 + 16)km/hr = 52 km/h

**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**

We have,

Number of articles = x

And, the total cost of articles = Rs 600

Then,

(i) Cost of one article = Rs 600/x

(ii) From the question we have,

⇒ x^{2} – 4x – 480 = 0

⇒ x^{2} – 24x – 20x – 480 = 0

⇒ x(x – 24) + 20(x – 24) = 0

⇒ (x – 24) (x + 20) = 0

⇒ x = 24 or -20

As the number of articles cannot be negative, x = 24.

**3. A hotel bill for a number of people for overnight stay is Rs 4800. If there were 4 people more, the bill each person had to pay, would have reduced by Rs 200. Find the number of people staying overnight.**

**Solution**

Let’s assume the number of people staying overnight as x.

Given, total hotel bill = Rs 4800

So, hotel bill for each person = Rs 4800/x

Then, according to the question

⇒ x^{2} + 4x – 96 = 0

⇒ x^{2} + 12x – 8x – 96 = 0

⇒ x(x +12) – 8(x + 12) = 0

⇒ (x – 8) (x + 12) = 0

So, x = 8 or -12

As, the number of people cannot be negative. We take x = 8.

∴ the number of people staying overnight is 8.

**4. An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for:**

**(i) the onward journey;**

**(ii) the return journey.**

**If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.**

**Solution**

Given,

Distance = 400 km

Average speed of the aeroplane = x km/hr

And, speed while returning = (x + 40) km/hr

We know that,

Time = Distance/ Speed

(i) Time taken for onward journey = 400/x hrs

(ii) Time take for return journey = 400/(x + 40) hrs

Then according to the question,

⇒ x^{2} + 40x – 32000 = 0

⇒ x^{2} + 200x – 160x – 32000 = 0

⇒ x(x + 200) – 160(x + 200) = 0

⇒ (x + 200) (x – 160) = 0

So, x = -200 or 160

As the speed cannot be negative, x = 160 is only valid.

**5. Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.**

**Solution**

Let’s take the original number of persons to be x.

Total money which was divided = Rs 6500

Each person’s share = Rs 6500/x

Then, according to the question

⇒ x^{2} + 15x – 3250 = 0

⇒ x^{2} + 65x – 50x – 3250 = 0

⇒ x(x + 65) – 50(x + 65) = 0

⇒ (x + 65) (x – 50) = 0

So, x = -65 or 50

As, the number of persons cannot be negative. x = 50

∴ the original number of persons are 50.

**6. A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.**

**Solution**

Let’s consider the usual speed of the plane to be x km/hr

The distance to travel = 1500km

We know that,

Time = Distance/ Speed

Then according to the question, we have

⇒ x^{2} + 250x – 750000 = 0

⇒ x^{2} + 1000x – 750x – 750000 = 0

⇒ x(x + 1000) – 750(x + 1000) = 0

⇒ (x + 1000) (x – 750) = 0

So, x = -1000 or 750

As, speed cannot be negative. We take x = 750 as the solution.

∴ the usual speed of the plan is 750km/hr.

**7. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.**

**Solution**

Let take the speed of the second train to be x km/hr.

Then, the speed of the first train is (x + 5) km/hr

Let O be the position of the railway station from which the two trains leave.

Distance travelled by the first train in 2 hours = OA = speed x time = 2(x + 5) km

Distance travelled by the second train in 2 hours = OB = speed x time = 2x km

(50)^{2} = [2(x + 5)]^{2} + (2x)^{2}

⇒ 2500 = 4(x^{2} + 10x + 25) + 4x^{2}

⇒ 2500 = 8x^{2} + 40x + 100

⇒ x^{2} + 5x – 300 = 0

⇒ x^{2} + 20x – 15x – 300 = 0

⇒ (x + 20) (x – 15) = 0

So, x = -20 or x = 15

As x cannot be negative, we have x = 15

Thus, the speed of the second train is 15 km/hr and the speed of the first train is 20 km/hr.

**8. The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.**

**Solution**

Given relation, S = n(n + 1)

And, S = 420

So, n(n + 1) = 420

⇒ n^{2} + n – 420 = 0

⇒ n^{2} + 21n – 20n – 420 = 0

⇒ n(n + 21) – 20(n + 21) = 0

⇒ (n + 21) (n – 20) = 0

⇒ n = -21, 20

As, n cannot be negative.

∴ n = 20.

**9. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.**

**Solution**

Let’s assume the present ages of father and his son to be x years and (45 – x) years respectively.

So five years ago,

Father’s age = (x – 5) years

Son’s age = (45 – x – 5) years = (40 – x) years

From the question, the below equation can be formed

(x – 5) (40 – x) = 124

⇒ 40x – x^{2} – 200 + 5x = 124

⇒ x^{2} – 45x +324 = 0

⇒ x^{2} – 36x – 9x +324 = 0

⇒ x(x – 36) – 9(x – 36) = 0

⇒ (x – 36) (x – 9) = 0

⇒ x = 36, 9

So, if x = 9,

The father’s age = 9 years and the son’s age = (45 – x) = 36 years

This is not possible.

Hence, x = 36

∴ The father’s age = 36 years

The son’s age = (45 – 36) years = 9 years