## Selina ICSE Solutions for Class 10 Maths – Solving Problems (Based on Quadratic Equations)

**Selina ICSE Solutions for Class 10 Maths Chapter 8 Solving Problems (Based on Quadratic Equations)**

**Exercise 8(A)**

**Solution 1:**

Let the two consecutive integers be x and x + 1.

From the given information,

x(x + 1) = 56

x^{2} + x – 56 = 0

(x + 8) (x – 7) = 0

x = -8 or 7

Thus, the required integers are – 8 and -7; 7 and 8.

**Solution 2:**

Let the numbers be x and x + 1.

From the given information,

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

But, -5 is not a natural number. So, x = 4.

Thus, the numbers are 4 and 5.

**Solution 3:**

Let the two numbers be x and x + 5.

From the given information,

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

Since, -9 is not a natural number. So, x = 4.

Thus, the numbers are 4 and 9.

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

Let the two numbers be x and y, y being the bigger number. From the given information,

x^{2 }+ y^{2 }= 208 ….. (i)

y^{2} = 18x ….. (ii)

From (i), we get y^{2}=208 – x^{2}. Putting this in (ii), we get,

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

⇒ x can’t be a negative number , hence x = 8

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

⇒ y = 12, since y is a positive integer

Hence, the two numbers are 8 and 12.

**Solution 8:**

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

From the given information,

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

Since, the numbers are positive, so x = 4.

Thus, the numbers are 4 and 6.

**Solution 9:**

Let the consecutive positive odd numbers be x and x + 2.

From the given information,

x^{2} + (x + 2)^{2} = 74

2x^{2} + 4x + 4 = 74

2x^{2} + 4x – 70 = 0

x^{2} + 2x – 35 = 0

(x + 7)(x – 5) = 0

x = -7, 5

Since, the numbers are positive, so, x = 5.

Thus, the numbers are 5 and 7.

**Solution 10:**

**Solution 11:**

Given, three positive numbers are in the ratio 1/2 : 1/3 : 1/4 = 6 : 4 : 3

Let the numbers be 6x, 4x and 3x.

From the given information,

(6x)^{2} + (4x)^{2} + (3x)^{2} = 244

36x^{2} + 16x^{2} + 9x^{2} = 244

61x^{2} = 244

x^{2} = 4

x = ± 2

Since, the numbers are positive, so x = 2.

Thus, the numbers are 12, 8 and 6.

**Solution 12:**

Let the two parts be x and y.

From the given information,

x + y = 20 ⇒ y = 20 – x

3x^{2} = (20 – x) + 10

3x^{2} = 30 – x

3x^{2} + x – 30 = 0

3x^{2} – 9x + 10x – 30 = 0

3x(x – 3) + 10(x – 3) = 0

(x – 3) (3x + 10) = 0

x = 3, -10/3

Since, x cannot be equal to -10/3, so, x = 3.

Thus, one part is 3 and other part is 20 – 3 = 17.

**Solution 13:**

Let the numbers be x – 1, x and x + 1.

From the given information,

x^{2} = (x + 1)^{2} – (x – 1)^{2} + 60

x^{2} = x^{2} + 1 + 2x – x^{2} – 1 + 2x + 60

x^{2} = 4x + 60

x^{2} – 4x – 60 = 0

(x – 10) (x + 6) = 0

x = 10, -6

Since, x is a natural number, so x = 10.

Thus, the three numbers are 9, 10 and 11.

**Solution 14:**

Let the numbers be p – 1, p and p + 1.

From the given information,

3(p + 1)^{2} = (p – 1)^{2} + p^{2} + 67

3p^{2} + 6p + 3 = p^{2} + 1 – 2p + p^{2} + 67

p^{2} + 8p – 65 = 0

(p + 13)(p – 5) = 0

p = -13, 5

Since, the numbers are positive so p cannot be equal to -13.

Thus, p = 5.

**Solution 15:**

**Solution 16:**

Let one pipe fill the cistern in x hours and the other fills it in (x – 3) hours.

Given that the two pipes together can fill the cistern in 6 hours 40 minutes, i.e.,

So, x = 15.

Thus, one pipe fill the cistern in 15 hours and the other fills in (x – 3) = 15 – 3 = 12 hours.

**Solution 17:**

**Exercise 8(B)**

**Solution 1:**

**Solution 2:**

Hypotenuse = 26 cm

The sum of other two sides is 34 cm.

So, let the other two sides be x cm and (34 – x) cm.

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

x = 10, 24

When x = 10, (34 – x) = 24

When x = 24, (34 – x) = 10

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

**Solution 3:**

Longer side = Hypotenuse = (3x + 1) cm

Lengths of other two sides are (x – 1) cm and 3x cm.

Using Pythagoras theorem,

(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

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

So, x = 8

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

Area of the triangle = ½ × 7 cm × 24 cm = 84 cm²

**Solution 4:**

Let one hypotenuse of the triangle be x cm.

From the given information,

Length of one side = (x – 1) cm

Length of other side = (x – 18) cm

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

When x = 13, x – 18 = 13 – 18 = -5, which being negative, is not possible.

So, x = 25

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

**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} = 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

But, x cannot be negative. So, x = 90.

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

**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 m^{2}

∴ 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

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.

**Solution 7:**

Let w be the width of the footpath.

Area of the path = Area of outer rectangle – Area of inner rectangle

∴ 208 = (32)(24) – (32 – 2w)(24 – 2w)

208 = 768 – 768 + 64w + 48w – 4w^{2}

4w^{2} – 112w + 208 = 0

w^{2} – 28w + 52 = 0

w^{2} – 26w – 2w + 52 = 0

w(w – 26) – 2(w – 26) = 0

(w – 26) (w – 2) = 0

w = 26, 2

If w = 26, then breadth of inner rectangle = (24 – 52) m = -28 m, which is not possible.

Hence, the width of the footpath is 2 m.

**Solution 8:**

Given that, two squares have sides x cm and (x + 4) cm.

Sum of their area = 656 cm^{2}

∴ x^{2} + (x + 4)^{2} = 656

x^{2} + x^{2} + 16 + 8x = 656

2x^{2} + 8x – 640 = 0

x^{2} + 4x – 320 = 0

x^{2} + 20x – 16x – 320 = 0

x(x + 20) – 16(x + 20) = 0

(x + 20) (x – 16) = 0

x = -20, 16

But, x being side, cannot be negative.

So, x = 16

Thus, the sides of the two squares are 16 cm and 20 cm.

**Solution 9:**

Let the width of the gravel path be w m.

Length of the rectangular field = 50 m

Breadth of the rectangular field = 40 m

Let the length and breadth of the flower bed be x m and y m respectively.

Therefore, we have:

x + 2w = 50 … (1)

y + 2w = 40 … (2)

Also, area of rectangular field = 50 m 40 m = 2000 m^{2}

Area of the flower bed = xy m^{2}

Area of gravel path = Area of rectangular field – Area of flower bed = (2000 – xy) m^{2}

Cost of laying flower bed + Gravel path = Area x cost of laying per sq. m

52000 = 30 xy + 20 (2000 – xy)

52000 = 10xy + 40000

xy = 1200

Using (1) and (2), we have:

(50 – 2w) (40 – 2w) = 1200

2000 – 180w + 4w^{2} = 1200

4w^{2} – 180w + 800 = 0

w^{2} – 45w + 200 = 0

w^{2} – 5w – 40w + 200 = 0

w(w – 5) – 40(w – 5) = 0

(w – 5) (w – 40) = 0

w = 5, 40

If w = 40, then x = 50 – 2w = -30, which is not possible.

Thus, the width of the gravel path is 5 m.

**Solution 10:**

Let the size of the larger tiles be x cm.

Area of larger tiles = x^{2} cm^{2}

Number of larger tiles required to pave an area is 128.

So, the area needed to be paved = 128 x^{2} cm^{2} …. (1)

Size of smaller tiles = (x – 2)cm

Area of smaller tiles = (x – 2)^{2} cm^{2}

Number of larger tiles required to pave an area is 200.

So, the area needed to be paved = 200 (x – 2)^{2} cm^{2} …. (2)

Therefore, from (1) and (2), we have:

128 x^{2} = 200 (x – 2)^{2}

128 x^{2} = 200x^{2} + 800 – 800x

72x^{2} – 800x + 800 = 0

9x^{2} – 100x + 100 = 0

9x^{2} – 90x – 10x + 100 = 0

9x(x – 10) – 10(x – 10) = 0

(x – 10)(9x – 10) = 0

Hence, the size of the larger tiles is 10 cm.

**Solution 11:**

Let the length and breadth of the rectangular sheep pen be x and y respectively.

From the given information,

x + y + x = 70

2x + y = 70 … (1)

Also, area = xy = 600

Using (1), we have:

x (70 – 2x) = 600

70x – 2x^{2} = 600

2x^{2} – 70x + 600 = 0

x^{2} – 35x + 300 = 0

x^{2} – 15x – 20x + 300 = 0

x(x – 15) – 20(x – 15) = 0

(x – 15)(x – 20) = 0

x = 15, 20

If x = 15, then y = 70 – 2x = 70 – 30 = 40

If x = 20, then y = 70 – 2x = 70 – 40 = 30

Thus, the length of the shorter side is 15 m when the longer side is 40 m. The length of the shorter side is 20 m when the longer side is 30 m.

**Solution 12:**

Let the side of the square lawn be x m.

Area of the square lawn = x^{2} m^{2}

The square lawn is bounded on three sides by a path which is 4 m wide.

Area of outer rectangle = (x + 4) (x + 8) = x^{2} + 12x + 32

Area of path = x^{2} + 12x + 32 – x^{2} = 12x + 32

From the given information, we have:

Since, x cannot be negative. So, x = 16 m.

Thus, each side of the square lawn is 16 m.

**Solution 13:**

Let the original length and breadth of the rectangular room be x m and y m respectively.

Area of the rectangular room = xy = 300

⇒ y = …..(1)

New length = (x – 5) m

New breadth = (y + 5) m

New area = (x – 5) (y + 5) = 300 (given)

Using (1), we have:

But, x cannot be negative. So, x = 20.

Thus, the length of the room is 20 m.

**Exercise 8(C)**

**Solution 1:**

(i) Speed of ordinary train = x km/hr

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

Distance = 300 km

(ii) Given that the ordinary train takes 2 hours more than the express train to cover the distance.

Therefore,

But, speed cannot be negative. So, x = 50.

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

**Solution 2:**

Let the speed of the car be x km/hr.

Distance = 36 km

From the given information, we have:

But, speed cannot be negative. So, x = 30.

Hence, the original speed of the car is 30 km/hr.

**Solution 3:**

Let the original speed of the aeroplane be x km/hr.

But, speed cannot be negative. So, x = 400.

Thus, the original speed of the aeroplane is 400 km/hr.

**Solution 4:**

Let x km/h be the original speed of the car.

We know that,

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 =

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.

**Solution 5:**

**Solution 6:**

**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

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 before the goods train.

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.

**Solution 8:**

C.P. of the article = Rs x

S.P. of the article = Rs 16

Loss = Rs (x – 16)

We know:

Thus, the cost price of the article is Rs 20 or Rs 80.

**Solution 9:**

C.P. of the article = Rs x

S.P. of the article = Rs 52

Profit = Rs (52 – x)

We know:

Since, C.P. cannot be negative. So, x = 40.

Thus, the cost price of the article is Rs 40.

**Solution 10:**

Let the C.P. of the chair be Rs x

S.P. of chair = Rs 75

Profit = Rs (75 – x)

We know:

But, C.P. cannot be negative. So, x = 50.

Hence, the cost of the chair is Rs 50.

**Exercise 8(D)**

**Solution 1:**

From the given information, we have:

n(n + 2) = 168

n² + 2n – 168 = 0

n² + 14n – 12n – 168 = 0

n(n + 14) – 12(n + 14) = 0

(n + 14) (n – 12) = 0

n = -14, 12

But, n cannot be negative.

Therefore, n = 12.

**Solution 2:**

From the given information,

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, 5

But, time cannot be negative.

Thus, the required time taken is 5 seconds.

**Solution 3:**

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

From the given information,

**Solution 4:**

The ages of two sisters are 11 years and 14 years.

Let in x number of years the product of their ages be 304.

But, the number of years cannot be negative. So, x = 5.

Hence, the required number of years is 5 years.

**Solution 5:**

Let the present age of the son be x years.

∴ Present age of man = x^{2} years

One year ago,

Son’s age = (x – 1) years

Man’s age = (x^{2} – 1) years

It is given that one year ago; a 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

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

So, x = 7.

Present age of son = x years = 7 years

Present age of man = x^{2} years = 49 years

**Solution 6:**

Let the present age of the son be x years.

Present age of father = 2x^{2} years

Eight years hence,

Son’s age = (x + 8) years

Father’s age = (2x^{2} + 8) years

It is 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

But, the age cannot be negative, so, x = 4.

Present age of son = 4 years

Present age of father = 2(4)^{2} years = 32 years

**Solution 7:**

Let the speed of the stream be x km/hr.

∴ Speed of the boat downstream = (15 + x) km/hr

Speed of the boat upstream = (15 – x) km/hr

But, x cannot be negative, so, x = 5.

Thus, the speed of the stream is 5 km/hr.

**Solution 8:**

**Solution 9:**

Let the number of children be x.

It is given that Rs 250 is divided amongst x students.

Since, the number of students cannot be negative, so, x = 100.

Hence, the number of students is 100.

**Solution 10:**

Original weekly wage of each worker = Rs x

Original weekly wage bill of employer = Rs 3150

Since, wage cannot be negative, x = 45.

Thus, the original weekly wage of each worker is Rs 45.

**Solution 11:**

Number of articles bought by the trader = x

It is given that the trader bought the articles for Rs 1200.

Number of articles cannot be negative. So, x = 100.

**Solution 12:**

Let the number of articles bought be x.

Total cost price of x articles = Rs 4800

Cost price of one article = Rs

Selling price of each article = Rs 100

Selling price of x articles = Rs 100x

Given, Profit = C.P. of 15 articles

∴ 100x – 4800 = 15 ×

100x^{2} – 4800x = 15 4800

x^{2} – 48x – 720 = 0

x^{2} – 60x + 12x – 720 = 0

x(x – 60) + 12(x – 60) = 0

(x – 60) (x + 12) = 0

x = 60, -12

Since, number of articles cannot be negative. So, x = 60.

Thus, the number of articles bought is 60.

**Exercise 8(E)**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**Solution 8:**

S = n(n + 1)

Given, S = 420

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

Since, n cannot be negative.

Hence, n = 20.

**Solution 9:**

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

Five years ago,

Father’s age = (x – 5) years

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

From the given information, we have:

(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

If x = 9,

Father’s age = 9 years, Son’s age = (45 – x) = 36 years

This is not possible.

Hence, x = 36

Father’s age = 36 years

Son’s age = (45 – 36) years = 9 years

**Solution 10:**

Let the number of rows in the original arrangement be x.

Then, the number of seats in each row in original arrangement = x

Total number of seats = x × x = x²

From the given information,

2x(x – 10) = x^{2} + 300

2x^{2} – 20x = x^{2} + 300

x^{2} – 20x – 300 = 0

(x – 30) (x + 10) = 0

x = 30, -10

Since, the number of rows or seats cannot be negative. So, x = 30.

(i) The number of rows in the original arrangement = x = 30

(ii) The number of seats after re-arrangement = x^{2} + 300 = 900 + 300 = 1200

**Solution 11:**

**Solution 12:**

Let the age of son 2 years ago be x years.

Then, father’s age 2 years ago = 3x^{2} years

Present age of son = (x + 2) years

Present age of father = (3x^{2} + 2) years

3 years hence:

Son’s age = (x + 2 + 3) years = (x + 5) years

Father’s age = (3x^{2} + 2 + 3) years = (3x^{2} + 5) years

From the given information,

3x^{2} + 5 = 4(x + 5)

3x^{2} – 4x – 15 = 0

3x^{2} – 9x + 5x – 15 = 0

3x(x – 3) + 5(x – 3) = 0

(x – 3) (3x + 5) = 0

x = 3,

Since, age cannot be negative. So, x = 3.

Present age of son = (x + 2) years = 5 years

Present age of father = (3x^{2} + 2) years = 29 years

**Solution 13:**

**Solution 14:**

Given, the difference between two digits is 6 and the ten’s digit is bigger than the unit’s digit.

So, let the unit’s digit be x and ten’s digit be (x + 6).

From the given condition, we have:

x(x + 6) = 27

x² + 6x – 27 = 0

x² + 9x – 3x – 27 = 0

x(x + 9) – 3(x + 9) = 0

(x + 9) (x – 3) = 0

x = -9, 3

Since, the digits of a number cannot be negative. So, x = 3.

Unit’s digit = 3

Ten’s digit = 9

Thus, the number is 93.

**Solution 15:**

**Solution 16:**