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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations)

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## Selina Textbook Solutions Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations).

All Selina textbook questions of Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

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## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations) Page/Excercise 6(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,

16t2 + 4t = 420

4t2 + t - 105 = 0

4t2 - 20t + 21t - 105 = 0

4t(t - 5) + 21(t - 5) = 0

(4t + 21)(t - 5) = 0

t = , 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,

The digit of a number cannot be negative, so, x = 3.

Thus, the required number is 38.

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 = x2 years

One year ago,

Son's age = (x - 1) years

Man's age = (x2 - 1) years

It is given that one year ago; a man was 8 times as old as his son.

(x2 - 1) = 8(x - 1)

x2 - 8x - 1 + 8 = 0

x2 - 8x + 7 = 0

(x - 7) (x - 1) = 0

x = 7, 1

If x = 1, then x2 = 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 = x2 years = 49 years

Solution 6

Let the present age of the son be x years.

Present age of father = 2x2 years

Eight years hence,

Son's age = (x + 8) years

Father's age = (2x2 + 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.

2x2 + 8 = 3(x + 8) +4

2x2 + 8 = 3x + 24 +4

2x2 - 3x - 20 = 0

2x2 - 8x + 5x - 20 = 0

2x(x - 4) + 5(x - 4) = 0

(x - 4) (2x + 5) = 0

x = 4,

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

Time taken to go 30 km downstream =

Time taken to come back =

From the given information,

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

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

Solution 8

Number of oranges = y

Cost of one orange =

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 =

=

Now, Mr. Mehra pays a total of Rs 15.

But, the number of oranges cannot be negative. So, x = 12.

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

If there were 25 children more, then

Money received by each child = Rs

From the given information,

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

Number of workers =

New weekly wage of each worker = Rs (x + 5)

New weekly wage bill of employer = Rs 3250

Number of workers =

From the given condition,

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.

So, cost of one article = Rs

Ten articles were damaged. So, number of articles left = x - 10

Selling price of each of (x - 10) articles = Rs

Selling price of (x - 10) articles = Rs (x - 10)

Profit = Rs 60

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

100x2 - 4800x = 15 4800

x2 - 48x - 720 = 0

x2 - 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.

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations) Page/Excercise 6(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

n2 + n - 420 = 0

n2 + 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 - x2 - 200 + 5x = 124

x2 - 45x +324 = 0

x2 - 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 =

From the given information,

2x(x - 10) = x2 + 300

2x2 - 20x = x2 + 300

x2 - 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 originalarrangement = x = 30

(ii) The number of seats after re-arrangement = x2 + 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 = 3x2 years

Present age of son = (x + 2) years

Present age of father = (3x2 + 2) years

3 years hence:

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

Father's age = (3x2 + 2 + 3) years = (3x2 + 5) years

From the given information,

3x2 + 5 = 4(x + 5)

3x2 - 4x - 15 = 0

3x2 - 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 = (3x2 + 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

Solution 17

Time taken by bus to cover total distance with speed x km/h =

Time taken by bus to cover total distance with speed (x - 10) km/h =

According to the given condition, we have

Since the speed cannot be negative, we have x = 40 km/h

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations) Page/Excercise 6(A)

Solution 1

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

From the given information,

x(x + 1) = 56

x2 + 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,

x2 + (x + 1)2 = 41

2x2 + 2x + 1 - 41 = 0

x2 + 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,

x2 + (x + 5)2 = 97

2x2 + 10x + 25 - 97 = 0

2x2 + 10x - 72 = 0

x2 + 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

Let the numbers be x and

From the given information,

x = 4

x = x = 4

Thus, the numbers are 4 and

Solution 5

Let the numbers be x and x + 3.

From the given information,

Since, x is a natural number, so x = 2.

Thus, the numbers are 2 and 5.

Solution 6

Let the two parts be x and x - 15.

x = 5 One part = 5 and other part = 10

x = 10 One part = 10 and other part = 5

Thus, the required two parts are 5 and 10.

Solution 7

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

x2 + y2 = 208 ..... (i)

y2 = 18x ..... (ii)

From (i), we get y2=208 - x2. Putting this in (ii), we get,

208 - x2 = 18x

x2 + 18x - 208 = 0

x2 + 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 y2 = 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,

x2 + (x + 2)2 = 52

2x2 + 4x + 4 = 52

2x2 + 4x - 48 = 0

x2 + 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,

x2 + (x + 2)2 = 74

2x2 + 4x + 4 = 74

2x2 + 4x - 70 = 0

x2 + 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

Let the required fraction be.

From the given information,

Thus, the required fraction is .

Solution 11

Given, three positive numbers are in the ratio

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

From the given information,

(6x)2 + (4x)2 + (3x)2 = 244

36x2 + 16x2 + 9x2 = 244

61x2 = 244

x2 = 4

x =

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

3x2 = (20 - x) + 10

3x2 = 30 - x

3x2 + x - 30 = 0

3x2 - 9x + 10x - 30 = 0

3x(x - 3) + 10(x - 3) = 0

(x - 3) (3x + 10) = 0

x = 3,

Since, x cannot be equal to, 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,

x2 = (x + 1)2 - (x - 1)2 + 60

x2 = x2 + 1 + 2x - x2 - 1 + 2x + 60

x2 = 4x + 60

x2 - 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 + p2 + 67

3p2 + 6p + 3 = p2 + 1 - 2p + p2 + 67

p2 + 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

Work done by A in one day =

Work done by B in one day =

Together A and B can do the work in 15 days. Therefore, we have:

Since, x cannot be negative.

Thus, x = 24.

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.,

If x =, then x - 3 =, which is not possible.

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

Let the smaller part be x.

Then, (larger part)2 = 8x

\ larger part =

Now, the sum of the squares of both the terms is given to be 208

Thus, the required number is 2 + 4 = 6.

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations) Page/Excercise 6(B)

Solution 1

Area of triangle = 30 cm2

But, x cannot be negative, so x = 3.

Thus, we have:

AB = 4 3 cm = 12 cm

BC = (2 3 - 1) cm = 5 cm

CA = (Using Pythagoras theorem)

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 = x2 + (34 - x)2

676 = x2 + x2 + 1156 - 68x

2x2 - 68x + 480 = 0

x2 - 34x + 240 = 0

x2 - 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

9x2 + 1 + 6x = x2 + 1 - 2x + 9x2

x2 - 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 =

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,

x2 = (x - 1)2 + (x - 18)2

x2 = x2 + 1 - 2x + x2 + 324 - 36x

x2 - 38x + 325 = 0

x2 - 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 = 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.

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 m and 20 m.

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 - 4w2

4w2 - 112w + 208 = 0

w2 - 28w + 52 = 0

w2 - 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 cm2

x2 + (x + 4)2 = 656

x2 + x2 + 16 + 8x = 656

2x2 + 8x - 640 = 0

x2 + 4x - 320 = 0

x2 + 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 m2

Area of the flower bed = xy m2

Area of gravel path = Area of rectangular field - Area of flower bed = (2000 - xy) m2

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 + 4w2 = 1200

4w2 - 180w + 800 = 0

w2 - 45w + 200 = 0

w2 - 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 = x2 cm2

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

So, the area needed to be paved = 128 x2 cm2 .... (1)

Size of smaller tiles = (x - 2)cm

Area of smaller tiles = (x - 2)2 cm2

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

So, the area needed to be paved = 200 (x - 2)2 cm2 .... (2)

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

128 x2 = 200 (x - 2)2

128 x2 = 200x2 + 800 - 800x

72x2 - 800x + 800 = 0

9x2 - 100x + 100 = 0

9x2 - 90x - 10x + 100 = 0

9x(x - 10) - 10(x - 10) = 0

(x - 10)(9x - 10) = 0

x = 10,

If , then , which is not possible.

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 - 2x2 = 600

2x2 - 70x + 600 = 0

x2 - 35x + 300 = 0

x2 - 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 = x2 m2

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

Area of outer rectangle = (x + 4) (x + 8) = x2 + 12x + 32

Area of path = x2 + 12x + 32 - x2 = 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

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.

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations) Page/Excercise 6(C)

Solution 1

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

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

Distance = 300 km

We know:

Time taken by ordinary train to cover 300 km =

Time taken by express train to cover 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

Time taken to cover a distance of 36 km =

New speed of the car = (x + 10) km/hr

New time taken by the car to cover a distance of 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.

Time taken to cover a distance of 1200 km =

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

Time taken to cover a distance of 1200 km =

From the given information, we have:

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

Now, the speed is increased by 12 km.

Also given that, increasing the speed of the car will decrease the time taken by 1 hour 40 minutes.

Hence,

Solution 5

We know:

Given, the girl covers a distance of 6 km at a speed x km/ hr.

Time taken to cover first 6 km =

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

Time taken to cover next 6 km =

Total time taken to cover the whole distance = 2 hrs 30 mins =

Since, speed cannot be negative. Therefore, x = 4.

Solution 6

Let the original speed of the car be y km/hr.

We know:

New speed of the car = (y + 4) km/hr

New time taken by the car to cover 390 km =

From the given information,

Since, time cannot be negative, so y = 26.

From (1), we have:

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 taken by goods train to cover a distance of 1040 km =

Time taken by express train to cover a distance of 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.

Therefore, we have:

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.

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TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations)  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.

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