જવાબ : y = 0 and y = -5 are Parallel lines, hence no solution.

જવાબ :

જવાબ :

જવાબ :

Area of triangle

= 12 × base × corresponding altitude

= 12 × 10 × 10 = 50 cm^{2}

3x + 2y = 8 6x – 4y = 9

જવાબ :

Therefore, given pair of linear equations is con-sistent.

3x = y + 5

5x – y = 11

જવાબ : We have, 3x = y + 5, and 5x – y = 11

Putting the value of x in (i), we get

3x – y = 5 ⇒ 3(3) – y = 5

9 – 5 = y ⇒ y = 4

∴ x = 3, y = 4

3x – y – 7

2x + 5y + 1 = 0

જવાબ : 3 x – y = 7 …(i)

2x + 5y = -1 -00

Multiplying equation (i) by 5 & (ii) by 1,

⇒ x = 2

Putting the value of x in (i), we have

3(2)-y = 7 ⇒ 6 – 7 = y

∴ y = -1 ∴ x = 2, y = -1

જવાબ : We have, 4x – 3y = 9 and 2x + ky = 11

જવાબ : Let wife’s monthly income = ₹x

Then man’s monthly income = ₹(x + 600)

According to the question,

110 (x + 600) + 16 (x) = ₹1,500

3(x+600)+5x30 = ₹1,500

3x + 1,800 + 5x = ₹45,000

8x = ₹45,000 – ₹1,800

x = ₹343,2008 = ₹5,400

Wife’s income = ₹x = ₹5,400

Man’s income = ₹(x + 600) = ₹6,000

ax + by = a

જવાબ :

Putting the value of x in (i), we get

b(a) – ay = 0 ⇒ ba = ay

baa = y ∴ b = y

∴ x = a, y = b

27x + 31y = 85;

31x + 2 7y = 89

જવાબ :

Putting the value of ‘x’ in (i), we get

2 + y = 3 ⇒ y = 3 – 2 = 1

∴ x = 2, y = 1

2x + 3y = 7;

αx + (α + β)y = 28 (2013)

જવાબ : We have, 2x + 3y = 7 and αx + (α + β)y = 28

y -4x= 1

6x- 5y= 9

જવાબ :

x + 2y – 3= 0

3x – 2y + 7 = 0

જવાબ :

જવાબ :

x+y=1

2x – 3y = 11

જવાબ :

જવાબ :

(ii) coincident lines

(iii) parallel lines

જવાબ :

જવાબ :

જવાબ :

For a journey of 13 km, the charge paid is? 129 and for journey of 22 km, the charge paid is ^ 210.

(i) What will a person have to pay for travelling a distance of 32 km?

(ii) Why did he decide to use CNG for his taxi as a fuel?

જવાબ :

જવાબ :

Why it is necessary to conserve these animals?

જવાબ :

6x + 2y = 3 and kx + y = 2 has a unique solution, find the value of k.

જવાબ :

જવાબ :

4x + 5y = 2;

(2p + 7q)x + (p + 8q)y = 2q-p + 1

જવાબ :

જવાબ :

જવાબ :

જવાબ :

3x + y = 1

(2k-l)x+ (k-l)y = 2k+l

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

2x + 3y = 9;

6x + (k – 2)y = (3k – 2).

જવાબ :

(p – 3)x + 3y = p;

px + py = 12

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

x + 2y – 8 = 0

2x + 4y = 16

જવાબ :

x +y = 14,

x-y = 4

જવાબ :

kx + 3y = k – 2 and 12x + ky = k has no solution.

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ : Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36 (l + b) = 36 ……….(1)Given, the length is 4 m more than its width. Let width = x And length = x + 4 Substituting this in eq(1), we get; x + x + 4 = 36

2x + 4 = 36

2x = 32

x = 16 Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

જવાબ : (i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0

and 2x – 3y = 7 or 2x – 3y – 7 = 0

And a

We get,

a

a

Since, a

Therefore,

a

a

જવાબ : 2x + 3y = 11…………………………..(i)

2x – 4y = -24………………………… (ii)

From equation (ii), we get;

x = (11 – 3y)/2 ……….…………………………..(iii)

11 – 3y – 4y = -24

-7y = -35

y = 5……………………………………..(iv)

Putting the value of y in equation (iii), we get; x = (11 – 15)/2 = -4/2 = −2

Hence, x = -2, y = 5

Also,

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore, the value of m is -1.

જવાબ : (i) Given,

x + y = 14 and x – y = 4 are the two equations.

From 1st equation, we get,

x = 14 – y

Now, put the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Or y = 5

By the value of y, we can now find the value of x;

∵ x = 14 – y

∴ x = 14 – 5

Or x = 9

Hence, x = 9 and y = 5.

3x – y = 3 and 9x – 3y = 9 are the two equations.

From 1st equation, we get,

x = (3 + y)/3

Now, substitute the value of x in the given second equation to get,

9[(3 + y)/3] – 3y = 9 ⇒ 3(3+y) – 3y = 9

⇒ 9 + 3y – 3y = 9

⇒ 9 = 9

Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

જવાબ : Let the cost of a bat be x and the cost of a ball be y.

According to the question,

7x + 6y = 3800 ………………. (i)

3x + 5y = 1750 ………………. (ii)

From (i), we get;

y = (3800 – 7x)/6 …………………… (iii)

Substituting (iii) in (ii). we get,

3x + 5[(3800 – 7x)/6] = 1750

⇒3x + (9500/3) – (35x/6) = 1750

3x – (35x/6) = 1750 – (9500/3)

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500

Putting the value of x in (iii), we get; y = (3800 – 7 × 500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

જવાબ : Let the fraction be x/y.

According to the question,(x + 2)/(y + 2) = 9/11 11x + 22 = 9y + 18 11x – 9y = -4 …………….. (1)

(x + 3)/(y + 3) = 5/6 6x + 18 = 5y +15

6x – 5y = -3 ………………. (2)

From (1), we get x = (-4 + 9y)/11 …………….. (3) Substituting the value of x in (2), we get

6[(-4 + 9y)/11] – 5y = -3 -24 + 54y – 55y = -33

-y = -9

y = 9 ………………… (4)

Substituting the value of y in (3), we get

x = (-4 + 81)/11 = 77/11 = 7 Hence, the fraction is 7/9.

જવાબ : (i)

Let us assume, the present age of Nuri is x.And the present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

Now,

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3) Substituting the value of y in eq.1, we get,

x – 3(20) = -10

x – 60 = -10

x = 50

Therefore,

Age of Nuri is 50 years

Age of Sonu is 20 years.

(ii) Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs. B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs. 15.

And the Additional charge per day is Rs. 3.

જવાબ : 8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get;

Substituting this value in equation 1, we get

8[(4 – 2y)/3] + 5y = 9

32 – 16y + 15y = 27

-y = -5

y = 5 ……………………………….(4)

Substituting this value in equation (2), we get

3x + 10 = 4 3x = -6

x = -2

Thus, x = -2 and y = 5. Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0 x/(-20 + 18) = y/(-27 + 32 ) = 1/(16 – 15)

-x/2 = y/5 = 1/1

∴ x = -2 and y =5.

જવાબ : (i) Let us consider,

Speed of boat is still water = x km/hr

Speed of current = y km/hr

Now, speed of Ritu, during,

Downstream = x + y km/hr

Upstream = x – y km/hr

As per the question given,

2(x + y) = 20

Or x + y = 10……………………….(1)

And, 2(x – y) = 4

Or x – y = 2………………………(2)

Adding both the eq.1 and 2, we get,

2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of current = 4 km/hr

જવાબ : Let unit’s place digit be x and ten’s place digit bey.

∴ Original number = x + 10y Reversed number = 10x + y

According to the Question,

10x + y = 3(x + 10y) – 9

⇒ 10x + y = 3x + 30y – 9

⇒ 10x + y – 3x – 30y = -9

⇒ 7x – 29y = -9 …(i)

10x + y – (x + 10y) = 45

⇒ 9x – 9y = 45

⇒ x – y = 5 …[Dividing both sides by 9

⇒ x – 5 + y …(ii)

Solving (i),

7x – 29y = -9

7(5 + y) – 29y = -9 …[From (ii)

35+ 7y – 29y = -9

-22y = -9 – 35

-22y = -44 ⇒ y = 4422 = 2

Putting the value of y in (ii),

x = 5 + 2 = 7

∴ Original number = x + 10y

= 7 + 10(2) = 27

જવાબ : Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,

x + 100 = 2(y – 100)

x + 100 = 2y – 200

x – 2y = 300 … (1)

Again, 6(x – 10) = (y + 10)

6x – 60 = y + 10

6x – y = 70 … (2)

Multiplying equation (2) by 2, we obtain:

12x – 2y = 140 … (3)

Subtracting equation (1) from equation (3), we obtain:

11x = 140 + 300

11x = 440

x = 40

Putting the value of x in equation (1), we obtain:

40 – 2y = –300

40 + 300 = 2y

2y = 340

y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

જવાબ : Let the number of rows be x and number of students in a row be y.

Total number of students in the class = Number of rows x Number of students in a row = xy

According to the question,

Total number of students = (x – 1) (y + 3)

xy = (x – 1) (y + 3)

xy= xy – y + 3x – 3

3x – y – 3 = 0

3x – y = 3 … (1)

Total number of students = (x + 2) (y – 3)

xy = xy + 2y – 3x – 6

3x – 2y = –6 … (2)

Subtracting equation (2) from (1), we obtain:

y = 9

Substituting the value of y in equation (1), we obtain:

3x – 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Hence, Total number of students in a class = xy = 4 x 9 = 36

જવાબ : The given system of equations may be written as

ax + by – (a – b) = 0

bx – ay – (a + b) = 0

By cross-multiplication, we have

Hence, the solution of the given system of equations is x = 1, y = -1

જવાબ : Let the length and breadth of a rectangle be x and y respectively.

Then area of the rectangle = xy

According to question, we have

(x – 5) 6 + 3) = xy – 9 ⇒ xy + 3x – 5y – 15 = xy – 9

⇒ 3x – 5y = 15 – 9 = 6 ⇒ 3x – 5y = 6 …(i)

Again, we have

(x + 3) 6 + 2) = xy + 67 ⇒ xy + 2x + 3y + 6 = xy + 67

⇒ 2x + 3y = 67 – 6 = 61 ⇒ 2x + 3y = 61…. (ii)

Now, from equation (i), we express the value of x in terms of y as

3x – 5 x 9 = 6 ⇒ 3x = 6 + 45 = 51

∴ x = 513 = 17

Hence, the length of rectangle = 17 units and breadth of rectangle = 9 units.

જવાબ : Let the fixed charge be *x and the cost of food per day be by.

Therefore, according to question,

x + 20y = 1000 …(i)

x + 26y = 1180 …(ii)

Now, subtracting equation (ii) from (i), we have

Putting the value of y in equation (i), we have

x + 20 x 30 = 1000 ⇒ x + 600 = 1000 ⇒ x = 1000 – 600 = 400

Hence, fixed charge is ₹400 and cost of food per day is ₹30.

જવાબ : x – y = 2 … (i)

x + y = 4 … (ii)

On adding (i) and (ii), we get 2x = 6 or x = 3

From (i), 3 – y ⇒ 2 = y = 1

a = 3, b = 1.

On comparing the ratios a1a2,b1b2 , and, c1c2 find out whether the following pair of linear equations consistent or inconsistent. is consistent or inconsistent. (Q. 5 to 6)

9x – 10y = 14

જવાબ :

જવાબ : Let cost price of table be ₹x and the cost price of the chair be ₹y.

From equation (i) and (ii) we get

110x + 125y = 105000

and 125x + 110y = 106500

On adding and subtracting these equations, we get

235x + 235y = 211500

and 15x – 15y = 1500

i.e., x + y = 900 …(iii)

x – y = 100 …… (iv)

Solving equation (iii) and (iv) we get

x = 500, y = 400

So, the cost price of the table is ₹500 and the cost price of the chair is ₹400.

જવાબ : Let the digits at unit and tens places be x and y respectively.

Then, number = 10y + x …(i)

Number formed by interchanging the digits = 10x + y

According to the given condition, we have

(10y + x) + (10x + y) = 110

⇒ 11x + 11y = 110

⇒ x + y – 10 = 0

Again, according to question, we have

(10y + x) – 10 = 5 (x + y) + 4

⇒ 10y + x – 10 = 5x + 5y + 4

⇒ 10y + x – 5x – 5y = 4 + 10

5y – 4x = 14 or 4x – 5y + 14 = 0

By using cross-multiplication, we have .

Putting the values of x and y in equation (i), we get

Number 10 × 6 + 4 = 64.

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) Roobi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by bus and the remaining by train. If she travels 100 km by bus and the remaining by train, she takes 10 minutes longer. Find the speed of the train and the bus separately.

જવાબ : (i) Let her speed of rowing in still water be x km/h and the speed of the current be y km/h.

Case I: When Ritu rows downstream

Her speed (downstream) = (x + y) km/h

Putting the value of x in equation (i), we have

6 + y = 10 ⇒ y = 10 – 6 = 4

Hence, speed of Ritu in still water = 6 km/h.

and speed of current = 4 km/h.

According to question, we have

જવાબ : Let the length and breadth of a rectangle be x and y respectively.

Then area of the rectangle = xy

According to question, we have

(x – 5) 6 + 3) = xy – 9 ⇒ xy + 3x – 5y – 15 = xy – 9

⇒ 3x – 5y = 15 – 9 = 6 ⇒ 3x – 5y = 6 …(i)

Again, we have

(x + 3) 6 + 2) = xy + 67 ⇒ xy + 2x + 3y + 6 = xy + 67

⇒ 2x + 3y = 67 – 6 = 61 ⇒ 2x + 3y = 61…. (ii)

Now, from equation (i), we express the value of x in terms of y as

Putting the value of y in equation (i), we have

3x – 5 x 9 = 6 ⇒ 3x = 6 + 45 = 51

∴ x = 513 = 17

Hence, the length of rectangle = 17 units and breadth of rectangle = 9 units.

જવાબ : Let x be the number of questions of right answer and y be the number of questions of wrong answer.

According to question,

3x – y = 40 … (i)

and 4x – 2y = 50

or 2x – y = 25 …(ii)

Subtracting (ii) from (i), we have

Putting the value of x in equation (i), we have

3 x 15 – y = 40 ⇒45 – y = 40

∴ y = 45 – 40 = 5

Hence, total number of questions is x + yi.e., 5 + 15 = 20.

Therefore, according to question,

x + 20y = 1000 …(i)

x + 26y = 1180 …(ii)

Now, subtracting equation (ii) from (i), we have

Putting the value of y in equation (i), we have

x + 20 x 30 = 1000 ⇒ x + 600 = 1000 ⇒ x = 1000 – 600 = 400

Hence, fixed charge is ₹400 and cost of food per day is ₹30.

જવાબ : Let x be the number of girls and y be the number of boys.

According to question, we have

x = y + 4

⇒ x – y = 4 ……(i)

Again, total number of students = 10

Therefore, x + y = 10 …(ii)

Hence, we have following system of equations

x – y = 4

and x + y = 10

From equation (i), we have the following table:

Here, the two lines intersect at point (7,3) i.e., x = 7, y = 3.

So, the number of girls = 7

and number of boys = 3.

જવાબ :

3 x 1 + y = 4⇒ y = 4 – 3 = 1

Hence, the solution of given system of equations is x = 1, y = 1.

2x + 3y = 7

(a – b)x + (a + b)y = 3a + b – 2

(ii) for which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

જવાબ : (i) We have, 2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2 … (ii)

Here, a1 = 2, b1 = 3, c1 = 7 and

a2 = a – b, b2 = a + b, c2 = 3a + b – 2

For infinite number of solutions, we have

⇒ 9a – 7a + 3b – 75 -6 = 0 ⇒ 2a – 45 – 6 = 0 => 2a – 4b = 6

⇒ a – 2b = 3 …(iv)

Putting a = 5b in equation (iv), we get

56 – 2b = 3 or 3b = 3 i.e., b = 33 =1

Putting the value of b in equation (ii), we get a = 5(1) = 5

Hence, the given system of equations will have an infinite number of solutions for a = 5 and b = 1.

(2k – 1) x + (k – 1) y = 2k + 1

⇒ (2k – 1) x + (k – 1) y – (2k + 1) = 0 ……(ii)

Here, a1 = 3, b1 = 1, C1 = -1

a2 = 2k – 1, b2 = k – 1, C2 = -(2k + 1)

For no solution, we must have

⇒ 3k – 2k = 3 – 1 ⇒ k = 2

Hence, the given system of equations will have no solutions for k = 2.

7x – 4y = 49 and 5x – y = 57

જવાબ : We have, 7x – 4y = 49 ……..(i)

and 5x – 6y = 57 ……..(ii)

Put y = -7 in equation (ii)

5x – 6(-7)57 ⇒ 5x = 57 – 42 ⇒ x = 3

hence, x = 3 and y = -7.

જવાબ :

Putting the value of y in (ii), we get

x + ab = 2ab ⇒ x = 2ab – ab ⇒ x = ab

∴ x = ab, y = ab

152x – 378y = -74 and -378x + 152y = -604

જવાબ : We have, 152x – 378y = -74 …(i)

-378x + 152y = -604 ……(ii)

Putting the value of x in (iii), we get

2 + y = 3 ⇒ y = 1

Hence, the solution of given system of equations is x = 2, y = 1.

gseb maths textbook std 10

- Math Book for GSEB ધોરણ ૧૦
- Chemistry Book for GSEB ધોરણ ૧૦
- Biology Book for GSEB ધોરણ ૧૦
- Physics Book for GSEB ધોરણ ૧૦
- History Book for GSEB ધોરણ ૧૦
- Geography Book for GSEB ધોરણ ૧૦
- Economics Book for GSEB ધોરણ ૧૦
- Political Science Book for GSEB ધોરણ ૧૦

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.