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જવાબ : Infinite Solutions
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જવાબ : One Solution
જવાબ : One Solution
જવાબ : One Solution
જવાબ : One Solution
જવાબ : No Solutions
જવાબ : x=2y−300 (1)
6x−y−70=0(2)
Substituting value from (1) to (2)
6(2y−300)−y−70=06(2y−300)−y−70=0
=> y=170y=170
Putting this in (1)
x=40
જવાબ : 5x−y=5 (1)
3x−y=3 (2)
Substituting value from (1) to (2)
3x + 5 5x = 3
> x=1
Putting this in (1)
> y=0
જવાબ : x+y−40=0 (A)
7x+3y=180 (B)
Multiplying equation (A) by 7
7x+7y−280=0 (C)
Subtracting equation (B) from equation (C)
We get
4y=100 => y=25
Substituting this in (A) ,we get x=15
જવાબ : True
જવાબ : False
જવાબ : False
જવાબ : True
જવાબ : False
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : 64
જવાબ : Let x be math’s students
y be bio students
Then
x+y=10
y=x+4
Solving these linear pair through any method we get
x=3 and y=7
જવાબ : Let x and y are the number
x+y=100x+y=100
y=4xy=4x
Solving them we get x=20 and y=80
જવાબ : x = 3, y = 2
જવાબ : x=3, y=4
જવાબ : x=2, y=1
જવાબ : 0
જવાબ : 3
જવાબ : 6
જવાબ : 10
જવાબ : 15/4
જવાબ : k ≠ 6
જવાબ : no
જવાબ : no
જવાબ : k=6
જવાબ : k≠6
જવાબ : Wrong
જવાબ : Correct
જવાબ : Correct
જવાબ : Correct
જવાબ : parallel
જવાબ : intersecting exactly at one point
જવાબ : parallel
જવાબ : 40 years
જવાબ : 78
જવાબ : 80°
જવાબ : 20°
જવાબ : parallel
જવાબ : intersecting or coincident
જવાબ : Let unit’s place digit be x and ten’s place digit be y.
Then original number = x + 10y
and reversed number = 10x + y
According to the Question,
x + 10y = 7(x + y)
x + 10y = 7x + 7y
⇒ 10y – 7y = 7x – x
⇒ 3y = 6x ⇒ y = 2x …(A)
(x + 10y) – (10x + y) = 18
x + 10y – 10x – y = 18
⇒ 9y – 9x = 180
⇒ y – x = 2 …[Dividing by 9
⇒ 2x – x = 2 …[From (A)
∴ x = 2
Putting the value of ‘x’ in (A), we get y = 2(2) = 4
∴ Required number = x + 10y
= 2 + 10(4) = 42
જવાબ : Let the present ages of his children be x years and y years.
Then the present age of the father = 2(x + y) …(A)
After 20 years, his children’s ages will be
(x + 20) and (y + 20) years
After 20 years, father’s age will be 2(x + y) + 20
According to the Question,
⇒ 2(x + y) + 20 = x + 20 + y + 20
⇒ 2x + 2y + 20 = x + y + 40
⇒ 2x + 2y – x – y = 40 – 20
⇒ x + y = 20 …[From (A)
∴ Present age of father = 2(20) = 40 years
જવાબ : Let, length of rectangular pond = x
breadth of rectangular pond = y
Area of rectangular pond = xy
According to Question,
=> Length of rectangular pond = 7 ft.
Breadth of rectangular pond = 4 ft.
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 …(A)
10x + y – (x + 10y) = 45
⇒ 9x – 9y = 45
⇒ x – y = 5 …[Dividing both sides by 9
⇒ x – 5 + y …(B)
Solving (A),
7x – 29y = 9
7(5 + y) – 29y = 9 …[From (B)
35+ 7y – 29y = 9
22y = 9 – 35
22y = 44 ⇒ y = 44/22 = 2
Putting the value of y in (B),
x = 5 + 2 = 7
∴ Original number = x + 10y
= 7 + 10(2) = 27
જવાબ : Let, speed of stream = x km/hr
Speed of boat in still water = 15 km/hr
then, the speed of the boat upstream = (15 – x) km/hr
and the speed of the boat downstream = (15 + x) km/hr
=> Speed of stream = 5 km/hr
જવાબ : Let the fixed charges = 7x
and the charge per km = ₹y
According to the Question,
Putting the value of y in (A), we get
x + 12(7) = 89
x + 84 = 89 ⇒ x = 89 – 84 = 5
Total fare for 30 km = x + 30y = 5 + 30(7)
= 5 + 210 = ₹215
જવાબ : Let the speed of the stream = y km/hr
Let the speed of boat in still water = x km/hr
then, the speed of the boat in downstream = (x + y) km/hr
and, the speed of the boat in upstream = (x – y) km/hr
From (A), x = 11 – 3 = 8
∴ Speed of the stream, y =3 km/hr
Speed of the boat in still water, x = 8 km/hr
જવાબ : Let, the speed of train = x km/hr
the speed of car = y km/ hr
=> Speed of the train = 60 km/hr
and Speed of the car = 80 kn/hr
જવાબ : Let fixed charge be ₹x and the charge for the distance = ₹y per km
According to the Question,
For a journey of 13 km,
x + 13y = 129 ⇒ x = 129 – 13y …(A)
For a journey of 22 km, x + 22y = 210 …(B)
⇒ 129 – 13y + 22y = 210 …[From (A)
⇒ 9y = 210 – 129 = 81
⇒ 9y = 81 ⇒ y = 9
From (A), x = 129 – 13(9)
= 129 – 117 = 12
∴ Fixed charge, x = ₹12
∴ The charge for the distance, y = ₹9 per km
To pay for travelling a distance of 32 km
= x + 32y = 12 + 32(9) = 12 + 288 = ₹300
જવાબ :
By plotting points and joining them, the lines intersesct at A(6, 0)જવાબ :
By plotting the points and joining the lines, they intersect at A(3,5).
=> x = 3, y = 5
∆ABC is the required triangle.
જવાબ :
Lines intersect at (2, 3) => x = 2, y = 3જવાબ : Let the price of one pencil = ₹x and the price of one chocolate = ₹y.
As per the Question,
Lines intersect at (1, 3).
=> x = 1, y = 3
Therefore the price of one pencil = ₹1
જવાબ : 7x – 5y – 4 = 0
જવાબ : Let the cost of 1 kg of oranges be ‘Rs. x’.
And, let the cost of 1 kg of grapes be ‘Rs. y’.
According to the question, the algebraic representation is
2x + y = 160
And 4x + 2y = 300
X  50  60  70 
Y  60  40  20 
X  70  80  75 
Y  10  10  0 
જવાબ : Given, half the perimeter of a rectangular garden = 36 m
so, 2(l + b)/2 = 36 (l + b) = 36 ……….(1)જવાબ : (i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0
and 2x – 3y = 7 or 2x – 3y – 7 = 0
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.
જવાબ : 2x + 3y = 11…………………………..(i)
2x – 4y = 24………………………… (ii)
From equation (ii), we get;
x = (11 – 3y)/2 ……….…………………………..(iii)
જવાબ : Let the cost of a bat be x and the cost of a pair of pad 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)
Coordinates of points on line x+y=0

Coordintes of x 

Coordinates of y 
1 
1 
A 
7 
2 
3 
B 
1 
3 
5 
C 
3 
4 
7 
D 
5 
જવાબ :
1B, 2C, 3D, 4A
Coordinates of points on line xy=0

Coordintes of x 

Coordinates of y 
1 
1 
A 
7 
2 
3 
B 
5 
3 
5 
C 
3 
4 
7 
D 
1 
જવાબ :
1D, 2C, 3B, 4A
Coordinates of points on line 2x+y=0

Coordintes of x 

Coordinates of y 
1 
4 
A 
4 
2 
6 
B 
2 
3 
8 
C 
5 
4 
10 
D 
3 
જવાબ :
1B,2D,3A, 4C
Coordinates of points on line x+2y=0

Coordintes of x 

Coordinates of y 
1 
8 
A 
1 
2 
2 
B 
2 
3 
6 
C 
3 
4 
4 
D 
4 
જવાબ :
1D, 2A, 3C, 4B
Coordinates of points on line 2xy=0

Coordintes of x 

Coordinates of y 
1 
2 
A 
6 
2 
3 
B 
10 
3 
4 
C 
4 
4 
5 
D 
8 
જવાબ :
1C, 2A, 3D, 4B
Coordinates of points on line 2x= 3y

Coordintes of x 

Coordinates of y 
1 
9 
A 
2 
2 
3 
B 
4 
3 
12 
C 
6 
4 
6 
D 
8 
જવાબ :
1C, 2A, 3D, 4B
Coordinates of points on line x3y=0

Coordintes of x 

Coordinates of y 
1 
6 
A 
1 
2 
9 
B 
2 
3 
12 
C 
3 
4 
3 
D 
4 
જવાબ :
1B, 2C, 3D, 4A
Coordinates of points on line 2x3y=0

Coordintes of x 

Coordinates of y 
1 
3 
A 
8 
2 
6 
B 
6 
3 
9 
C 
4 
4 
12 
D 
2 
જવાબ :
1D, 2C, 3B, 4A
Coordinates of points on line 2x3y=0

Coordintes of x 

Coordinates of y 
1 
3 
A 
8 
2 
6 
B 
6 
3 
9 
C 
4 
4 
12 
D 
2 
જવાબ :
1D, 2C, 3B, 4A
Coordinates of points on line x5y=0

Coordintes of x 

Coordinates of y 
1 
40 
A 
5 
2 
25 
B 
6 
3 
35 
C 
7 
4 
30 
D 
8 
જવાબ :
1D, 2A, 3C, 4B
Chapter 03 : Pair of Linear Equations in Two Variables
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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.
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