જવાબ : Infinite Solutions
જવાબ : One Solution
જવાબ : No Solutions
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
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
1 |
A |
-7 |
2 |
3 |
B |
-1 |
3 |
5 |
C |
-3 |
4 |
7 |
D |
-5 |
જવાબ :
1-B, 2-C, 3-D, 4-A
Coordinates of points on line x-y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
1 |
A |
7 |
2 |
3 |
B |
5 |
3 |
5 |
C |
3 |
4 |
7 |
D |
1 |
જવાબ :
1-D, 2-C, 3-B, 4-A
Coordinates of points on line 2x+y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
4 |
A |
-4 |
2 |
-6 |
B |
-2 |
3 |
8 |
C |
5 |
4 |
-10 |
D |
3 |
જવાબ :
1-B,2-D,3-A, 4-C
Coordinates of points on line x+2y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
-8 |
A |
1 |
2 |
-2 |
B |
-2 |
3 |
6 |
C |
-3 |
4 |
4 |
D |
4 |
જવાબ :
1-D, 2-A, 3-C, 4-B
Coordinates of points on line 2x-y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
2 |
A |
6 |
2 |
3 |
B |
10 |
3 |
4 |
C |
4 |
4 |
5 |
D |
8 |
જવાબ :
1-C, 2-A, 3-D, 4-B
Coordinates of points on line 2x= -3y
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
-9 |
A |
2 |
2 |
-3 |
B |
4 |
3 |
-12 |
C |
6 |
4 |
-6 |
D |
8 |
જવાબ :
1-C, 2-A, 3-D, 4-B
Coordinates of points on line x-3y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
6 |
A |
1 |
2 |
9 |
B |
2 |
3 |
12 |
C |
3 |
4 |
3 |
D |
4 |
જવાબ :
1-B, 2-C, 3-D, 4-A
Coordinates of points on line 2x-3y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
3 |
A |
8 |
2 |
6 |
B |
6 |
3 |
9 |
C |
4 |
4 |
12 |
D |
2 |
જવાબ :
1-D, 2-C, 3-B, 4-A
Coordinates of points on line 2x-3y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
3 |
A |
8 |
2 |
6 |
B |
6 |
3 |
9 |
C |
4 |
4 |
12 |
D |
2 |
જવાબ :
1-D, 2-C, 3-B, 4-A
Coordinates of points on line x-5y=0
|
Co-ordintes of x |
|
Co-ordinates of y |
1 |
40 |
A |
5 |
2 |
25 |
B |
6 |
3 |
35 |
C |
7 |
4 |
30 |
D |
8 |
જવાબ :
1-D, 2-A, 3-C, 4-B
Math
Chapter 03 : Pair of Linear Equations in Two Variables
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