Test: Pair of Linear Equations in Two Variables (Hard) - Class 10 MCQ

⇒ .04x + .02y = 5....eq (1)

⇒ x = 5−.0.042y ⇒ .5x − .4y = 30....eq (2)

Substitute the value of x in eq(2)

⇒ - .4y = 30

⇒ 2.5 − .01y − .016y = 1.2

⇒−.026y = −1.3

⇒ y = 50

Substitute the value of y in eq(2)

⇒ .5x − .4 (50) = 30

⇒.5x = 50

⇒ x = 100

⇒ x = 100, y = 50

Perimeter = 60

2(x + x - 6)= 60

4x = 72

x = Length = 18m

Breadth = 18 - 6 = 12m

= = So,from equation (1) and (2) we can write the value of a,b and (c) = = = = = = = = = = 16 y = 2

Therefore, x =

a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

Hence, 3 / 2 = 2k / 5

k = 15 / 4

a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

For option a,

a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂ = 1 / 2

x = 2 + y

Substituting the value of x in the second equation we get;

2 + y + y = 4

2 + 2y = 4

2y = 2

y = 1

Now putting the value of y, we get;

x = 2 + 1 = 3

Hence, the solutions are x = 3 and y = 1.

So, as per the question given,

(x - 1) / y = 1 / 3 => 3x – y = 3…………………(1)

x / (y + 8) = 1 / 4 => 4x – y = 8 ………………..(2)

Subtracting equation (1) from (2), we get

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

4 × 5 – y = 8

y = 12

Therefore, the fraction is 5 / 12.

4m + 3y = 14

3m – 4y = 23

By cross multiplication we get;

m / (-69 - 56) = y / (-42 - (-92)) = 1 / (-16 - 9)

m / -125 = y / 50 = -1 / 25

m / -125 = -1 / 25 and y / 50 = -1 / 25

m = 5 and y = -2

m = 1 / x or x = 1 / m = 1 / 5

Speed of Stream = y km/hr

Now, speed of Ritu, during,

Downstream = x + y km/h

Upstream = x – y km/h

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 equations, we get,

2x = 12

x = 6

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

y = 4

Therefore, Speed of Ritu is still water = 6 km/hr

Speed of Stream = 4 km/hr

Hence,

A + C = 180°

6x + 10 + x + y = 180

=> 7x + y = 170°

And B + D = 180°

5x + 3y - 10 = 180

=> 5x + 3y = 190°

By solving the above two equations we get;

x = 20° and y = 30°.

From question, l = 6 + b… (1)

×2(l + b) = 46 ⟹ l + b = 46… (2)

Using Substitution method: Substituting the value of l from (1) in (2), we get

6 + b + b = 46

⇒ 6 + 2b = 46

⟹ 2b = 40

⟹ b = 20 m.

Thus, l = 26 m

3x + 2y = 12… (2)

Multiply equation (1) by 2, we get:

2(2x + y) = 2(7)

⇒ 4x + 2y = 14… (3)

Subtracting (2) from (3) we get,

x = 2

Substituting the value of x in (1) we get,

2(2) + y = 7 ⟹ y = 3

Thus, the solution for the given pair of linear equations is (2, 3).

Solve the pair equations 4x + y = 3, 3x + 2y = 1

let 4x + y = 3…..(1)

and 3x + 2y = 1 …..(2)

y = 3 − x

[ From (1)] Substituting value if y in (2)

3x + 2y = 1

3x + 2(3 − 4x) = 1

3x + 6 − 8x = 1

−5x = −5 ⇒ x = 1

Substituting x = 1 in (1),

4(1) + y = 3 ⇒ y = −1

⇒ (1, -1) is the solution of pair of equation.

∴ Pair of equations which satisfy the point (1, -1)

x + y = 54…. (i)

And 10x + 22y = 780 ——————– (ii)

Multiply (i) by 10, we get

Substituting y = 20 in x + y = 54, we have x + 20 = 54; x = 34

Hence, x = 34 and y = 20.

⇒ 0.5x− y + √7= 0

The general form of an equation is ax + by + c = 0.

Here, on comparing, we get

a = 0.5, b = −1 and c = √7

7x + 8y = 61 = 7(3) + 8(4) = 53.

Substituting the value of x = 2 and y = 5,

7x + 8y = 61 = 7(2) + 8(5) = 54.

Substituting the value of x = -3 and y = 7,

7x + 8y = 61 = 7(−3) + 8(7) = 35.

Substituting the value of x = 3 and y = 5,

7x + 8y = 61 = 7(3) + 8(5) = 61

Hence, (3, 5) lies on the given line.

L.H.S = 2x + 3y

L.H.S = 2(−3) + 3(2)

L.H.S = 0 = R.H.S

Hence x = −3, y = 2 is the solution of the equation 2x + 3y = 0

y = 1 / 2 (3x + 7)

Simplifying the equation we get:

2y − 3x − 7 = 0

⇒ −3x + 2y − 7 = 0 (1)

Thus, the value of a, b and c is -3, 2 and -7 respectively.

The equation can also be written as, 3x − 2y + 7 = 0

Thus, the value of a, b and c is +3, -2 and +7 respectively.

The option -3, 2 and -7 is correct [Since +3, -2 and +7 is not an option]

(As x ≠ 0, y ≠ 0)

Then, the given equations become

a + 3b = 1 … (1)

6a − 12b = 2 … (2)

Multiplying equation (1) by 4, we get

4a + 12b = 4 … (3)

On adding equation (2) and equation (3), we get 10a = 6

⇒ a = 3 / 5

Putting a = 3 / 5 in equation (1), we get (3 / 5) + 3b = 1

⇒ b = 2 / 15

Hence, x = 5 / 3 and y = 15 / 2