Math Olympiad Test: Linear Equations in Two Variables- 1 - Class 10 MCQ

Let

then 57a + 6b = 5 ...(1)
38a + 21b = 9 ...(2)
Solving (1) and (2) we get a = 1/19, b = 1/3
Hence
x + y = 19 ...(4)
x - y = 3 ...(5)
Solving (4) and (5), we get x = 11, y = 8

Let the numbers be x and y
x + y = 16 ...(1)
and

Now


⇒ x - y = 8 ...(2)
From (1) and (2), we get x = 12, y = 4.

L et ∠A = x°, ∠B = y°, ∠C = 3∠B = 3y°
In ΔABC, ∠A + ∠B + ∠C = 180°
x° + y° +3y° = 180°
x° + 4y° = 180 ...(1)
∠C = 2(∠A + ∠B) ⇒ 3y = 2 (x° + y°)
2x° - y° = 0 ...(2)
From (1) and (2)
x = 20 , y = 40
∴ ∠A = 20°, ∠B = 40°, ∠C = 3 × 40 = 120°

For no solution, it must have

We have
2ax - 2by = - a - 4b ...(1)
and 2bx + 2ay = 4a - b ...(2)
2a²x - 2aby = -a² - 4ab ...(3)
Now 2b²x + 2aby = 4ab - b²
2x (a² + b²) = - (a² + b²) ⇒ x = -1/2
Now putting the value of 2aby from eq. (3), we get

Let the breadth of the field be x m and length be y m.
Then, area = xy m²
Now y = 3 + x
⇒ y - x = 3 ...(1)
New length = y + 3
breadth = x -2
∴ area =  (x - 2) (y + 3) = xy
⇒ xy + 3x - 2y - 6 = xy
⇒ 3x -2y = 6 ...(2)
From (1) and (2), we get
x = 12 m, y = 15 m

The given system of equations are 
2x + 3y – 5 = 0 
kx - 6y - 8 = 0 
This system is of the form: 
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 
where, a₁ = 2, b₁ = 3, c₁ = -5 and a₂ = k, b₂ = -6, c₂ = -8 
Now, for the given system of equations to have a unique solution, we must have:

Let Shashi’s present age be x and Ravi’s present age be y
y - 5 = 3 (x - 5) ⇒ 3x - y = 10 ...(1)
y + 10 = 2 (x + 10) ⇒ 2x - y = 10 ...(2)
Solving eqns (1) and (2), we get
x = 20, y = 50

In a cyclic quadrilateral,
∠A + ∠C = 180°, ∠B + ∠D = 180°
Here 2x - 1 + 2y + 15 = 180°
⇒ 2x + 2y = 166
x + y = 83 ...(1)
and 4x - 7 + y + 5 = 180°
⇒ 4x + y = 182 ...(2)
Solving eq. (1) and (2), we get x = 33, y = 50
∠A = 2x - 1 = 2 × 33 - 1 = 65°
∠B = y + 5 = 50 + 5 = 55°
∠C = 2y + 15 = 2 × 50 + 15 = 115°
∠D = 4x - 7 = 4 × 33 - 7 = 125°

Let the number of right answer be x and the number of wrong answer be y
3x - y = 40 ...(1)
4x - 2y = 50 ⇒ 2x - y = 25 ...(2)
Solving (1) and (2) x = 15, y = 5
No. of questions = 15 + 5 = 20