Math Olympiad Test: Linear Equations in Two Variables- 3 - Class 10 MCQ

Let the unit’s and ten’s digits in the number be y and x respectively
So, the number be 10x + y.
According to the question, x + y = 12
Also, 10x + y + 18 = 10y + x
⇒ 9x – 9y = –18 ⇒ x – y = –2
Solving (1) and (2), we get x = 5 and y = 7
∴ Required number is 57.

Let the two numbers be x and y
According to the given conditions, x + y = 8...(1)
 and  ...(2)
Putting value of x = 8 – y in (2), we get

⇒ y² – 8y + 15 = 0 ⇒ y² – 5y – 3y + 15 = 0
⇒ (y – 5) (y – 3) = 0 ⇒ y = 5 or y = 3
From (1), x = 3 or x = 5
Thus, the numbers are 5 and 3.

We have, ax + by = c and lx + my = n
Now, 
∴ The given system of equations has no solution.

Let the number be x/y
According to the question,
y = x + 3 ⇒ x = y − 3 ... (1)
Also,  ⇒ 5x – 15 = y + 2
⇒ 5x = y + 17 ...(2)
Solving (1) and (2), we get x = 5 and y = 8
∴ Required number = x/y = 5/8.

Given equations are
2x + 3y = 7 and 2ax + (a + b)y = 28
For infinitely many solutions, we have

Taking first two members, we get 2a + 2b = 6a
⇒ 4a = 2b ⇒ 2a = b ...(1)
Also,  ...(2)
From (1) and (2), we have
2(4) = b ⇒ b = 8

It is given that, kx – 3y + 6 = 0 and 4x – 6y + 15 = 0 are two parallel lines.
i.e., The given lines has no solution or

⇒ 

ax + by + c = 0
When a and b are equal to zero, then the above linear equation does not represent an equation of line.

In a triangle, sum of angles is 180°
∴ ∠A + ∠B + ∠C = 180° ...(1)
∠C = 3∠B ...(2)
and 3∠B = 2∠A + 2∠B
∴ ∠A = ∠B/2 ...(3)
From (1), (2) and (3), we get
∠B/2 + ∠B + 3∠B = 180^°
⇒ (9/2)∠B = 180° ⇒ ∠B = 40°
∴ ∠C = 120° and ∠A = 20°

Given equations are
6x + 3y = c – 3 and 12x + cy = c
For infinitely many solutions, 
Taking first two members, we have 
Thus, only one value of c exists.

Let the digit at unit place be x and the digit at tens place be y, then the number = 10y + x
Now, according to the question,
 
⇒ 10y + x = 4y + 4x
⇒ 6y = 3x ⇒ x = 2y ...(1)
Also, x = 3 + y ⇒ 2y = 3 + y [From (1)]
⇒ y = 3 and x = 6
∴ Number = 36