Linear Equation in 2 or more Variables MCQ Quiz - Objective Question with Answer for Linear Equation in 2 or more Variables - Download Free PDF

Last updated on Jun 20, 2025

Solve these Linear Equations in Two or More Variables MCQs Quiz to truly understand the complex concepts behind the seemingly easy questions. Linear Equation in Two or More Variables objective questions are commonly asked in competitive examinations. Therefore candidates should thoroughly understand the basics of it. Get solutions and their explanations so that there are no doubts about Linear Equations in Two or More Variables question answers.

Latest Linear Equation in 2 or more Variables MCQ Objective Questions

Linear Equation in 2 or more Variables Question 1:

Find the value of y.

x + y - z = 6

2x + y - z = 7

3x + y - 2z = 11

  1. 2
  2. -3
  3. 4
  4. -5
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 2

Linear Equation in 2 or more Variables Question 1 Detailed Solution

Given:

x + y - z = 6

2x + y - z = 7

3x + y - 2z = 11

Concept Used:

Solving a system of linear equations using elimination method to find the value of y.

Calculation:

We have,

⇒ x + y - z = 6     (1)

⇒ 2x + y - z = 7     (2)

⇒ 3x + y - 2z = 11     (3)

Subtract equation (1) from equation (2):

⇒ (2x + y - z) - (x + y - z) = 7 - 6

⇒ x = 1     (4)

Subtract equation (2) from equation (3):

⇒ (3x + y - 2z) - (2x + y - z) = 11 - 7

⇒ x - z = 4

From equation (4), substitute x = 1:

⇒ 1 - z = 4

⇒ z = -3     (5)

Substitute x = 1 and z = -3 into equation (1):

⇒ 1 + y - (-3) = 6

⇒ 1 + y + 3 = 6

⇒ y + 4 = 6

⇒ y = 2

∴ The value of y is 2.

Linear Equation in 2 or more Variables Question 2:

 The cost of two pencils and three erasers is  ₹ 18 whereas the cost of one pencil and two erasers is ₹11. The cost of each pencil is

  1. 8
  2. 4
  3. 3
  4. 6

Answer (Detailed Solution Below)

Option 3 : 3

Linear Equation in 2 or more Variables Question 2 Detailed Solution

Given:

Cost of 2 pencils + 3 erasers = ₹18

Cost of 1 pencil + 2 erasers = ₹11

Formula used:

Let cost of 1 pencil = ₹x and cost of 1 eraser = ₹y.

Then, equations are:

2x + 3y = 18

x + 2y = 11

Calculation:

From x + 2y = 11:

⇒ x = 11 - 2y

Substitute x = 11 - 2y into 2x + 3y = 18:

⇒ 2(11 - 2y) + 3y = 18

⇒ 22 - 4y + 3y = 18

⇒ 22 - y = 18

⇒ y = 4

Substitute y = 4 into x = 11 - 2y:

⇒ x = 11 - 2×4

⇒ x = 11 - 8

⇒ x = 3

∴ The cost of each pencil is ₹3.

The correct answer is option (3).

Linear Equation in 2 or more Variables Question 3:

If a pair of linear equations is given by a1x + b1y + c1  = 0 and a2x + b2y c2= 0 where a1a2=b1b2c1c2 then,

  1. The pair of linear equation is consistent
  2. The pair of linear equation is inconsistent
  3. The pair of linear equation is independent
  4. The pair of linear equation is dependent

Answer (Detailed Solution Below)

Option 2 : The pair of linear equation is inconsistent

Linear Equation in 2 or more Variables Question 3 Detailed Solution

Given:

The pair of linear equations is:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Condition:

(a1/a2) = (b1/b2) ≠ (c1/c2)

Formula Used:

To determine the nature of the pair of linear equations:

If (a1/a2) = (b1/b2) ≠ (c1/c2), the system of equations is inconsistent.

Calculation:

Given:

(a1/a2) = (b1/b2) ≠ (c1/c2)

From the formula, when the above condition holds, the system of equations does not have a solution. This means the two lines are parallel and will never intersect.

Hence, the pair of linear equations is inconsistent.

The correct option is Option 2.

Linear Equation in 2 or more Variables Question 4:

Two students appeared for an examination. One of them secured 9 marks more than the other and his marks has 56% of the sum of their marks. What are the marks obtained by them? 

  1. 35, 44
  2. 35, 42
  3. 33, 44
  4. 33, 42
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : 33, 42

Linear Equation in 2 or more Variables Question 4 Detailed Solution

Given:

Two students appeared for an examination.

One of them secured 9 marks more than the other.

His marks are 56% of the sum of their marks.

Concept Used:

Use algebraic equations to represent the given conditions and solve for the unknowns.

Calculation:

Let the marks of the student who scored less be x.

Then, the marks of the student who scored more are x + 9.

The sum of their marks = x + (x + 9) = 2x + 9.

According to the problem, the marks of the student who scored more are 56% of the sum of their marks.

⇒ x + 9 = 56/100 × (2x + 9)

⇒ x + 9 = 0.56(2x + 9)

⇒ x + 9 = 1.12x + 5.04

⇒ x + 9 - 1.12x = 5.04

⇒ -0.12x + 9 = 5.04

⇒ -0.12x = 5.04 - 9

⇒ -0.12x = -3.96

⇒ x = -3.96 / -0.12

⇒ x = 33

Marks of the student who scored less = 33

Marks of the student who scored more = 33 + 9 = 42

∴ The marks obtained by the students are 33 and 42.

Linear Equation in 2 or more Variables Question 5:

The simplified value of following

(315a5b6c3×59ab5c4)÷1027a2bc3

  1. 9a2bc410
  2. 3ab4c310
  3. 3a4b10c410
  4. 1a4b4c410
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 3a4b10c410

Linear Equation in 2 or more Variables Question 5 Detailed Solution

Calculation

15(a6b11c7)/135/10(a2bc3)/27

⇒ 15 × 27(a4b10c4)/(135 × 10)

3(a4b10c4)/10

Top Linear Equation in 2 or more Variables MCQ Objective Questions

Three-fifths of my current age is the same as five-sixths of that of one of my cousins’. My age ten years ago will be his age four years hence. My current age is ______ years.

  1. 55
  2. 45
  3. 60
  4. 50

Answer (Detailed Solution Below)

Option 4 : 50

Linear Equation in 2 or more Variables Question 6 Detailed Solution

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Let my current age = x years and my cousin’s age = y years.

Three-fifths of my current age is the same as five-sixths of that of one of my cousins’,

⇒ 3x/5 = 5y/6

⇒ 18x = 25y

My age ten years ago will be his age four years hence,

⇒ x – 10 = y + 4

⇒ y = x – 14,

⇒ 18x = 25(x – 14)

⇒ 18x = 25x – 350

⇒ 7x = 350

∴ x = 50 years

In a four-digit number, the second digit and the last digit places are interchanged. The new number thus formed is greater than the original number by 297. What is the difference between the second and last digit of that number?

  1. 2
  2. 3
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 2 : 3

Linear Equation in 2 or more Variables Question 7 Detailed Solution

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Given:

Interchange number – original number = 297

Calculation:

Let the number be ‘pqrs’.

⇒ pqrs = 1000p + 100q + 10r + s

⇒ psrq – pqrs = 297

⇒ 1000p + 100s + 10r + q – (1000p + 100q + 10r + s) = 297

⇒ 1000p + 100s + 10r + q – 1000p – 100q – 10r – s = 297

⇒ 100s + q – 100q – s = 297

⇒ 99s – 99q = 297

⇒ 99(s – q) = 297

⇒ s – q = 3

second digit – last digit = 3

Alternate Method

The original number is a four-digit number. Let's represent it as ABCD (where A, B, C, D are its digits).

When the second digit and the last digit are interchanged, the new number becomes ADCB.

The problem states that ADCB = ABCD + 297.

In a four-digit number:
- The thousands place contributes its value times 1000
- The hundreds place contributes its value times 100
- The tens place contributes its value times 10
- The ones place contributes its value times 1

So, we can write the original number (ABCD) as 1000A + 100B + 10C + D.

Similarly, the new number (ADCB) can be written as 1000A + 100D + 10C + B.

Setting up the equation as given in the problem:

1000A + 100D + 10C + B = 1000A + 100B + 10C + D + 297

Solving the equation, we find that 99B - 99D = 297, or B - D = 3.

Therefore, the difference between the second digit (B) and the last digit (D) of the number is 3.

If two mixers and one T.V cost Rs. 700. While two T.V s and one mixer cost Rs. 980. The value of one T.V is: 

  1. Rs. 420
  2. Rs. 400
  3. Rs. 450
  4. Rs. 480

Answer (Detailed Solution Below)

Option 1 : Rs. 420

Linear Equation in 2 or more Variables Question 8 Detailed Solution

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Given

2 mixers + 1 TV = Rs. 700

2 TVs + 1 mixer = Rs. 980

Concept:

This problem can be solved using a system of equations.

Solution:

2M + T = 700

2T + M = 980

Add both equations:

 2T + M + (2M + T) = 980 + 700 ⇒ T + M = 1680/3 = 560

2T + M = 980

T + T + M = 980

T + 560 = 980

T = 420

Therefore, the value of one TV is Rs. 420.

If x + y = 12, y + z = 15 and x + z = 18, then find x + y + z =?

  1. 18
  2. 12
  3. 15
  4. 22.5

Answer (Detailed Solution Below)

Option 4 : 22.5

Linear Equation in 2 or more Variables Question 9 Detailed Solution

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Given:

x + y = 12, y + z = 15, x + z = 18

Calculation:

x + y = 12      ----(1)

y + z = 15      ----(2)

x + z = 18      ----(3)

By solving equations (1) and (2)

⇒ x – z = -3      ----(4)

By solving equations (3) and (4)

⇒ x = 7.5

Put the value of x in equation (1)

⇒ y = 4.5

Put the value of y in equation (2)

⇒ z = 10.5

x + y + z

⇒ 7.5 + 4.5 + 10.5

⇒ 22.5

∴ The value of x + y + z is 22.5.

Shortcut Trick

Add (1), (2) and (3)

⇒ 2(x + y + z) = 45     

⇒ (x + y + z) = 45/2 = 22.5

∴ The value of x + y + z is 22.5.

If (x + y) : (y + z) : (z + x) = 11 : 13 : 16, and x + y + z = 200, then find the value of z.

  1. 50
  2. 60
  3. 90
  4. 80

Answer (Detailed Solution Below)

Option 3 : 90

Linear Equation in 2 or more Variables Question 10 Detailed Solution

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Given:

(x + y) : (y + z) : (z + x) = 11 : 13 : 16, and x + y + z = 200

Calculation:

The value of (x + y) = 11A

The value of (y + z) = 13A

The value of (z + x) = 16A

Add all three equations.

⇒ x + y + y + z + z + x = 40A

⇒ 2(x + y + z) = 40A

⇒ (x + y + z) = 20A

According to the question,

⇒ 20A = 200

⇒ A = 10

Now, 

The value of (x + y) = 11 × 10 = 110

According to the question,

⇒ (x + y + z) - (x + y) = 200 - 110

⇒ z = 90

∴ The value of the variable 'z' is 90.

Three cups of ice cream, two burgers and four soft drinks together cost Rs. 128. Two cups of ice cream, one burger and two soft drinks together cost Rs. 74. What is the cost of five burgers and ten soft drinks?

  1. Rs. 160
  2. Rs. 128
  3. Rs. 170
  4. Cannot be determined

Answer (Detailed Solution Below)

Option 3 : Rs. 170

Linear Equation in 2 or more Variables Question 11 Detailed Solution

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Let cost of each ice cream, burger and soft drink is x, y and z respectively.

3x + 2y + 4z = 128      ---- (i)

2x + y + 2z = 74      ---- (ii)

Multiply 3 × (ii) and 2 × (i), we get

6x + 3y + 6z = 222      ----(iii)

6x + 4y + 8z = 256      ----(iv)

substract equation (iv) to equation (iii)

y + 2z = 34

Multiply the above equation by 5 

we get,

5 (y + 2z) = 5 × 34

5y + 10z = 170 

∴ cost of 5 burgers and 10 soft drinks = 34 × 5 = 170

If the equations 14x + 8y + 5 = 0 and 21x - ky - 7 = 0 have no solution, then the value of k is:

  1. 12
  2. -12
  3. 8
  4. -16

Answer (Detailed Solution Below)

Option 2 : -12

Linear Equation in 2 or more Variables Question 12 Detailed Solution

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⇒ The equations have no solution when their slopes are same

⇒ Slope of equation 1 = - 14/8 = - 7/4

⇒ Slope of equation 2 = 21/k

⇒ So, 21/k = - 7/4

∴ The value of k is - 12.

If the equations 6x – 5y + 11 = 0 and 15x + ky – 9 = 0 have no solution, then the value of k is:

  1. -12.5
  2. 12.5
  3. -18
  4. 18

Answer (Detailed Solution Below)

Option 1 : -12.5

Linear Equation in 2 or more Variables Question 13 Detailed Solution

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Concept used:

Consider the pair of linear equations in two variables x and y.
a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Here a1, b1, c1, a2, b2, c2 are all real numbers.

Note that, a12 + b12 ≠ 0, a22 + b22 ≠ 0

If (a1/a2) = (b1/b2) ≠ (c1/c2), then there will be no solution.

Calculation:

When two equation have no solution then, Using parallel system of equations,

Then,

⇒ 6/15 = -5/k

⇒ k = -25/2

⇒ k = -12.5

If a(a + b + c) = 126, b(a + b + c) = 147 and c(a + b + c) = 168, then value of (a + b + c) = ?

  1. 23
  2. 21
  3. 41
  4. 22

Answer (Detailed Solution Below)

Option 2 : 21

Linear Equation in 2 or more Variables Question 14 Detailed Solution

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Given:

3 equations, a(a + b + c) = 126, b(a + b + c) = 147 and c(a + b + c) = 168

Calculation:

Adding all, we get (a + b + c) (a + b + c) = 126 + 147 + 168

⇒ (a + b + c)2 = 441

⇒ (a + b + c) = 21

If x + y + z = 0, then find the value of [(x + y)3 + (x + z)3 + (y + z)3 – 9xyz]/[4(x + y)(x + z)(z + y)].

  1. 0
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 3 : 3

Linear Equation in 2 or more Variables Question 15 Detailed Solution

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Calculation:

x + y + z = 0

⇒ x + y = -z ---- (1)

⇒ x + z = -y ---- (2)

⇒ y + z = -x ---- (3)

Now,

⇒ [(x + y)3 + (x + z)3 + (y + z)3 – 9xyz]/[4(x + y)(x + z)(z + y)],

⇒ [(-z)3 + (-y)3 + (-x)3 – 9xyz]/[4(-z)(-y)(-x)], (By equation 1, 2 and 3)

⇒ [(z)3 + (y)3 + (x)3 + 9xyz]/[4(z)(y)(x)],

⇒ [(3xyz + 9xyz]/[4(z)(y)(x)], (∵ if a + b + c = 0 then, a3 + b3 + c3 = 3abc)

⇒ 12/4 = 3

∴ [(x + y)3 + (x + z)3 + (y + z)3 – 9xyz]/[4(x + y)(x + z)(z + y)] = 3

Shortcut Trick

Put x = 1, y = 1 and z = -2

So, [(x + y)3 + (x + z)3 + (y + z)3 – 9xyz] / [4(x + y)(x + z)(z + y)]

⇒ [(1 + 1)3 + (1 - 2)3 + (1 - 2)3 - 9(1)(1)(-2)] / [4(1 + 1)(1 - 2)(-2 + 1)]

⇒ [23 + (-1)3 + (-1)3 + 18] / [4 × 2 × (-1) × (-1)]

⇒ [8 - 1 - 1 + 18] / [8]

⇒ 24/8 = 3

Hence, the correct answer is option (3).

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