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

⇒ x – y = K; x + y = 7K; xy= 6K

⇒ (x+y)² – (x – y)² = (7K)² – K²

⇒ x²+ y² + 2xy – x² –y² + 2xy = 49K² – K²

⇒ 4xy = 48K²

Putting value of xy

⇒ 4 × 6K = 48K²

⇒ 24K = 48K²

⇒ K = 0.5

Now,

xy = 6 × 0.5

xy = 3

Thus, the value of xy is 3

Let the ten’s digit be denoted by Y and unit’s digit be denoted by X. Then,

The original number = 10 × Y + X

= 10Y + X

A.T.Q: ⇒ Sum of the digits = 12

∴ Y + X = 12 … (1)

On adding 36, the new number = 10Y + X + 36

Given that on adding 36 the digits gets reversed, thus

⇒ New number = 10X + Y

Hence, ⇒ 10Y + X + 36 = 10X + Y

⇒ Y - X = -4 …. (2)

Solving equations (1) & (2), we get

⇒ Y = 4 & X = 8

Hence, the original number = 4 × 10 + 8

= 48

Given:

Cost of 5 shirts + 7 trousers = ₹6,150

Cost of 3 shirts = Cost of 4 trousers

Formula used:

Let cost of 1 shirt = ₹S and cost of 1 trouser = ₹T

Calculation:

⇒ 5S + 7T = 6150 …(1)

⇒ 3S = 4T ⇒ S = 4T/3 …(2)

Substitute equation (2) in equation (1):

⇒ 5 × (4T/3) + 7T = 6150

⇒ (20T/3) + 7T = 6150

⇒ (20T + 21T)/3 = 6150

⇒ 41T/3 = 6150

⇒ 41T = 6150 × 3 = 18,450

⇒ T = 18450 ÷ 41 = 450

⇒ S = (4 × 450)/3 = 600

Cost of 2 shirts and 2 trousers = 2S + 2T = 2 × 600 + 2 × 450 = 1200 + 900 = 2100

∴ The cost of 2 shirts and 2 trousers is ₹2100.

Given:

Pair of linear equations:

kx + 3y - (k - 3) = 0

12x + ky - k = 0

Formula Used:

For a pair of linear equations to have infinitely many solutions, their ratios must be equal:

Calculation:

Given equations are:

kx + 3y - (k - 3) = 0

12x + ky - k = 0

Rewrite in the form ax + by + c = 0:

kx + 3y + (-k + 3) = 0

12x + ky + (-k) = 0

Here:

a₁ = k , b₁ = 3 , c₁ = -k + 3

a₂ = 12 , b₂ = k , c₂ = -k

For infinitely many solutions:

⇒

From :

⇒ k² = 36

⇒ k = 6 or k = -6

From :

⇒

⇒ k = 6

Therefore, the value of k that satisfies both conditions is:

k = 6

The correct answer is option 4.

Given:

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

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.

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.

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.

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.

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

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).

⇒ 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.

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

Calculation:

x + 1/y = 3      ----(1) 

y + 1/z = 2      ----(2) 

z + 1/x = 4      ----(3)

Add the equation. (1), (2) and (3).

⇒ x + y + z + 1/x + 1/y + 1/z = 9      ----(4)

Now multiply the eq. (1), (2) and (3)

⇒ (x + 1/y) × (y + 1/z) × (z + 1/x) = 3 × 2 × 4

⇒ (xy + x/z + 1 + 1/zy)(z + 1/x) = 24

⇒ (xyz + y + x + 1/z + z + 1/x + 1/y + 1/xyz) = 24

⇒ [xyz + (1/xyz) + x + y + z + 1/x + 1/y + 1/z] = 24

⇒ xyz + 1/xyz + 9 = 24 

⇒ xyz + 1/xyz = 24 – 9 = 15

∴ The answer is 15