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

Calculation

15(a⁶b¹¹c⁷)/135/10(a²bc³)/27

⇒ 15 × 27(a⁴b¹⁰c⁴)/(135 × 10)

⇒ 3(a⁴b¹⁰c⁴)/10

Given:

Cost of 2 mixers and 1 T.V = ₹7000

Cost of 2 T.Vs and 1 mixer = ₹9800

Formula used:

Let the cost of 1 mixer be M and the cost of 1 T.V be T

2M + T = 7000

2T + M = 9800

Calculation:

2M + T = 7000      ...(i)

2T + M = 9800      ...(ii)

From equation (i), multiply by 2:

⇒ 4M + 2T = 14000      ...(iii)

From equation (ii), multiply by 1:

⇒ 2T + M = 9800

Subtract equation (ii) from equation (iii):

⇒ 4M + 2T - (2T + M) = 14000 - 9800

⇒ 4M + 2T - 2T - M = 4200

⇒ 3M = 4200

⇒ M = 1400

Substitute M = 1400 in equation (i):

⇒ 2 × 1400 + T = 7000

⇒ 2800 + T = 7000

⇒ T = 4200

⇒ 2T = 2 × 4200 = 8400

∴ The correct answer is option (2).

Given:

The total monthly salary of 4 men and 2 women is ₹46,000.

A woman earns ₹500 more than a man.

Formula used:

Let the monthly salary of a man be ₹x.

Then, the monthly salary of a woman = ₹(x + 500).

Total salary of 4 men and 2 women = 4x + 2(x + 500).

Calculation:

4x + 2(x + 500) = 46,000

⇒ 4x + 2x + 1000 = 46,000

⇒ 6x + 1000 = 46,000

⇒ 6x = 46,000 - 1000

⇒ 6x = 45,000

⇒ x = 7,500

Monthly salary of a woman = x + 500

⇒ 7,500 + 500 = 8,000

∴ The correct answer is option (3).

Ans. (3)

Solution 

a + 7b + 3c + 5d = 0 → (1)

8a + 4b + 6c + 2d = -16 → (2)

2a + 6b + 4c + 8d = 16 → (3)

5a + 3b + 7c + d = -16 → (4)

Adding (1) and (4), we have

6(a + d) + 10(b + c) = -16

adding (2) and (3), we get 10(a + d) + 10(b + c) = 0.

Therefore, a + d = 4. And (b + c) = -4.

Therefore, (a + d) (b + c) = -16.

Ans. (4)

Solution 

\(\frac{b x-a y}{b}=\frac{c y-b z}{c}=\frac{a z-c x}{a}=k\)

Therefore, c(bx - ay) + a(cy - bz) + b(az - cx) = k(ab + bc + ca)

⇒ k(ab + bc + ca) = 0

⇒ ab + bc + ca = 0.

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

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

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 = 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:

(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 

⇒ 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

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

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

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