Linear Equation in 2 Variable MCQ Quiz - Objective Question with Answer for Linear Equation in 2 Variable - Download Free PDF

Given:

A 3-digit number such that:

Units digit is 2 more than the tens digit

If 311 is subtracted from the number, result is 13

Formula used:

Let the number be represented as:
Number = 100a + 10b + c

Where:

a = hundreds digit

b = tens digit

c = units digit

Also, c = b + 2

Calculation:

(100a + 10b + c) − 311 = 13

⇒ 100a + 10b + c = 324

⇒ 100a + 10b + (b + 2) = 324

⇒ 100a + 11b + 2 = 324

⇒ 100a + 11b = 322

Trying values for a:

Let a = 3

⇒ 100 x 3 + 11b = 322

⇒ 300 + 11b = 322

⇒ 11b = 22

⇒ b = 2

⇒ c = b + 2 = 4

Therefore, Number = 100a + 10b + c = 100 x 3 + 10 x 2 + 4 = 324

Check: 324 − 311 = 13 

Sum of digits = 3 + 2 + 4 = 9

∴ The correct answer is 9.

Given:

7x + 3y = 4 and 2x + y = 2

Calculation:

7x + 3y = 4      ----(1)

2x + y = 2     ----(2)

By solving,

equation (1) - 3 × equation(2)

⇒ 7x + 3y - 3(2x + y) = 4 - (3 × 2)

⇒ x = -2

Putting the value of x in equation (1)

7x + 3y = 4

⇒ 7 × (-2) + 3y = 4

⇒ y = 6

∴ The value of x and y is -2 and 6.

Given:

30 apples + 23 guavas = ₹392

15 apples + 7 guavas = ₹178

Calculation:

Let cost of 1 apple = x, cost of 1 guava = y

Equation 1: 30x + 23y = 392

Equation 2: 15x + 7y = 178

Multiply Eq(2) by 2:

⇒ 30x + 14y = 356

Now subtract:

(30x + 23y) - (30x + 14y) = 392 - 356

⇒ 9y = 36

⇒ y = 4

Substitute y in Eq(2):

15x + 7×4 = 178

⇒ 15x + 28 = 178

⇒ 15x = 150

⇒ x = 10

∴ The cost of an apple is ₹10

Given:

Equation 1: 2x – ky + 5 = 0

Equation 2: 3x + 2y – 7 = 0

Formula used:

For a unique solution: a₁/a₂ ≠ b₁/b₂

Equation 1: a₁ = 2, b₁ = –k

Equation 2: a₂ = 3, b₂ = 2

Calculation:

2/3 ≠ –k/2

⇒ Cross multiply: 2 × 2 ≠ –k × 3

⇒ 4 ≠ –3k

⇒ k ≠ –4/3

∴ The correct answer is k ≠ –4/3

Given:

Simultaneous linear equations:

Equation 1: 2x – 3y + 4 = 0

Equation 2: 3x – 2y – 4 = 0

Formula Used:

To determine the nature of solutions for simultaneous linear equations:

Compare the ratios of coefficients: \(a_1/a_2\), \(b_1/b_2\), and \(c_1/c_2\)

Calculation:

For Equation 1: 2x – 3y + 4 = 0 → a₁ = 2, b₁ = -3, c₁ = 4

For Equation 2: 3x – 2y – 4 = 0 → a₂ = 3, b₂ = -2, c₂ = -4

Compare the ratios:

\(a_1/a_2 = 2/3\)

\(b_1/b_2 = (-3)/(-2) = 3/2\)

\(c_1/c_2 = 4/(-4) = -1\)

⇒ \(a_1/a_2 ≠ b_1/b_2\)

Conclusion:

Since \(a_1/a_2 ≠ b_1/b_2\), the given equations have a unique solution.

The correct answer is Option 2: Unique solution.

Given:

8k6 + 15k3 – 2 = 0

Calculation:

Let, k3 = x

So, 8x² + 15x - 2 = 0

⇒ 8x2 + 16x - x - 2 = 0

⇒ 8x (x + 2) - 1 (x + 2) = 0

⇒ (8x - 1) (x + 2) = 0

⇒ 8x - 1 = 0 ⇒ x = 1/8

⇒ x + 2 = 0 ⇒ x = - 2 [Not possible because of negative value]

Now, k3 = 1/8

⇒ k = 1/2 ⇒ 1/k = 2

Then, (k + 1/k) = (1/2 + 2) = 5/2 = \(2\frac{1}{2}\)

∴ The value of (k + 1/k) is \(2\frac{1}{2}\)

Calculation

let the number of toffee with A be x and with B be y.

If A gives one toffee to B, then:

⇒ x - 1 = y + 1

⇒ x = y + 2  .........(1)

Now when B gives one toffee to A, then the toffees with A are double with B:

⇒ x + 1 = 2 (y - 1)  ......(2)

Putting the value of eq.(1) in eq. (2).

⇒ y + 3 = 2y - 2

⇒ y = 5

If y = 5 then x = 7.

⇒ x + y = 12

The total number of toffees with A and B are 12.

Given:

Difference between the two numbers = 5

Ratio If 25 is subtracted from the smaller number and 20 is added to the greater number = 1 : 2

Calculation:

Let the greater number and smaller number be x and (x – 5) respectively

Now, according to the question,

(x – 5 – 25) : (x + 20) = 1 : 2

⇒ (x –  30)/(x + 20) = 1/2

⇒ 2x – 60= x + 20

⇒ x = 80

∴ The greater number is 80

Calculation-

Let the cost of 1 table be 'x' and 1 chair be 'y'

Than according to the given condition 

2x + 4y = 16,000 and x = 6y 

Now, 2x + 4y = 16,000

⇒ 2(6y) + 4y = 16,000

⇒ 16y = 16,000

⇒ y = 1,000

∴ cost of 9 chairs will be 9y = 9,000

Given:

x + y + 3 = 0

Formula used:

(a + b)3 = a3 + b3 + 3ab (a + b)

Calculation:

x + y + 3 = 0

⇒ x + y = - 3   .....(1)

⇒ (x + y)³ = (- 3)³  [Taking cube of both sides]

⇒ x3 + y3 + 3xy (x + y) = - 27

⇒ x3 + y3 + 3xy × (- 3) = - 27  [∵ x + y = - 3]

⇒ x3 + y3 - 9xy = - 27

⇒ x3 + y3 - 9xy + 9 = - 27 + 9  [Adding 9 in both sides]

⇒ x3 + y3 - 9xy + 9 = - 18

∴ The value of x3 + y3 - 9xy + 9 is (- 18)

Given equation are x + 2y = 8 and 2x + 4y = 16 or x + 2y = 8,

Both the given equations are same

Let the price of single pencil, pen, and eraser be x, y, and z respectively

According to question,

8x + 5y + 3z = Rs. 111      ----(1)

9x + 6y + 5z = Rs. 130      ----(2)

16x + 11y + 3z = Rs. 221      ----(3)

Subtracting equation (1) from (3)

⇒ (16x + 11y + 3z) - (8x + 5y + 3z) = 221 - 111

⇒ 8x + 6y = 110

⇒ 4x + 3y = 55      ----(4)

Multiply the equation (2) by 3 and (3) by 5 and then subtracting equation (2) from (3)

⇒ (16x + 11y + 3z) × 5 - (9x + 6y + 5z) × 3 = 221 × 5 - 130 × 3

⇒ 80x + 55y + 15z - 27x - 18y - 15z = 1105 - 390

⇒ 53x + 37y = 715      ----(5)

Multiply the equation (4) by 53 and (5) by 4 and then subtracting equation (4) from (5)

⇒ 212x + 159y - 212x - 148y = 2915 - 2860

⇒ 11y = 55

⇒ y = 5

By putting the value of y = 5 in equation (4)

⇒ 4x + 3 × 5 = 55

⇒ x = 10

By putting the value of y = 5 and x = 10 in equation (1)

⇒ 8 × 10 + 5 × 5 + 3z = 111

⇒ 80 + 25 + 3z = 111

⇒ z = 2

∴ Cost of 39 pencils, 26 pens and 13 erasers is 39x + 26y + 13z =39 × 10 + 26 × 5 + 13 × 2 = Rs. 546

Shortcut Trick

Let, price of 1 pencil = x, price of 1 pen = y and price of one eraser = z

Then, 8x + 5y + 3z = 111      ----(1)

9x + 6y + 5z = 130      ----(2)

16x + 11y + 3z = 221      ----(3)

Adding (1), (2) and (3), we get

33x + 22y + 11z = 462

⇒ 3x + 2y + z = 42

⇒ 39x + 26y + 13z = 546      (multiplying with 13) 

Let the price of one pen be Rs. P, one notebook be Rs. N and one file be Rs. F

According to the problem statement

⇒ 4P + 6N + 9F = 305     ---- (i)

⇒ 3P + 4N + 2F = 145     ---- (ii)

Now 2 × (i) – (ii)

⇒ (8 – 3)P + (12 – 4)N + (18 – 2)F = 5P + 8N + 16F = 2 × 305 – 145 = 465

Given:

The price of an article is reduced by ₹ then more articles can be purchased for ₹ 288.

Calculation:

Let the original price of each article = y

Number of articles sold = x

Total price = xy = 288

⇒ x = 288/y --(i)

New price of each article = y - 4

No. of new articles sold = x + 12

∴ According to the question,

⇒ (x + 12) (y - 4) = xy

⇒ xy - 4x + 12y - 48 = xy

⇒ 4x - 12y = - 48

From (i),

⇒ 4(288/y) - 12y = - 48

⇒ 1152 - 12y² + 48y = 0

⇒ 12y² - 48y - 1152 = 0

⇒ y² - 4y - 96 = 0

⇒ (y - 12) (y + 8) = 0

⇒ y = 12, y = -8

Since the price can't be negative so y = -8 is not possible.

∴ The original price of the new article is Rs 12.

Alternate MethodCalculation:

According to the question:

⇒ 288/(x - 4) - 288/x = 12

⇒ x - x + 4/(x - 4) x = 12/288

⇒ 4/(x - 4) x = 1/24

⇒ x (x - 4) = 96

So from option we can put the value of x.

If we put x = 12

⇒ 12 × 8 = 96

⇒ 96 = 96 (Equation satisfied)

∴ The correct answer is Rs.12

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