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

Given:

The cost of 4 pens and 3 notebooks is ₹ 150.

The cost of 5 notebooks is ₹ 41 more than the cost of 6 pens.

Formula Used:

Let the cost of one pen be ₹ x and the cost of one notebook be ₹ y.

From the given information:

4x + 3y = 150     ...(i)

5y = 6x + 41     ...(ii)

Calculation:

From equation (ii), express y in terms of x:

5y = 6x + 41

y = (6x + 41) / 5

Substitute y in equation (i):

4x + 3((6x + 41) / 5) = 150

Multiply through by 5 to clear the fraction:

20x + 3(6x + 41) = 750

20x + 18x + 123 = 750

38x + 123 = 750

38x = 750 - 123

38x = 627

x = 627 / 38

x = 16.5

Now, find y:

y = (6 × 16.5 + 41) / 5

y = (99 + 41) / 5

y = 140 / 5

y = 28

Now, find the cost of 3 pens and 2 notebooks:

3 pens cost 3x = 3 × 16.5 = 49.5

2 notebooks cost 2y = 2 × 28 = 56

Total cost = 49.5 + 56 = 105.5

The cost of 3 pens and 2 notebooks is ₹ 105.50.

Given:

The cost of 2 tables and 3 chairs is Rs. 540.

The cost of 2 tables and 1 chair is Rs. 470.

Formula used:

Let the cost of one table be T and the cost of one chair be C.

Calculations:

2T + 3C = 540       ......(1)

2T + 1C = 470       ......(2)

Subtract equation (2) from equation (1):

⇒ (2T + 3C) - (2T + 1C) = 540 - 470

⇒ 2C = 70 ⇒ C = 35

Substitute the value of C into equation (2):

⇒ 2T + 35 = 470

⇒ 2T = 470 - 35

⇒ T = 217.5

Now, calculate the cost of 2 tables and 2 chairs:

⇒ 2T + 2C = 2 × 217.5 + 2 × 35

⇒ 2T + 2C = 435 + 70

⇒ 2T + 2C = 505

∴ The correct answer is option 1.

Concept:

System of equations

a1x + b1y = c1

a2x + b2y = c2

For Infinite solution

\(\frac {a_1}{a_2}= \frac {b_1}{b_2}= \frac {c_1}{c_2}\)

Calculation:

We have 

ax + by + c = 0       ----(1)

lx + my + n = 0       ----(2)

For infinite solutions, 

\(\frac{a}{l} = \frac{b}{m} = \frac{c}{n}\)

Important Points

For unique solution

\(\frac {a_1}{a_2}≠ \frac {b_1}{b_2}\)

For inconsistent solution

\(\frac {a_1}{a_2}=\frac {b_1}{b_2}≠ \frac {c_1}{c_2}\)

Calculation:

Let the 2 digits be x,y. respectively

2 digit number =10x + y

2 digit number obtained by using the digit =10y + x

∴10x + y + 10y + x = 99

⇒ 11x + 11y = 99

⇒ x + y = 9

⇒ x − y = 3

On adding :

⇒ 2x = 12

⇒ x = 6

putting x in ⟶(i)

⇒ 6 + y=9

⇒ y = 3

∴ Number =10x + y = 63, 10y + x = 36.

∴ The correct answer will be 36 or 63.

Given:

The system of linear equations 2x + 3y = 7 and

2ax + (a + b) y = 8.

Concept Used:

If a₁x + b₁y + c₁ = 0 and a₂x +b₂y + c₂ = 0 be two equations than,

For infinitely many solutions, 

\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)

Solution:

We have,

2x + 3y + 7 i.e. a₁ = 2, b₁ = 3 and c₁ = 7

And 2ax + (a + b) y + 8 i.e. a₂ = 2a, b₂ = (a + b) and c₂ = 8

We know that, 

For infinitely many solutions,

\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)

⇒ \(\frac{2}{2a} = \frac{3}{(a +b)} = \frac{7}{8}\)

\(\frac{1}{a} = \frac{3}{(a + b)}\)

⇒ a + b = 3a

⇒ b = 2a

\(\therefore\) option 2 is correct.

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:

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 \(2\frac{1}{2}\)

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

Given:

The system of equations 17x + my + 102 = 0 and 23x + 299y + 138 = 0 has infinite number of solutions.

Concept used:

When Ax + By = C and Px + Qy = R form a system of linear equation, then it has infinitely many solutions if,

A/P = B/Q = C/R

Calculation:

Since the given system of linear equation is said to have an infinite number of solutions, so,

According to the concept,

17/23 = m/299 = 102/138

Hence,

17/23 = m/299

⇒ m = (17 × 299) ÷ 23

⇒ m = 221

∴ The value of m is 221.