Test: Linear Equations in Two Variables - 2 - SAT MCQ

8a²b = 27, ab² = 216

Multiplying these two equations, we get,

8a³b³    = 27*216

a³b³    = 27*27

ab = 9

(b² – 4b + 4) + (a² – 6a + 9) = 0

(b – 2)² + (a – 3)² = 0

a = 3, b = 2

a + b = 5

⇒ 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.

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) 

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 posiible because of negative value]

Now, k³ = 1/8

⇒ k = 1/2 ⇒ 1/k = 2

Then, (k + 1/k) = (1/2 + 2) = 5/2 = 

∴ The value of (k + 1/k) is 

Let the wrong questions attempt by Mr. X be a then

Right question = (100 – a)

According to the question

⇒ (100 – a) × 4 – a × 1 = 340

⇒ 400 – 4a – a = 340

⇒ 5a = 400 – 340 = 60

⇒ a = 60/5 = 12

Let, length of the piece of cloth = x m

Cost of cloth = Rs. 35

∴ Cost of 1 m cloth = Rs. 35/x

According to the question,

⇒ (x + 4) (35/x - 1) = 35

⇒ 35 - x + 140/x - 4 = 35

⇒ 35x - x² + 140 - 4x = 35x

⇒ x² + 4x - 140 = 0

⇒ x² + 14x - 10x - 140 = 0

⇒ x(x + 14) - 10(x + 14) = 0

⇒ (x + 14) (x - 10) = 0

⇒ x = - 14 or 10

∴ The piece of cloth is 10 m long
Alternate Method:
We can tabulate the given data according to the following table:

As the total remains constant at Rs.35, we get:

xy = 35     

⇒ x = 35/y      ----(i)

Also, (x – 1) × (y + 4) = 35      ----(ii)

On substituting the value of x from equation (i) into equation (ii), we get:

[(35/y) – 1] × (y + 4) = 35 

On solving, we get:

y = -14 and 10

∴ The piece of cloth is 10 m long

Calculation:    

Let assume that numerator = x

Denominator = y 

So,

The fraction is = x/y

Now,

y - x = 2                          ............................. (1)

⇒ 4x + 12 = 3y + 9

⇒ 4x - 3y = -3                   ..............................  (2)

From equation (1) and (2)

x = 3

y = 5

∴ The fraction is 3/5

The correct option is 4 i.e. 3/5

Assume a + b = x

x² – 2x – 80 = 0

x = 10

ab = 16

a = 8, b = 2

3 × a – 19b = −14

x +  y = 14      … (i)

x² – y² = 56

→ x – y = 4     … (ii)

x = 9, y = 5

sum of their squares = 81 + 25 = 106

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

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

Given: 

System of equation:

2x + 3y = 5

4x + ky = 10

Concept:

System of equations

a₁x + b₁y = c₁

a₂x + b₂y = c₂

For Infinite solution

Calculation:

From the equations, it can be deduced that

a₁ = 2, b₁ = 3, c₁ = 5

a₂ = 4, b₂ = k, c₂ = 10

For infinite solutions, 2/4 = 3/k

⇒ k = 6

∴ The value of k is 6.
Important Points:
For unique solution

For inconsistent solution

Given:

The sum of two numbers = 184

Calculation:

Let the numbers be x and (184 − x)

According to the question,

x × (1/3) - (184 − x)/7 = 8

⇒ (7x - 552 + 3x)/21 = 8

⇒ 7x - 552 + 3x = 8 × 21

⇒ 10x = 168 + 552

⇒ x = 720/10 = 72

One number = 72

Other number = 184 − x = 184 - 72 = 112

∴ The smaller number is 72.

Given:

Number obtained by interchanging the positions of its digits = 36

Calculation:

Let the ten's digit be x and unit's digit be y respectively

Then,

⇒ (10x + y) – (10y + x) = 36

⇒ 9(x – y) = 36

⇒ x – y = 4

∴ The difference between the two digits of that number is 4