Test: Linear Equations in Two Variables - 1 - SAT MCQ

Given:

3 chairs and 2 tables cost Rs. 700

5 chairs and 3 tables cost Rs. 1100

Calculation:

Let the cost of 1 chair be Rs. x and 1 table be Rs. y

Now, According to the question

3x + 2y = 700      ----(1)

5x + 3y = 1100      ----(2)

By solving (1) and (2), we get 

x = 100 and y = 200

Now, we have to find the cost of 1 chair and 2 tables

∴ x + 2y = 100 + 400 = Rs. 500

(a + b)² – 2(a + b) = 80

⇒ (a + b)² – 2(a + b) + 1 = 81

⇒ (a + b – 1)² = 81

⇒ a + b – 1 = 9

⇒ a + b = 10

given ab = 16

⇒ a = 8 and b = 2

⇒ 3a – 19b = 24 – 38 = -14

⇒ Here, k₁/k₂ = k₃/k_4, where k₁ is the initial number of red balls, k₂ is the initial number of blue balls, k₃ is the new number of red balls and k₄ is the new number of blue balls

⇒ 68/36 = k₃/63

⇒ k₃ = 119

∴There should be 119 red balls in the bag.

Using algebraic identities,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

By putting the respective values given in question,

⇒ (9)² = a² + b² + c² + 2(ab + bc + ca) [∵ ab + bc + ca = 26]

⇒ (9)² = a² + b² + c² + 2(26)

⇒ a² + b² + c² = 81 - 52 = 29

Given equations,

a³ + b³ = 91      ----(1)

b³ + c³ = 72      ----(2)

c³ + a³ = 35      ----(3)

On adding (1), (2) and (3)

a³ + b³ + b³ + c³ + c³ + a³ = 91 + 72 + 35

⇒ 2(a³ + b³ + c³) = 198

⇒ a³ + b³ + c³ = 99

Using algebraic identities,

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

By putting the respective values,

⇒ 99 - 3abc = 9 (29 - 26) [∵ ab + bc + ca = 26 and a + b + c = 9]

⇒ 3abc = 99 - 27

⇒ abc = 72/3

∴ abc = 24

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)

The given system of equations is of the form:

⇒ a₁x + b₁y + c₁ = 0

⇒ a₂x + b₂y + c₂ = 0

Here, a₁ = p; b₁ = -3; c₁ = -5; a₂ = 6; b₂ = 2; c₂ = -12

For no solution

⇒ a₁/a₂ = b₁/b₂ ≠ c₁/c₂

⇒ p/6 = -3/2

⇒ p = (-3/2) × 6

⇒ p = -9

∴ The value of p is -9

Given:

(x + 6y) = 8

xy = 2

Formula used:

(a + b)³ = a³ + b³ + 3ab (a + b)

Calculation:

xy = 2

⇒ y = 2/x   .....(1)

Also, (x + 6y) = 8

⇒ x + (6 × 2/x) = 8

⇒ x + 12/x = 8   .....(2)

⇒ (x + 12/x)³ = 8³  [Taking cube of both sides]

⇒ x³ + (12/x)³ + 3 × x × (12/x) (x + 12/x) 512

⇒ x³ + (12/x)³ + 3 × 12 × 8 = 512  [∵ We get from equation (2), x + 12/x = 8]

⇒ x³ + (12/x)³ + 288 = 512

⇒ x³ + (12/x)³ = 512 - 288

⇒ x³ + [(2 × 6)/x]³ = 224

⇒ x³ + (6y)³ = 224  [∵ y = 2/x]

⇒ x³ + 216y³ = 224

∴ The value of x³ + 216y³ is 224

The cost of one dozen bananas = Rs. 5

One and a quarter dozen means 12 + 1/4 × 12 = 15

So, cost of one and a quarter dozen bananas = 15 × 5/12 = Rs. 6.25

The cost of one dozen oranges = Rs. 75

So, cost of three-fourth dozen oranges = 12 × 3/4 × 75/12 = Rs. 56.25

So, total cost = 6.25 + 56.25 = Rs. 62.50

Let the cost of 1 pencil, 1 pen and 1 eraser be x, y and z respectively.

2x + 4y + 8z = 12      ----(1)

10x + 8y + 4z = 36      ----(2)

Adding (1) and (2), 12x + 12y + 12z = 48

⇒ 3x + 3y + 3z = 48/4 = 12

∴ Cost of 3 pencils, 3 pens and 3 erasers is Rs. 12

Let the cost of a chair, a table and a fan be Rs. ‘x’, Rs. ‘y’ and Rs. ‘z’ respectively

⇒ Cost of 7 chairs, 2 tables and 5 fans = 7x + 2y + 5z = 9350        ----(1)

Also,

⇒ Cost of 3 chairs and a fan = 3x + z = 1950        ----(2)

Subtracting (2) from (1), we get,

⇒ 7x + 2y + 5z - 3x - z = 9350 - 1950

⇒ 4x + 2y + 4z = 7400

⇒ 2x + y + 2z = 3700

∴ Cost of 2 chairs, 1 table and 2 fans = Rs. 3700