Test: Linear Equations in Two Variables - 1 - Class 10 MCQ

# Test: Linear Equations in Two Variables - 1 - Class 10 MCQ

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## 10 Questions MCQ Test - Test: Linear Equations in Two Variables - 1

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Test: Linear Equations in Two Variables - 1 - Question 1

### 3 chairs and 2 tables cost Rs. 700 and 5 chairs and 3 tables cost Rs. 1100. What is the cost of 1 chair and 2 tables?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 1

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

Test: Linear Equations in Two Variables - 1 - Question 2

### If (a + b)2 – 2(a + b) = 80 and ab = 16, then what can be the value of 3a – 19b?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 2

(a + b)– 2(a + b) = 80

⇒ (a + b)– 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

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Test: Linear Equations in Two Variables - 1 - Question 3

### The ratio of the number of blue and red balls in a bag is constant. When there were 68 red balls, the number of blue balls was 36. If the number of blue balls was 63, how many red balls should be there in the bag?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 3

⇒ Here, k1/k2 = k3/k4, 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.

Test: Linear Equations in Two Variables - 1 - Question 4

If a + b + c = 9, ab + bc + ca = 26, a3 + b3 = 91, b3 + c3 = 72 and c3 + a3 = 35, then what is the value of abc?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 4

Using algebraic identities,

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

By putting the respective values given in question,

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

⇒ (9)2 = a+ b2 + c2 + 2(26)

⇒ a+ 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) (a+ 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

Test: Linear Equations in Two Variables - 1 - Question 5

If x + y + 3 = 0, then find the value of x3 + y- 9xy + 9.

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 5

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

⇒ x+ 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)

Test: Linear Equations in Two Variables - 1 - Question 6

Find the value of p for which the system of linear equations: px - 3y = 5 and 6x + 2y = 12 has no solutions?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 6

The given system of equations is of the form:

⇒ a1x + b1y + c1 = 0

⇒ a2x + b2y + c2 = 0

Here, a1 = p; b1 = -3; c= -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

Test: Linear Equations in Two Variables - 1 - Question 7

If (x + 6y) = 8, and xy = 2, where x > 0, what is the value of (x3 + 216y3)?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 7

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

Test: Linear Equations in Two Variables - 1 - Question 8

The cost of one dozen bananas is Rs. 5. The cost of one dozen oranges is Rs. 75. What will the cost of one and a quarter dozen bananas and three-fourth dozen oranges?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 8

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

Test: Linear Equations in Two Variables - 1 - Question 9

The cost of 2 pencil, 4 pens and 8 erasers is Rs. 12 and the cost of 8 pens, 10 pencils and 4 erasers is Rs. 36. How much will 3 pencils, 3 pens and 3 erasers cost?

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 9

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

Test: Linear Equations in Two Variables - 1 - Question 10

The cost of 7 chairs, 2 tables and 5 fans is Rs. 9350. If the cost of 3 chairs and a fan is Rs. 1950, find the cost of 2 chairs, 1 table and 2 fans.

Detailed Solution for Test: Linear Equations in Two Variables - 1 - Question 10

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

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