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Test: Linear Equations in Two Variables - 2 - Class 10 MCQ


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15 Questions MCQ Test The Complete SAT Course - Test: Linear Equations in Two Variables - 2

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

If 8a2b = 27, ab2 = 216, find the value of ab.

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

8a2b = 27, ab2 = 216

Multiplying these two equations, we get,

8a3b3    = 27*216

a3b3    = 27*27

ab = 9

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

If a2 + b2 = 4b + 6a – 13, then what is the value of a + b?

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

(b2 – 4b + 4) + (a2 – 6a + 9) = 0

(b – 2)2 + (a – 3)2 = 0

a = 3, b = 2

a + b = 5

Test: Linear Equations in Two Variables - 2 - Question 3

If the equations 14x + 8y + 5 = 0 and 21x - ky - 7 = 0 have no solution, then the value of k is:

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 3

⇒ 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.

Test: Linear Equations in Two Variables - 2 - Question 4

Cost of 8 pencils, 5 pens and 3 erasers is Rs. 111. Cost of 9 pencils, 6 pens and 5 erasers is Rs. 130. Cost of 16 pencils, 11 pens and 3 erasers is Rs. 221. What is the cost (in Rs) of 39 pencils, 26 pens and 13 erasers?

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 4

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) 

Test: Linear Equations in Two Variables - 2 - Question 5

If 8k6 + 15k3 – 2 = 0, then the positive value ofis:

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 5

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, k= 1/8

⇒ k = 1/2 ⇒ 1/k = 2

Then, (k + 1/k) = (1/2 + 2) = 5/2 = 

∴ The value of (k + 1/k) is 

Test: Linear Equations in Two Variables - 2 - Question 6

X attempts 100 questions and gets 340 marks. If for every correct answer is 4 marks and wrong answer is negative one mark, then the number of questions wrongly answered by Mr. X is:

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 6

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

Test: Linear Equations in Two Variables - 2 - Question 7

A piece of cloth costs Rs. 35. If the piece were 4 m longer and each meter was to cost Rs. 1 lesser, then the total cost would remain unchanged. How long is the piece of cloth?

A. 10 m

B. 14 m

C. 12 m

D. 8 m

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 7

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

Test: Linear Equations in Two Variables - 2 - Question 8

The denominator of a fraction is 2 more than its numerator. If 3 is added to the numerator and denominator, the fraction becomes 3/4. What was the original fraction?

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 8

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

Test: Linear Equations in Two Variables - 2 - Question 9

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 - 2 - Question 9

Assume a + b = x

x2 – 2x – 80 = 0

x = 10

ab = 16

a = 8, b = 2

3 × a – 19b = −14

Test: Linear Equations in Two Variables - 2 - Question 10

The sum of two positive numbers is 14 and difference between their squares is 56. What is the sum of their squares?

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 10

x +  y = 14      … (i)

x2 – y2 = 56

→ x – y = 4     … (ii)

x = 9, y = 5

sum of their squares = 81 + 25 = 106

Test: Linear Equations in Two Variables - 2 - Question 11

Three cups of ice cream, two burgers and four soft drinks together cost Rs. 128. Two cups of ice cream, one burger and two soft drinks together cost Rs. 74. What is the cost of five burgers and ten soft drinks?

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 11

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

Test: Linear Equations in Two Variables - 2 - Question 12

The difference between two numbers is 5. If 25 is subtracted from the smaller number and 20 is added to the greater number the ratio becomes 1 : 2. What is the greater number?

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

Test: Linear Equations in Two Variables - 2 - Question 13

If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions then k =?

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 13

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

Test: Linear Equations in Two Variables - 2 - Question 14

The sum of two numbers is 184. If one-third of one exceeds one-seventh of the other by 8,  find the smaller number

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 14

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.

Test: Linear Equations in Two Variables - 2 - Question 15

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

Detailed Solution for Test: Linear Equations in Two Variables - 2 - Question 15

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

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