## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself.

### RS Aggarwal Solutions Class 10 Chapter 3

**MCQ****Question 1.****Solution:****(a)**

**Question 2.****Solution:****(d)**

**Question 3.****Solution:****(a)**

**Question 4.****Solution:****(d)**

**Short-Answer Questions****Question 5.****Solution:**

**Question 6.****Solution:**

**Question 7.****Solution:**

**Question 8.****Solution:**

**Question 9.****Solution:**

**Question 10.****Solution:**

Let first number = x

and second number = y

According to the conditions, x – y = 26 …(i)

and x = 3y …..(ii)

From (i),

3y – y = 26

⇒ 2y = 26

⇒ y = 13

and x = 3 x 13 = 39

Numbers are 39 and 13

**Short-Answer Questions (3 marks)****Question 11.****Solution:**

23x + 29y = 98 …..(i)

29x + 23y = 110 …..(ii)

Adding, we get 52x + 52y = 208

x + y = 4 …..(iii) (Dividing by 52)

and subtracting,

-6x + 6y = -12

x – y = 2. …..(iv) (Dividing by -6)

Adding (iii) and (iv),

2x = 6 ⇒ x = 3

Subtracting,

2x = 2 ⇒ y = 1

Hence, x = 3, y = 1

**Question 12.****Solution:**

x = 1, y = \\(\\frac { 3 }{ 2 })

**Question 13.****Solution:**

**Question 14.****Solution:**

**Question 15.****Solution:**

Let cost of one pencil = ₹ x

and cost of one pen = ₹ y

According to the condition,

5x + 7y = 195 …(i)

7x + 5y= 153 …(ii)

Adding, (i) and (ii)

12x + 12y = 348

x + y = 29 ….(iii) (Dividing by 12)

and subtracting,

-2x + 2y = 42

-x + y = 21 …..(iv) (Dividing by -2)

Now, Adding (iii) and (iv),

2y = 50 ⇒ y = 25

and from (iv),

-x + 25 = 21 ⇒ -x = 21 – 25 = -4

x = 4

Cost of one pencil = ₹ 4

and cost of one pen = ₹ 25

**Question 16.****Solution:**

2x – 3y = 1, 4x – 3y + 1 = 0

2x – 3y = 1

2x = 1 + 3y

x = \\(\\frac { 1 + 3y }{ 2 })

Giving some different values to y, we get corresponding values of x as given below

Now plot the points (2, 1), (5, 3) and (-1, -1) on the graph and join them to get a line.

Similarly,

4x – 3y + 1 = 0

⇒ 4x = 3y – 1

⇒ x = \\(\\frac { 3y – 1 }{ 4 })

Now plot the points (-1, -1), (-4, -5) and (2, 3) on the graph and join them to get another line which intersects the first line at the point (-1, -1).

Hence, x = -1, y = -1

**Long-Answer Questions****Question 17.****Solution:**

We know that opposite angles of a cyclic quadrilateral are supplementary.

∠A + ∠C = 180° and ∠B + ∠D = 180°

Now, ∠A = 4x° + 20°, ∠B = 3x° – 5°, ∠C = 4y° and ∠D = 7y° + 5°

But ∠A + ∠C = 180°

4x + 20° + 4y° = 180°

⇒ 4x + 4y = 180° – 20 = 160°

x + y = 40° …(i) (Dividing by 4)

and ∠B + ∠D = 180°

⇒ 3x – 5 + 7y + 5 = 180°

⇒ 3x + 7y = 180° …(ii)

From (i), x = 40° – y

Substituting the value of x in (ii),

3(40° – y) + 7y = 180°

⇒ 120° – 3y + 7y = 180°

⇒ 4y = 180°- 120° = 60°

y = 15°

and x = 40° – y = 40° – 15° = 25°

∠A = 4x + 20 = 4 x 25 + 20 = 100 + 20= 120°

∠B = 3x – 5 = 3 x 25 – 5 = 75 – 5 = 70°

∠C = 4y = 4 x 15 = 60°

∠D = 7y + 5 = 7 x 15 + 5 = 105 + 5 = 110°

**Question 18.****Solution:**

**Question 19.****Solution:**

Let numerator of a fraction = x

and denominator = y

Fraction = \\(\\frac { x }{ y })

According to the conditions,

**Question 20.****Solution:**

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself are helpful to complete your math homework.

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