Miscellaneous applications of linear equations are the key focus of this exercise. The RD Sharma Solutions Class 10 developed by experts at BYJU’S is a prime resource for students to clarify their doubts and help them analyse their weaker areas. Further, students can also utilise the** RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.11** PDF given below.

## RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.11 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.11

**1. If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.**

**Solution:**

Let’s assume the length and breadth of the rectangle be x units and y units, respectively.

Hence, the area of rectangle = xy sq. units

From the question we have the following cases,

Case 1:

Length is increased by 2 units ⇒ now, the new length is x+2 units

Breadth is reduced by 2 units ⇒ now, the new breadth is y-2 units

And it’s given that the area is reduced by 28 square units i.e., = xy – 28

So, the equation becomes

(x+2)(y−2) = xy − 28

⇒ xy − 2x + 2y – 4 = xy − 28

⇒ −2x + 2y – 4 + 28 = 0

⇒ −2x + 2y + 24 = 0

⇒ 2x − 2y – 24 = 0 ……… (i)

Case 2:

Length is reduced by 1 unit ⇒ now, the new length is x-1 units

Breadth is increased by 2 units ⇒ now, the new breadth is y+2 units

And, it’s given that the area is increased by 33 square units i.e. = i.e. = xy + 33

So, the equation becomes

(x−1)(y+2) = xy + 33

⇒ xy + 2x – y – 2 = x + 33

⇒ 2x – y − 2 − 33 = 0

⇒ 2x – y −35 = 0 ……….. (ii)

Solving (i) and (ii),

By using cross multiplication, we get

x = 46/2

x = 23

And,

y = 22/2

y = 11

Hence,

The length of the rectangle is 23 units.

The breadth of the rectangle is 11 units.

So, the area of the actual rectangle = length x breadth,

= x×y

= 23 x 11

= 253 sq. units

Therefore, the area of rectangle is 253 sq. units.

**2. The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and the breadth is increased by 5 metres. Find the dimensions of the rectangle.**

**Solution:**

Let’s assume the length and breadth of the rectangle be x units and y units, respectively.

Hence, the area of rectangle = xy sq.units

From the question we have the following cases,

Case 1

Length is increased by 7 metres ⇒ now, the new length is x+7

Breadth is decreased by 3 metres ⇒ now, the new breadth is y-3

And it’s given, the area of the rectangle remains the same i.e. = xy.

So, the equation becomes

xy = (x+7)(y−3)

xy = xy + 7y − 3x − 21

3x – 7y + 21 = 0 ………. (i)

Case 2:

Length is decreased by 7 metres ⇒ now, the new length is x-7

Breadth is increased by 5 metres ⇒ now, the new breadth is y+5

And it’s given that, the area of the rectangle still remains the same i.e. = xy.

So, the equation becomes

xy = (x−7)(y+5)

xy = xy − 7y + 5x − 35

5x – 7y – 35 = 0 ………. (ii)

Solving (i) and (ii),

By using cross-multiplication, we get,

x = 392/14

x = 28

And,

y = 210/14

y = 15

Therefore, the length of the rectangle is 28 m. and the breadth of the actual rectangle is 15 m.

**3. In a rectangle, if the length is increased by 3 metres and breadth is decreased by 4 metres, the area of the triangle is reduced by 67 square metres. If length is reduced by 1 metre and breadth is increased by 4 metres, the area is increased by 89 sq. metres. Find the dimension of the rectangle.**

**Solution:**

Let’s assume the length and breadth of the rectangle be x units and y units, respectively.

Hence, the area of rectangle = xy sq.units

From the question we have the following cases,

According to the question,

Case 1:

Length is increased by 3 metres ⇒ now, the new length is x+3

Breadth is reduced by 4 metres ⇒ now, the new breadth is y-4

And it’s given, the area of the rectangle is reduced by 67 m^{2} = xy – 67.

So, the equation becomes

xy – 67 = (x + 3)(y – 4)

xy – 67 = xy + 3y – 4x – 12

4xy – 3y – 67 + 12 = 0

4x – 3y – 55 = 0 —— (i)

Case 2:

Length is reduced by 1 m ⇒ now, the new length is x-1

Breadth is increased by 4 metre ⇒ now, the new breadth is y+4

And it’s given, the area of the rectangle is increased by 89 m^{2} = xy + 89.

Then, the equation becomes

xy + 89 = (x -1)(y + 4)

4x – y – 93 = 0 —— (ii)

Solving (i) and (ii),

Using cross multiplication, we get

x = 224/8

x = 28

And,

y = 152/8

y = 19

Therefore, the length of rectangle is 28 m and the breadth of rectangle is 19 m.

**4. The income of X and Y are in the ratio of 8: 7 and their expenditures are in the ratio 19: 16. If each saves ₹ 1250, find their incomes.**

**Solution:**

Let the income be denoted by x and the expenditure be denoted by y.

Then, from the question we have

The income of X is ₹ 8x and the expenditure of X is 19y.

The income of Y is ₹ 7x and the expenditure of Y is 16y.

So, on calculating the savings, we get

Saving of X = 8x – 19y = 1250

Saving of Y = 7x – 16y = 1250

Hence, the system of equations formed are

8x – 19y – 1250 = 0 —– (i)

7x – 16y – 1250 = 0 —– (ii)

Using cross-multiplication method, we have

x = 3750/5

x = 750

If, x = 750, then

The income of X = 8x

= 8 x 750

= 6000

The income of Y = 7x

= 7 x 750

= 5250

Therefore, the income of X is ₹ 6000 and the income of Y is ₹ 5250

**5. A and B each has some money. If A gives ₹ 30 to B, then B will have twice the money left with A. But, if B gives ₹ 10 to A, then A will have thrice as much as is left with B. How much money does each have?**

**Solution:**

Let’s assume the money with A be ₹ x and the money with B be ₹ y.

Then, from the question we have the following cases

Case 1: If A gives ₹ 30 to B, then B will have twice the money left with A.

So, the equation becomes

y + 30 = 2(x – 30)

y + 30 = 2x – 60

2x – y – 60 – 30 = 0

2x – y – 90 = 0 —— (i)

Case 2: If B gives ₹ 10 to A, then A will have thrice as much as is left with B.

x + 10 = 3(y – 10)

x + 10 = 3y – 10

x – 3y + 10 + 30 = 0

x – 3y + 40 = 0 —— (ii)

Solving (i) and (ii),

On multiplying equation (ii) with 2, we get,

2x – 6y + 80 = 0

Subtract equation (ii) from (i), we get,

2x – y – 90 – (2x – 6y + 80) = 0

5y – 170 =0

y = 34

Now, on using y = 34 in equation (i), we find,

x = 62

Hence, the money with A is ₹ 62 and the money with B be ₹ 34

** **

**7. 2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it?**

**Solution:**

Assuming that the time required for a man alone to finish the work be x days and also the time required for a boy alone to finish the work be y days.

Then, we know

The work done by a man in one day = 1/x

The work done by a boy in one day = 1/y

Similarly,

The work done by 2 men in one day = 2/x

The work done by 7 boys in one day = 7/y

So, the condition given in the question states that,

2 men and 7 boys together can finish the work in 4 days

4(2/x + 7/y) = 1

8/x + 28/y = 1 ——–(i)

And, the second condition from the question states that,

4 men and 4 boys can finish the work in 3 days

For this, the equation so formed is

3(4/x + 4/y) = 1

12/x + 12/y = 1 ——–(ii)

Hence, solving (i) and (ii) ⇒

Taking, 1/x = u and 1/y = v

So, the equations (i) and (ii) becomes,

8u + 28v = 1

12u + 12v = 1

8u + 28v – 1 = 0 —— (iii)

12u + 12v – 1 = 0 —— (iv)

By using cross multiplication, we get,

u = 1/15

1/x = 1/15

x = 15

And,

v = 1/60

1/y = 1/60

y = 60

Therefore,

The time required for a man alone to finish the work is 15 days and the time required for a boy alone to finish the work is 60 days.

**8. In a **Δ**ABC, **∠**A = x ^{o}, **∠

**B = (3x – 2)**∠

^{o},**C = y**∠

^{o}. Also,**C –**∠

**B = 9**

^{o}. Find the three angles.**Solution:**

It’s given that,

∠A = x^{o},

∠B = (3x – 2)^{o},

∠C = y^{o}

Also given that,

∠C – ∠B = 9^{o}

⇒ ∠C = 9^{∘} + ∠B

⇒ ∠C = 9^{∘} + 3x^{∘} − 2^{∘}

⇒ ∠C = 7^{∘} + 3x^{∘}

Substituting the value for

∠C = y^{o} in above equation we get,

y^{o} = 7^{o} + 3x^{o}

We know that, ∠A + ∠B + ∠C = 180^{o }(Angle sum property of a triangle)

⇒ x^{∘} + (3x^{∘ }− 2^{∘}) + (7^{∘} + 3x^{∘}) = 180^{∘}

⇒ 7x^{∘} + 5^{∘} = 180^{∘}

⇒ 7x^{∘} = 175^{∘}

⇒ x^{∘} = 25^{∘}

Hence, calculating for the individual angles we get,

∠A = x^{o} = 25^{o}

∠B = (3x – 2)^{o} = 73^{o}

∠C = (7 + 3x)^{o} = 82^{o}

Therefore,

∠A = 25^{o}, ∠B = 73^{o} and ∠C = 82^{o }.

**9. In a cyclic quadrilateral ABCD, **∠**A = (2x + 4) ^{o}, **∠

**B = (y + 3)**∠

^{o},**C = (2y + 10)**∠

^{o},**D = (4x – 5)**

^{o}. Find the four angles.**Solution:**

We know that,

The sum of the opposite angles of cyclic quadrilateral should be 180^{o}.

And, in the cyclic quadrilateral ABCD,

Angles ∠A and ∠C & angles ∠B and ∠D are the pairs of opposite angles.

So,

∠A + ∠C = 180^{o} and

∠B + ∠D = 180^{o}

Substituting the values given to the above two equations, we have

For ∠A + ∠C = 180^{o}

⇒ ∠A = (2x + 4)^{o} and ∠C = (2y + 10)^{o}

2x + 4 + 2y + 10 = 180^{o}

2x + 2y + 14 = 180^{o}

2x + 2y = 180^{o }– 14^{o}

2x + 2y = 166 —— (i)

And for, ∠B + ∠D = 180^{o}, we have

⇒ ∠B = (y+3)^{o} and ∠D = (4x – 5)^{o}

y + 3 + 4x – 5 = 180^{o}

4x + y – 5 + 3 = 180^{o}

4x + y – 2 = 180^{o}

4x + y = 180^{o} + 2^{o}

4x + y = 182^{o} ——- (ii)

Now for solving (i) and (ii), we perform

Multiplying equation (ii) by 2 to get,

8x + 2y = 364 —— (iii)

And now, subtract equation (iii) from (i) to get

-6x = -198

x = −198/ −6

⇒ x = 33^{o}

Now, substituting the value of x = 33^{o} in equation (ii) to find y

4x + y = 182

132 + y = 182

y = 182 – 132

⇒ y = 50

Thus, calculating the angles of a cyclic quadrilateral we get:

∠A = 2x + 4

= 66 + 4

= 70^{o}

∠B = y + 3

= 50 + 3

= 53^{o}

∠C = 2y + 10

= 100 + 10

= 110^{o}

∠D = 4x – 5

= 132 – 5

= 127^{o}

Therefore, the angles of the cyclic quadrilateral ABCD are

∠A = 70^{o}, ∠B = 53^{o}, ∠C = 110^{o} and ∠D = 127^{o}

** **

**10. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test? **

**Solution: **

Let’s assume that the total number of correct answers be x and the total number of incorrect answers be y.

Hence, their sum will give the total number of questions in the test i.e. x + y

Further from the question, we have two type of marking scheme:

1) When 3 marks is awarded for every right answer and 1 mark deducted for every wrong answer.

According to this type, the total marks scored by Yash is 40. (Given)

So, the equation formed will be

3x – 1y = 40 ….. (i)

Next,

2) When 4 marks is awarded for every right answer and 2 marks deducted for every wrong answer.

According to this type, the total marks scored by Yash is 50. (Given)

So, the equation formed will be

4x – 2y = 50 …… (ii)

Thus, by solving (i) and (ii) we obtained the values of x and y.

From (i), we get

y = 3x – 40 …….. (iii)

Using (iii) in (ii) we get,

4x – 2(3x – 40) = 50

4x – 6x + 80 = 50

2x = 30

x = 15

Putting x = 14 in (iii) we get,

y = 3(15) – 40

y = 5

So, x + y = 15 + 5 = 20

Therefore, the number of questions in the test were 20.

**11. In a **Δ**ABC, **∠**A = x ^{o}, **∠

**B = 3x**∠

^{o},**C = y**

^{o}. If 3y – 5x = 30, prove that the triangle is right-angled.**Solution:**

We need to prove that ΔABC is right-angled.

Given:

∠A = x^{o}, ∠B = 3x^{o} and ∠C = y^{o}

Sum of the three angles in a triangle is 180^{o }(Angle sum property of a triangle)

i.e., ∠A + ∠B + ∠C = 180^{o}

x + 3x + y = 180^{o}

4x + y = 180 —— (i)

From question it’s given that, 3y – 5x = 30 —– (ii)

To solve (i) and (ii), we perform

Multiplying equation (i) by 3 to get,

12x + 36y = 540 —– (iii)

Now, subtracting equation (ii) from equation (iii) we get

17x = 510

x = 510/17

⇒ x = 30^{o}

Substituting the value of x = 30^{o} in equation (i) to find y

4x + y = 180

120 + y = 180

y = 180 – 120

⇒ y = 60^{o}

Thus the angles ∠A, ∠B and ∠C are calculated to be

∠A = x^{o} = 30^{o}

∠B = 3x^{o} = 90^{o}

∠C = y^{o} = 60^{o}

A right angled triangle is a triangle with any one side right angled to other, i.e., 90^{o} to other.

And here we have,

∠B = 90^{o}.

Therefore, the triangle ABC is right angled. Hence proved.

**12. The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is ₹ 89 and for a journey of 20 km, the charge paid is ₹ 145. What will a person have to pay for travelling a distance of 30 km?**

**Solution:**

Let the fixed charge of the car be ₹ x and,

Let the variable charges of the car be ₹ y per km.

So according to the question, we get 2 equations

x + 12y = 89 —— (i) and,

x + 20y = 145 —— (ii)

Now, by solving (i) and (ii) we can find the charges.

On subtraction of (i) from (ii), we get,

-8y = -56

y = −56 − 8

⇒ y = 7

So, substituting the value of y = 7 in equation (i) we get

x + 12y = 89

x + 84 = 89

x = 89 – 84

⇒ x = 5

Thus, the total charges for travelling a distance of 30 km can be calculated as: x + 30y

⇒ x + 30y = 5 + 210 = ₹ 215

Therefore, a person has to pay ₹ 215 for travelling a distance of 30 km by the car.