Miscellaneous applications of linear equations are the key focus in this exercise. The RD Sharma Solutions Class 10 developed by experts at BYJUâ€™S is a prime weapon 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 Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.11 Download PDF

### Access RD Sharma Solutions for Class 10 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:**

Letâ€™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 â‡’ now, the new length is x+2 units

Breadth is reduced by 2 units â‡’ now, the new breadth is y-2 units

And itâ€™s given that the area is reduced by 28 square units i.e. = xy – 28

So, the equation becomes

(x+2)(yâˆ’2) = xy âˆ’ 28

â‡’ xy âˆ’ 2x + 2y â€“ 4 = xy âˆ’ 28

â‡’ âˆ’2x + 2y â€“ 4 + 28 = 0

â‡’ âˆ’2x + 2y + 24 = 0

â‡’ 2x âˆ’ 2y â€“ 24 = 0 â€¦â€¦â€¦ (i)

Case 2:

Length is reduced by 1 unit â‡’ now, the new length is x-1 units

Breadth is increased by 2 units â‡’ now, the new breadth is y+2 units

And, itâ€™s given that the area is increased by 33 square units i.e. = i.e. = xy + 33

So, the equation becomes

(xâˆ’1)(y+2) = xy + 33

â‡’ xy + 2x â€“ y â€“ 2 = x + 33

â‡’ 2x â€“ y âˆ’ 2 âˆ’ 33 = 0

â‡’ 2x â€“ y âˆ’35 = 0 â€¦â€¦â€¦.. (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,

= xÃ—y

= 23 x 11

= 253 sq. units

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

Â

**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:**

Letâ€™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 â‡’ now, the new length is x+7

Breadth is decreased by 3 metres â‡’ now, the new breadth is y-3

And itâ€™s given, the area of the rectangle remains same i.e. = xy.

So, the equation becomes

xy = (x+7)(yâˆ’3)

xy = xy + 7y âˆ’ 3x âˆ’ 21

3x â€“ 7y + 21 = 0 â€¦â€¦â€¦. (i)

Case 2:

Length is decreased by 7 metres â‡’ now, the new length is x-7

Breadth is increased by 5 metres â‡’ now, the new breadth is y+5

And itâ€™s given that, the area of the rectangle still remains same i.e. = xy.

So, the equation becomes

xy = (xâˆ’7)(y+5)

xy = xy âˆ’ 7y + 5x âˆ’ 35

5x â€“ 7y â€“ 35 = 0 â€¦â€¦â€¦. (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.

Â

**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:**

Letâ€™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 â‡’ now, the new length is x+3

Breadth is reduced by 4 metres â‡’ now, the new breadth is y-4

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

So, the equation becomes

xy â€“ 67 = (x + 3)(y â€“ 4)

xy â€“ 67 = xy + 3y â€“ 4x â€“ 12

4xy â€“ 3y â€“ 67 + 12 = 0

4x â€“ 3y â€“ 55 = 0 â€”â€” (i)

Case 2:

Length is reduced by 1 m â‡’ now, the new length is x-1

Breadth is increased by 4 metre â‡’ now, the new breadth is y+4

And itâ€™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 â€“ y â€“ 93 = 0 â€”â€” (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.

Â

**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 â‚¹ 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 â‚¹ 8x and the expenditure of X is 19y.

The income of Y is â‚¹ 7x and the expenditure of Y is 16y.

So, on calculating the savings, we get

Saving of X = 8x â€“ 19y = 1250

Saving of Y = 7x â€“ 16y = 1250

Hence, the system of equations formed are

8x â€“ 19y â€“ 1250 = 0 â€”â€“ (i)

7x â€“ 16y â€“ 1250 = 0 â€”â€“ (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 â‚¹ 6000 and the income of Y is â‚¹ 5250

Â

**5. A and B each has some money. If A gives â‚¹ 30 to B, then B will have twice the money left with A. But, if B gives â‚¹ 10 to A, then A will have thrice as much as is left with B. How much money does each have?**

**Solution:**

Letâ€™s assume the money with A be â‚¹ x and the money with B be â‚¹ y.

Then, from the question we have the following cases

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

So, the equation becomes

y + 30 = 2(x â€“ 30)

y + 30 = 2x â€“ 60

2x â€“ y â€“ 60 â€“ 30 = 0

2x â€“ y â€“ 90 = 0 â€”â€” (i)

Case 2: If B gives â‚¹ 10 to A, then A will have thrice as much as is left with B.

x + 10 = 3(y â€“ 10)

x + 10 = 3y â€“ 10

x â€“ 3y + 10 + 30 = 0

x â€“ 3y + 40 = 0 â€”â€” (ii)

Solving (i) and (ii),

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

2x â€“ 6y + 80 = 0

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

2x â€“ y â€“ 90 – (2x â€“ 6y + 80) = 0

5y â€“ 170 =0

y = 34

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

x = 62

Hence, the money with A is â‚¹ 62 and the money with BÂ be â‚¹ 34

**Â **

**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 + 2/8y = 1 â€”â€”â€“(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 â€”â€”â€“(ii)

Hence, solving (i) and (ii) â‡’

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

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

8u + 28v = 1

12u + 12v = 1

8u + 28v â€“ 1 = 0 â€”â€” (iii)

12u + 12v â€“ 1 = 0 â€”â€” (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.

Â

**8. In aÂ **Î”**ABC, **âˆ **A = x ^{o}, **âˆ

**B = (3x â€“ 2)**âˆ

^{o},**C = y**âˆ

^{o}. Also,**C â€“**âˆ

**B = 9**

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

Itâ€™s given that,

âˆ A = x^{o},

âˆ B = (3x â€“ 2)^{o},

âˆ C = y^{o}

Also given that,

âˆ C â€“ âˆ B = 9^{o}

â‡’ âˆ C = 9^{âˆ˜} + âˆ B

â‡’ âˆ C = 9^{âˆ˜} + 3x^{âˆ˜} âˆ’ 2^{âˆ˜}

â‡’ âˆ C = 7^{âˆ˜} + 3x^{âˆ˜}

Substituting the value for

âˆ C = y^{o}Â in above equation we get,

y^{o}Â = 7^{o}Â + 3x^{o}

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

â‡’ x^{âˆ˜} + (3x^{âˆ˜ }âˆ’ 2^{âˆ˜}) + (7^{âˆ˜} + 3x^{âˆ˜}) = 180^{âˆ˜}

â‡’ 7x^{âˆ˜} + 5^{âˆ˜} = 180^{âˆ˜}

â‡’ 7x^{âˆ˜} = 175^{âˆ˜}

â‡’ x^{âˆ˜} = 25^{âˆ˜}

Hence, calculating for the individual angles we get,

âˆ A = x^{o}Â = 25^{o}

âˆ B = (3x â€“ 2)^{o}Â = 73^{o}

âˆ C = (7 + 3x)^{o}Â = 82^{o}

Therefore,

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

Â

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

**B = (y + 3)**âˆ

^{o},**C = (2y + 10)**âˆ

^{o},**D = (4x â€“ 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 âˆ A and âˆ C & angles âˆ B and âˆ D are the pairs of opposite angles.

So,

âˆ A + âˆ C = 180^{o}Â and

âˆ B + âˆ D = 180^{o}

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

For âˆ A + âˆ C = 180^{o}Â

â‡’ âˆ A = (2x + 4)^{o}Â and âˆ C = (2y + 10)^{o}

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

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

2x + 2y = 180^{oÂ }â€“ 14^{o}

2x + 2y = 166 â€”â€” (i)

And for, âˆ B + âˆ D = 180^{o}, we have

â‡’ âˆ B = (y+3)^{o}Â and âˆ D = (4x â€“ 5)^{o}

y + 3 + 4x â€“ 5 = 180^{o}

4x + y â€“ 5 + 3 = 180^{o}

4x + y â€“ 2 = 180^{o}

4x + y = 180^{o}Â + 2^{o}

4x + y = 182^{o}Â â€”â€”- (ii)

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

Multiplying equation (ii) by 2 to get,

8x + 2y = 364 â€”â€” (iii)

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

-6x = -198

x = âˆ’198/ âˆ’6

â‡’ x = 33^{o}

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

4x + y = 182

132 + y = 182

y = 182 â€“ 132

â‡’ y = 50

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

âˆ A = 2x + 4

= 66 + 4

= 70^{o}

âˆ B = y + 3

= 50 + 3

= 53^{o}

âˆ C = 2y + 10

= 100 + 10

= 110^{o}

âˆ D = 4x â€“ 5

= 132 â€“ 5

= 127^{o}

Therefore, the angles of the cyclic quadrilateral ABCD are

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

**Â **

**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: **

Letâ€™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 â€¦.. (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 â€¦â€¦ (ii)

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

From (i), we get

y = 3x – 40 â€¦â€¦.. (iii)

Using (iii) in (ii) we get,

4x â€“ 2(3x â€“ 40) = 50

4x â€“ 6x + 80 = 50

2x = 30

x = 15

Putting x = 14 in (iii) we get,

y = 3(15) â€“ 40

y = 5

So, x + y = 15 + 5 = 20

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

**11. In aÂ **Î”**ABC, **âˆ **A = x ^{o}, **âˆ

**B = 3x**âˆ

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

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

We need to prove that Î”ABC is right angled.

Given:

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

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

i.e., âˆ A + âˆ B + âˆ C = 180^{o}

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

4x + y = 180 â€”â€” (i)

From question itâ€™s given that, 3y â€“ 5x = 30 â€”â€“ (ii)

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

Multiplying equation (i) by 3 to get,

12x + 36y = 540 â€”â€“ (iii)

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

17x = 510

x = 510/17

â‡’ x = 30^{o}

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

4x + y = 180

120 + y = 180

y = 180 â€“ 120

â‡’ y = 60^{o}

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

âˆ A = x^{o} = 30^{o}

âˆ B = 3x^{o} = 90^{o}

âˆ C = y^{o} = 60^{o}

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

And here we have,

âˆ B = 90^{o}.

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

Â

**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 â‚¹ 89 and for a journey of 20 km, the charge paid is â‚¹ 145. What will a person have to pay for travelling a distance of 30 km?**

**Solution:**

Let the fixed charge of the car be â‚¹ x and,

Let the variable charges of the car be â‚¹ y per km.

So according to the question, we get 2 equations

x + 12y = 89 â€”â€” (i) and,

x + 20y = 145 â€”â€” (ii)

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

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

-8y = -56

y = âˆ’56 âˆ’ 8

â‡’ y = 7

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

x + 12y = 89

x + 84 = 89

x = 89 â€“ 84

â‡’ x = 5

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

â‡’ x + 30y = 5 + 210 = â‚¹ 215

Therefore, a person has to pay â‚¹ 215 for travelling a distance of 30 km by the car.