# Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.11 | Set 1

**Question 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 unit and y unit.

Therefore, the area of rectangle = x * y sq. units

Given that,

Case 1:

Length is increased by 2 unit = New length is x+2 unit.

Breadth is reduced by 2 unit = New breadth is y-2 unit.

Area is reduced by 28 square units i.e. = (x * y) â€“ 28

therefore the equation becomes,

= (x+2)(yâˆ’2) = xy âˆ’ 28

= xy âˆ’ 2x + 2y â€“ 4 = xy âˆ’ 28

= âˆ’2x + 2y â€“ 4 + 28 = 0

= 2x âˆ’ 2y â€“ 24 = 0 —————(i)

Case 2:

Length is reduced by 1 unit = New length is x-1 unit.

Breadth is increased by 2 unit = New breadth is y+2 unit.

Area is increased by 33 square units i.e. = (x * y) + 33

therefore the equation becomes,

(xâˆ’1)(y+2) = xy + 33

= xy + 2x â€“ y â€“ 2 = x + 33

= 2x â€“ y âˆ’ 2 âˆ’ 33 = 0

= 2x â€“ y âˆ’35 = 0 ———————(ii)

Now by solving equation (i) and (ii) we get,

x = 46/2 = 23

and,

y = 22/2 = 11

Hence,

The length of the rectangle is 23 unit.

The breadth of the rectangle is 11 unit.

therefore, 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.

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

**Solution: **

Letâ€™s assume the length and breadth of the rectangle be x unit and y unit.

therefore, the area of rectangle = x * y sq. units

Given that,

Case 1:

Length is increased by 7 meter = New length is x+7

Breadth is decreased by 3 meter = New breadth is y-3

Area of the rectangle remains same i.e. = x * y.

therefore, the equation becomes,

xy = (x+7)(yâˆ’3)

xy = xy + 7y âˆ’ 3x âˆ’ 21

3x â€“ 7y + 21 = 0 —————-(i)

Case 2:

Length is decreased by 7 meter = New length is x-7

Breadth is increased by 5 meter = New breadth is y+5

Area of the rectangle still remains same i.e. = x * y.

therefore, the equation becomes

xy = (xâˆ’7)(y+5)

xy = xy âˆ’ 7y + 5x âˆ’ 35

5x â€“ 7y â€“ 35 = 0 —————–(ii)

Now by solving equation (i) and (ii) we get,

x = 392/14 = 28

And,

y = 210/14 = 15

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

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

**Solution: **

Letâ€™s assume the length and breadth of the rectangle be x units and y units respectively.

therefore, the area of rectangle = x * y sq. units

Given that,

Case 1:

Length is increased by 3 meter = New length is x+3

Breadth is reduced by 4 meter = New breadth is y-4

Area of the rectangle is reduced by 67 m2 = (x * y) â€“ 67.

therefore, 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 —————-(i)

Case 2:

Length is reduced by 1 meter = New length is x-1

Breadth is increased by 4 meter = New breadth is y+4

Area of the rectangle is increased by 89 m2 = (x * y) + 89.

hence, the equation becomes

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

4x â€“ y â€“ 93 = 0 —————–(ii)

Now by solving equation (i) and (ii) we get,

x = 224/8 = 28

y = 152/8 = 19

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

**Question 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 respectively.

Given that,

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, by calculating the savings, we get

Saving of X = 8x â€“ 19y = 1250

Saving of Y = 7x â€“ 16y = 1250

Hence, the equations are:

8x â€“ 19y â€“ 1250 = 0 —————-(i)

7x â€“ 16y â€“ 1250 = 0 —————(ii)

Now by solving equation (i) and (ii) we get,

x = 3750/5 = 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

**Question 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.

Given that,

Case 1:

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

According to the case 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 —————(i)

Case 2:

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

According to the case the equation becomes,

x + 10 = 3(y â€“ 10)

x + 10 = 3y â€“ 10

x â€“ 3y + 10 + 30 = 0

x â€“ 3y + 40 = 0 —————(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 â€“ (2x â€“ 6y + 80) = 0

5y â€“ 170 =0

y = 34

Now, put the value y = 34 in equation (i), and we get,

x = 62

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

**Question 6. ABCD is a cyclic quadrilateral such that âˆ A = (4y + 20)Â°, âˆ B = (3y â€“ 5)Â°, âˆ C = (+4x)Â° and âˆ D = (7x + 5)Â°. Find the four angles. [NCERT]**

**Solution: **

Given that,

âˆ A = 4y + 20,

âˆ B = 3y-5,

âˆ C = – 4x,

âˆ D = 7x+5.

As we know that opposite angles of a cyclic quadrilateral are supplementary,

âˆ A + âˆ C = 180Â°

4y+20-4x = 180

-4x+4y = 160

(-x + y) = 40 ——————-(i)

and,

âˆ B+âˆ D = 180Â°

3y-5+7x+5 = 180

7x+3y = 180 —————-(ii)

-7x+7y = 280 (multiplying eq

^{n}(i) by 7 and we get)10y = 460

y = 46

put the value of y in eq

^{n}(i) and we get,-x+46 = 40

-x = -6

= 6

Hence the value of x = 6 and y = 46.

**Question 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:**

Let’s assume 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.

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

therefore according to the condition given in question,

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

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

12/x + 12/y = 1 —————-(ii)

Now by solving equation (i) and (ii) we get,

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

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

8u + 28v = 1

12u + 12v = 1

8u + 28v â€“ 1 = 0 —————-(iii)

12u + 12v â€“ 1 = 0 —————-(iv)

Now by solving equation (iii) and (iv) 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.

**Question 8. In a **Î”**ABC, âˆ A = x**^{o}, âˆ B = (3x â€“ 2)^{o}, âˆ C = y^{o}. Also, âˆ C â€“ âˆ B = 9^{o}. Find the three angles.

^{o}, âˆ B = (3x â€“ 2)

^{o}, âˆ C = y

^{o}. Also, âˆ C â€“ âˆ B = 9

^{o}. Find the three angles.

**Solution: **

Given that,

âˆ A = x

^{o},âˆ B = (3x â€“ 2)

^{o},âˆ C = y

^{o},âˆ C â€“ âˆ B = 9

^{o}âˆ C = 9

^{âˆ˜}+ âˆ Bâˆ C = 9 + 3x âˆ’ 2

âˆ C = 7

^{o}+ 3x^{o}Substituting the value for

âˆ C = y

^{o}in above equation we get,yo = 7

^{o}+ 3x^{o}As we know that, âˆ A + âˆ B + âˆ C = 180

^{o}(Angle sum property of a triangle)= x

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

^{âˆ˜}+ 5^{âˆ˜}= 180^{âˆ˜}= 7x

^{âˆ˜}= 175^{âˆ˜}= x = 25

^{âˆ˜}Hence, calculating for the individual angles we get,

âˆ A = x

^{o}= 25^{o}âˆ B = (3x â€“ 2)

^{o}= 73^{o}âˆ C = (7 + 3x)

^{o}= 82Â°

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

**Question 9. In a cyclic quadrilateral ABCD, âˆ A = (2x + 4)**^{o}, âˆ B = (y + 3)^{o}, âˆ C = (2y + 10)^{o}, âˆ D = (4x â€“ 5)^{o}. Find the four angles.

^{o}, âˆ B = (y + 3)

^{o}, âˆ C = (2y + 10)

^{o}, âˆ D = (4x â€“ 5)

^{o}. Find the four angles.

**Solution: **

As we know that,

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

And, in the cyclic quadrilateral ABCD,

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

therefore,

âˆ A + âˆ C = 180

^{o}andâˆ B + âˆ D = 180

^{o}Substituting the values given to the above two equations, we have

For âˆ A + âˆ C = 180

^{o}= âˆ A = (2x + 4)

^{o}and âˆ C = (2y + 10)^{o}2x + 4 + 2y + 10 = 180

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

^{o}2x + 2y = 180

^{o}â€“ 14^{o}2x + 2y = 166 —————(i)

And for, âˆ B + âˆ D = 180

^{o}, we have= âˆ B = (y+3)o and âˆ D = (4x â€“ 5)o

y + 3 + 4x â€“ 5 = 180

^{o}4x + y â€“ 5 + 3 = 180

4x + y â€“ 2 = 180

4x + y = 180 + 2

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) and we get,

-6x = -198

x = âˆ’198/ âˆ’6 = 33

Now, substituting the value of x = 33o in equation (ii) and we get,

4x + y = 182

132 + y = 182

y = 182 â€“ 132 = 50

Thus, calculating the angles of a cyclic quadrilateral and 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}

**Question 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

Given that,

Case 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)

therefore, the equation will be

3x â€“ 1y = 40 —————–(i)

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

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

therefore, the equation will be

4x â€“ 2y = 50 ————–(ii)

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

From (i), we get

y = 3x â€“ 40 ————–(iii)

put this value of y in (ii) and we get,

4x â€“ 2(3x â€“ 40) = 50

4x â€“ 6x + 80 = 50

2x = 30

x = 15

Putting x = 14 in (iii) and we get,

y = 3(15) â€“ 40 = 5

therefore, x + y = 15 + 5 = 20

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

**Question 11. In a **Î”**ABC, âˆ A = x**^{o}, âˆ B = 3x^{o}, âˆ C = y^{o}. If 3y â€“ 5x = 30, prove that the triangle is right angled.

^{o}, âˆ B = 3x

^{o}, âˆ 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 that,

âˆ A = x

^{o}, âˆ B = 3x^{o}and âˆ C = y^{o}As we know that,

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

âˆ A + âˆ B + âˆ C = 180

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

4x + y = 180 —————-(i)

3y â€“ 5x = 30 ————-(ii) (Given)

To solve (i) and (ii),

Multiplying equation (i) by 3 and we get,

12x + 36y = 540 ————-(iii)

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

17x = 510

x = 510/17 = 30

Substituting the value of x = 30o in equation (i) and we get,

4x + y = 180

120 + y = 180

y = 180 â€“ 120 = 60

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

âˆ A = xo = 30

^{o}âˆ B = 3xo = 90

^{o}âˆ C = yo = 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 âˆ B = 90

^{o}.

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

**Question 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.

therefore the equations become,

x + 12y = 89 ————–(i)

x + 20y = 145 ————–(ii)

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

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

-8y = -56

y = âˆ’56 âˆ’ 8 = 7

now, substitute the value of y in equation (i) and we get,

x + 12y = 89

x + 84 = 89

x = 89 â€“ 84 = 5

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

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.

## Please

Loginto comment...