## Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.11

**Exercise 3.11**

**Solution**

Let the length and breadth of the rectangle be x and y units respectively

Then, area of rectangle = xy square units

If length is increased and breadth reduced each by 2 units, then the area is reduced by 28 square units

(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

Therefore, 2x – 2y – 24 = 0 …(i)

Then the length is reduced by 1 unit and breadth is increased by 2 units then the area is increased by 33 square units

(x-1) (y+2) = xy + 33

⇒ xy + 2x – y – 2 = xy + 33

⇒ 2x – y – 2 - 33 = 0

⇒ 2x – y – 35 = 0

Therefore , 2x – y – 35 = 0 …(ii)

Thus we get the following , we have

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

**Solution**

Then, area of rectangle = xy square units

If length is increased by 7 meters and breadth is decreased by 3 meters when the area of a rectangle remains the same

Therefore,

xy = (x+7)(y-3)

xy = xy + 7y – 3x – 21

xy = xy + 7y – 3x – 21

3x – 7y + 21 = 0 …(i)

If the length is decreased by 7 meters and breadth is increased by 5 meters when the area remains unaffected, then

xy = (x-7)(y+5)

xy = xy - 7y + 5x – 35

xy = xy – 7y + 5x – 35

0 = 5x – 7y – 35 …(ii)

Thus we get the following system of linear equaton

3x – 7y + 21 = 0

5x – 7y – 35 = 0

By using cross – multiplication, we have

3. In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 meters, the area of the rectangle is reduced by 67 square meters. If length is reduced by 1 meter and breadth is increased by 4 meters, the area is increased by 89 Sq. meters. Find the dimensions of the rectangle.

**Solution**

Let the length and breadth of the rectangle be x and y units respectively

Then, area of rectangle = xy square units

If the length is increased by 3 meters and the breath is reduced each by 4 square meters the area is reduced by 67 square units

Therefore ,

xy – 67 = (x+3)(y-4)

xy – 67 = xy + 3y – 4x – 12

xy – 67 = xy + 3y – 4x – 12

4x – 3y – 67 + 12 = 0

4x – 3y – 55 = 0 …(i)

Then the length is reduced by 1 meter and breadth is increased by 4 meter then the area is increased by 89 square units

Therefore, 0 = 4x – y – 93 …(ii)

Thus, we get the following system of linear equation

4x - 3y - 55 = 0

4x – 3y – 93 = 0

By using cross multiplication we have

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

**Solution**

Let the income of x be Rs 8x and the income of y be Rs 7x . Further let the expenditure of x be

19y and the expenditure of y be 16y respectively then,

Saving of x = 8x - 19y

Saving of y = 7x – 16y

8x – 19y = 1250

7x – 16y = 1250

8x – 19y – 1250 = 0 …(i)

7x – 16y – 1250 = 0 …(ii)

Solving equation (i) and (ii) by cross – multiplication , we have

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

**Solution**

Let the money with A be Rs x and the money with B be Rs y.

If A gives Rs 30 to B, Then B will have twice the money left with A, According to the condition we have,

y+30 = 2(x-30)

y+30 = 2x-60

0 = 2x – y – 60 – 30

0 = 2x – y – 90 …(i)

If B gives Rs 10 to A, then a will have thrice as much as is left with B,

x + 10 = 3(y-10)

x + 10 = 3y – 30

x – 3y + 10 + 30 = 0

x – 3y + 40 = 0 …(ii)

By multiplying equation (ii) with 2 we get, 2x – 6y + 80 = 0

By subtracting (ii) from (i) we get ,

By substituting y = 34 in equation (i) we get

x = 124/2

x = 62

Hence the money with A be Rs . 62 and the money with B be Rs. 34 .

6. ABCD is a cyclic quadrilateral such that ∠A=(4y+20)°, ∠B=(3y-5)°, ∠C=(4x)° and ∠D=(7x+5)°. Find the four angles.

**Solution**

We know that the sum of the opposite angles of cyclic quadrilateral is 180°. In the cyclic quadrilateral ABCD angles A and C and angles B and D pairs of opposite angles

Therefore

∠A + ∠C = 180° and ∠B + ∠D = 180°

∠A + ∠C = 180°

By substituting ∠A = (4y + 20)° and ∠C = (-4x)° we get

4y + 20 – 4x = 180°

-4x + 4y + 20 = 180°

By substituting ∠A = (4y + 20)° and ∠C = (-4x)° we get

4y + 20 – 4x = 180°

-4x + 4y + 20 = 180°

-4x + 4y = 180° - 20

-4x + 4y = 160°

4x – 4y = -160°

Dividing both sides of equations by 4 we get

x – y = -40°

x – y + 40° = 0 …(i)

∠B + ∠D = 180°

By substituting ∠B = (3y – 5)° and ∠D = (7x+5)° we get

3y – 5 + 7x + 5 = 180°

7x + 3y = 180

7x + 3y – 180 = 0 …(ii)

By multiplying equation (i) by 3 we get

3x – 3y + 120° = 0 …(iii)

By subtracting equation (iii) from (ii) we get

3x – 3y + 120 = 0

7x + 3y – 180 = 0

10x = 60

x = 60/10

x = 6

By substituting x = 6° in equation (i) we get

x – y + 40°= 0

6 – y + 40 = 0

-1y = -40 -6

-1y = -46

y = -46/-1

By substituting x = 6° in equation (i) we get

x – y + 40° = 0

6 – y + 40 = 0

-1y = - 40 – 6

-1y = - 46

y = 46

The angles of a cyclic quadrilateral are

∠A = 4y + 20

= 4 × 46 + 20

= 184 + 20

= 204°

∠B = 3y – 5

= 3 × 46 + 5

= 138 – 5

= 133°

∠C = -4x°

= - 4(6)

= - 24°

∠D = 7x + 5

= 7 × 6 + 5

= 42 + 5

= 47°

Hence the angles of quadrilateral are ∠A = 204° , ∠B = 133° , ∠C = -24° , ∠D = 47°.

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?

A man can also finish the work in x days and one boy alone can finish it in y days then

**Solution**
8. In a △ABC, ∠A = X° , ∠B = (3x – 2)° , ∠C = y°. Also, ∠C - ∠B = 9° . Find the three angles.

Let ∠A = X° , ∠B = (3x – 2)° , ∠C = y° and

∠C - ∠B = 9°

⇒ ∠C = 9° + ∠B

⇒ ∠C = 9° + 3x° - 2°

⇒ ∠C = 7° + 3x°

Substitute ∠C = y° in above equation we get ,

y° = 7° + 3x°

∠A + ∠B + ∠C = 180°

⇒ x° + (3x° - 2° ) + (7° + 3x° ) = 180°

⇒ 7x° = 180° - 5° = 175°

⇒ x° = 175°/7° = 25°

∠A = x° = 25°

∠B = (3x – 2)° = 3(25°) - 2° = 73°

∠C = (7° + 3x° ) = 7° + 3(25)° = 82°

∠A = 25°, ∠B = 73°, ∠C = 82°

Hence, the answer

**Solution**

∠C - ∠B = 9°

⇒ ∠C = 9° + ∠B

⇒ ∠C = 9° + 3x° - 2°

⇒ ∠C = 7° + 3x°

Substitute ∠C = y° in above equation we get ,

y° = 7° + 3x°

∠A + ∠B + ∠C = 180°

⇒ x° + (3x° - 2° ) + (7° + 3x° ) = 180°

⇒ 7x° = 180° - 5° = 175°

⇒ x° = 175°/7° = 25°

∠A = x° = 25°

∠B = (3x – 2)° = 3(25°) - 2° = 73°

∠C = (7° + 3x° ) = 7° + 3(25)° = 82°

∠A = 25°, ∠B = 73°, ∠C = 82°

Hence, the answer

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 awanded for each correct answer and 2 marks been deducted for each incorrect answer, the Yash would have scored 50 marks. How many question were there in the test ?

**Solution**

Let take right answer will be x and wrong answer will be y.

Hence total number of questions will be x + y …(i)

If yash scored 40 marks in atleast getting 3 marks for each right answer and losing 1 mark for each wrong answer then

3x – 1y = 40 …(ii)

If 4 marks awarded for each right answer and 2 marks deduced for each wrong answer the he scored 50 marks

4x – 2y = 50 …(iii)

By multiplying equation (i) by 2 we get

6x – 2y = 80 …(iv)

By subtracting (iii) from (iv) we get

6x – 2y = 80

-4x – 2y = 50

2x = 30

x = 30/2

x = 15

Putting x = 15 in equation (ii) we have

3x – 1y = 40

3 × 15 – 1y = 40

45 – 1y = 40

-1y = 40 – 45

-y = -5

Total number question will be

= x+y

= 15 + 5

= 20

Hence, the total number of question is 20 .

11. In a △ABC , ∠A = X°, ∠B = 3X° and ∠C = y° . If 3y – 5x =30 , prove that the triangle is right angled .

**Solution**

We have to prove that the triangle is right

Given ∠A = x°, ∠B = 3x° and ∠C = y°

Sum of three angles in triangle are ∠A + ∠B + ∠C = 180°

∠A + ∠B + ∠C = 180°

x + 3x + y = 180

Sum of three angles in triangles are ∠A + ∠B + ∠C = 180°

∠A + ∠B + ∠C = 180°

x + 3x + y = 180

4x + y = 180 … (i)

By solving 4x + y = 180 with 3y – 5x = 30 we get,

4x + y = 180

-5x + 3y = 30 … (ii)

Multiplying equation (i) by 3 we get

12x + 3y = 540 …(iii)

Subtracting equation (ii) from (iii)

12x + 3y = 540

-5x + 3y = 30

17x = 510

x = 510/17

x = 30°

Substituting x = 30° in equtaion (i) we get

4x + y = 180

4 × 30 + y = 180

120 + y = 180

y = 180 – 120

y = 60°

Angles ∠A, ∠B and ∠C are

∠A = x°

= 30°

∠B = 3x°

= 3 × 30°

= 90°

∠C = y°

= 60°

A right angled triangle is a triangle in which one side should has a right angle that is 90° in it .

Hence, ∠B = 90° The triangle ABC is right angled.

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

**Solution**

Let the fixed charges of car be Rs x per km and the running charges be Rs. y km/hr

According to the given condition we have

x + 12 = 89 …(i)

x + 20y = 145 …(ii)

x + 12y = 89

x + 20y = 145

-8y = - 56

y = -56/-8

y = 7

Putting y = 7 in equation (i) we get

x + 12y = 89

x + 12 × 7 = 89

x + 84 = 89

x = 89 – 54

x = 5

Therefore, total charges for travelling distance of 30 km

= x + 30y

= 5 + 30 × 7

= 5 + 210

= Rs 215

Hence, A person have to pay Rs . 215 for travelling a distance of 30 km .

13. A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes foods for 20 days, he has to pay Rs 1000 as hostel charges whereas a students B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.

**Solution**

According to the given condition we have ,

x + 20y = 1000 …(i)

x + 26y = 1180 …(ii)

Subtracting equation (ii) from equation (i) we get

x + 20y = 1000

-x – 26y = -1180

-6 = -180

y = -180/-6

y = 30

Putting y = 30 in equation (i) we get

x + 20y = 1000

x + 20 × 30 = 1000

x + 600 = 1000

x = 1000 – 600

x = 400

Hence, the fixed charges of hostel is Rs 400

The cost of food per day is Rs 30.

14. Half the perimeter of a garden , whose length is 4 , more than its width is 36 m . Find the dimension of the garden.

**Solution**

*l*+b) . If half the perimeter of a garden will be 36 m .

1/2 × 2

*(l+b)*= 36*(l+b)*= 36

2b = 36 – 4

b = 32/2

b = 16

Putting b = 16 in equation (i) we get

*(l+b)*= 36

l + 16 = 36

l = 36 – 16

l = 20

Hence, the dimensions of rectangular garden are width = 16 m and length = 20 m .

15. The larger of two supplementary angles exceeds the smaller by 18 degrees . Find them .

**Solution**

We know that the sum of supplementary angles will be 180° .

Let the longer supplementary angles will be ‘y’ .

Then, x + y = 180° …(i)

If larger of supplementary angles exceeds the smaller by 18 degree , According to he given condition .

We have ,

x = y + 18 …(ii)

Substitute x = y + 18 in equation (i) , we get

x + y = 180°

y + 18 + y = 180°

2y + 18 = 180°

2y = 180° - 18°

2y = 162°/2

y = 81°

Put y = 81° equation (ii), we get

x = y + 18

x = 81 + 18

x = 99°

Hence , the larger supplementary angle is 99° ,

The smaller supplementary angle is 81° .

16. 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery, and that taken by 1 man alone.

**Solution**

1 women alone can finish the work in x days and 1 man alone can finish it in y days . Then

One woman one day work = 1/x

One man one days work = 1/y

2 women’s one day work = 2/x

5 man’s one days work = 5/y

Since 2 women and 5 men can finish the work in 4 days

**Solution**

Let Rs x be the notes of Rs . 50 and Rs . 100 notes wii be Rs. y

If Meena ask for Rs. 50 and Rs. 100 notes only, then the equation will be,

50x + 100y = 2000
Dividing both sides by 50 then we get,

x + 2y = 40 …(i)

If Meena got 25 notes in all then the equation will be ,

x + y = 25 …(ii)

By subtracting the equation (ii) from (i) we get ,

x + 2y = 40

-x – y = -25

1y = 15

y = 15/1

y = 15

Substituting y = 15 in equation (ii) , we get

x + y = 25

x + 15 = 25

x = 25 -15

x = 10

Therefore x = 10 and y = 15

Hence , Meena has 10 notes of Rs. 50 and 15 notes of Rs. 100

18. There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.

**Solution**

Let us take the A examination room will be x and the B examination room wiil be y

If 10 candidates are sent from A to B , the number of students in each room is same . According to the above condition will be

y + 10 = x – 10

0 = x – y – 10 – 10

x – y – 20 = 0 …(i)

If 20 candidates are sent from B to A, the number of students in A is double the number of students in B, then equation will be

x + 20 = 2(y – 20)

x + 20 = 2y – 40

x + 20 – 2y + 40 = 0

x – 2y + 20 + 40 = 0

By subtracting the equation (i) from (ii) we get , y = 80

Substituting y = 80 in equation (i) , we get

Hence 100 candidaes are in A examination Room,

80 candidates are in B examination Room .

19. A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs 216 and one full and one half reserved first class tickets cost Rs 327. What is the basic first class full fare and what is the reservation charge ?

**Solution**

20. A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that'. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the the stake of money each of the cock-owners have.

**Solution**

21. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of student in the class.

**Solution**

Let the number of students be x and the number of row be y. Then ,

Number of students in each row = x/y When three students is extra in each row , there are one row less that is when each row has

**Solution**

Let the money with first person be Rs x and the money with the second person be Rs y . Then, (x+100) = 2(y-100)

(y+10) = 6(x-10)

If first person give Rs 100 to second person then the second persson will become twice as rich as first person, According to the given condition, we have,

(x+100) = 2 (y -100)

x + 100 = 2y – 200

x – 2y + 100 + 200 = 0

x – 2y + 300 = 0 …(i)

If second person gives Rs 10 to first person then the first person will becomes six time as rich as second person , According to given condition , we have ,

(y+10) = 6 (x-10)

y + 10 = 6x – 60

0 = 6x – 60 – y – 10

0 = 6x – y – 70 …(ii)

Multiplying (ii) equation by 2 we get,

12x – 2y – 140 = 0 …(iii)

By subtracting (iii) from (i) , we get

x – 2y + 300 = 0

-12x + 2y + 140 = 0

-11x + 440 = 0

-11x = -440

x = -440/-11

x = 40

Putting x = 40 in equation (i) , we get ,

x – 2y + 300 = 0

40 – 2y + 300 = 0

-2y + 340 = 0

-2y = -340

y = -340/-2

y = 170

Hence , first person’s capital will be Rs. 40,

Second person’s capital will be Rs. 170.

23. A shopkeeper sells a saree at 8% profit and the sweater at 10% discount , there by getting a sum of 1008 . If she had sold the saree at 10% profit and the sweater at 8% discount , she would have got 1028 . Find the cost price of the sarees and the list price (price before discount ) of the sweater .

**Solution**

Let the CP of saree be ₹x and the list price of sweater be ₹y.

Case I: When saree is sold at 8% profit and sweater at 10% discount.

24. In a competitive examination , one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer . Jayanti answered 120 questions and got 90 marks . How many questions did she answer correctly .

**Solution**

Let the number of correct answer be x and the number of wrong answers be y.

Total questions jayanthi answered = 120

So, x + y = 120 …(i)

Now, marks obtained for answering corectly = 1 × x = x

25. A shopkeeper gives books on rent for getting . She takes a fixed charge for the first two days, and an additional charge for each day thereafter . Latika Paid 22 for a book kept for 6 days , while Anand paid 16 for the book kept for four days . Find the fixed charges and charge and charge for each extraday .

**Solution**

Let the fixed charge for first two days be ₹x and the additional charge for each day extra be ₹y.

It is given that Latika kept the book for 6 days and paid ₹22.

So,

Fixed charge for the first 2 days + Additional charge for 4 days = ₹22

∴ x + 4y = 22 .....(i)

Anand kept the book for 4 days and paid ₹16. So,

Fixed charge for the first 2 days + Additional charge for 2 days = ₹16

∴ x + 2y = 16 .....(ii)

Subtracting (ii) from (i), we get

2y = 6

⇒ y = 3

Putting y = 3 in (i), we get

x + 12 = 22

⇒ x = 10

Thus, the fixed charge is ₹10 and the additional charge for each extraday is ₹ 3.