RS Aggarwal Class 10 Solutions Chapter 3 - Linear Equations In Two Variables - Ex 3K(3.11)

RS Aggarwal Class 10 Chapter 3 - Linear Equations In Two Variables - Ex 3K(3.11) Solutions Free PDF

Mathematics is an important subject for Class 10 students and to score good marks in this subject you need to prepare effectively. For this purpose, you can refer to the RS Aggarwal Class 10 solutions Chapter 3 which are prepared by subject experts following the latest CBSE syllabus. By practicing these solutions you can draw meaning out of every problem quite effectively and master the subject.

By solving these solutions you can grasp each topic in quick time and develop a confidence that surely will help you approach the exam in the best possible way as possible. Also, using these solutions you can greatly improve your question-solving skills and tackle difficult questions with full confidence. The RS Aggarwal Solutions Class 10 Chapter 3 – Linear Equations In Two Variables help students in clearing the concepts of the subject and at the same time provide a good platform for practicing the questions.

Download PDF of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3K(3.11)

Question 1: Abdul travelled 250 km by train and 120 km by car taking hours. But, if he travels 125 km by train and 60 km by car, he takes 2 minutes longer. Find the speed of the train and that of the car.

Solution:

Let the speed of train and car be x km/hr and y km/hr respectively.

Then,

\( \frac{250}{x} + \frac{120}{y} = 4 \)

\( \Rightarrow \frac{125}{x} + \frac{60}{y} = 2 \)

When, \( \frac{1}{x} = u and \frac{1}{y} = v \)

125u + 60v = 2 ……..(1)

And, \( \frac{130}{x} + \frac{240}{y} = 4 + \frac{18}{60} = 4 + \frac{3}{10} = \frac{43}{10} \)

\( \Rightarrow \frac{1300}{x} + \frac{2400}{y} = 43 \)

\( \Rightarrow \) 1300u + 2400v = 43 ……..(2)

Multiplying (1) by 40 and (2) by 1, we get

5000u + 2400v = 80 ……..(3)

1300u + 2400v = 43 ……..(4)

Subtracting (4) from (3), we get

3700u = 37

\( \Rightarrow u = \frac{1}{100} \)

Putting \( u = \frac{1}{100} \) in (1), we get

\( 125 \times \frac{1}{100} + 60v = 2 \Rightarrow 6000v = 200 – 125 \Rightarrow v = \frac{1}{80} \)

\( u = \frac{1}{100} \Rightarrow \frac{1}{x} = \frac{1}{100} \Rightarrow x = 100 \)

\( v = \frac{1}{80} \Rightarrow \frac{1}{y} = \frac{1}{80} \Rightarrow y = 80 \)

Hence, speeds of the train and the cars are 100 km/hr and 80 km/hr respectively.

Question 2: A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

Solution:

Let the speed of the boat in still water be x km/hr and speed of the stream by y km/hr

Then,

Speed upstream = (x – y) km/hr

Speed downstream = (x + y) km/hr

Time taken to cover 12 km upstream = \( \frac{12}{x – y} hrs \)

Time taken to cover 40 km downstream = \( \frac{40}{x + y} hrs \)

Total time taken = 8 hrs

\( \frac{12}{x – y} + \frac{40}{x + y} = 8 \)

Again, time taken to cover 16 km upstream = \( \frac{16}{x – y} \)

Times taken to cover 32 km downstream = \( \frac{32}{x + y} \)

Total time taken = 8 hrs

\( \frac{16}{x – y} + \frac{32}{x + y} = 8 \)

Putting \( \frac{1}{x – y} = u and \frac{1}{x + y} = v \) , we get,

12u + 40v = 8

3u + 10v = 2 ……..(1)

And,

16u + 32v = 8

2u + 4v = 1 ……..(2)

Multiplying (1) by 4 and (2) by 10, we get

12u + 40v = 8 ……..(3)

20u + 40v = 10 ……..(4)

Subtracting (3) from (4), we get

8u = 2

\( u = \frac{1}{4} \)

Putting \( u = \frac{1}{4} \) in (3), we get

\( 3 \times \frac{1}{4} + 10v = 2 \Rightarrow 10v = \frac{5}{4} \Rightarrow v = \frac{1}{8} \)

\( u = \frac{1}{4} \Rightarrow \frac{1}{x – y} = \frac{1}{4} \Rightarrow x – y = 4 \) ……..(5)

\( v = \frac{1}{8} \Rightarrow \frac{1}{x + y} = \frac{1}{8} \Rightarrow x + y = 8 \) ……..(6)

On adding (5) and (6), we get:

2x = 12

x = 6

Putting x = 6 in (6), we get

6 + y = 8

y = 8 – 6 = 2

x = 6, y = 2

Hence, the speed of the boat in still water = 6 km/hr and speed of the stream = 2 km/hr

Question 3: Taxi charges in a city consisting of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 110 km, he pays Rs.1130, and travelling 200 km, he pays Rs.1850. Find the fixed charges and rate per km.

Solution:

Let the fixed charges of taxi per day be Rs. x and charges for travelling for 1 km be Rs. y.

For travelling 110 km, he pays

Rs. x + Rs. 110y = Rs. 1130

x + 110y = 1130 ……..(1)

For travelling 200 km, he pays

Rs. x + Rs. 200y = Rs. 1850

x + 200y = 1850 ……..(2)

Subtracting (1) from (2), we get

90y = 1850 – 1130 = 720

\( y = \frac{720}{90} = 8 \)

Putting y = 8 in (1),

x + 110(8) = 1130

x = 1130 – 880 = 250

Hence, fixed charges = Rs. 250

And charges for travelling 1 km = Rs. 8

Question 4: A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs.3500, whereas a student B who takes food for 28 days, has to pay Rs.3800. Find the fixed charges per month and the cost of the food per day.

Solution:

Let the fixed charges be Rs. x and food charges per day be Rs. y respectively.

For student A:

Student takes food for 25 days and he has to pay: Rs. 3500

x + 25y = 3500 ……..(1)

For student B:

Student takes food for 28 days and he has to pay: Rs. 3800

x + 28y = 3800 ……..(2)

Subtracting (1) from (2), we get

3y = 3800 – 3500

3y = 300

y = 100

Putting y = 100 in (1), we get

x + 25(100) = 3500

Or, x = 1000

Thus, fixed charges for hostel = Rs. 1000 and charges for food per day = Rs. 100

Question 5: The length of a room exceeds its breadth by 3 metres. If the length is increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.

Solution:

Let the length = x metres and breadth = y metres

Then,

x = y + 3

x – y = 3 ……..(1)

Also,

(x + 3)(y – 2) = xy

3y – 2y = 6 ……..(2)

Multiplying (1) by 2 and (2) by 1

-2y + 2x = 6 ……..(3)

3y – 2x = 6 ……..(4)

Adding (3) and (4), we get

y = 12

Putting y = 12 in (1), we get

x – 12 = 3

x = 15

x = 15, y = 12

Hence, length = 15 metres and breadth = 12 metres.

Question 6: The area of a rectangle gets reduced by 8 m2, when its length is reduced by 5 m and its breadth is increased by 3 m. If we increase the length by 3m and breadth by 2m, the area is increased by 74 m2. Find the length and breadth of the rectangle.

Solution:

Let the length of a rectangle be x metres and breadth be y metres.

Then,

Area = xy sq.m

Now,

xy – (x – 5)(y + 3) = 8

xy [xy * 5y + 3x – 15] = 8

(xy)(xy) + (5y)(3x) + 15 = 8

3x – 5y = 7 ……..(1)

And, (x + 3)(y + 2) – xy = 74

xy + 3y + 2x + 6(xy) = 74

2x + 3y = 68 ……..(2)

Multiplying (1) by 3 and (2) by 5, we get

9x – 15y = 21 ……..(3)

10x + 15y = 340 ……..(4)

Adding (3) and (4), we get

\( 19x = 361 \Rightarrow x = \frac{361}{19} = 19 \)

Putting x = 19 in (3) we get

\( 9 \times 19 – 15y = 21 \Rightarrow 171 – 15y = 21 \Rightarrow y = \frac{150}{15} = 10 \)

x = 19 metres, y = 10 metres

Hence, length = 19m and breadth = 10m

Question 7: 2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.

Solution:

Let man’s 1 day’s work be \( \frac{1}{x} \) and 1 boy’s day’s work be \( \frac{1}{y} \)

Also,

Let \( \frac{1}{x} = u \, and \,  \frac{1}{y} = v \)

Then, \( \frac{2}{x} + \frac{5}{y} = \frac{1}{4} \Rightarrow 2u + 5v = \frac{1}{4} \) ……..(1)

And, \( \frac{3}{x} + \frac{6}{y} = \frac{1}{3} \Rightarrow 3u + 6v = \frac{1}{3} \) ……..(2)

Multiplying (1) by 6 and (2) by 5, we get

\( 12u + 30v = \frac{6}{4} \) ……..(3)

\( 15u + 30v = \frac{5}{3} \) ……..(4)

Subtracting (3) and (4), we get

\( 3u = \frac{5}{3} – \frac{6}{4} \)

\( \Rightarrow 3u = \frac{20 – 18}{12} \)

\( \Rightarrow 3u = \frac{2}{12} \)

\( \Rightarrow 3u = \frac{1}{6} \)

\( u = \frac{1}{18} \)

Putting \( u = \frac{1}{18} \) , we get

\( 2 \times \frac{1}{18} + 5v = \frac{1}{4} \Rightarrow \frac{1}{9} + 5v = \frac{1}{4} \Rightarrow \frac{1}{4} = \frac{1}{9} \)

\( \Rightarrow 5v = \frac{5}{36} \Rightarrow v = \frac{1}{36} \)

Now, \( u = \frac{1}{18} \Rightarrow x = \frac{1}{u} = 18 \)

And, v = \(\frac{1}{36} \Rightarrow y = \frac{1}{v} = 36 \)

x = 18, y = 36

The man will finish the work in 18 days and the boy will finish the work in 36 days when they work alone.

Question 8: In a \(\Delta ABC,\angle A=x^{\circ},\angle B=\left ( 3x-2 \right )^{\circ},\angle C=y^{\circ}\:and\:\angle C-\angle B=9^{\circ}\). Find the three angles.

Solution:

\( \angle A + \angle B + \angle C = 180^{\circ} \)

x + 3x + y = 180

4x + y = 180 ……..(1)

Also,

3y – 5x = 30

-5y + 3x = 30 ……..(2)

Multiplying (1) by 3 and (2) by 1, we get

12x + 3y = 540 ……..(3)

-5x + 3y = 30 ……..(4)

Subtracting (4) from (3), we get

17x = 510

x = 30

Putting x = 30 in (1), we get

4(30) + y = 180

y = 60

Hence, \( \angle A = 30^{\circ} , \, \angle B = 3 \times 30^{\circ} = 90^{\circ} , \, \angle C = 60^{\circ} \)

Therefore, the triangle is right angled.

Question 9: In a cycle quadrilateral ABCD, it is given that \(\angle A=\left ( 2x+4 \right )^{\circ},\angle B=\left ( y+3 \right )^{\circ},\angle C=\left ( 2y+10 \right )^{\circ}\:and\:\angle D=\left ( 4x-5 \right )^{\circ}\). Find the four angles.

Solutions:

In a cyclic quadrilateral ABCD:

\( \angle A = (x + y + 10)^{\circ} \)

\( \angle B = (y + 20)^{\circ} \)

\( \angle C = (x + y – 30)^{\circ} \)

\( \angle D = (x + y)^{\circ} \)

We have, \( \angle A + \angle C = 180^{\circ} and \angle B + \angle D = 180^{\circ} \)

(because, ABCD is a quadrilateral)

Now,

\( \angle A + \angle C = (x + y + 10)^{\circ} + (x + y – 30)^{\circ} = 180^{\circ} \)

2x + 2y – 20 = 180

x + y – 10 = 90

x + y = 100 ……..(1)

Also,

\( \angle B + \angle D = (y + 20)^{\circ} + (x + y)^{\circ} = 180^{\circ} \)

x + 2y + 20 = 180

x + 2y = 160

Subtracting (1) from (2) we get

y = 160 – 100 = 60

Putting y = 60 in (1), we get

x = 100 – y

x = 100 – 60

x = 40

Therefore,

\( \angle A = (x + y + 10)^{\circ} = (100 + 10)^{\circ} = 110^{\circ} \)

\( \angle B = (y + 20)^{\circ} = (60 + 20)^{\circ} \)

\( \angle C = (x + y – 30)^{\circ} = (60 + 40 – 30)^{\circ} = (100 – 30)^{\circ} = 70^{\circ} \)

\( \angle D = (x + y)^{\circ} = (60 + 40)^{\circ} = 100^{\circ} \)<

Key Features of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3K(3.11)

  • The solutions explains each problem in a simple and understandable language.
  • You can refer to the RS Aggarwal maths solutions whenever you have any doubt.
  • It covers the entire syllabus so it is the best study material for exam preparation.
  • It boosts your confidence level while writing the final question paper.

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