# RS Aggarwal Solutions Class 10 Ex 3K

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:

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

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

$\angle C = (x + y – 30)^{\circ}$

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

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

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,

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

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

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

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

#### Practise This Question

State whether the following statement is True or False:
Incomplete combustion of carbon is due to insufficient amount of air (oxygen) which leads to the formation of carbon monoxide (CO).