NCERT Solutions for Exercise 3.7 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Q2 One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

[Hint : $x + 100 = 2(y - 100), y + 10 = 6(x - 10)$ ]

Answer:

Let the amount of money the first person and the second person having is x and y respectively

Noe, According to the question.

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

$\Rightarrow x - 2y =-300...........(1)$

Also

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

$\Rightarrow y - 6x =-70..........(2)$

Multiplying (2) by 2 we get,

$2y - 12x =-140..........(3)$

Now adding (1) and (3), we get

$-11x=-140-300$

$\Rightarrow 11x=440$

$\Rightarrow x=40$

Putting this value in (1)

$40-2y=-300$

$\Rightarrow 2y=340$

$\Rightarrow y=170$

Thus two friends had 40 Rs and 170 Rs respectively.

Q3 A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer:

Let the speed of the train be v km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

Now As we Know,

$speed=\frac{distance }{time}$

$\Rightarrow v=\frac{d}{t}$

$\Rightarrow d=vt..........(1)$
Now, According to the question,
$(v+10)=\frac{d}{t-2}$

$\Rightarrow (v+10){t-2}=d$

$\Rightarrow vt +10t-2v-20=d$

Now, Using equation (1), we have

$\Rightarrow -2v+10t=20............(2)$
Also,

$(v-10)=\frac{d}{t+3}$

$\Rightarrow (v-10)({t+3})=d$

$\Rightarrow vt+3v-10t-30=d$
$\Rightarrow3v-10t=30..........(3)$

Adding equations (2) and (3), we obtain:
$v=50.$
Substituting the value of x in equation (2), we obtain:
$(-2)(50)+10t=20$

$\Rightarrow -100+10t=20$

$\Rightarrow 10t=120$

$\Rightarrow t=12$
Putting this value in (1) we get,

$d=vt=(50)(12)=600$

Hence the distance covered by train is 600km.

Q4 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 students in the class.

Answer:

Let the number of rows is x and the number of students in a row is y.
Total number of students in the class = Number of rows * Number of students in a row
$=xy$

Now, According to the question,

$\\xy = (x - 1) (y + 3) \\\Rightarrow xy= xy - y + 3x - 3 \\\Rightarrow 3x - y - 3 = 0 \\ \Rightarrow 3x - y = 3 ...... ... (1)$

Also,

$\\xy=(x+2)(y-3)\\\Rightarrow xy = xy + 2y - 3x - 6 \\ \Rightarrow 3x - 2y = -6 ... (2)$
Subtracting equation (2) from (1), we get:
$y=9$
Substituting the value of y in equation (1), we obtain:
$\\3x - 9 = 3 \\\Rightarrow 3x = 9 + 3 = 12 \\\Rightarrow x = 4$
Hence,
The number of rows is 4 and the Number of students in a row is 9.

Total number of students in a class

: $xy=(4)(9)=36$

Hence there are 36 students in the class.

Q5 In a $\Delta\textup{ABC}$, $\angle C = 3 \angle B = 2( \angle A + \angle B)$ . Find the three angles.

Answer:

Given,

Also, As we know that the sum of angles of a triangle is 180, so

$\angle A +\angle B+ \angle C=180$

$\angle A +\angle B+ 3\angle B=180^0$

$\angle A + 4\angle B=180^0..........(2)$

Now From (1) we have

$\angle B = 2 \angle A.......(3)$

Putting this value in (2) we have

$\angle A + 4(2\angle A)=180^0.$

$\Rightarrow 9\angle A=180^0.$

$\Rightarrow \angle A=20^0.$

Putting this in (3)

$\angle B = 2 (20)=40^0$

And

$\angle C = 3 \angle B =3(40)=120^0$

Hence three angles of triangles $20^0,40^0\:and\:120^0.$

Q6 Draw the graphs of the equations $5x - y =5$and $3x - 7 = 3$ . Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.

Answer:

Given two equations,

$5x - y =5.........(1)$

And

$3x - y = 3........(2)$

Points(x,y) which satisfies equation (1) are:

Points(x,y) which satisfies equation (1) are:

GRAPH:

As we can see from the graph, the three points of the triangle are, (0,-3),(0,-5) and (1,0).

Q7 (i) Solve the following pair of linear equations:$\\px + qy = p - q\\ qx - py = p + q$

Answer:

Given Equations,

$\\px + qy = p - q.........(1)\\ qx - py = p + q.........(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(q)(-p-q)-(q-p)(-p)}=\frac{y}{(q-p)(q)-(p)(-p-q)}=\frac{1}{(p)(-p)-(q)(q)}$

$\frac{x}{-p^2-q^2}=\frac{y}{p^2+q^2}=\frac{1}{-p^2-q^2}$

$x=1,\:and\:y=-1$

Q7 (ii) Solve the following pair of linear equations:$\\ax + by = c\\ bx +ay = 1 + c$

Answer:

Given two equations,

$\\ax + by = c.........(1)\\ bx +ay = 1 + c.........(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(b)(-1-c)-(-c)(a)}=\frac{y}{(-c)(b)-(a)(-1-c)}=\frac{1}{(a)(a)-(b)(b)}$

$\frac{x}{-b-bc+ac}=\frac{y}{-cb+a+ac}=\frac{1}{a^2-b^2}$

$x=\frac{-b-bc+ac}{a^2-b^2},\:and\:y=\frac{a-bc+ac}{a^2-b^2}$

Q7(iii) Solve the following pair of linear equations: $\\\frac{x}{a} - \frac{y}{b} = 0\\ ax + by = a^2 +b^2$

Answer:

Given equation,

$\\\frac{x}{a} - \frac{y}{b} = 0\\\Rightarrow bx-ay=0...........(1)\\ ax + by = a^2 +b^2...............(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-a)(-a^2-b^2)-(b)(0)}=\frac{y}{(0)(a)-(b)(-a^2-b^2)}=\frac{1}{(b)(b)-(-a)(a)}$

$\frac{x}{a(a^2+b^2)}=\frac{y}{b(a^2+b^2)}=\frac{1}{a^2+b^2}$

$x=a,\:and\:y=b$

Q7(iv) Solve the following pair of linear equations: $\\(a-b)x + (a+b)y = a^2 -2ab - b^2\\ (a+b)(x+y) = a^2 +b^2$

Answer:

Given,

$\\(a-b)x + (a+b)y = a^2 -2ab - b^2..........(1)$

And

$\\ (a+b)(x+y) = a^2 +b^2\\\Rightarrow (a+b)x+(a+b)y=a^2+b^2...........(2)$

Now, Subtracting (1) from (2), we get

$(a+b)x-(a-b)x=a^2+b^2-a^2+2ab+b^2$

$\Rightarrow(a+b-a+b)x=2b^2+2ab$

$\Rightarrow 2bx=2b(b+2a)$

$\Rightarrow x=(a+b)$

Substituting this in (1), we get,

$(a-b)(a+b)+(a+b)y=a^2-2ab-b^2$

$\Rightarrow a^2-b^2+(a+b)y=a^2-2ab-b^2$

$\Rightarrow (a+b)y=-2ab$

$\Rightarrow y=\frac{-2ab}{a+b}$ .

Hence,

$x=(a+b),\:and\:y=\frac{-2ab}{a+b}$

Q7(v) Solve the following pair of linear equations: $\\152x - 378y = -74\\ -378x + 152y = -604$

Answer:

Given Equations,

$\\152x - 378y = -74............(1)\\ -378x + 152y = -604............(2)$

As we can see by adding and subtracting both equations we can make our equations simple to solve.

So,

Adding (1) and )2) we get,

$-226x-226y=-678$

$\Rightarrow x+y=3...........(3)$

Subtracting (2) from (1) we get,

$530x-530y=530$

$\Rightarrow x-y=1...........(4)$

Now, Adding (3) and (4) we get,

$2x=4$

$\Rightarrow x=2$

Putting this value in (3)

$2+y=3$

$\Rightarrow y=1$

Hence,

$x=2\:and\:y=1$

Q8 ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Answer:

As we know that in a quadrilateral the sum of opposite angles is 180 degrees.

So, From Here,

$4y+20-4x=180$

$\Rightarrow 4y-4x=160$

$\Rightarrow y-x=40............(1)$

Also,

$3y-5-7x+5=180$

$\Rightarrow 3y-7x=180........(2)$

Multiplying (1) by 3 we get,

$\Rightarrow 3y-3x=120........(3)$

Now,

Subtracting, (2) from (3) we get,

$4x=-60$

$\Rightarrow x=-15$

Substituting this value in (1) we get,

$y-(-15)=40$

$\Rightarrow y=40-15$

$\Rightarrow y=25$

Hence four angles of a quadrilateral are :

$\angle A =4y+20=4(25)+20=100+20=120^0$

$\angle B =3y-5=3(25)-5=75-5=70^0$

$\angle C =-4x=-4(-15)=60^0$

$\angle D =-7x+5=-7(-15)+5=105+5=110^0$