# Ncert Solutions For Class 10 Maths Ex 3.2

## Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.2

Question 1:

Find the graphical solution for the given problems.

(a).In Class 10th, 10 students participated in a Maths test. If there are 4 more girls than the total number of boys, find how many girls and boys participated in the test.

(b)The total cost of 5 erasers and 7 chocolates is Rs.50, but the total cost of 7 erasers and 5 chocolates is  Rs. 46. Now, calculate the cost of one eraser and the cost of one chocolate.

Solution:

(a)Let there are x number of girls and y number of boys. As per  the given question,  it is represented as follows.

x+y=10xy=4Nowx+y=10,x=10y$x \:+\: y = \:10\\ x\:-\: y = \:4 \\ Now\: x +\: y = \:10, x \:= \:10-\:y$

 X 5 4 6 Y 5 6 4

Forxy=4,x=4+y$For\: x- y = 4,\:x= \:4 +\: y$
 X 4 5 3 Y 0 1 -1

Hence, the table is represented in graphical form as follows.

From the figure, it is seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class .

(b)Let 1 pencil costs Rs.x and 1 pen costs Rs.y. According to the question, the algebraic representation is

5x+7y=507x+5y=46For5x+7y=50,$5x\: + \:7y = \:50\\ 7x\: + \:5y = \:46\\ For\: 5x +\: 7y = \:50,$ x=507y5$x=\frac{50-7y}{5}$

 x 3 10 -4 y 5 0 10

7x+5y=46$7x\: +\: 5y =\: 46$

 X 8 3 -2 Y -2 5 12

Hence, the graphic representation is as follows.

From the figure, it is seen that the given  lines cross each other at point (3, 5).

So, the cost of a eraser is 3/- and cost of a chocolate is 5/-.

Question 2:

Comparing the given ratios, a1a2,b1b2andc1c2$\frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}}\;and\;\frac{c_{1}}{c_{2}}$,Figure out whether the lines are parallel or coincident:

(a) 5x4y+8=0$5x-4y+8=0$

7x+6y9=0$7x+6y-9=0$

(b) 9x+3y+12=018x+6y+24=0$9x+3y+12=0\\ 18x+6y+24=0$

(c) 6x3y+10=02xy+9=0$6x-3y+10=0\\ 2x-y+9=0$

Solution:

(a) 5x4y+8=0$5x-4y+8=0$

7x+6y9=0$7x+6y-9=0$

Comparing these equations with a1x+b1y+c1=0$a_{1}x+b_{1}y+c_{1}=0$

and a2x+b2y+c2=0$a_{2}x+b_{2}y+c_{2}=0$

We get,

a1=5,b1=4,c1=8a2=7,b2=6,c2=9$a_{1}=5,\; b_{1}=-4, \;c_{1}=8\\ a_{2}=7,\; b_{2}=6, \;c_{2}=-9$ a1a2=57$\frac{a_{1}}{a_{2}}=\frac{5}{7}$ b1b2=46=23$\frac{b_{1}}{b_{2}}=\frac{-4}{6}=\frac{-2}{3}$

Since a1a2b1b2$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$

So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(b) 9x+3y+12=018x+6y+24=0$9x+3y+12=0\\ 18x+6y+24=0$

Comparing these equations with a1x+b1y+c1=0$a_{1}x+b_{1}y+c_{1}=0$

and a2x+b2y+c2=0$a_{2}x+b_{2}y+c_{2}=0$

We get,

a1=9,b1=3,c1=12a2=18,b2=6,c2=24$a_{1}=9,\; b_{1}=3, \;c_{1}=12\\ a_{2}=18,\; b_{2}=6, \;c_{2}=-24$ a1a2=918=12$\frac{a_{1}}{a_{2}}=\frac{9}{18}=\frac{1}{2}$ b1b2=36=12$\frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}$ c1c2=1224=12$\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}$

Since a1a2=b1b2=c1c2$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(c) 6x3y+10=02xy+9=0$6x-3y+10=0\\ 2x-y+9=0$

Solution: Comparing these equations with a1x+b1y+c1=0$a_{1}x+b_{1}y+c_{1}=0$

and a2x+b2y+c2=0$a_{2}x+b_{2}y+c_{2}=0$

We get,

a1=6,b1=3,c1=10a2=2,b2=1,c2=9$a_{1}=6,\; b_{1}=-3, \;c_{1}=10\\ a_{2}=2,\; b_{2}=-1, \;c_{2}=9$ a1a2=62=31$\frac{a_{1}}{a_{2}}=\frac{6}{2}=\frac{3}{1}$ b1b2=31=31$\frac{b_{1}}{b_{2}}=\frac{-3}{-1}=\frac{3}{1}$ c1c2=109$\frac{c_{1}}{c_{2}}=\frac{10}{9}$

Since a1a2=b1b2c1c2$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

Question 3:

Compare the given ratios, a1a2,b1b2andc1c2$\frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}}\;and\;\frac{c_{1}}{c_{2}}$ find out whether the following pair of linear equations are consistent, or inconsistent.

(a) 3x+2y=5;2x3y=7$3x+2y=5;\\ 2x-3y=7$

(b) 2x3y=8;4x6y=9$2x-3y=8;\\ 4x-6y=9$

(c) 32x+53y=7;9x10y=14$\frac{3}{2}x+\frac{5}{3}y=7;\\ 9x-10y=14$

Solution:

(a) 3x+2y=5;2x3y=7$3x+2y=5;\\ 2x-3y=7$

a1a2=32$\frac{a_{1}}{a_{2}}=\frac{3}{2}$ b1b2=23$\frac{b_{1}}{b_{2}}=\frac{-2}{3}$ c1c2=57$\frac{c_{1}}{c_{2}}=\frac{5}{7}$

Since a1a2b1b2$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$

So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(b) 2x3y=8;4x6y=9$2x-3y=8;\\ 4x-6y=9$

a1a2=24$\frac{a_{1}}{a_{2}}=\frac{2}{4}$ b1b2=36$\frac{b_{1}}{b_{2}}=\frac{-3}{-6}$ c1c2=89$\frac{c_{1}}{c_{2}}=\frac{8}{9}$

Since a1a2=b1b2c1c2$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

So, the equations are parallel to each other  and they have no possible solution. So, the equations are inconsistent.

(c) 32x+53y=7;9x10y=14$\frac{3}{2}x+\frac{5}{3}y=7;\\ 9x-10y=14$

a1a2=329=16$\frac{a_{1}}{a_{2}}=\frac{\frac{3}{2}}{9}=\frac{1}{6}$ b1b2= frac5310=16$\frac{b_{1}}{b_{2}}=\frac{\ frac{5}{3}}{-10}=\frac{-1}{6}$ c1c2=714$\frac{c_{1}}{c_{2}}=\frac{7}{14}$

Since a1a2b1b2$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$

So, the equations are intersecting  each other at one point and they have only one possible solution. So, the equations are consistent.

Question 4:

Find whether the following pairs of linear equations are consistent or  inconsistent. If consistent, find the solution graphically:

(a)x + y = 5, 2x + 2y = 10

(b)2x + y – 6 = 0,4x – 2y – 4 = 0

Solution:

(a)x + y = 5, 2x + 2y = 10

a1a2=12$\frac{a_{1}}{a_{2}}=\frac{1}{2}$ b1b2=12$\frac{b_{1}}{b_{2}}=\frac{1}{2}$ c1c2=510=12$\frac{c_{1}}{c_{2}}=\frac{5}{10}=\frac{1}{2}$

Since a1a2=b1b2=c1c2$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}}$

∴ the equations are coincident  and they have infinite number of possible solutions.

So, the equations are consistent.

x + y = 5, 2x + 2y = 10

 x 4 3 2 y 1 2 3

And, 2x + 2y = 10

x=102y2$x=\frac{10-2y}{2}$
 x 4 3 2 y 1 2 3

So, the equations are represented as follows:

From the figure, it is seen that the lines are overlapping each other.

Therefore, the equations have infinite possible solutions.

Question 5:

Consider the two equations x − y + 1 = 0 and 3x + 2y − 12 = 0. Draw a graph for both. Find out the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution:

xy+1=0$x\:-\: y \:+\: 1\: = \:0$

Or

x=y1$x\:=y\:-\:1$
 x 0 1 2 y 1 2 3

3x+2y12=0x=122y3$3x +2y-12=0\\ x=\frac{12-2y}{3}$

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).