RS Aggarwal Solutions Class 10 Ex 3B

Question 1: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

2x – 3y – 17 = 0, 4x + y – 13 =0

Solution:

Given equations are 2x – 3y – 17 = 0 and 4x + y – 13 =0

Graph of 2x – 3y – 17 = 0:

2x – 3y – 17 = 0 \( \Rightarrow y = \frac{2x – 17}{3} \)

Putting x = 1, we get y = -5

Putting x = 4, we get y = -3

Putting x = 7, we get y = -1

Hence, the table is,

x 1 4 7
y -5 -3 -1

Plot the points A(1, -5), B(4, -3) and C(7, -1) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 2x – 3y – 17 = 0.

Graph of 4x + y – 13 =0 :

4x + y – 13 =0 \( \Rightarrow y = -4x + 13 \)

Putting x = 4, we get y = -3

Putting x = 2, we get y = 5

Putting x = 3, we get y = 1

Hence, the table is,

x 4 2 3
y -3 5 1

Now, on the same graph paper as above plot the points P(2, 5) and Q(3, 1). The point B(4, -3) has already been plotted. Join PQ and QB to get PB. The line PB is the graph of the equation 4x + y – 13 =0 is the solution of the given system of equations.

These graph lines intersect the x-axis at R and S. the region bounded by these lines and the x-axis has been shaded.

https://lh5.googleusercontent.com/OTWuKbZrM_pEvFKLn82P7CONsudtM6ivxCHCM0Am-YPxVKhUhNenNuxQY0aBr1kvz0neVIkHpjiihhaThutFnWWhYMly1a_0arCZYlj_3b50BNvZQId8DIH21zaMKyWjZc_dzY5f

Question 2: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

4x – y = 4, 3x + 2y = 14

Solution:

Given equations are 4x – y = 4 and 3x + 2y = 14

Graph of 4x – y = 4:

4x – y = 4 \( \Rightarrow y = 4x – 4 \)

Putting x = 0, we get y = -4

Putting x = 1, we get y = 0

Putting x = 2, we get y = 4

Hence, the table is,

x 0 1 2
y -4 0 4

Plot the points A(0, -4), B(1, 0) and C(2, 4) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x – y = 4.

Graph of 3x + 2y = 14 :

3x + 2y = 14  \( \Rightarrow y =\frac{14 – 3x}{2} \)

Putting x = 0, we get y = 7

Putting x = 2, we get y = 4

Putting x = 4, we get y = 1

Hence, the table is,

x 0 2 4
y 7 4 1

Now, on the same graph paper as above plot the points P(0, 7) and Q(4, 1). The point C(2, 4) has already been plotted. Join PC and CQ to get PQ.

https://lh5.googleusercontent.com/iPndWSUd6SKRd4_9ZkoQ8ZEOlQxGGLoVl0WduhrF4Ty5LsLHVBaUvnnrFXAi1BJxIsiDPlDPXS_2zonF6Cp4joXLTravhotnskMisgm2lGCjwdxaa7ulVxBgEi9gUl7lijuhFFwb

The line PQ is the graph of the equation 3x + 2y = 14. The two graph lines intersect at point C(2, 4)

Therefore, x = 2, y = 4 is the solution of the given system of equations. The region bounded by these lines and y-axis has been shown by the shaded area.

Question 3: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

2x – y = 1, x – y = -1

Solution:

Given equations are 2x – y = 1 and x – y = -1

Graph of 2x – y = 1:

2x – y = 1 \( \Rightarrow y = 2x – 1 \)

Putting x = 1, we get y = 1

Putting x = 2, we get y = 3

Putting x = 0, we get y = -1

Hence, the table is,

x 1 2 0
y 1 3 -1

Plot the points A(1, 1), B(2, 3) and C(0, -1) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 2x – y = 1.

Graph of x – y = -1:

x – y = -1 \( \Rightarrow y =x + 1 \)

Putting x = 1, we get y = 2

Putting x = 2, we get y = 3

Putting x = 0, we get y = 1

Hence, the table is:

x 1 2 0
y 2 3 1

Now, on the same graph paper as above plot the points P(1, 2) and Q(0, 1). The point B(2, 3) has already been plotted. Join PB and PQ to get BQ.

https://lh5.googleusercontent.com/G2dP08T3y1_C2nCkajkKk7MLFcCmaEZPYOSHfk9K5u9ST7F8lKzo67etZ1w3-RHDYs2yjHiI0rN4z4W1ARxC-Jrs6MAvSwRSX1brYYbjqLcAV9LVK8w362X55Q1-qahHCbvhbu9l

The line BQ is the graph of the equation x – y = -1. The two graph lines intersect at point B(2, 3)

Therefore, x = 2, y = 3 is the solution of the given system of equations. The region bounded by these lines and y-axis has been shown by the shaded area.

Question 4: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

2x – 5y + 4 = 0,

2x + y – 8 = 0

Solution:

Given equations are 2x – 5y + 4 = 0 and 2x + y – 8 = 0

Graph of 2x – 5y + 4 = 0:

2x – 5y + 4 = 0 \( \Rightarrow y = \frac{2x + 4}{5} \)

Putting x = 3, we get y = 2

Putting x = -2, we get y = 0

Putting x = 8, we get y = 4

Hence, the table is,

x 3 -2 8
y 2 0 4

Plot the points A(3, 2), B(-2, 0) and C(8, 4) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 2x – 5y + 4 = 0.

Graph of 2x + y – 8 = 0:

2x + y – 8 = 0 \( \Rightarrow y = -2x + 8 \)

Putting x = 1, we get y = 6

Putting x = 2, we get y = 4

Putting x = 3, we get y = 4

Hence, the table is,

x 1 2 3
y 6 4 2

Now, on the same graph paper as above plot the points P(1, 6) and Q(2, 4). The point A(3, 2) has already been plotted. Join PA.

Thus, line PA is the graph of 2x + y – 8 = 0. On extending the graph lines on both sides, we find that these graph lines intersect the y-axis at the point R(0, 8) and S(0, 0.8).

Question 5: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

4x – 5y – 20 = 0 and 3x + 5y – 15 = 0

Solution:

Given equations are 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0

Graph of 4x – 5y – 20 = 0:

4x – 5y – 20 = 0 \( \Rightarrow y = \frac{4x – 20}{5} \)

Putting x = 0, we get y = -4

Putting x = 5, we get y = 0

Putting x = 10, we get y = 4

Hence, the table is:

x 0 5 10
y -4 0 4

Plot the points A(0, -4), B(5, 0) and C(10, 4) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x – 5y – 20 = 0.

Graph of 3x + 5y – 15 = 0:

3x + 5y – 15 = 0 \( \Rightarrow y = \frac{-3x + 15}{5} \)

Putting x = 5, we get y = 0

Putting x = 0, we get y = 3

Putting x = -5, we get y = 6

Hence, the table is:

x 5 0 -5
y 0 3 6

Now, on the same graph paper as above plot the points P(0, 3) and Q(-5, 6). The point B(5, 0) has already been plotted. Join PQ and PB to get the line QB.

https://lh4.googleusercontent.com/cX1guxswgqyV35OnynQbmfhmX5L93PwSPAbsJsKU4i4ns1kn3vWuRGCeSro-ur5kjLoRCJTm13FkvsD6s3I4glgnKsqeP28m_3bwd60tzpxAR3ozFIQZN1JVv-bk-FO-Jb4niFUu

Thus, line QB is the graph of 3x + 5y – 15 = 0. The two graph lines intersect at B(5, 0).

Hence, x = 5, y = 0 is the solution of the given system of equations.

Clearly, the vertices of \( \Delta PBA \) formed by these lines and the y-axis are A(0, -4), B(5, 0) and P(0, 3).

Question 6: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

4x – 3y + 4 = 0,

4x + 3y – 20 = 0

Solution:

Given equations are 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0

Graph of 4x – 3y + 4 = 0:

4x – 3y + 4 = 0 \( \Rightarrow y = \frac{4x + 4}{3} \)

Putting x = -1, we get y = 0

Putting x = 2, we get y = 4

Putting x = 5, we get y = 8

Hence, the table is:

x -1 2 5
y 0 4 8

Plot the points A(-1, 0), B(2, 4) and C(5, 8) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x – 3y + 4 = 0.

Graph of 4x + 3y – 20 = 0:

4x + 3y – 20 = 0 \( \Rightarrow y = \frac{-4x + 20}{3} \)

Putting x = 2, we get y = 4

Putting x = -1, we get y = 8

Putting x = 5, we get y = 0

Hence, the table is:

x 2 -1 5
y 4 8 0

Now, on the same graph paper as above plot the points P(-1, 8) and Q(5, 0). The point B(2, 4) has already been plotted. Join QB and PB to get the line PQ.

https://lh3.googleusercontent.com/5sGDnpwqxah5eqGseOdZuj9HVWUV4ZlI579va3PXLX2gF-iW7XOIOSwgInJXD9djdQxbj1KO81pth2IRAP3F1W-rqKo648gncHP3MFTE9zibOXfUqpvqD6EjhOcOk2ut2lOx4CIw

Thus, line PQ is the graph of 4x + 3y – 20 = 0. The two graph lines intersect at B(2, 4).

Hence, x = 2, y = 4 is the solution of the given system of equations.

Clearly, the vertices of \( \Delta ABQ \) formed by these lines and the x-axis are A(-1, 0), B(2, 4) and Q(5, 0).

Consider the triangle \( \Delta ABQ \) :

Height of the triangle = 4 units and base(AQ) = 6 units

Area of the triangle \( \Delta ABQ \):

\( Area = \frac{1}{2} \times Base \times height = \frac{1}{2} \times 4 \times 6 \) sq. units

Area of \( \Delta ABQ \) = 12 sq. units

Question 7: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

x – y + 1 = 0, 3x + 2y – 12 = 0

Solution:

Given equations are x – y + 1 = 0 and 3x + 2y – 12 = 0

Graph of x – y + 1 = 0:

x – y + 1 = 0 \( \Rightarrow y = x + 1 \)

Putting x = -1, we get y = 0

Putting x = 1, we get y = 2

Putting x = 2, we get y = 3

Hence, the table is,

x -1 1 2
y 0 2 3

Plot the points A(-1, 0), B(1, 2) and C(2, 3) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of x – y + 1 = 0.

Graph of 3x + 2y – 12 = 0:

3x + 2y – 12 = 0 \( \Rightarrow y = \frac{-3x + 12}{2} \)

Putting x = 0, we get y = 6

Putting x = 2, we get y = 3

Putting x = 4, we get y = 0

Hence, the table is:

x 0 2 4
y 6 3 0

Now, on the same graph paper as above plot the points P(0, 6) and Q(4, 0). The point C(2, 3) has already been plotted. Join PC and CQ to get the line PQ.

https://lh5.googleusercontent.com/m9ePQtpSjQZybPedralVqgDds6iQrjBmbCRen6YcWqNC5s2x371QYqHhZ4xlB0ToxVhMDuAyPoxf_uH3Fmml3koAwA3x4xwXvkuF_2gieaudo4DUDaUmoK7IxddV-FQO_WxoVTVN

Thus, line PQ is the graph of 3x + 2y – 12 = 0. The two graph lines intersect at C(2, 3).

Hence, x = 2, y = 3 is the solution of the given system of equations.

Clearly, the vertices of \( \Delta ACQ \) formed by these lines and the x-axis are A(-1, 0), B(2, 3) and Q(4, 0).

Consider the triangle \( \Delta ACQ \) :

Height of the triangle = 3 units and base(AQ) = 5 units

Area of the triangle \( \Delta ACQ \):

\( Area = \frac{1}{2} \times Base \times height = \frac{1}{2} \times 3 \times 5 \) sq. units

Area of \( \Delta ACQ \) = 7.5 sq. units


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Which of the following metal will melt on touching?