Here you get the CBSE Class 10 Mathematics chapter 3, Pair of Linear Equations in Two Variables: NCERT Exemplar Problems and Solutions (Part-IVA). This part of the chapter includes solutions of Question Number 1 to 6 from Exercise 3.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Pair of Linear Equations in Two Variables. This exercise comprises only of the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
NCERT Exemplar Solution for CBSE Class 10 Mathematics: Pair of Linear Equations (Part-I)
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
NCERT Exemplar Solution for CBSE Class 10 Mathematics: Pair of Linear Equations (Part-II)
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Pair of Linear Equations in Two Variables:
Exercise 3.4
Long Answer Type Questions (Q. No. 1 to 6)
Quesntion1. Graphically, solve the following pair of equations
2x + y = 6 and 2x - y + 2 = 0
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the X-axis and the lines with the Y-axis.
Solution:
First equation is: 2x + y = 6
⟹ y = 6 – 2x
If x = 0, y = 6
And if x = 3, y = 0
| x |
| 0 |
| 3 |
| y |
| 6 |
| 0 |
| Points |
| B |
| A |
Second equation is: 2x - y + 2= 0,
⟹ y = 6 – 2x
If x = 0, y = 2
And if x = −1, y = 0
| x | 0 | -1 |
| y | 2 | 0 |
| Points | D | C |
Plotting 2x + y = 6 and 2x - y + 2= 0, as shown below, we obtain two lines AB and CD respectively intersecting at point, E (1, 4).
Hence, the ratio of the areas of the two triangles DACE and DBDE is 4:1
Quesntion2. Determine graphically, the vertices of the triangle formed by the lines
y = x, 3y = x and x + y = 8.
Solution:
First equation is: y = x
If x = 1, y = 1
If x = 0, y = 0
If x = 2, y = 2
| x | 0 | 1 | 2 |
| y | 0 | 1 | 2 |
| Points | 0 | A | 8 |
Second equation is: x = 3y,
If x = 0, y = 0
If x = 3, y = 1
If x = 6, y = 2
| x | 0 | 3 | 6 |
| y | 0 | 1 | 2 |
| Points | 0 | C | D |
Third equation is: x + y = 8
⟹ y = 8 -x
If x = 0, y = 8
If x = 8, y = 0
If x = 4, y = 4
| x | 0 | 4 | 8 |
| y | 8 | 4 | 0 |
| Points | P | Q | R |
Plotting y = x, 3y = x and x + y = 8, we get three lines OQ, OD and PR respectively, as shown below.
We see that lines OQ and OD intersect the line PR on Q and D, respectively.
So, DOQD is formed by these lines. Hence, the vertices of the DOQD formed by the given lines are O(0, 0), Q(4, 4) and D(6,2).
Quesntion3. Draw the graphs of the equations x = 3, x = 5 and 2x - y -4 = 0. Also find the area of the quadrilateral formed by the lines and the X-axis.
Solution:
First equation is: 2x - y- 4 = 0
⟹ y = 2x - 4
If x = 0, y = -4
If x = 2, y = 0
If x = 4, y = 4
| x | 0 | 2 |
| y | -4 | 0 |
| Points | P | Q |
Plotting x = 3 and x = 5 and 2x - y- 4 = 0, we obtain three lines I, II and III respectively forming a quadrilateral ABCD with X-axis as shown below:
Quadrilateral ABCD is a trapezium with AB = OB - OA = 5 - 3 = 2, AD = 2 and BC = 6
Now, Area of trapezium ABCD = 1/2 × distance between parallel lines (AB) × (AD + BC) = 1/2 × 2 × (6 + 2) = 8sq units
Quesntion4. The cost of 4 pens and 4 pencils boxes is Rs.100. Three times the cost of a pen is Rs.15 more than the cost of a pencil box. Form the pair of Linear equations for the above situation. Find the cost of a pen and a pencil box.
Solution:
Let cost of a pen be Rs. x
And cost of a pencil box be Rs. y.
According to question,
4x + 4y = 100
⟹ x + y = 25 ...(i)
Also, 3x = y + 15
⟹ 3x - y = 15 ...(ii)
On adding equations (i) and (ii), we get
4x = 40
⟹ x = 10
By substituting x = 10, in equation (i) we get
y = 25 -10 = 15
Hence, the cost of a pen is Rs. 10 and that of a pencil box is Rs.15.
Quesntion5. Determine, algebraically, the vertices of the triangle formed by the lines 3 x - y = 3; 2x - 3y = 2 and x+ 2y = 8.
Solution:
Given equations are:
3x - 2y = 3 ...(i)
2x - 3y = 2 ...(ii)
x + 2y = 8 ...(iii)
Let lines (i), (ii) and (iii) represent the sides of a DABC i.e., AB, BC and CA, respectively.
On solving lines (i) and (ii), we will get the intersecting point B, solving (ii) and (iii) we will get C and solving (iii) and (i), will give point A.
⟹ y = 6 - 3
⟹ y = 3
So, the coordinate of point or vertex A is (2, 3).
Hence, the vertices of the ΔABC formed by the given lines are A (2, 3), B (1, 0) and C (4, 2)
Quesntion6. Ankita travels 14km to her home partly by rickshaw and partly by bus She takes half an hour, if she travels 2km by rickshaw and the remaining distance by bus. On the other hand, if she travels 4km by rickshaw and the remaining distance by bus, she takes 9min longer. Find the speed of the rickshaw and of the bus.
Solution:
Let the speed of the rickshaw and the bus are x and y km/h, respectively.
Hence, the speed of rickshaw is 10 km/h and that of bus is 40km/h.
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