Here you get the CBSE Class 10 Mathematics chapter 3, Pair of Linear Equations in Two Variables: NCERT Exemplar Problems and Solutions (PartIIIB). This part of the chapter includes solutions of Question Number 12 to 22 from Exercise 3.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Pair of Linear Equations in Two Variables. This exercise comprises of only the Short 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 (PartI)
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 20172018 as well as other competitive exams.
NCERT Exemplar Solution for CBSE Class 10 Mathematics: Pair of Linear Equations (PartII)
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Pair of Linear Equations in Two Variables:
Exercise 3.3
Short Answer Type Questions (Q. No. 12 to 22)
Quesntion12. Draw the graph of the pair of equations 2x + y = 4 and 2x  y = 4. Write the vertices of the triangle formed by these Lines and the Yaxis, find the area of this triangle?
Solution:
First equation is:
2x + y = 4
⟹ y = 4 − 2x
If x = 0, y = 4
And x = 2, y = 0
x 
2 

y 
4 

Points 
A 
B 
Second equation is:
2x  y = 4,
⟹ y = 2x − 4
If x = 0, y = −4
And x = 2, y = 0
x 
2 

y 
−4 

Points 
C 
B 
Plotting graph for 2x + y = 4 and 2x  y = 4, using the respective points given in above two tables, we get two lines AB and CB respectively, both intersecting at B.
Tringle formed by the lines with y–axis is ΔABC. Coordinates of vertices are A(0, 4), B(2, 0) and C(0, −4)
NCERT Exemplar Solution for CBSE Class 10 Mathematics: Pair of Linear Equations (PartIIIA)
Quesntion13. Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x  y = 1, How many such lines can we find?
Solution:
Given pair of linear equations is x + y  2 = 0 and 2x  y 1 = 0
On comparing with equations a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0, we have:
So, the pair of equations has a unique solution.
Now first equation is:
x + y = 2
⟹ y = 2  x
If x = 0, y = 2
And x = 2, y = 0
x 
2 

y 
2 

Points 
A 
B 
Second equation is:
2x  y  1 = 0
⟹ y = 2x  1
If x = 0, y =  1
If x = 1/2, y = 0
And x = 1, y = 1
x 
1/2 
1 

y 
1 
1 

Points 
C 
D 
E 
Plotting x + y = 2 and 2x  y  1 = 0 we get two lines AB and CD respectively intersecting at a point E (1, 1).
We can make infinite linear equations passing through E (1, 1). Some of the linear equations are given below:
(i) y = x,
(ii) 2x + y = 3,
(iii) x + 2y = 3 and so on.
Quesntion14. If (x + 1) is a factor of 2x^{3} + ax^{2} + 2bx + 1, then find the value of a and b given that 2a  3b = 4.
Solution:
Let f(x) = 2x^{3} + ax^{2} + 2bx + 1
Since (x + 1) is a factor of f(x) then by factor theorem, f( 1) = 0.
⟹ 2( 1)^{3} + a( 1)^{2} + 2b(  1) + 1 = 0
⟹ 2 + a2b + 1 = 0
⟹ a 2b  1 = 0 ...(i)
Hence, the required values of a and b are 5 and 2, respectively.
Quesntion15. If the angles of a triangle are x, y and 40° and the difference between the two angles x and y is 30°. Then, find the value of x and y.
Solution:
Given, x, y and 40° are the measure of interior angles of a triangle.
Therefore, x + y + 40° = 180° [Sum of interior angles of a triangle is 180°]
⟹ x + y= 140° ... (i)
Also, x  y = 30° ... (ii)
On adding equations (i) and (ii), we get
2x = 170°
⟹ x = 85°
On putting x = 85° in equation (i), we get
85° + y = 140°
⟹ y = 55°
Hence, the required values of x and y are 85° and 55°, respectively.
Quesntion16. Two years ago, Salim was thrice as old as his daughter and six years Later, he will be four year older than twice her age. How old are they now?
Solution:
Let the present ago of Salim be x years.
And the present age of his daughter be y years.
Age of Salim 2 years ago = (x – 2) years
Age of Salim’s daughter 2 years ago = (y – 2) years
According to the question, we have
x  2 = 3(y  2)
⟹ x  2 = 3y  6
⟹ x 3y =  4 ... (i)
Now, age of Salim 6 years later = (x + 6) years
And age of Salim’s daughter 6 years later = (y + 6) years
According to the second condition given in question, we have:
x + 6 = 2 (y+ 6) + 4
⟹ x + 6 = 2y +12 + 4
⟹ x  2y = 16  6
⟹ x  2y = 10 ...(ii)
On subtracting equation (i) from (ii), we get
Put the value of y in equation (ii), we get
x  2 x 14 = 10
⟹ x = 10 + 28
⟹ x = 38
Hence, present age of Salim = 38 yrs
And present age of his daughter = 14 yr
Quesntion17. The age of the father is twice the sum of the ages of his two children. After 20 year, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the present age of father be x year
And sum of the ages of his two children be y year.
According to the first condition, we have
x = 2y ... (i)
After 20 years father’s age will be = x + 20
And after 20 years sum of the ages of two children = y + 40
Using the second condition given in question, we have:
x + 20 = y + 40
⟹ y = x  20 ….(ii)
On putting the value of y from (ii) in equatoion (i) we get:
x = 2(x 20)
⟹ x = 2x  40
⟹ x = 40
Hence, the father's age is 40 years.
Quesntion18. Two numbers are in the ratio 5:6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5, then find the numbers.
Solution:
Quesntion19. There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B but, if 20 students are sent from B to A, the number of students in A becomes double the number of students in B, then find the number of students in the both halls.
Solution:
Let the number of students in hall A be x.
And the number of students in hall B be y.
By the first condition given in question, we get:
x  10 = y + 10
⟹ x  y = 20 ...(i)
Again, by using the second condition given in question we get:
(x + 20) = 2(y  20)
⟹ x  2y =  60
⟹ (x  y) – y =  60 ...(ii)
Using (i) and (ii) we get:
20 – y =  60
⟹ y = 80
On putting y = 80 in (i), we get
x  80 = 20
⟹ x = 100
and y = 80
Hence, the number of students in hall A and B are 100 and 80 respectively .
Quesntion20. A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days and an additional charge for each day thereafter. Latika paid `22 for a book kept for six days, while Anand paid `16 for the book kept for four days. Find the fixed charges and the charge for each extra day.
Solution:
Let Latika takes a fixed charge for the first two day be Rs. x and additional charge for each day thereafter be Rs. y.
Now by first condition.
Latika paid Rs. 22 for a book kept for six days i.e.,
x + 4y = 22 ...(i)
And by second condition,
Anand paid Rs. 16 for a book kept for four days i.e.,
x + 2y = 16 ...(ii)
Now, subtracting equation (ii) from equation (i), we get
2y = 6
⟹ y = 3
On putting the value of y in eqaution (ii), we get
x + 2 ´ 3= 16
⟹ x = 16  6 = 10
So, the fixed charges for first 2 days = Rs.10
And the additional charges after 2 days = Rs.3 per day
Quesntion21. In a competitive examination, 1 mark is awarded for each correct answer. while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
Solution:
Let the number of questions attempted correctly = x Number of questions answered = 120 So, wrong answer attempted = (120 – x)
Then, by given condition,
Hence, Jayanti answered 100 questions correctly.
Quesntion22. The angles of a cyclic quadrilateral ABCD are ∠A = (6x + 10)°, ∠B = (5x)°, ∠C = (x + y)° and ∠D = (3y  10)°.
Find x and y and hence the values of the four angles.
Solution:
The sum of opposite angles of a cyclic cyclic quadrilateral is 180°.
⟹ ∠A + ∠C = 180°
⟹ (6x + 10)° + (x + y)° = 180°
⟹ 7x + y = 170 ...(i)
Also, ∠B + ∠D = 180°
⟹ (5x)° + (3y  10)° = 180°
⟹ 5x + 3y = 190° ...(ii)
Now, 3 × (i) – (ii) gives:
3 × (7x + y)  (5x + 3y) = 510° 190°
⟹ 21x + 3y  5x 3y = 320°
⟹ 16x = 320°
⟹ x = 20°
Putting x = 20° in equation (i), we get
7 × 20 + y = 170°
⟹ y = 170°140°
⟹ y = 30°
Thus, ∠A = (6x + 10)° = 6 × 20° + 10° = 120° + 10° = 130°
∠B = (5x)° = 5 × 20° = 100
∠C = (x + y)° =20° + 30° = 50°
∠D = (3y10)° = 3 × 30° = 10° = 90° 10° = 80°
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