Important Questions for CBSE Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables 3 Mark Question

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CBSE Class 10 Maths Chapter-3 Pair of Linear Equations in Two Variables Important Questions

CBSE Class 10 Maths Important Questions Chapter 3 – Pair of Linear Equations in Two Variables

3 Mark Questions

1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Ans. Let the present age of Aftab and his daughter be x and y respectively.
Seven years ago, Age of Aftab = x – 7 and Age of his daughter = y – 7
According to the given condition,

 
 
Again, Three years hence, Age of Aftab = x + 3 and Age of his daughter = y + 3
According to the given condition,

 
 
Thus, the given conditions can be algebraically represented as:
x – 7y = –42
 x = –42 + 7y
Three solutions of this equation can be written in a table as follows:

And x – 3y = 6   x = 6 + 3y
Three solutions of this equation can be written in a table as follows:

Concept insight: In order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.

2. Form the pair of linear equations in the following problems, and find their solutions graphically. 
(i) 10 students of class X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Ans. (i) Let number of boys who took part in the quiz = x
Let number of girls who took part in the quiz = y
According to given conditions, we have
x+y=10 … (1)
And, y=x+4
⇒ x– y=−4 … (2)
For equation x + y = 10, we have following points which lie on the line

For equation x – y = –4, we have following points which lie on the line

We plot the points for both of the equations to find the solution.

We can clearly see that the intersection point of two lines is (3, 7).
Therefore, number of boys who took park in the quiz = 3 and, number of girls who took part in the quiz = 7.
(ii) Let cost of one pencil = Rs x and Let cost of one pen = Rs y
According to given conditions, we have
5x+7y=50 … (1)
7x+5y=46 … (2)
For equation 5x + 7y = 50, we have following points which lie on the line

For equation 7x + 5y = 46, we have following points which lie on the line

We can clearly see that the intersection point of two lines is (3, 5).
Therefore, cost of pencil = Rs 3 and, cost of pen = Rs 5

3. On comparing the ratios , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x − 4y + 8 = 0
7x+ 6y – 9 = 0
(ii) 9x+ 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x − 3y + 10 = 0
2x– y + 9 = 0
Ans. (i) 5x−4y+8= 0, 7x+6y– 9=0
Comparing equation 5x−4y+8=0 with a₁x+b₁y+c₁=0and 7x+6y– 9=0 with a₂x+b₂y+c₂=0,
We get a₁=5, b₁=−4, c₁=8, a₂=7, b₂=6, c₂=−9
We have because 
Hence, lines have unique solution which means they intersect at one point.
(ii) 9x+3y+12=0, 18x+6y+24=0
Comparing equation 9x+3y+12=0 with a₁x+b₁y+c₁=0and 18x+6y+24=0 with a₂x+b₂y+c₂=0,
We get, a₁= 9,b₁= 3,c₁= 12, a₂= 18,b₂=6,c₂= 24
We have  because 
Hence, lines are coincident.
(iii) 6x−3y+10=0, 2x– y+9=0
Comparing equation 6x−3y+10=0 with a₁x+b₁y+c₁=0and 2x– y+9=0 with a₂x+b₂y+c₂=0,
We get, a₁= 6,b₁= –3,c₁= 10, a₂= 2,b₂= –1,c₂= 9
We have  because 
Hence, lines are parallel to each other.

4. Given the linear equation (2x+3y – 8=0), write another linear equation in two variables such that the geometrical representation of the pair so formed is: 
(i) Intersecting lines
(ii) Parallel lines
(iii) Coincident lines
Ans. (i) Let the second line be equal to a₂x+b₂y+c₂=0
Comparing given line 2x+3y – 8=0 with a₁x+b₁y+c₁=0,
We geta₁=2,b₁=3 and c₁=−8
Two lines intersect with each other if 
So, second equation can be x+2y=3 because 
(ii) Let the second line be equal to a₂x+b₂y+c₂=0
Comparing given line 2x+3y – 8=0 with a₁x+b₁y+c₁=0,
We geta₁=2,b₁=3 and c₁=−8
Two lines are parallel to each other if 
So, second equation can be 2x+3y – 2=0 because 
(iii) Let the second line be equal to a₂x+b₂y+c₂=0
Comparing given line 2x+3y – 8=0 with a₁x+b₁y+c₁=0,
We get a₁=2,b₁=3 and c₁=−8
Two lines are coincident if 
So, second equation can be 4x+6y – 16=0 because 

5. Solve 2x+3y=11 and 2x−4y=−24 and hence find the value of ‘m’ for which
 y=mx+3. 
Ans. 2x+3y=11 … (1)
2x−4y=−24 … (2)
Using equation (2), we can say that
2x=−24+4y
⇒ x=−12+2y
Putting this in equation (1), we get
2(−12+2y)+3y=11
⇒ −24+4y+3y=11
⇒ 7y=35
⇒ y=5
Putting value of y in equation (1), we get
2x+3(5)=11
⇒ 2x+15=11
⇒ 2x=11 – 15=−4
⇒ x=−2
Therefore, x=−2 and y=5
Putting values of x and y in y=mx+3, we get
5=m(−2)+3
⇒ 5=−2m+3
⇒ −2m=2
⇒ m=−1

6. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 
2x + 3y = 7
(a− b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k − 1) x + (k − 1) y = 2k + 1
Ans. (i) Comparing equation 2x + 3y – 7 = 0 with a₁x + b₁y + c₁ = 0 and (a − b) x + (a + b)
y− 3a – b + 2 = 0 with a₂x + b₂y + c₂ = 0
We get a₁ = 2, b₁=3 and c₁ = −7, a₂ = (a − b), b₂ = (a + b) and c₂ = 2 – b − 3a
Linear equations have infinite many solutions if 
⇒ 
⇒ 
⇒ 2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b
⇒ a = 5b  … (1) and −2a = −4b – 6  … (2)
Putting (1) in (2), we get
−2 (5b) = −4b – 6
⇒ −10b + 4b = −6
⇒ −6b = –6   ⇒ b = 1
Putting value of b in (1), we get
a = 5b = 5 (1) = 5
Therefore, a = 5 and b = 1
(ii) Comparing (3x + y – 1 = 0) with a₁x + b₁y + c₁ = 0 and (2k − 1)x+ (k − 1)y −2k – 1 = 0) with a₂x + b₂y + c₂=0,
We get a₁ = 3,b₁ = 1 and c₁ = −1, a₂ = (2k − 1), b₂ = (k − 1) and c₂ = −2k − 1
Linear equations have no solution if 
⇒ 
⇒ 
⇒ 3 (k − 1) = 2k – 1
⇒ 3k – 3 = 2k − 1
⇒ k = 2

7. 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.
Ans. Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Since Speed = 
 
   … (1)
According to the question

 
 
  ……(2)  [Using eq. (1)]
Again, 
 
 
  ……(3)  [Using eq. (1)]
Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(–2) x (50) + 10t = 20
 –100 + 10t = 20
 10t = 120    t = 12
From equation (1), we obtain:
d = xt = 50 x 12 = 600
Thus, the distance covered by the train is 600 km.

8. Draw the graphs of the equations   and  Determine the co-ordinate of the vertices of the triangle formed by these lines and the  axis.
Ans. 
 
Three solutions of this equation can be written in a table as follows:

It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, –3), C (0, –5).

9. ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Ans. We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
 A + C = 180°
 
 
  ………(1)
Also B + D = 180°
 
  ………(2)
Multiplying equation (1) by 3, we obtain:
 ……….(3)
Adding equations (2) and (3), we obtain:

 
Substituting the value of in equation (1), we obtain:

 
 A = 
B = 
C = 
D = 

10. Draw graphs of the equations on the same graph paper 2x + 3y = 12, x – y = 1. Find the area and co-ordinate of the vertices of the triangle formed by the two straight lines and the y-axis.
Ans. 

11. Solve:  and 
Ans. 

12. The sum of a two-digit number and the number obtained by reversing the order of digits is 99. If the digits differ by 3, find the number. 
Ans. 

13. In a cyclic quadrilateral ABCD, , and  Find the four angles.
Ans. 

14. A two-digit number is obtained by either multiplying the sum of the digits by 8 and adding 1 or by multiplying number. How many such numbers are there?
Ans. 

15. A leading library has a fixed charge for the first three days and an additional charge for each day thereafter Sarika paid Rs. 27 for a book kept for seven days, while Sury paid Rs. 21 for the book she kept for five days, find the fixed charge and the charge for each extra day.
Ans. 

16. If 2 is added to the numerator of a fraction, it reduces to  and if 1 is subtracted from the denominator, it reduces to . Find the fraction.
Ans. 

17. Solve the following system of equation graphically.
x + 2y = 1, x – 2y = -7, also read the points from the graph where the lines meet the x-axis and y-axis.
 Ans. 

18. Solve 23x – 29y = 98 and 29x – 23y = 110.
Ans. 

19. A man has only 20 paise coins and 25 paisea coins in his purse. If he has 50 coins in all totaling Rs 11.25. How many coins of each kind does he have? 
Ans. 

20. A says to B my present age is five times your that age when I was an old as you are now. It the sum of their present ages is 48 years, find their present ages.
Ans. 

21. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km down stream. Determine the speed of the stream and that of the boat in still water.
Ans. 

22. Determine graphically the coordinates of the vertices of the triangle, the equation of whose sides are y = x, 3y = x and x + y = 8.
Ans. 

23. Father’s age is three times the sum of ages of his two children. After 5 years, his age will be twice the sum of ages of two children. Find the age of father.
Ans. 

24. On selling a T.V. at 5% gain and a fridge at 10% gain shop keeper gains Rs 2000. But if he sells the T.V at 10% gain and the Fridge at 5% loss, he gains Rs 1500 on the transaction. Find the actual Price of TV and Fridge.
Ans. Let the selling price of TV = Rs x

25. A taken 3 hours more than B to walk a distance of 30 km. But if A doubles his speed, he is ahead of B by  hours. Find their original speed.
Ans.