CBSE Class 10 Pair of Linear Equations in Two Variables Sure Shot Questions Set 02

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1. Find the value of k, so that the following system of equations has no solution:

3x – y – 5 = 0;    6x – 2y – k = 0.

2. Find the value of k, so that the following system of equations has a non-zero solution:

3x + 5y = 0;     kx + 10y = 0.  

Find the value of k, so that the following system of equations has no solution:

3. 3x+y= 1;     (2k- 1)x+ (k- 1)y= (2k- 1). 

4. 3x+y= 1;     (2k- 1)x+ (k- 1)y= (2k+ 1).

5. x-2y =3;     3x ky =1 .

6. x+ 2y= 5;   3x +ky +15 = 0.

7. kx+2y=5;   3x-4y =10.

8. x +2y=3;    5x+ky+7=0.

9. 8x+5y= 9;   kx+10y=15. 

10. (3k+ 1)x+ 3y-2= 0;     ( k2+1)x+ (k- 2)y- 5= 0. 

11. kx+3y=3 ;    12x+ky=6.

Find the value of k, so that the following system of equations has a unique solution:

12. x-2y=3;   3x+ky=1. 

13. x +2y= 5;    3x+ky+15=0 .

14. kx +2y =5;    3x-4y =10.

15. x +2y=3;     5x+ ky+7=0.

16. 8x+ 5y= 9;    kx+10y=15.

17. kx+3y =(k-3);    12x+ ky= k.

18. kx+2y=5;     3x+ y=1.

19. x-2y=3;     3x+ ky=1.

20. 4x- 5y=k ;      2x- 3y= 12.

For what value of k, the following pair of linear equations has infinite number of solutions:

21. kx + 3y = (2k+1); 2(k+1)x + 9y = (7k +1).

22. 2x + 3y = 2; (k + 2)x = (2k +1)y = 2(k -1).

23. x + (2k -1)y = 4; kx + 6y = k + 6.

24. (k +1)x-y = 5; (k +1)x + (1-k) y = (3k +1).

25. x + (k +1)y =5; (k +1)x + 9y = (8k-1).

26. 2x + 3y = 7; (k-1)x + (k + 2) y = 3k.

27. 2x + (k-2) y = k; 6x + (2k-1)y = (2k + 5).

Find the value of a and b for which each of the following systems of linear equations has a infinite number of solutions:

28. (a-1)x + 3y = 2; 6x + (1-2b) y = 6.

29. 2x +3y = 7; (a + b)x - (a + b-3) y = 4a + b.

30. 2x + 3y = 7; (a + b +1)x + (a + 2b + 2) y = 4(a + b) +1.

31. 2x + 3y = 7; a(x + y) - b(x - y) = 3a + b - 2

32. (2a -1)x + 3y = 5; 3x + (b -1)y = 2.

33. Find the value of k, so that the following system of equations has a non-zero solution:
5x – 3y = 0; 2x + ky = 0.

Show that the following system of the equations has a unique solution and hence find the solution of the given system of equations.
34. x/3 + y/2 =3;   x-2y = 2

35. 3x + 5y =12; 5x + 3y = 4.

Very Short Answer 

Question. How many solutions does the pair of equations \( y = 0 \) and \( y = -5 \) have?
Answer: Sol. \( y = 0 \) and \( y = -5 \) are Parallel lines, hence no solution.

Question. If \( ax + by = a^2 - b^2 \) and \( bx + ay = 0 \), find the value of \( (x + y) \).
Answer: Sol. \( ax + by = a^2 - b^2 \)
\( ay + bx = 0 \)
\( a(x + y) + b(x + y) = a^2 - b^2 \) ...[adding
\( (x + y)(a + b) = a^2 - b^2 \)
\( (x + y) = \frac{(a + b)(a - b)}{(a + b)} \)
\( \therefore x + y = a - b \)

Question. For what value of \( k \), the pair of equations \( 4x - 3y = 9, 2x + ky = 11 \) has no solution?
Answer: Sol. We have, \( 4x - 3y = 9 \) and \( 2x + ky = 11 \)
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) (No solution)
\( \frac{4}{2} = \frac{-3}{k} \neq \frac{9}{11} \)
\( \Rightarrow 2 = \frac{-3}{k} \)
\( 2k = -3 \Rightarrow k = \frac{-3}{2} \)

Short Answer-I 

Question. Find whether the following pair of linear equations is consistent or inconsistent:
\( 3x + 2y = 8 \)
\( 6x - 4y = 9 \)
Answer: Sol. Here, \( \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{2}{-4} = \frac{-1}{2} \)
Here, \( \frac{1}{2} \neq -\frac{1}{2} \therefore \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (unique solution)
Therefore, given pair of linear equations is consistent.

Short Answer-II 

Question. Solve the following pair of equations for x and y:
\( \frac{a^2}{x} - \frac{b^2}{y} = 0; \frac{a^2b}{x} + \frac{b^2a}{y} = a + b, x \neq 0; y \neq 0 \)
Answer: Sol. \( \frac{a^2}{x} - \frac{b^2}{y} = 0 \) ...(i)
\( \frac{a^2b}{x} + \frac{b^2a}{y} = a + b \) ...(ii)
Multiplying (i) by \( a \) and (ii) by 1,
\( \frac{a^3}{x} - \frac{b^2a}{y} = 0 \)
\( \frac{a^2b}{x} + \frac{b^2a}{y} = a + b \)
\( \frac{a^3 + a^2b}{x} = a + b \) ...[By adding
\( \frac{a^2(a + b)}{x} = \frac{a + b}{1} \)
\( x(a + b) = a^2(a + b) \therefore x = a^2 \)
Putting the value of \( x \) in (i), we get \( \frac{a^2}{a^2} - \frac{b^2}{y} = 0 \Rightarrow 1 = \frac{b^2}{y} \Rightarrow y = b^2 \)

Question. Solve for x and y:
\( \frac{10}{x + y} + \frac{2}{x - y} = 4; \frac{15}{x + y} - \frac{5}{x - y} = -2; x + y \neq 0, x - y \neq 0 \)
Answer: Sol. \( \frac{10}{x + y} + \frac{2}{x - y} = 4 \) ...(i)
\( \frac{15}{x + y} - \frac{5}{x - y} = -2 \) ...(ii)
Multiplying (i) by 5 and (ii) by 2, we get
\( \frac{50}{x + y} + \frac{10}{x - y} = 20 \)
\( \frac{30}{x + y} - \frac{10}{x - y} = -4 \)
\( \frac{80}{x + y} = 16 \) ...[By adding
\( 16(x + y) = 80 \Rightarrow x + y = 5 \) ...(iii)
Putting the value of \( (x + y) \) in (i), we get \( \frac{10}{5} + \frac{2}{x - y} = 4 \Rightarrow 2 + \frac{2}{x - y} = 4 \)
\( \frac{2}{x - y} = 4 - 2 = 2 \Rightarrow 2(x - y) = 2 \therefore x - y = 1 \) ...(iv)
Solving (iii) and (iv), we get \( 2x = 6 \Rightarrow x = 3 \)
Putting the value of \( x \) in (iii), we get \( 3 + y = 5 \Rightarrow y = 2 \)
\( \therefore x = 3, y = 2 \)

Question. Solve the following pair of linear equations for x and y:
\( 141x + 93y = 189; \)
\( 93x + 141y = 45 \)
Answer: Sol. \( 141x + 93y = 189 \)
\( 93x + 141y = 45 \)
\( 234x + 234y = 234 \) ...[By adding
\( \Rightarrow x + y = 1 \) ...(i) (\( \div \) by 234)
Again
\( 141x + 93y = 189 \)
\( 93x + 141y = 45 \)
\( 48x - 48y = 144 \) ...[By subtracting
\( \Rightarrow x - y = 3 \) ...(ii) (\( \div \) by 48)
By adding (i) and (ii), we get \( 2x = 4 \Rightarrow x = 2 \)
Putting the value of \( x \) in (i), we get \( 2 + y = 1 \Rightarrow y = -1 \)
\( \therefore x = 2, y = -1 \)

Question. Solve the following pair of linear equations for x and y:
\( \frac{b}{a}x + \frac{a}{b}y = a^2 + b^2; x + y = 2ab \)
Answer: Sol. \( \frac{b}{a}x + \frac{a}{b}y = a^2 + b^2 \) ...(i)
\( x + y = 2ab \) ...(ii)
Multiplying (ii) by \( b/a \), we get
\( \frac{b}{a}x + \frac{b}{a}y = 2b^2 \)
Subtracting this from (i):
\( (\frac{a}{b} - \frac{b}{a})y = a^2 + b^2 - 2b^2 \)
\( (\frac{a^2 - b^2}{ab})y = (a^2 - b^2) \)
\( y = ab \)
Putting the value of \( y \) in (ii), we get \( x + ab = 2ab \Rightarrow x = ab \)
\( \therefore x = ab, y = ab \)

Question. Solve by elimination:
\( 3x = y + 5 \)
\( 5x - y = 11 \)
Answer: Sol. We have, \( 3x - y = 5 \) ...(i) and \( 5x - y = 11 \) ...(ii)
Subtracting (i) from (ii): \( 2x = 6 \Rightarrow x = 3 \)
Putting \( x = 3 \) in (i): \( 3(3) - y = 5 \Rightarrow 9 - 5 = y \therefore y = 4 \)
\( \therefore x = 3, y = 4 \)

Question. Solve by elimination:
\( 3x - y = 7 \)
\( 2x + 5y + 1 = 0 \)
Answer: Sol. \( 3x - y = 7 \) ...(i) and \( 2x + 5y = -1 \) ...(ii)
Multiplying equation (i) by 5: \( 15x - 5y = 35 \)
Adding this to (ii): \( 17x = 34 \Rightarrow x = 2 \)
Putting \( x = 2 \) in (i): \( 3(2) - y = 7 \Rightarrow 6 - 7 = y \therefore y = -1 \)
\( \therefore x = 2, y = -1 \)

Question. Solve for x and y:
\( 27x + 31y = 85; \)
\( 31x + 27y = 89 \)
Answer: Sol. Adding equations: \( 58x + 58y = 174 \Rightarrow x + y = 3 \) ...(i)
Subtracting equations: \( -4x + 4y = -4 \Rightarrow x - y = 1 \) ...(ii)
Adding (i) and (ii): \( 2x = 4 \Rightarrow x = 2 \)
Putting \( x = 2 \) in (i): \( 2 + y = 3 \Rightarrow y = 1 \)
\( \therefore x = 2, y = 1 \)

Question. Solve for x and y: \( \frac{x}{a} = \frac{y}{b}; ax + by = a^2 + b^2 \)
Answer: Sol. \( \frac{x}{a} - \frac{y}{b} = 0 \Rightarrow bx - ay = 0 \) ...(i)
\( ax + by = a^2 + b^2 \) ...(ii)
Multiplying (i) by \( b \) and (ii) by \( a \),
\( b^2x - aby = 0 \)
\( a^2x + aby = a(a^2 + b^2) \)
Adding: \( x(b^2 + a^2) = a(a^2 + b^2) \Rightarrow x = a \)
Putting \( x = a \) in (i): \( b(a) - ay = 0 \Rightarrow ay = ab \therefore y = b \)
\( \therefore x = a, y = b \)

Question. Solve: \( \frac{x}{a} + \frac{y}{b} = a + b; \frac{x}{a^2} + \frac{y}{b^2} = 2, a, b \neq 0 \)
Answer: Sol. \( \frac{bx + ay}{ab} = a + b \Rightarrow bx + ay = a^2b + ab^2 \) ...(i)
\( \frac{b^2x + a^2y}{a^2b^2} = 2 \Rightarrow b^2x + a^2y = 2a^2b^2 \) ...(ii)
Multiplying (i) by \( b \) and subtracting (ii) from it:
\( (b^2x + aby) - (b^2x + a^2y) = (a^2b^2 + ab^3) - 2a^2b^2 \)
\( ay(b - a) = ab^2(b - a) \therefore y = b^2 \)
Putting \( y = b^2 \) in (ii): \( \frac{x}{a^2} + 1 = 2 \Rightarrow \frac{x}{a^2} = 1 \therefore x = a^2 \)
\( \therefore x = a^2, y = b^2 \)

Question. Solve the following pair of equations:
\( 49x + 51y = 499 \)
\( 51x + 49y = 501 \)
Answer: Sol. Adding: \( 100x + 100y = 1000 \Rightarrow x + y = 10 \) ...(i)
Subtracting: \( -2x + 2y = -2 \Rightarrow -x + y = -1 \) ...(ii)
Solving (i) and (ii): \( 2y = 9 \Rightarrow y = \frac{9}{2} \)
Putting \( y = \frac{9}{2} \) in (i): \( x = 10 - \frac{9}{2} = \frac{11}{2} \)
\( \therefore x = \frac{11}{2}, y = \frac{9}{2} \)

Question. Find the two numbers whose sum is 75 and difference is 15.
Answer: Sol. Let the two numbers be \( x \) and \( y \).
\( x + y = 75 \) ...(i)
\( x - y = 15 \) ...(ii)
Adding: \( 2x = 90 \Rightarrow x = 45 \)
Putting \( x = 45 \) in (i): \( 45 + y = 75 \Rightarrow y = 30 \)
\( \therefore \) Numbers are 45, 30.

Question. Find the value of \( \alpha \) and \( \beta \) for which the following pair of linear equations has infinite number of solutions:
\( 2x + 3y = 7; \alpha x + (\alpha + \beta)y = 28 \)
Answer: Sol. For infinite solutions: \( \frac{2}{\alpha} = \frac{3}{\alpha + \beta} = \frac{7}{28} = \frac{1}{4} \)
\( \frac{2}{\alpha} = \frac{1}{4} \Rightarrow \alpha = 8 \)
\( \frac{3}{\alpha + \beta} = \frac{1}{4} \Rightarrow \alpha + \beta = 12 \)
\( 8 + \beta = 12 \Rightarrow \beta = 4 \)
\( \therefore \alpha = 8, \beta = 4 \)

Question. Solve the following pair of linear equations by the cross multiplication method:
\( x + 2y = 2; x - 3y = 7 \)
Answer: Sol. \( x + 2y - 2 = 0 \) and \( x - 3y - 7 = 0 \)
\( \frac{x}{2(-7) - (-3)(-2)} = \frac{y}{(-2)(1) - (-7)(1)} = \frac{1}{1(-3) - (1)(2)} \)
\( \frac{x}{-14 - 6} = \frac{y}{-2 + 7} = \frac{1}{-3 - 2} \)
\( \frac{x}{-20} = \frac{y}{5} = \frac{1}{-5} \)
\( x = \frac{-20}{-5} = 4 \) and \( y = \frac{5}{-5} = -1 \)
\( \therefore x = 4, y = -1 \)

Question. A man earns Rs 600 per month more than his wife. One-tenth of the man’s salary and 1/6th of the wife’s salary amount to Rs 1,500, which is saved every month. Find their incomes.
Answer: Sol. Let wife’s monthly income \( = \text{Rs } x \)
Then man’s monthly income \( = \text{Rs }(x + 600) \)
According to question: \( \frac{1}{10}(x + 600) + \frac{1}{6}(x) = 1500 \)
\( \frac{3(x + 600) + 5x}{30} = 1500 \)
\( 3x + 1800 + 5x = 45000 \Rightarrow 8x = 43200 \)
\( x = \frac{43200}{8} = 5400 \)
Wife’s income \( = \text{Rs } 5,400 \)
Man’s income \( = \text{Rs } (5400 + 600) = \text{Rs } 6,000 \)

Long Answer 

Question. The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.
Answer: Sol. Let unit and tens digit be \( x \) and \( y \).
Original number \( = x + 10y \)
Reversed number \( = 10x + y \)
According to question: \( x + y = 8 \Rightarrow y = 8 - x \) ...(ii)
Also, \( (1x + 10y) - (10x + y) = 18 \Rightarrow 9y - 9x = 18 \Rightarrow y - x = 2 \)
Substituting \( y \): \( (8 - x) - x = 2 \Rightarrow 8 - 2 = 2x \Rightarrow 2x = 6 \therefore x = 3 \)
\( y = 8 - 3 = 5 \)
Original number \( = 3 + 10(5) = 53 \)

Question. The age of the father is twice the sum of the ages of his 2 children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Answer: Sol. Let present ages of children be \( x \) and \( y \). Present age of father \( = 2(x + y) \) ...(i)
After 20 years: Father's age \( = 2(x + y) + 20 \). Children's ages sum \( = (x + 20) + (y + 20) \)
According to Question: \( 2(x + y) + 20 = x + y + 40 \)
\( x + y = 20 \)
Present age of father \( = 2(20) = 40 \text{ years} \)

Question. A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.
Answer: Sol. Let unit’s digit be \( x \) and ten’s digit \( y \). Original number \( = x + 10y \)
\( x + 10y = 7(x + y) \Rightarrow 3y = 6x \Rightarrow y = 2x \) ...(i)
\( (x + 10y) - (10x + y) = 18 \Rightarrow 9y - 9x = 18 \Rightarrow y - x = 2 \)
Substituting \( y \): \( 2x - x = 2 \Rightarrow x = 2 \). Then \( y = 4 \).
Original number \( = 2 + 10(4) = 42 \)

Question. Sita Devi wants to make a rectangular pond... The area of the pond will be decreased by 3 square feet if its length is decreased by 2 ft. and breadth is increased by 1 ft. Its area will be increased by 4 square feet if the length is increased by 1 ft. and breadth remains same. Find the dimensions of the pond.
Answer: Sol. Let length \( = x \), breadth \( = y \). Area \( = xy \)
According to Question: \( (x - 2)(y + 1) = xy - 3 \Rightarrow xy + x - 2y - 2 = xy - 3 \Rightarrow x - 2y = -1 \) ...(i)
Also, \( (x + 1)y = xy + 4 \Rightarrow xy + y = xy + 4 \therefore y = 4 \)
Putting \( y = 4 \) in (i): \( x - 2(4) = -1 \Rightarrow x = 7 \)
Length \( = 7 \text{ ft} \), Breadth \( = 4 \text{ ft} \)

Question. On reversing the digits of a two digit number, number obtained is 9 less than three times the original number. If difference of these two numbers is 45, find the original number.
Answer: Sol. Let unit digit be \( x \), tens digit \( y \). Original number \( = x + 10y \)
\( 10x + y = 3(x + 10y) - 9 \Rightarrow 7x - 29y = -9 \) ...(i)
\( (10x + y) - (x + 10y) = 45 \Rightarrow 9x - 9y = 45 \Rightarrow x - y = 5 \Rightarrow x = 5 + y \) ...(ii)
Solving (i) and (ii): \( 7(5 + y) - 29y = -9 \Rightarrow 35 + 7y - 29y = -9 \Rightarrow -22y = -44 \therefore y = 2 \)
Then \( x = 5 + 2 = 7 \). Original number \( = 27 \)

Question. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Answer: Sol. Let speed of stream \( = x \text{ km/hr} \). Speed upstream \( = 15 - x \), downstream \( = 15 + x \)
According to question: \( \frac{30}{15 - x} + \frac{30}{15 + x} = 4.5 \text{ hrs} \)
\( 30 \left[ \frac{15 + x + 15 - x}{225 - x^2} \right] = \frac{9}{2} \Rightarrow \frac{30 \times 30}{225 - x^2} = \frac{9}{2} \)
\( 1800 = 9(225 - x^2) \Rightarrow 200 = 225 - x^2 \Rightarrow x^2 = 25 \therefore x = 5 \text{ km/hr} \)

Question. Taxi charges in city comprises of fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and for journey of 20 km, the charge paid is Rs 145. What will a person have to pay for travelling a distance of 30 km?
Answer: Sol. Let fixed charges \( = \text{Rs } x \), charge per km \( = \text{Rs } y \)
\( x + 12y = 89 \) ...(i)
\( x + 20y = 145 \) ...(ii)
Subtracting (i) from (ii): \( 8y = 56 \Rightarrow y = 7 \)
\( x + 12(7) = 89 \Rightarrow x + 84 = 89 \therefore x = 5 \)
Total fare for 30 km \( = x + 30y = 5 + 30(7) = 5 + 210 = \text{Rs } 215 \)

Question. A boat takes 4 hours to go 44 km downstream and it can go 20 km upstream in the same time. Find the speed of the stream and that of the boat in still water.
Answer: Sol. Let boat speed \( = x \), stream speed \( = y \)
\( \frac{44}{x + y} = 4 \Rightarrow x + y = 11 \) ...(i)
\( \frac{20}{x - y} = 4 \Rightarrow x - y = 5 \)
Solving: \( 2x = 16 \Rightarrow x = 8 \text{ km/hr} \). Then \( y = 3 \text{ km/hr} \).
Boat speed \( = 8 \text{ km/hr} \), Stream speed \( = 3 \text{ km/hr} \)

Question. A man travels 300 km partly by train and partly by car. He takes 4 hours if he travels 60 km by train and the rest by car. If he travels 100 km by train and the remaining by car, he takes 10 minutes longer. Find the speeds of the train and the car separately.
Answer: Sol. Let train speed \( = x \text{ km/hr} \), car speed \( = y \text{ km/hr} \)
\( \frac{60}{x} + \frac{240}{y} = 4 \) ...(i)
\( \frac{100}{x} + \frac{200}{y} = 4 \text{ hr } 10 \text{ min} = \frac{25}{6} \) ...(ii)
Solving: \( x = 60 \text{ km/hr}, y = 80 \text{ km/hr} \)

Please click the link below to download CBSE Class 10 Pair of Linear Equations Sure Shot Questions Set B.