Question 1

If x = 1 and y = 2 is a solution to the system of linear equations 2x - 3y + a = 0 and 2x + 3y - b = 0, which of the following relationships between a and b is correct?

Accepted Answer

Answer: 2a = b. Substituting x=1 and y=2 into the first equation gives 2(1) - 3(2) + a = 0, which simplifies to a = 4. Substituting into the second equation gives 2(1) + 3(2) - b = 0, which simplifies to b = 8. The relationship 2a = b holds true since 2(4) = 8.

Question 2

Determine the value of 'k' for which the pair of linear equations 6x + y = 3k and 36x + 6y = 3 has an infinite number of solutions.

Accepted Answer

Answer: 1/6. For a system to have infinitely many solutions, the ratios of the corresponding coefficients must be equal (a₁/a₂ = b₁/b₂ = c₁/c₂). Here, 6/36 = 1/6 and 3k/3 = k. Setting 1/6 = k gives the required value.

Question 3

Determine the nature of the solution for the pair of linear equations x + 2y + 5 = 0 and -3x = 6y - 1.

Accepted Answer

Answer: no solutions. Rewriting the second equation as -3x - 6y + 1 = 0, we compare the coefficient ratios. a₁/a₂ = 1/(-3), b₁/b₂ = 2/(-6) = -1/3, and c₁/c₂ = 5/1. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and have no solution.

Question 4

Find the value of k for which the pair of linear equations 5x + 2y - 7 = 0 and 2x + ky + 1 = 0 has no solution.

Accepted Answer

Answer: 4/5. For a pair of linear equations to have no solution, the condition is a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Applying this, we get 5/2 = 2/k. Solving for k gives 5k = 4, so k = 4/5.

Question 5

If the pair of linear equations x - y = 1 and x + ky = 5 has a unique solution at x = 2, y = 1, find the value of k.

Accepted Answer

Answer: 3. Substituting the given solution x = 2 and y = 1 into the second equation, x + ky = 5, we get 2 + k(1) = 5. Solving for k gives k = 5 - 2 = 3.

Question 6

A father's current age is three times his son's current age. In 12 years, the father will be twice as old as his son. What is the sum of their present ages?

Accepted Answer

Answer: 48 years. Let the father's age be x and the son's age be y. The equations are x = 3y and x + 12 = 2(y + 12). Solving these gives x = 36 and y = 12, so their sum is 36 + 12 = 48.

Question 7

Given the system of linear equations 17x - 19y = 53 and 19x - 17y = 55, find the value of (x + y).

Accepted Answer

Answer: 1. Subtracting the first equation from the second gives (19x - 17x) - (17y - 19y) = 55 - 53, which simplifies to 2x + 2y = 2, and therefore x + y = 1.

Question 8

Determine the solution for the pair of linear equations given by x = -5 and y = 6.

Accepted Answer

Answer: (-5, 6). The equations explicitly state the values x = -5 and y = 6, so the solution is the coordinate point (-5, 6).

Question 9

Find the values of x and y that satisfy the pair of linear equations 32x + 33y = 34 and 33x + 32y = 31.

Accepted Answer

Answer: x = -1, y = 2. Adding the two equations gives 65x + 65y = 65, simplifying to x + y = 1. Subtracting the second equation from the first gives -x + y = 3. Solving these new equations yields y = 2 and x = -1.

Question 10

Find the value of k for which the linear equations 3x + 5y = 8 and kx + 15y = 24 have an infinite number of solutions.

Accepted Answer

Answer: 9. For infinitely many solutions, the condition is a₁/a₂ = b₁/b₂ = c₁/c₂. This gives 3/k = 5/15 = 8/24. From 5/15 = 1/3, we get 3/k = 1/3, which implies k = 9.

Question 11

One of two parallel lines is represented by the equation 3x - 2y = 5. Which of the following could be the equation of the second line?

Accepted Answer

Answer: -12x + 8y = 7. For parallel lines, a₁/a₂ = b₁/b₂ ≠ c₁/c₂. For the line 3x - 2y = 5 and -12x + 8y = 7, we have 3/(-12) = -1/4 and -2/8 = -1/4, so a₁/a₂ = b₁/b₂. Also, 5/7 ≠ -1/4. Thus, the lines are parallel.

Question 12

A bookstore's lending library has a fixed charge for the first two days and an additional charge for each day thereafter. Amruta paid ₹22 for a book kept for 6 days, while Radhika paid ₹16 for a book kept for 4 days. Assuming the fixed charge is ₹x and the additional charge per day is ₹y, which of the following equations algebraically represents the amount paid by Radhika?

Accepted Answer

Answer: x + 2y = 16. The fixed charge ₹x is for the first 2 days. Radhika kept the book for 4 days, so she pays an additional charge for (4 - 2) = 2 days, which is 2y. The total charge is the sum of the fixed charge and the additional charges, leading to the equation x + 2y = 16.

Question 13

A bookstore's lending library has a fixed charge for the first two days and an additional charge for each day thereafter. Amruta paid ₹22 for a book kept for 6 days, while Radhika paid ₹16 for a book kept for 4 days. Assuming the fixed charge is ₹x and the additional charge per day is ₹y, which of the following equations algebraically represents the amount paid by Amruta?

Accepted Answer

Answer: x + 4y = 22. The fixed charge ₹x covers the first 2 days. Amruta kept the book for 6 days, incurring an additional charge for (6 - 2) = 4 days, which amounts to 4y. The total payment is the sum of the fixed and additional charges, resulting in the equation x + 4y = 22.

Question 14

A bookstore's lending library has a fixed charge for the first two days and an additional charge for each day thereafter. Amruta paid ₹22 for a book kept for 6 days, and Radhika paid ₹16 for a book kept for 4 days. What is the fixed charge for a book?

Accepted Answer

Answer: ₹ 10. The two linear equations representing the situation are x + 4y = 22 and x + 2y = 16. Solving this system by subtracting the second equation from the first gives 2y = 6, so y = 3. Substituting y=3 into the second equation gives x + 2(3) = 16, which simplifies to x = 10. The fixed charge (x) is ₹10.

Question 15

A bookstore's lending library has a fixed charge for the first two days and an additional charge for each day thereafter. Amruta paid ₹22 for a book kept for 6 days, and Radhika paid ₹16 for a book kept for 4 days. What would be the total amount paid by both if each of them had kept their book for 2 more days?

Accepted Answer

Answer: ₹ 50. The fixed charge is ₹10 and the additional daily charge is ₹3. Amruta's new rental period is 6+2=8 days, costing 10 + (8-2)*3 = ₹28. Radhika's new rental period is 4+2=6 days, costing 10 + (6-2)*3 = ₹22. The total amount is ₹28 + ₹22 = ₹50.

Class 10 Maths: Algebraic Methods of Solving a Pair of Linear Equations — Practice Questions with Answers

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