RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 27 marks
One equation in two unknowns has endless solutions; pair it with a second and you usually pin the answer down to a single point. Here you learn to read a pair of equations from its coefficients alone — telling at a glance whether the lines cross, run parallel or lie on top of each other — and to solve them by substitution and elimination. Most board questions arrive dressed as word problems about ages, digits, boats and bills.
For which value of k does the pair kx + 3y = 5 and 2x + 6y = 11 have no solution?
Answer
k = 1. There is no solution when a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Here a₁/a₂ = k/2, b₁/b₂ = 3/6 = 1/2 and c₁/c₂ = 5/11. Setting k/2 = 1/2 gives k = 1. We must also check the inequality: 1/2 ≠ 5/11, which is true. So k = 1 makes the lines parallel and the system inconsistent.
The pair of equations 5x - y = 7 and 10x - 2y = 9 represents:
Answer
They represent parallel lines. Compare the coefficients: a₁/a₂ = 5/10 = 1/2, b₁/b₂ = (-1)/(-2) = 1/2 and c₁/c₂ = 7/9. Since a₁/a₂ = b₁/b₂ but this common ratio 1/2 is not equal to 7/9, the lines are parallel. A parallel pair never meets, so the system has no solution and is inconsistent.
The pair of equations 3x - 2y = 5 and 6x - 4y = 10 has:
Answer
It has infinitely many solutions. Here a₁/a₂ = 3/6 = 1/2, b₁/b₂ = (-2)/(-4) = 1/2 and c₁/c₂ = 5/10 = 1/2. All three ratios are equal, so the second equation is just the first multiplied by 2 and both equations describe the same line. Every point on that line satisfies both equations, so there are infinitely many solutions and the pair is consistent and dependent.
The lines x = 4 and y = -3 intersect at the point:
Answer
They intersect at (4, -3). The equation x = 4 is a vertical line on which every point has x-coordinate 4. The equation y = -3 is a horizontal line on which every point has y-coordinate -3. The only point satisfying both conditions has x = 4 and y = -3, i.e. (4, -3). Note that (4, 0) lies on x = 4 only and (0, -3) lies on y = -3 only.
Assertion (A): The pair of equations 2x + 3y = 9 and 4x + 6y = 18 has infinitely many solutions. Reason (R): If a₁/a₂ = b₁/b₂ = c₁/c₂, the two lines are coincident and the pair of equations is consistent and dependent.
Answer
Both A and R are true and R is the correct explanation of A. Reason R correctly states the condition for coincident lines. Applying it: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2 and c₁/c₂ = 9/18 = 1/2. All three ratios are equal, so the lines coincide and the pair has infinitely many solutions, which is precisely Assertion A. Hence R explains A.
Solve the pair of equations 3x + 4y = 10 and 2x - 2y = 2 by the method of elimination.
Answer
Divide the second equation by 2 to simplify: x - y = 1. ...(ii) Multiply (ii) by 4: 4x - 4y = 4. ...(iii) Add (iii) to the first equation 3x + 4y = 10: (3x + 4y) + (4x - 4y) = 10 + 4, so 7x = 14 and x = 2. Put x = 2 into (ii): 2 - y = 1, so y = 1. Solution: x = 2, y = 1. Check: 3(2) + 4(1) = 10 ✔ and 2(2) - 2(1) = 2 ✔
Find the value of k for which the pair 4x + ky = 8 and 12x + 9y = 24 has infinitely many solutions.
Answer
For infinitely many solutions we need a₁/a₂ = b₁/b₂ = c₁/c₂. a₁/a₂ = 4/12 = 1/3 and c₁/c₂ = 8/24 = 1/3, so these two already agree. Now set b₁/b₂ = 1/3: k/9 = 1/3, so k = 3. With k = 3 the first equation is 4x + 3y = 8, and multiplying it by 3 gives exactly 12x + 9y = 24. The lines coincide, so k = 3.
Five pens and three notebooks together cost ₹ 390, while three pens and five notebooks together cost ₹ 410. Find the cost of one pen and one notebook.
Answer
Let the cost of one pen be ₹ p and of one notebook be ₹ n. 5p + 3n = 390 ...(i) 3p + 5n = 410 ...(ii) Add (i) and (ii): 8p + 8n = 800, so p + n = 100. ...(iii) Subtract (i) from (ii): (3p + 5n) - (5p + 3n) = 410 - 390, so -2p + 2n = 20, giving n - p = 10. ...(iv) Add (iii) and (iv): 2n = 110, so n = 55, and then p = 100 - 55 = 45. A pen costs ₹ 45 and a notebook costs ₹ 55. Check: 5(45) + 3(55) = 225 + 165 = 390 ✔ and 3(45) + 5(55) = 135 + 275 = 410 ✔
A motorboat covers 24 km upstream and 36 km downstream in 6 hours. The same boat covers 36 km upstream and 24 km downstream in 6 hours 30 minutes. Find the speed of the boat in still water and the speed of the stream.
Answer
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h. Then speed upstream = (x - y) km/h and speed downstream = (x + y) km/h. Using time = distance/speed: 24/(x - y) + 36/(x + y) = 6 ...(i) 36/(x - y) + 24/(x + y) = 6.5 ...(ii) Step 1 — Substitute p = 1/(x - y) and q = 1/(x + y): 24p + 36q = 6 ...(iii) 36p + 24q = 6.5 ...(iv) Step 2 — Multiply (iii) by 3 and (iv) by 2: 72p + 108q = 18 and 72p + 48q = 13. Subtracting, 60q = 5, so q = 1/12. Step 3 — Put q = 1/12 in (iii): 24p + 36(1/12) = 6, so 24p + 3 = 6, giving 24p = 3 and p = 1/8. Step 4 — Convert back: 1/(x - y) = 1/8 gives x - y = 8, and 1/(x + y) = 1/12 gives x + y = 12. Step 5 — Adding, 2x = 20 so x = 10; subtracting, 2y = 4 so y = 2. The boat's speed in still water is 10 km/h and the stream flows at 2 km/h. Check with (ii): 36/8 + 24/12 = 4.5 + 2 = 6.5 hours ✔
(a) The sum of the digits of a two-digit number is 11. If the digits are reversed, the new number is 27 less than the original number. Find the original number. (b) For what value of m will the pair mx - 5y = 8 and 6x - 15y = 20 have a unique solution?
Answer
(a) Let the tens digit be x and the units digit be y, so the number is 10x + y. Given x + y = 11. ...(i) The reversed number is 10y + x, and it is 27 less than the original: (10x + y) - (10y + x) = 27 9x - 9y = 27, so x - y = 3. ...(ii) Add (i) and (ii): 2x = 14, so x = 7. Then y = 11 - 7 = 4. The original number is 10(7) + 4 = 74. Check: 7 + 4 = 11 ✔ and 74 - 47 = 27 ✔ (b) A unique solution requires a₁/a₂ ≠ b₁/b₂. Here a₁/a₂ = m/6 and b₁/b₂ = (-5)/(-15) = 1/3. We need m/6 ≠ 1/3, i.e. m ≠ 2. So the pair has a unique solution for every real value of m except m = 2.
Read the following and answer the questions that follow: A neighbourhood gym charges each member a fixed monthly registration amount plus a fixed charge for every training session attended in that month. In one month Ritu attended 8 sessions and her bill came to ₹ 1400, while Sameer attended 14 sessions in the same month and his bill came to ₹ 2000. (a) Taking the fixed monthly amount as ₹ x and the charge per session as ₹ y, write the pair of linear equations for the situation. (b) Find the fixed monthly amount and the charge per session. (c) What would a member pay in a month in which she attends 20 sessions? (d) A member's bill for a month was ₹ 3100. How many sessions did he attend?
Answer
(a) Ritu: x + 8y = 1400 ...(i) Sameer: x + 14y = 2000 ...(ii) (b) Subtract (i) from (ii): (x + 14y) - (x + 8y) = 2000 - 1400, so 6y = 600 and y = 100. Put y = 100 in (i): x + 800 = 1400, so x = 600. The fixed monthly amount is ₹ 600 and each session costs ₹ 100. (c) Bill for 20 sessions = 600 + 20 × 100 = 600 + 2000 = ₹ 2600. (d) Let the number of sessions be n. Then 600 + 100n = 3100. 100n = 2500, so n = 25. He attended 25 sessions.
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