IMO Maths Olympiad Class 10 Chapter 03 Pair of Linear Equations in Two Variables MCQs with Answers

Answer: (a) 57

Explanation:
1. Let the tens digit be x and the units digit be y; the number is 10x + y and x + y = 12 …(1)
2. (10y + x) − (10x + y) = 18 ⇒ 9y − 9x = 18 ⇒ y − x = 2 …(2)
3. From (1) and (2), y = 7 and x = 5, so the number is 57.

Teacher's Note:
1. Interchanging the digits changes the number by 9 × (difference of the digits).
2. Check: 75 − 57 = 18 and 5 + 7 = 12.

Answer: (b) Has no solution.

Explanation:
1. am = bl gives \( \dfrac{a}{l} = \dfrac{b}{m} \).
2. bn ≠ cm gives \( \dfrac{b}{m} \neq \dfrac{c}{n} \).
3. So \( \dfrac{a}{l} = \dfrac{b}{m} \neq \dfrac{c}{n} \): the lines are parallel and the system has no solution.

Teacher's Note:
1. Equal ratios of the x- and y-coefficients mean parallel or coincident lines.
2. The ratio of the constants decides which of the two it is.

Answer: (b) 2

Explanation:
1. For parallel lines, \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \).
2. \( \dfrac{k}{4} = \dfrac{-3}{-6} = \dfrac{1}{2} \) ⇒ k = 2.
3. \( \dfrac{c_1}{c_2} = \dfrac{6}{15} = \dfrac{2}{5} \neq \dfrac{1}{2} \), so the lines are parallel for k = 2.

Teacher's Note:
1. Always check that the constants are not in the same ratio, or the lines would coincide.
2. Watch the signs of the coefficients.

Answer: (a) 20°, 40°, 120°

Explanation:
1. ∠A + ∠B + ∠C = 180° and ∠C = 3∠B.
2. 3∠B = 2∠A + 2∠B ⇒ ∠A = \( \dfrac{\angle B}{2} \).
3. \( \dfrac{\angle B}{2} + \angle B + 3\angle B = 180^\circ \) ⇒ ∠B = 40°, so ∠A = 20° and ∠C = 120°: 20°, 40°, 120°.

Teacher's Note:
1. Express every angle in terms of one angle.
2. Then use the angle sum property of a triangle.

Answer: (b) 36

Explanation:
1. Let the tens digit be x and the units digit be y. Then 10x + y = 4(x + y) ⇒ 6x = 3y ⇒ y = 2x.
2. Also y = x + 3, so 2x = x + 3 ⇒ x = 3 and y = 6.
3. The number is 36.

Teacher's Note:
1. Check: 36 : (3 + 6) = 36 : 9 = 4 : 1.
2. Be careful which digit is in the tens place.

Answer: (a) 1

Explanation:
1. For infinitely many solutions, \( \dfrac{6}{12} = \dfrac{3}{c} = \dfrac{c-3}{c} \).
2. \( \dfrac{1}{2} = \dfrac{3}{c} \) ⇒ c = 6; check: \( \dfrac{c-3}{c} = \dfrac{3}{6} = \dfrac{1}{2} \). ✓
3. So there is only one value of c.

Teacher's Note:
1. All three ratios must be equal at the same time.
2. Find c from one pair of ratios and check it in the third.

Answer: (d) Intersecting or coincident

Explanation:
1. A consistent pair has at least one solution.
2. Intersecting lines give exactly one solution and coincident lines give infinitely many; parallel lines give none.
3. So the lines are intersecting or coincident.

Teacher's Note:
1. Consistent means at least one solution, not exactly one.
2. Only parallel lines give an inconsistent pair.

Answer: (a) \( \dfrac{33}{2} \)

Explanation:
1. Dividing by 6: 6x + 4y = 117 …(1) and 4x + 6y = 93 …(2)
2. 3 × (1) − 2 × (2): 18x − 8x = 351 − 186 ⇒ 10x = 165.
3. \( x = \dfrac{165}{10} = \dfrac{33}{2} \).

Teacher's Note:
1. Simplify by a common factor before eliminating.
2. Make the y-coefficients equal (12 and 12) and subtract.

Answer: (d) 12x + 15y = 45

Explanation:
1. For coincident lines, \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \).
2. For (d): \( \dfrac{4}{12} = \dfrac{5}{15} = \dfrac{15}{45} = \dfrac{1}{3} \).
3. So 12x + 15y = 45 (3 times the given equation) coincides with it. In (a), 25 ≠ 2 × 15.

Teacher's Note:
1. A coincident line is the same equation multiplied by a non-zero number.
2. Check the constant term as well as the coefficients.

Answer: (b) 40 km/hr, 30 km/hr

Explanation:
1. Let the speeds be x km/hr (from M) and y km/hr (from N), x > y.
2. Same direction: the gap of 90 km closes at (x − y) km/hr, so 9(x − y) = 90 ⇒ x − y = 10.
3. Opposite directions: \( \dfrac{9}{7}(x+y) = 90 \) ⇒ x + y = 70. So x = 40, y = 30: 40 km/hr, 30 km/hr.

Teacher's Note:
1. Moving in the same direction, use the difference of speeds; towards each other, use the sum.
2. Then add and subtract the two equations.

Answer: (c) 51 years

Explanation:
1. Let Ram be born in year x and his grandpa in year y. Then \( 2002 - x = \dfrac{2002 - y}{2} \) ⇒ 2x − y = 2002 …(1)
2. x + y = 3854 …(2). Adding: 3x = 5856 ⇒ x = 1952.
3. Ram's age at the end of 2003 = 2003 − 1952 = 51 years.

Teacher's Note:
1. Use the birth years as the unknowns, since their sum is given.
2. Read carefully: the question asks for the age in 2003, not 2002.

Answer: (a) ₹ 100, ₹ 80

Explanation:
1. Let the cost prices be ₹ x (tea-set) and ₹ y (lemon-set).
2. \( \dfrac{15y}{100} - \dfrac{5x}{100} = 7 \) ⇒ 3y − x = 140 …(1) and \( \dfrac{5x}{100} + \dfrac{10y}{100} = 13 \) ⇒ x + 2y = 260 …(2)
3. Adding: 5y = 400 ⇒ y = 80 and x = 100. The prices are ₹ 100 and ₹ 80.

Teacher's Note:
1. A loss is subtracted from the gain to find the net gain.
2. Clear the denominators first to keep the numbers small.

Answer: (b) (P) → (iii); (Q) → (i); (R) → (iv); (S) → (ii)

Explanation:
1. (P) x = \( \dfrac{5}{3} \); 3y = \( \dfrac{10}{3} \) + 15 ⇒ \( y = \dfrac{55}{9} \) → (iii).
2. (Q) y = 2x − 1, so 4x + 6x − 3 = 27 ⇒ x = 3, y = 5 → (i).
3. (R) Adding the equations: 4x + 4 = 0 ⇒ x = −1, y = 2 → (iv).
4. (S) The equations are 12x + y = 16 and x + y = 5; subtracting, 11x = 11 ⇒ x = 1, y = 4 → (ii).

Teacher's Note:
1. You can also substitute each solution into the equations to match quickly.
2. Clear fractions in (S) before solving.

Answer: (a)

Explanation:
1. 2x + 4y = 8 meets the axes at (4, 0) and (0, 2); it slopes downwards.
2. 3x − 4y = 12 meets the axes at (4, 0) and (0, −3); it slopes upwards.
3. Both lines pass through (4, 0), and only graph (a) shows one line through (0, 2) and the other through (0, −3).

Teacher's Note:
1. Plot a line quickly from its x- and y-intercepts.
2. The common intercept (4, 0) is the solution of the pair.

Answer: (d) (i) F, (ii) F, (iii) T

Explanation:
1. (i) 7 × (x + 2y = 5) − (7x + 3y = 13) gives 11y = 22, so y = 2 and x = 1. The solution is unique but it is x = 1, y = 2, not x = 2, y = 1: F.
2. (ii) \( \dfrac{\sqrt{2}}{\sqrt{3}} \neq \dfrac{\sqrt{3}}{-\sqrt{8}} \), so there is a unique solution (x = 0, y = 0): F.
3. (iii) With p = 1, q = 5 the second equation is 6x − 3y = 15, i.e. 3(2x − y = 5), so there are infinitely many solutions: T.

Teacher's Note:
1. Always substitute a claimed solution back into both equations.
2. A pair with both constants zero always has at least the solution (0, 0).