CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 19

Answer: (a) all real values except 10

Explanation:
1. For parallel lines \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
2. Here \( \frac{3}{6} = \frac{-1}{-2} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-5}{-p} = \frac{5}{p} \).
3. We need \( \frac{5}{p} \neq \frac{1}{2} \), so \( p \neq 10 \).

Teacher's Note:
1. Parallel lines need the third ratio to be different.
2. If p = 10 the lines coincide.

Answer: (c) \( k = 6 \)

Explanation:
1. For infinitely many solutions \( \frac{k}{12} = \frac{3}{k} = \frac{k-3}{k} \).
2. From the first two, \( k^2 = 36 \), so \( k = \pm 6 \).
3. From \( \frac{3}{k} = \frac{k-3}{k} \), \( k = 6 \). Both conditions hold only for \( k = 6 \).

Teacher's Note:
1. All three ratios must be equal.
2. Take the value common to every condition.

Answer: (c) \( m \neq -\frac{11}{4} \)

Explanation:
1. Write the first equation as \( 3x - 2y + 8 = 0 \). A unique solution needs \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
2. \( \frac{3}{2m-5} \neq \frac{-2}{7} \), so \( 21 \neq -2(2m - 5) = -4m + 10 \).
3. Hence \( 4m \neq -11 \), i.e. \( m \neq -\frac{11}{4} \).

Teacher's Note:
1. Unique solution means the ratios of x and y coefficients differ.
2. Cross-multiply carefully, keeping the signs.

Answer: (b) \( l = 12 \) cm, \( b = 8 \) cm

Explanation:
1. Let the length be \( x \) cm and the breadth \( y \) cm. Then \( 2(x + y) = 40 \), so \( x + y = 20 \).
2. The ratio of breadth to length is 2 : 3, so \( 3y = 2x \).
3. Solving, \( x = 12 \) and \( y = 8 \). Length = 12 cm and breadth = 8 cm.

Teacher's Note:
1. The larger side is the length.
2. Keep the units in cm, as in the question.

Answer: (b) Simultaneous equations

Explanation:
1. Two linear equations in the same two variables are solved together, so they are called simultaneous equations.
2. Consistent, inconsistent and dependent describe the nature of the solution, not the pair itself.

Teacher's Note:
1. Simultaneous means solved at the same time.
2. Consistent and inconsistent say whether solutions exist.

Answer: (a) 39, 13

Explanation:
1. Let the numbers be \( x > y \). Then \( x - y = 26 \) and \( x = 3y \).
2. \( 3y - y = 26 \), so \( y = 13 \) and \( x = 39 \).

Teacher's Note:
1. Substitute one equation into the other.
2. Check: 39 − 13 = 26 and 39 = 3 × 13.

Answer: (b) Rs. 3900, Rs. 150

Explanation:
1. Let the initial salary be Rs. \( x \) and the annual increment Rs. \( y \).
2. After 4 years: \( x + 4y = 4500 \). After 10 years: \( x + 10y = 5400 \).
3. Subtracting, \( 6y = 900 \), so \( y = 150 \) and \( x = 3900 \).

Teacher's Note:
1. After n years the salary is initial + n × increment.
2. Subtract the equations to remove x.

Answer: (a) a unique solution

Explanation:
1. \( \frac{a_1}{a_2} = \frac{1}{3} \) and \( \frac{b_1}{b_2} = \frac{2}{12} = \frac{1}{6} \).
2. Since \( \frac{1}{3} \neq \frac{1}{6} \), the lines intersect at exactly one point.

Teacher's Note:
1. Compare only a1/a2 and b1/b2 first.
2. Different ratios mean a unique solution.

Answer: (c) 104

Explanation:
1. Let the girls be \( x \) and boys \( y \). Then \( x + y = 150 \) and \( 0.50x + 0.25y = 49 \), i.e. \( 2x + y = 196 \).
2. Subtracting, \( x = 46 \), so \( y = 104 \).

Teacher's Note:
1. Convert paise to rupees (50 paise = Rs. 0.50).
2. Multiply the money equation by 4 to clear decimals.

Answer: (a) 45 years

Explanation:
1. Let the father be \( x \) and the son \( y \). Then \( x + y = 65 \) and \( 2(x - y) = 50 \), so \( x - y = 25 \).
2. Adding, \( 2x = 90 \), so \( x = 45 \).

Teacher's Note:
1. Halve the second equation before adding.
2. Check: son = 20, sum = 65.

Answer: (d) 10

Explanation:
1. Write as \( x + 2y - 3 = 0 \) and \( 5x + ky + 7 = 0 \).
2. Inconsistent: \( \frac{1}{5} = \frac{2}{k} \neq \frac{-3}{7} \).
3. So \( k = 10 \), and \( \frac{1}{5} \neq \frac{-3}{7} \) holds.

Teacher's Note:
1. Inconsistent means parallel lines.
2. Always check that the third ratio is different.

Answer: (d) 30

Explanation:
1. B is 60% more efficient, so the ratio of their work rates is \( 100 : 160 = 5 : 8 \).
2. Time taken is in the inverse ratio: \( A : B = 8 : 5 \), so B takes \( 24 \times \frac{5}{8} = 15 \) days for the earlier work.
3. For twice the work, B needs \( 2 \times 15 = 30 \) days.

Teacher's Note:
1. More efficient means less time for the same work.
2. Do not forget the factor of 2 for the larger work.

Answer: (b) \( 33\frac{1}{3} \)

Explanation:
1. Let Y's salary be \( A \); then X's is \( \frac{A}{2} \) and the total is \( \frac{3A}{2} \).
2. New salaries: X = \( \frac{3A}{4} \), Y = \( \frac{5A}{4} \), total \( 2A \).
3. Increase = \( 2A - \frac{3A}{2} = \frac{A}{2} \); percentage = \( \frac{A/2}{3A/2} \times 100 = 33\frac{1}{3} \).

Teacher's Note:
1. Take Y's salary as a symbol and find everything in it.
2. Percentage increase = increase / original × 100.

Answer: (b) \( 4y = x + 1 \)

Explanation:
1. For \( 4y = x + 1 \): at \( (7, 2) \), \( 8 = 8 \); at \( (-1, 0) \), \( 0 = 0 \). Both satisfy it.
2. Option (a) fails at \( (-1, 0) \) (\( 0 \neq -10 \)), (c) fails at \( (7,2) \), and (d) fails at \( (7, 2) \) (7 ≠ 9).

Teacher's Note:
1. A point lies on a line if its coordinates satisfy the equation.
2. Test the easier point (−1, 0) first to eliminate options.

Answer: (c) 70

Explanation:
1. Let the present ages be \( 7x \) and \( 3x \).
2. After 10 years \( \frac{7x + 10}{3x + 10} = \frac{2}{1} \), so \( 7x + 10 = 6x + 20 \) and \( x = 10 \).
3. Father's age = \( 7x = 70 \) years.

Teacher's Note:
1. Use a common multiplier x for a ratio.
2. Check: 70 and 30 become 80 and 40, ratio 2 : 1.

Answer: (d) 15

Explanation:
1. Let the fraction be \( \frac{x}{y} \). Then \( \frac{x+1}{y+1} = 4 \), so \( x = 4y + 3 \).
2. Also \( \frac{x-1}{y-1} = 7 \), so \( x = 7y - 6 \).
3. Equating, \( 3y = 9 \), \( y = 3 \) and \( x = 15 \).

Teacher's Note:
1. Form two equations by clearing the denominators.
2. Check: 16/4 = 4 and 14/2 = 7.

Answer: (a) 3, 1.5

Explanation:
1. Let the speeds of the boat and the current be \( x \) and \( y \) km/h.
2. Downstream \( x + y = \frac{9}{2} = 4.5 \) and upstream \( x - y = \frac{9}{6} = 1.5 \).
3. Adding, \( x = 3 \); then \( y = 1.5 \).

Teacher's Note:
1. Downstream speed = boat + current; upstream = boat − current.
2. Speed = distance ÷ time.

Answer: (b) 54

Explanation:
1. Let the number be \( 10x + y \). Then \( \frac{5}{6}(10x + y) = 10y + x \), so \( 44x = 55y \) and \( x : y = 5 : 4 \).
2. With digit difference 1, \( x = 5 \) and \( y = 4 \), so the number is 54.
3. Check: reversed 45 = \( \frac{5}{6} \times 54 \).

Teacher's Note:
1. Use the ratio of digits together with their difference.
2. Verify by reversing the digits.

Answer: (b) 11

Explanation:
1. The numbers are \( 10x + y \) and \( 10y + x \); their sum is \( 11(x + y) \).
2. For \( 11(x+y) \) to be a perfect square, \( x + y \) must be 11 times a square; as \( x + y \le 17 \), \( x + y = 11 \) (sum 121).

Teacher's Note:
1. Add the two numbers to see the common factor 11.
2. A perfect square divisible by 11 is divisible by 121.

Answer: (c) 7 : 1

Explanation:
1. Cross-multiplying, \( 4(3x + 4y) = 9(x + 2y) \), so \( 3x = 2y \), i.e. \( x = \frac{2}{3}y \).
2. \( 3x + 5y = 2y + 5y = 7y \) and \( 3x - y = 2y - y = y \).
3. The ratio is \( 7 : 1 \).

Teacher's Note:
1. Find 3x in terms of y and substitute.
2. Ratios can be written as fractions and cross-multiplied.

Answer: (c) 48

Explanation:
1. Let the ten's digit be \( y \) and the unit's digit \( x \), so \( x = 2y \).
2. \( 10y + x + 36 = 10x + y \) gives \( x - y = 4 \).
3. So \( 2y - y = 4 \), \( y = 4 \), \( x = 8 \) and the number is 48.

Teacher's Note:
1. Reversing the digits swaps the place values.
2. Check: 48 + 36 = 84.

Answer: (b) \( \frac{16}{3} \) km/hr

Explanation:
1. Upstream speed \( = 6 - 2 = 4 \) km/h and downstream speed \( = 6 + 2 = 8 \) km/h.
2. If upstream takes \( t \) hours, downstream takes \( \frac{t}{2} \) hours for the same distance \( 4t \).
3. Average speed \( = \frac{4t + 4t}{t + t/2} = \frac{8t}{3t/2} = \frac{16}{3} \) km/h.

Teacher's Note:
1. Average speed = total distance ÷ total time.
2. It is not the mean of the two speeds.

Answer: (c) 8 : 9

Explanation:
1. Downstream speed = 20 km/h and upstream speed = 10 km/h. For a distance \( d \) each way:
2. Average speed \( = \frac{2d}{\frac{d}{20} + \frac{d}{10}} = \frac{40}{3} \) km/h.
3. Ratio to 15 \( = \frac{40}{3} : 15 = 8 : 9 \).

Teacher's Note:
1. Average speed over equal distances is the harmonic mean.
2. Convert 40/3 : 15 to whole numbers.

Answer: (d) 15

Explanation:
1. \( x + y = 5 + \frac{y}{2} \) (from \( x = 5 - \frac{y}{2} \)), with \( 0 \le y \le 10 \) because \( x \ge 0 \).
2. Maximum at \( y = 10 \): 10. Minimum at \( y = 0 \): 5.
3. Sum = 15.

Teacher's Note:
1. Write x + y in terms of one variable and use its range.
2. Non-negative x limits y to at most 10.

Answer: (d) lying between −1 and \( \frac{3}{2} \)

Explanation:
1. Adding the equations, \( (k + 1)x = 5 \), so \( x = \frac{5}{k+1} \); then \( y = x - 2 = \frac{3 - 2k}{k + 1} \).
2. \( x > 0 \) needs \( k > -1 \). Then \( y > 0 \) needs \( 3 - 2k > 0 \), i.e. \( k < \frac{3}{2} \).
3. So \( -1 < k < \frac{3}{2} \).

Teacher's Note:
1. First quadrant means x > 0 and y > 0.
2. Both conditions must hold together.

Answer: (a) 6

Explanation:
1. Infinite solutions need \( \frac{p}{12} = \frac{3}{p} = \frac{p-3}{p} \).
2. \( p^2 = 36 \) gives \( p = \pm 6 \); \( \frac{3}{p} = \frac{p-3}{p} \) gives \( p = 6 \).
3. So \( p = 6 \).

Teacher's Note:
1. Check each root in all three ratios.
2. p = −6 satisfies only the first equality.

Answer: (b) 24

Explanation:
1. Let the boys be \( x \) and girls \( y \). Then \( \frac{4x/5}{y} = \frac{2}{3} \) and \( \frac{4x/5}{y - 44} = \frac{5}{2} \).
2. The first gives \( y = \frac{6x}{5} \). The second gives \( 8x = 25(y - 44) \), so \( 8x = 30x - 1100 \), \( x = 50 \) and \( y = 60 \).
3. Boys now \( = 40 \), girls now \( = 16 \). Boys to leave \( = 40 - 16 = 24 \).

Teacher's Note:
1. Track the boys and girls after each change.
2. Check: 40 : 60 = 2 : 3 and 40 : 16 = 5 : 2.

Answer: (a) 10, 5

Explanation:
1. Adding, \( \frac{2}{x} = 20 \), so \( \frac{1}{x} = 10 \). Then \( \frac{1}{y} = 15 - 10 = 5 \).
2. \( ax = 1 \) means \( a = \frac{1}{x} = 10 \) and \( by = 1 \) means \( b = \frac{1}{y} = 5 \).

Teacher's Note:
1. Treat 1/x and 1/y as the unknowns.
2. Then a and b are exactly those reciprocals.

Answer: (b) both I and II are false

Explanation:
1. Statement I is false: a consistent pair may also have infinitely many solutions (coincident lines).
2. Statement II is false: \( 13^2 + 14^2 = 169 + 196 = 365 \).
3. So both are false.

Teacher's Note:
1. Consistent means at least one solution, not exactly one.
2. Test the statement with actual numbers.

Answer: (a) 10

Explanation:
1. Let one woman take \( W \) hours and one girl \( G \) hours. Then \( \frac{8}{W} + \frac{12}{G} = \frac{1}{10} \) and \( \frac{6}{W} + \frac{8}{G} = \frac{1}{14} \).
2. Solving (with \( \frac{1}{W} = u \), \( \frac{1}{G} = v \)), \( W = 140 \) and \( G = 280 \).
3. 7 women and 14 girls do \( \frac{7}{140} + \frac{14}{280} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10} \) of the mural per hour, so they take 10 hours.

Teacher's Note:
1. Work done per hour adds up; use reciprocals as new variables.
2. Check the answer in the first equation: it is the same composition scaled.

Answer: (b) 8 km/hr

Explanation:
1. Downstream speed \( = \frac{30}{3} = 10 \) km/h and upstream speed \( = \frac{30}{5} = 6 \) km/h.
2. Speed in still water \( = \frac{10 + 6}{2} = 8 \) km/h (and the stream is 2 km/h).

Teacher's Note:
1. Still water speed is the mean of upstream and downstream speeds.
2. Check: 8 + 2 = 10 and 8 − 2 = 6.

Answer: (c) infinitely many solutions

Explanation:
1. \( \frac{a_1}{a_2} = \frac{5}{3} \), \( \frac{b_1}{b_2} = \frac{-15}{-9} = \frac{5}{3} \) and \( \frac{c_1}{c_2} = \frac{8}{24/5} = \frac{5}{3} \).
2. All three ratios are equal, so the lines coincide and there are infinitely many solutions.

Teacher's Note:
1. Equal ratios mean coincident lines.
2. A pair of lines never has exactly two solutions.

Answer: (d) 36

Explanation:
1. Let the ten's digit be \( y \) and the unit's digit \( x \): \( x + y = 9 \).
2. \( 10y + x + 27 = 10x + y \) gives \( x - y = 3 \).
3. So \( x = 6 \), \( y = 3 \) and the number is 36 (36 + 27 = 63).

Teacher's Note:
1. Reversal swaps the digits; subtract to get a difference of digits.
2. Verify: 36 + 27 = 63.

Answer: (d) no value

Explanation:
1. Write as \( cx - y - 2 = 0 \) and \( 6x - 2y - 3 = 0 \).
2. Infinite solutions need \( \frac{c}{6} = \frac{-1}{-2} = \frac{-2}{-3} \).
3. But \( \frac{1}{2} \neq \frac{2}{3} \), so no value of c works.

Teacher's Note:
1. Check the b and c ratios first; they do not depend on c.
2. If they differ, no value of c can give infinitely many solutions.

Answer: (c) 3 and 1

Explanation:
1. Adding the equations, \( 2x = 6 \), so \( x = 3 \). Then \( y = 4 - 3 = 1 \).
2. So \( a = 3 \) and \( b = 1 \).

Teacher's Note:
1. The solution is the pair (x, y) that satisfies both equations.
2. Elimination by addition works neatly here.

Answer: (c) III only

Explanation:
1. I: from \( x - y = 1 - x - y \), \( x = \frac{1}{2} \); then \( xy = x - y \) gives \( y = \frac{1}{3} \) and \( x + y = \frac{5}{6} \neq \frac{5}{3} \). False.
2. II: infinite solutions need \( \frac{3}{5} = \frac{2}{b} = \frac{a}{4} \), so \( b = \frac{10}{3} \) and \( a = \frac{12}{5} \), not 4 and 3. False.
3. III: \( x = a, y = b \) gives \( 1 + 1 = 2 \) and \( a^2 - b^2 = a^2 - b^2 \), and the pair has a unique solution since \( \frac{1/a}{a} \neq \frac{1/b}{-b} \). True.

Teacher's Note:
1. Test each statement separately by solving or substituting.
2. Substituting the claimed solution is the quickest check for III.

Answer: (c) I and II only

Explanation:
1. I: \( x - y = -1 \) gives \( y = x + 1 \); then \( 3x - 5x - 5 = -1 \), so \( x = -2 \) and \( y = -1 \). True.
2. II: eliminating, \( x = -21 \) and \( y = 17 \) (check: \( -42 + 51 = 9 \), \( -63 + 68 = 5 \)). True.
3. III: adding gives \( \frac{3x}{a} = 6 \), so \( x = 2a \), and then \( \frac{y}{b} = -2 \), so \( y = -2b \), not \( 2b \). False.
4. So I and II are true.

Teacher's Note:
1. Substitute the claimed values into both equations.
2. A single wrong sign makes the statement false.

Answer: (a) parallel

Explanation:
1. Inconsistent means there is no common solution.
2. Lines with no common point never meet, so they are parallel.

Teacher's Note:
1. Coincident lines have all points in common, so they are consistent.
2. Intersecting lines have exactly one common point.

Answer: (c) 2

Explanation:
1. Write the second as \( 6x - ky + 16 = 0 \).
2. Coincident lines: \( \frac{3}{6} = \frac{-1}{-k} = \frac{8}{16} \).
3. So \( \frac{1}{2} = \frac{1}{k} \) and \( k = 2 \).

Teacher's Note:
1. Bring both equations to the form ax + by + c = 0 first.
2. Coincident needs all three ratios equal.