CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 35

Answer: (d) the solution \( x = 2\frac{1}{8}, y = 1\frac{7}{8} \)

Explanation:
1. \( 3^{x+y} = 3^{4} \) gives \( x + y = 4 \).
2. \( 81^{x-y} = 3 \) gives \( 3^(4(x-y)) = 3^{1} \), so \( x - y = 1/4 \).
3. Adding: \( 2x = 17/4, x = 17/8 \) = 2⅛ and \( y = 4 - 17/8 = 15/8 \) = 1⅞.

Teacher's Note:
1. Convert both sides to the same base to get linear equations.
2. Then solve by adding and subtracting.

Answer: (b) 4y = x + 1

Explanation:
1. Slope = \( (2 - 0)/(7 + 1) = 1/4 \).
2. Line through (-1, 0): \( y = (x + 1)/4 \), i.e. \( 4y = x + 1 \).
3. Check (7, 2): \( 8 = 8 \).

Teacher's Note:
1. Find the slope first, then use a point.
2. Always check the second point.

Answer: (b) k ≠ 3

Explanation:
1. Unique solution needs \( a_{1}/a_{2} \ne b_{1}/b_{2} \).
2. \( k/6 \ne (-1)/(-2) = 1/2 \), so \( k \ne 3 \).

Teacher's Note:
1. Compare the ratios of the \( x \) and \( y \) coefficients.
2. If \( k = 3 \) the lines are parallel (and different).

Answer: (a) ad = bc

Explanation:
1. \( x = -b/a \) from the first and \( x = -d/c \) from the second.
2. They agree when \( b/a = d/c \), i.e. bc = ad.

Teacher's Note:
1. Consistent means both equations have the same solution.
2. Cross-multiply the two values of \( x \).

Answer: (d) x = y = 1/2 or x = y = -1/2

Explanation:
1. \( x - y = 0 \) gives \( x = y \), so \( |2x| = 1 \).
2. \( 2x = 1 \) or \( 2x = -1 \), so \( x = y = 1/2 \) or \( x = y = -1/2 \).

Teacher's Note:
1. Absolute value equations give two cases.
2. Both cases must be checked against the second equation.

Answer: (b) 3/2

Explanation:
1. Multiply the first by 12: \( 3x + 4y = 5 \). Multiply the second by 2: \( x + 2y = 2 \).
2. Subtract \( 2 \times (x + 2y = 2) \) from the first: \( x = 1 \), then \( y = 1/2 \).
3. \( x + y = 3/2 \).

Teacher's Note:
1. Clear fractions first.
2. Use elimination to solve.

Answer: (b) -1

Explanation:
1. For \( x \ge 3 \): \( 3x - 1 = 4 \) gives \( x = 5/3 \), which is not ≥ 3.
2. For \( -1 \le x \) < 3: \( (3 - x) + 2(x + 1) = x + 5 = 4 \) gives \( x = -1 \).
3. For \( x \) < \( -1 \): \( (3 - x) - 2(x + 1) = 1 - 3x = 4 \) gives \( x = -1 \), which is not < \( -1 \).
4. So \( x = -1 \) only; check: \( 4 + 0 = 4 \).

Teacher's Note:
1. Split into intervals at the points where each modulus becomes zero.
2. Reject a root that falls outside its interval.

Answer: (d) any number except 0

Explanation:
1. The right side equals \( (x + 1)/x = 1 + 1/x \) for every \( x \ne 0 \).
2. So the equation is an identity, true for every \( x \) except 0.

Teacher's Note:
1. Simplify one side to see if it equals the other.
2. \( x = 0 \) is excluded because \( 1/x \) is undefined.

Answer: (c) 75

Explanation:
1. Solving the three equations gives \( x = 8, y = 10, z = 15 \).
2. Check: \( 32 + 50 = 82, 24 + 30 = 54 \) and \( 50 + 60 = 110 \).
3. \( 5x + 2y + z = 40 + 20 + 15 = 75 \).

Teacher's Note:
1. Eliminate one variable at a time.
2. Always substitute back into all three equations.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( y + 8 = x + 8 \) simplifies to \( y = x \), i.e. \( x - y = 0 \), which has constant term 0 and passes through (0, 0). So \( A \) is true.
2. \( R \) is true: if the constant term is \( 0, x = 0 \) and \( y = 0 \) satisfy the equation.
3. \( R \) is exactly the reason \( A \) holds.

Teacher's Note:
1. Check whether (0, 0) satisfies the equation.
2. Simplify before deciding about the constant term.

Answer: (c) 2

Explanation:
1. xy \( - x - y = 0 \) gives \( (x - 1)(y - 1) = 1 \).
2. Integer factor pairs of 1 are (1, 1) and (-1, -1).
3. So \( (x, y) = (2, 2) \) or (0, 0): two pairs.

Teacher's Note:
1. Add 1 to both sides to factorise.
2. List all integer factor pairs, including negatives.

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.
3. As \( x + y \le 17, x + y = 11 \).

Teacher's Note:
1. The sum of a number and its reverse is always a multiple of 11.
2. A square divisible by the prime 11 is divisible by 121.

Answer: (c) 3 hours

Explanation:
1. Let the length be 1. After \( t \) hours the first has \( 1 - t/6 \) left and the second \( 1 - t/4 \).
2. \( 1 - t/6 = 2(1 - t/4) \) gives \( 1 - t/6 = 2 - t/2 \).
3. \( t/2 - t/6 = 1 \), so \( t/3 = 1 \) and \( t = 3 \).

Teacher's Note:
1. Express each remaining length as a fraction of the original.
2. Set up the given ratio as an equation.

Answer: (c) 8 : 9

Explanation:
1. Downstream speed = 20 km/h and upstream speed = 10 km/h.
2. For a distance \( d \) each way, average speed = \( 2d / (d/20 + d/10) = 40/3 \) km/h.
3. Ratio = (40/3) : \( 15 = 40 \) : \( 45 = 8 \) : 9.

Teacher's Note:
1. Average speed is total distance divided by total time, not the mean of the speeds.
2. The distance cancels out.

Answer: (c) 3x - 2y - 27 = 0

Explanation:
1. A parallel line has the form \( 3x - 2y + c = 0 \).
2. Substituting (5, -6): \( 15 + 12 + c = 0 \), so \( c = -27 \).
3. The line is \( 3x - 2y - 27 = 0 \).

Teacher's Note:
1. Parallel lines have the same coefficients of \( x \) and \( y \).
2. Only the constant term changes.

Answer: (a) parallel

Explanation:
1. Slope of the first line = \( (6 - 3)/(5 - 2) = 1 \).
2. Slope of the second line = \( (-2 + 5)/(9 - 6) = 1 \).
3. Equal slopes: parallel. They are not the same line, since the first is \( y = x + 1 \) and (9, -2) does not lie on it.

Teacher's Note:
1. Compare the slopes to test parallel or perpendicular.
2. Parallel lines also need different intercepts to be distinct.

Answer: (a) 3x - 4y = 12

Explanation:
1. Intercept form: \( x/4 + y/(-3) = 1 \).
2. Multiply by 12: \( 3x - 4y = 12 \).

Teacher's Note:
1. Use \( x/a + y/b = 1 \) with signs of the intercepts.
2. Check with (4, 0) and (0, -3).

Answer: (b) -1 < x < 11

Explanation:
1. \( |5 - x| \) < 6 means \( -6 \) < \( x - 5 \) < 6.
2. So \( -1 \) < \( x \) < 11.

Teacher's Note:
1. \( |A| \) < \( k \) means \( -k \) < \( A \) < \( k \).
2. Add 5 to all parts to isolate \( x \).

Answer: (d) if b ≠ 3

Explanation:
1. Rearrange: \( (3 - b)x = 6 - a \).
2. A unique solution exists exactly when the coefficient \( 3 - b \ne 0 \), i.e. \( b \ne 3 \).

Teacher's Note:
1. Collect the \( x \) terms on one side.
2. The value of a does not affect uniqueness.

Answer: (a) coincident lines

Explanation:
1. Ratios: \( 2/6 = 1/3 = (-5)/(-15) = 1/3 \).
2. All three ratios are equal, so the lines coincide and there are infinitely many solutions.

Teacher's Note:
1. Equal ratios of all three coefficients mean coincident lines.
2. The second equation is 3 times the first.

Answer: (a) (1) and (2) only

Explanation:
1. Multiplying \( p \) > \( q \) by a negative \( r \) reverses the sign: pr < qr, so (1) is true.
2. Adding \( r \) to both sides keeps the inequality: (2) is true.
3. Subtracting \( r \) keeps it: \( p - r \) > \( q - r \), so (3) is false. Doubling then adding \( r \): \( 2p + r \) > \( 2q + r \), so (4) is false.

Teacher's Note:
1. Multiplying by a negative number reverses an inequality.
2. Adding or subtracting the same number never does.

Answer: (b) (2) and (4)

Explanation:
1. On 0 < \( u \) < \( 1, u^{2} \) increases with \( u \), so (4) increases and (1) decreases.
2. \( u - 1 \) increases with \( u \), so (2) increases.
3. \( 1/u^{2} \) decreases as \( u \) increases, so (3) decreases.

Teacher's Note:
1. Test with \( u = 0.2 \) and \( u = 0.8 \).
2. A reciprocal reverses the direction of change.

Answer: (a) 0 and 1

Explanation:
1. If \( x \ge 1 \): \( x^{2} + x - 1 = 1 \) gives \( x^{2} + x - 2 = 0 \), so x = 1 (x = -2 rejected).
2. If \( x \) < 1: \( x^{2} - x + 1 = 1 \) gives \( x(x - 1) = 0 \), so x = 0 (x = 1 is not < 1 but is already found).
3. The solutions are 0 and 1.

Teacher's Note:
1. Split into cases for the modulus.
2. Check each root against its case.

Answer: (a) a unique solution if p ≠ 2, and infinitely many solutions if p = 2 and q = 4

Explanation:
1. Ratios of \( x \) coefficients: \( 2/p \); of \( y \) coefficients: \( -1/-1 = 1 \).
2. If \( p \ne 2 \) the ratios differ, so there is a unique solution.
3. If \( p = 2 \) and \( q = 4 \) the equations are identical: infinitely many solutions. If \( p = 2 \) and \( q \ne 4 \) the lines are parallel and there is no solution.

Teacher's Note:
1. Compare \( a_{1}/a_{2}, b_{1}/b_{2} \) and \( c_{1}/c_{2} \).
2. The constant term decides between 'no solution' and 'infinitely many'.

Answer: (a) x + y = 2

Explanation:
1. Slope = \( (3 - 5)/(-1 + 3) = -1 \).
2. \( y = 3 - (x + 1) = 2 - x \), i.e. \( x + y = 2 \).
3. Check with (1, 1): \( 1 + 1 = 2 \).

Teacher's Note:
1. Find the slope from two points, then the intercept.
2. Test the third point.

Answer: (a) 4

Explanation:
1. The relation is \( x + y = 2 \).
2. At \( x = -2, y = 2 - (-2) = 4 \).

Teacher's Note:
1. Use the relation found from the table.
2. Equal steps in \( x \) give equal steps in \( y \) for a linear function.

Answer: (c) -1

Explanation:
1. Write \( y = -x + 2 \).
2. The slope is the coefficient of \( x \), i.e. \( -1 \).

Teacher's Note:
1. In \( y \) = mx \( + c \) the slope is \( m \).
2. Rearrange the equation to this form.