CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 29

Answer: (a) Rs. 8 and Rs. 6

Explanation:
1. \(x+10y=68\) and \(x+15y=98\)
2. Subtracting: \(5y=30\), so \(y=6\)
3. \(x=68-60=8\)

Teacher's Note:
1. Subtract the equations to remove the fixed charge.
2. Check the answer by substituting it in both original equations.

Answer: (c) Rs. 255

Explanation:
1. \(x+10y=105\) and \(x+15y=155\), so \(y=10\) and \(x=5\)
2. For 25 km: \(5+25\times10=255\)

Teacher's Note:
1. Find the fixed charge and rate first, then evaluate for 25 km.
2. Check: 5 + 100 = 105.

Answer: (b) 60 km/h and 40 km/h

Explanation:
1. Same direction: \(6(x-y)=120\), so \(x-y=20\)
2. Opposite: \(\dfrac65(x+y)=120\), so \(x+y=100\)
3. \(x=60\), \(y=40\)

Teacher's Note:
1. Convert 1 hour 12 minutes to 6/5 hours.
2. Relative speed for opposite directions is the sum.

Answer: (d) 3 km/h and 8 km/h

Explanation:
1. Put \(p=\dfrac1{x-y},\ q=\dfrac1{x+y}\): \(30p+44q=10\) and \(40p+55q=13\)
2. Solving: \(p=\dfrac15,\ q=\dfrac1{11}\), so \(x-y=5\) and \(x+y=11\)
3. Boat \(x=8\) km/h, stream \(y=3\) km/h

Teacher's Note:
1. Read which speed the question asks for first: the stream here.
2. Check: 30/5 + 44/11 = 10.

Answer: (c) 10 km/h and 2 km/h

Explanation:
1. Downstream speed \(=\dfrac{8}{2/3}=12\) km/h; upstream speed \(=8\) km/h
2. \(x+y=12\) and \(x-y=8\)
3. \(x=10\), \(y=2\)

Teacher's Note:
1. Convert 40 minutes to 2/3 hour.
2. Still-water speed is the average of the two speeds.

Answer: (b) 6 km/h and 4 km/h

Explanation:
1. Downstream speed \(=10\), upstream speed \(=2\)
2. \(x+y=10\), \(x-y=2\)
3. \(x=6\), \(y=4\)

Teacher's Note:
1. Speed = distance / time for each direction.
2. The current is half the difference of the two speeds.

Answer: (a) 23 m and 11 m

Explanation:
1. \((x+2)(y-2)=xy-28\) gives \(x-y=12\)
2. \((x-1)(y+2)=xy+33\) gives \(2x-y=35\)
3. Subtracting: \(x=23\), then \(y=11\)

Teacher's Note:
1. Expand both area equations; xy cancels.
2. Check: 25 x 9 = 225 = 253 - 28.

Answer: (c) 99 degrees and 81 degrees

Explanation:
1. \(x+y=180\) and \(x=y+18\)
2. \(2y=162\), so \(y=81\) and \(x=99\)

Teacher's Note:
1. Supplementary angles add up to 180 degrees.
2. Check the answer by substituting it in both original equations.

Answer: (b) \(\angle A=20^\circ,\ \angle B=40^\circ,\ \angle C=120^\circ\)

Explanation:
1. \(C=2(A+B)\) and \(A+B+C=180^\circ\), so \(3(A+B)=180^\circ\), \(A+B=60^\circ\) and \(C=120^\circ\)
2. \(3B=C=120^\circ\), so \(B=40^\circ\)
3. \(A=60^\circ-40^\circ=20^\circ\)

Teacher's Note:
1. Use the angle sum property to find C first.
2. Check: 20 + 40 + 120 = 180.

Answer: (a) \(y=x\)

Explanation:
1. For each of the three points the x and y coordinates are equal
2. So every point of the line satisfies \(y=x\)
3. Check \((4,4)\): \(4=4\), so \((4,4)\) lies on the graph of \(y=x\)

Teacher's Note:
1. A line through the origin with equal coordinates is y = x.
2. Test one of the given points in each option.

Answer: (d) \((3,-2)\)

Explanation:
1. \(x=3\) is a vertical line (parallel to the y-axis) 3 units to the right of the y-axis
2. \(y=-2\) is a horizontal line (parallel to the x-axis) 2 units below the x-axis
3. They meet at \((3,-2)\)

Teacher's Note:
1. The first coordinate comes from the vertical line and the second from the horizontal line.
2. Lines x = c and y = d always meet at (c, d).

Answer: (c) \(x=3,\ y=1\)

Explanation:
1. The solution is the common point of the two lines, read from the graph
2. Algebraically: from \(x=1+2y\), \(2(1+2y)+y=7\) gives \(y=1\)
3. \(x=3\). Check: \(3-2=1\) and \(6+1=7\)

Teacher's Note:
1. The solution of a pair is the point of intersection.
2. Verify by substituting into both equations.

Answer: (b) \((1,0)\) and \(\left(\dfrac72,0\right)\)

Explanation:
1. On the x-axis \(y=0\)
2. \(x-2y=1\Rightarrow x=1\); \(2x+y=7\Rightarrow x=\dfrac72\)

Teacher's Note:
1. Put y = 0 to find where a line cuts the x-axis.
2. Do not confuse with the y-intercepts.

Answer: (a) \(x=1,\ y=3\); \(\left(0,\dfrac72\right)\) and \((0,1)\)

Explanation:
1. Common point: \(x=7-2y\) in the second gives \(14-4y-y=-1\), so \(y=3,\ x=1\)
2. y-axis (x = 0): first line \(y=\dfrac72\); second line \(y=1\)

Teacher's Note:
1. The y-intercept is found by putting x = 0.
2. Check: 1 + 6 - 7 = 0 and 2 - 3 + 1 = 0.

Answer: (d) \(x=3,\ y=2\); \((0,4)\) and \(\left(0,\dfrac12\right)\)

Explanation:
1. Put \(x=2y-1\) in \(2x+3y=12\): \(4y-2+3y=12\), so \(y=2\) and \(x=3\)
2. At \(x=0\): \(3y=12\Rightarrow y=4\); \(2y-1=0\Rightarrow y=\dfrac12\)

Teacher's Note:
1. Rewrite 2y - 1 = x as x - 2y = -1 to compare with the graph.
2. Check the answer by substituting it in both original equations.

Answer: (c) inconsistent, with no solution

Explanation:
1. \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=1\) but \(\dfrac{c_1}{c_2}=\dfrac{-12}{-6}=2\)
2. The lines are parallel, so they never meet: the pair is inconsistent

Teacher's Note:
1. Parallel lines mean no solution.
2. A pair of lines can never have exactly two common points.

Answer: (b) consistent (dependent), with infinitely many solutions

Explanation:
1. \(9x=15y-12\) is \(9x-15y+12=0\), which is 3 times \(3x-5y+4=0\)
2. \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=\dfrac13\): the lines coincide

Teacher's Note:
1. Divide the second equation by 3 and compare.
2. Coincident lines share every point.

Answer: (d) two lines that coincide

Explanation:
1. \(3x+6y=3900\) is \(3\) times \(x+2y=1300\)
2. \(\dfrac31=\dfrac62=\dfrac{3900}{1300}\), so the lines coincide (infinitely many possible prices)

Teacher's Note:
1. The second purchase is exactly one third of the first.
2. So the data cannot fix the individual prices.

Answer: (a) \((2,5)\)

Explanation:
1. Substitute \(y=7-x\): \(5x+14-2x=20\), so \(x=2\)
2. \(y=5\). Check: \(2+5=7\) and \(10+10=20\)

Teacher's Note:
1. Highways meet where both equations are satisfied.
2. Check the answer by substituting it in both original equations.

Answer: (b) 7.5 square units

Explanation:
1. Vertices: the lines meet at \((2,3)\); \(x-y+1=0\) meets the x-axis at \((-1,0)\); \(3x+2y-12=0\) meets it at \((4,0)\)
2. Base on the x-axis \(=4-(-1)=5\); height \(=3\)
3. Area \(=\dfrac12\times5\times3=7.5\) square units

Teacher's Note:
1. Find the three vertices before computing the area.
2. The height is the y-coordinate of the intersection point.

Answer: (d) 6 square units

Explanation:
1. Vertices: \(2x-3y=-6\) and \(y=2\) meet at \((0,2)\); \(2x+3y=18\) and \(y=2\) at \((6,2)\); the first two lines at \((3,4)\)
2. Base \(=6\), height \(=4-2=2\)
3. Area \(=\dfrac12\times6\times2=6\) square units

Teacher's Note:
1. The base is the part of the horizontal line y = 2 between the two vertices.
2. The height is the vertical distance of the third vertex from that line.

Answer: (c) none

Explanation:
1. \(\dfrac24=\dfrac36=\dfrac12\) but \(\dfrac{10}{12}\ne\dfrac12\)
2. The lines are parallel, so the pair has no solution

Teacher's Note:
1. Parallel lines never meet.
2. The second line cuts the y-axis at y = 2 and the first at y = 10/3.

Answer: (b) infinitely many solutions

Explanation:
1. Multiply \(2x+5y=16\) by \(\dfrac32\): \(3x+\dfrac{15}2y=24\)
2. The two equations are the same line, so every point of the line is a solution

Teacher's Note:
1. Multiply one equation by a constant to see if it gives the other.
2. Equal ratios of all three coefficients: 2/3 = 5/(15/2) = 16/24.