CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 27

Answer: (a) I, II, IV, III

Explanation:
1. Quadrants: I \((+,+)\), II \((-,+)\), III \((-,-)\), IV \((+,-)\)
2. \((3,3)\) is in I, \((-3,2)\) in II, \((2,-4)\) in IV, \((-1,-2)\) in III

Teacher's Note:
1. Look at the signs of the x and y coordinates.
2. Quadrants are numbered anticlockwise from the top right.

Answer: (d) \(x=\dfrac{6-4y}3\)

Explanation:
1. \(3x=6-4y\)
2. \(x=\dfrac{6-4y}3\)

Teacher's Note:
1. Isolate the term containing x and then divide by its coefficient.
2. Be careful to divide every term, including 6.

Answer: (c) No, because \(3(3)+4(2)=17\ne6\)

Explanation:
1. Put \(x=3,\ y=2\): \(9+8=17\)
2. \(17\ne6\), so the point does not lie on the line

Teacher's Note:
1. A point lies on a line if its coordinates satisfy the equation.
2. Put x first, then y.

Answer: (b) on the negative x-axis, on the positive y-axis, in quadrant III

Explanation:
1. \((-2,0)\) has \(y=0\) and \(x<0\): negative x-axis
2. \((0,1)\) has \(x=0\) and \(y>0\): positive y-axis
3. \((-2,-3)\) has both coordinates negative: quadrant III

Teacher's Note:
1. A point with y = 0 lies on the x-axis; with x = 0 it lies on the y-axis.
2. Points on an axis belong to no quadrant.

Answer: (a) \(x=\dfrac{15+3y}7\) and the statement is true

Explanation:
1. \(7x=15+3y\), so \(x=\dfrac{15+3y}7\)
2. For \(y=-5\): \(x=\dfrac{15-15}7=0\)
3. So \((0,-5)\) is on the line, i.e. the line meets the y-axis at \(y=-5\)

Teacher's Note:
1. The y-axis is where x = 0.
2. Substitute the given y to test.

Answer: (d) a line parallel to the y-axis through \((4,0)\)

Explanation:
1. \(1.5x=6\), so \(x=4\)
2. The equation \(x=4\) holds for every y, so the graph is a vertical line meeting the x-axis at \((4,0)\)

Teacher's Note:
1. An equation x = c always gives a vertical line.
2. A line parallel to the y-axis cuts the x-axis.

Answer: (c) \(a=7,\ b=5\)

Explanation:
1. At \(x=0\): \(y=b=5\)
2. At \(x=1\): \(a+5=12\), so \(a=7\)
3. Check: \(x=2\): \(14+5=19\); \(x=-3\): \(-21+5=-16\); \(x=5\): \(35+5=40\)

Teacher's Note:
1. The value of y at x = 0 gives b directly.
2. Check with the remaining table entries.

Answer: (b) All three points A, B and C

Explanation:
1. A: \(9(1)-7=2\) matches
2. B: \(9(-1)-7=-16\) matches
3. C: \(9(0)-7=-7\) matches

Teacher's Note:
1. Substitute x and compare with y for each point.
2. All three satisfy the equation, so the points are collinear on the line.

Answer: (b) a solution, because \(2(2)+3=7\)

Explanation:
1. Put \(x=3,\ y=2\) in \(2y+x\): \(4+3=7\)
2. This equals the right side, so the point lies on the graph

Teacher's Note:
1. Remember that the coefficient 2 multiplies y.
2. Check by reading the point off the graph as well.

Answer: (d) 8 and 12

Explanation:
1. Let the numbers be x and \(x+4\): \(x(x+4)=96\)
2. \(x^2+4x-96=0\), so \((x+12)(x-8)=0\)
3. Positive root \(x=8\), so the numbers are 8 and 12

Teacher's Note:
1. Reduce the product condition to a quadratic.
2. Check: 12 - 8 = 4 and 8 x 12 = 96.

Answer: (c) all real \(k\) except \(k=-\dfrac23\)

Explanation:
1. \(\dfrac{a_1}{a_2}=\dfrac13\) and \(\dfrac{b_1}{b_2}=\dfrac{-k}2\)
2. Unique solution needs \(\dfrac13\ne\dfrac{-k}2\), i.e. \(k\ne-\dfrac23\)

Teacher's Note:
1. Compare only the ratios of the x and y coefficients for a unique solution.
2. Exclude the single value that makes the lines parallel or coincident.

Answer: (d) \(k=2\)

Explanation:
1. No solution needs \(\dfrac3{2k-1}=\dfrac1{k-1}\ne\dfrac{1}{2k+1}\)
2. \(3k-3=2k-1\), so \(k=2\)
3. At \(k=2\): \(\dfrac33=\dfrac11=1\) but \(\dfrac{c_1}{c_2}=\dfrac{-1}{-5}=\dfrac15\ne1\), so the lines are parallel

Teacher's Note:
1. The third ratio must be different from the first two.
2. Always check it after finding k.

Answer: (c) all real \(k\) except \(k=-\dfrac{14}5\)

Explanation:
1. \(\dfrac25\ne\dfrac k{-7}\)
2. \(k\ne-\dfrac{14}5\)

Teacher's Note:
1. Compare a1/a2 with b1/b2.
2. Careful with the negative sign of the y-coefficient -7.

Answer: (b) \(k=4\)

Explanation:
1. \(\dfrac2{k+2}=\dfrac3{2k+1}=\dfrac2{2(k-1)}\)
2. \(\dfrac2{k+2}=\dfrac1{k-1}\) gives \(2k-2=k+2\), so \(k=4\)
3. Check: \(\dfrac26=\dfrac39=\dfrac13\)

Teacher's Note:
1. Simplify 2/(2(k - 1)) to 1/(k - 1) first.
2. Test with the middle ratio.

Answer: (c) \(a=-1,\ b=\dfrac52\)

Explanation:
1. \(\dfrac2{2b+1}=\dfrac{2a+5}9=\dfrac5{15}=\dfrac13\)
2. \(2b+1=6\), so \(b=\dfrac52\)
3. \(2a+5=3\), so \(a=-1\)

Teacher's Note:
1. The constants give the common ratio 1/3 immediately.
2. Use it with each of the other ratios.

Answer: (b) \(c=6\) and \(c=-6\)

Explanation:
1. No solution needs \(\dfrac c{12}=\dfrac3c\ne\dfrac35\)
2. \(c^2=36\), so \(c=\pm6\)
3. For \(c=6\) the ratio \(\dfrac12\ne\dfrac35\); for \(c=-6\) the ratio \(-\dfrac12\ne\dfrac35\), so both values give parallel lines

Teacher's Note:
1. Equal ratios of the coefficients give c squared = 36.
2. Check the constant ratio for each root.

Answer: (a) \(p=-6\)

Explanation:
1. Write as \(px+3y-(p-3)=0\) and \(12x+py-p=0\)
2. \(\dfrac p{12}=\dfrac3p\) gives \(p^2=36\), \(p=\pm6\)
3. Need \(\dfrac3p\ne\dfrac{p-3}p\), i.e. \(p\ne6\) (for \(p=6\) the lines coincide); so \(p=-6\)

Teacher's Note:
1. Check the third ratio for each candidate root.
2. For p = 6 all three ratios are equal, giving coincident lines.

Answer: (d) \(k=-1\)

Explanation:
1. No solution needs \(\dfrac{3k+1}{k^2+1}=\dfrac3{k-2}\ne\dfrac25\)
2. \((3k+1)(k-2)=3(k^2+1)\) gives \(3k^2-5k-2=3k^2+3\), so \(k=-1\)
3. At \(k=-1\): \(\dfrac3{k-2}=-1\ne\dfrac25\), so the lines are parallel

Teacher's Note:
1. The k squared terms cancel, leaving a linear equation in k.
2. After finding k verify the constant ratio.

Answer: (c) \(k=8\)

Explanation:
1. Both equations are homogeneous, so a non-trivial solution needs \(\dfrac4k=\dfrac5{10}\)
2. \(k=8\)

Teacher's Note:
1. For equations through the origin the constants are both zero, so the lines either coincide or meet only at the origin.
2. Compare the x and y ratios.

Answer: (b) \(x=\dfrac{54}{17},\ y=\dfrac{13}{17}\)

Explanation:
1. From the first: \(x=\dfrac{7y+1}2\). Put in the second: \(2(7y+1)+3y=15\)
2. \(17y=13\), so \(y=\dfrac{13}{17}\)
3. \(x=\dfrac{7\cdot13/17+1}{2}=\dfrac{54}{17}\)

Teacher's Note:
1. Substitute the whole expression for x in brackets.
2. Check the answer by substituting it in both original equations.

Answer: (d) \(x=5,\ y=7\)

Explanation:
1. \(y=17-2x\). Then \(17x-11(17-2x)=8\)
2. \(39x=195\), so \(x=5\)
3. \(y=17-10=7\)

Teacher's Note:
1. Substitute y = 17 - 2x into the second equation.
2. Check the answer by substituting it in both original equations.

Answer: (b) \(x=y=\dfrac{a^2+b^2}{2a}\)

Explanation:
1. Adding: \(2a(x+y)=2(a^2+b^2)\), so \(x+y=\dfrac{a^2+b^2}a\)
2. Subtracting: \(2b(x-y)=0\), so \(x=y\)
3. \(x=y=\dfrac{a^2+b^2}{2a}\)

Teacher's Note:
1. Adding and subtracting the equations is quicker than substitution here.
2. Symmetry suggests x = y.

Answer: (a) \(-1\)

Explanation:
1. Subtracting: \(7y=35\), so \(y=5\); then \(2x=11-15\), so \(x=-2\)
2. \(5=m(-2)+3\), so \(-2m=2\)
3. \(m=-1\)

Teacher's Note:
1. First find the point of intersection, then use it in the line y = mx + 3.
2. Check the answer by substituting it in both original equations.

Answer: (d) \(x=\dfrac{13}7,\ y=\dfrac3{14}\)

Explanation:
1. Multiply the second by 3: \(9x+6y=18\). Add to the first: \(14x=26\)
2. \(x=\dfrac{13}7\)
3. \(2y=6-\dfrac{39}7=\dfrac37\), so \(y=\dfrac3{14}\)

Teacher's Note:
1. Making the y coefficients 6 and -6 allows direct addition.
2. Check the answer by substituting it in both original equations.