CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 26

Answer: (b) \(x=-\dfrac1{11}\)

Explanation:
1. LCM of 5, 4, 7 is 140: \(28(2x-3)+35(x+3)=20(4x+1)\)
2. \(56x-84+35x+105=80x+20\), so \(91x+21=80x+20\)
3. \(11x=-1\), so \(x=-\dfrac1{11}\)

Teacher's Note:
1. Multiply by the LCM of all denominators to clear fractions.
2. Watch the signs while expanding the brackets.

Answer: (a) \(x=7\)

Explanation:
1. Simplify the bracket: \(2x-\dfrac{1-x}2=\dfrac{5x-1}2\)
2. Equation: \(\dfrac{7x-1}4-\dfrac{5x-1}6=\dfrac{19}3\). Multiply by 12: \(3(7x-1)-2(5x-1)=76\)
3. \(21x-3-10x+2=76\), so \(11x=77\) and \(x=7\)

Teacher's Note:
1. Work from the innermost bracket outwards.
2. Check the answer by substituting it in both original equations.

Answer: (d) \(x=-8\)

Explanation:
1. Multiply by 6: \(4x+5-4(2x+7)=9\)
2. \(4x+5-8x-28=9\), so \(-4x=32\)
3. \(x=-8\)

Teacher's Note:
1. Multiplying by 6 turns 2(2x + 7)/3 into 4(2x + 7).
2. Check the answer by substituting it in both original equations.

Answer: (b) \(x=\dfrac85\)

Explanation:
1. Multiply by \((x-1)(x-2)(x-3)\): \(3(x-2)(x-3)+4(x-1)(x-3)=7(x-1)(x-2)\)
2. \(3x^2-15x+18+4x^2-16x+12=7x^2-21x+14\)
3. \(-31x+30=-21x+14\), so \(10x=16\) and \(x=\dfrac85\)

Teacher's Note:
1. The x squared terms cancel, so the equation becomes linear.
2. Check that the answer is not 1, 2 or 3.

Answer: (a) \(x=1\)

Explanation:
1. LHS \(=2x^2+3x+1\); RHS \(=2x^2+9x+9-14=2x^2+9x-5\)
2. \(3x+1=9x-5\), so \(6x=6\)
3. \(x=1\)

Teacher's Note:
1. Quadratic terms cancel on both sides.
2. Check the answer by substituting it in both original equations.

Answer: (c) \(27\)

Explanation:
1. \(2^x-2^{x-1}=2^{x-1}(2-1)=2^{x-1}\)
2. \(2^{x-1}=4=2^2\), so \(x=3\)
3. \(x^x=3^3=27\)

Teacher's Note:
1. Factor out 2^{x-1}.
2. Evaluate x^x only after finding x.

Answer: (b) \(a=-11,\ b=5\)

Explanation:
1. Expand the right side: \(2x^2+xy-3y^2+(b-4)x-(b+6)y-2b\)
2. Compare constants: \(-2b=-10\), so \(b=5\); compare x: \(b-4=1\), consistent
3. Compare y: \(a=-(b+6)=-11\)

Teacher's Note:
1. Compare the coefficients of like terms on both sides.
2. The x coefficient is a useful check on b.

Answer: (a) \(x^2-45x+324=0\)

Explanation:
1. A now has \(x-5\) and B now has \(45-x-5=40-x\)
2. \((x-5)(40-x)=124\), so \(-x^2+45x-200=124\)
3. \(x^2-45x+324=0\)

Teacher's Note:
1. B's original coins are 45 - x.
2. Move everything to one side and make the x squared term positive.

Answer: (d) 16 and 48

Explanation:
1. Let the parts be x and y: \(x+y=64\), \(y=3x\)
2. \(4x=64\), so \(x=16\) and \(y=48\)

Teacher's Note:
1. Check the answer by substituting it in both original equations.
2. The ratio of the parts is 1 : 3.

Answer: (c) \(k=5\)

Explanation:
1. We need \(\dfrac2{k+1}=\dfrac{-3}{1-2k}=\dfrac7{5k-4}\)
2. \(2(1-2k)=-3(k+1)\) gives \(k=5\)
3. Check: \(\dfrac26=\dfrac{-3}{-9}=\dfrac7{21}=\dfrac13\)

Teacher's Note:
1. Use two of the ratios to find k, then confirm with the third.
2. Equal ratios of all three coefficients give infinitely many solutions.

Answer: (b) \(k\ne3\)

Explanation:
1. \(\dfrac{a_1}{a_2}=\dfrac k6\) and \(\dfrac{b_1}{b_2}=\dfrac{-1}{-2}=\dfrac12\)
2. Unique solution needs \(\dfrac k6\ne\dfrac12\), i.e. \(k\ne3\)

Teacher's Note:
1. A unique solution needs the ratio of x-coefficients different from the ratio of y-coefficients.
2. Do not mix this with the condition for no solution.

Answer: (d) none

Explanation:
1. Infinitely many solutions need \(\dfrac k6=\dfrac12=\dfrac23\)
2. But \(\dfrac12\ne\dfrac23\) whatever k is, so no value of k works

Teacher's Note:
1. The ratio of the y-coefficients and the ratio of the constants are already different.
2. So the lines can only intersect or be parallel.

Answer: (c) 96

Explanation:
1. Let there be r rows of s students: \((s+4)(r-2)=rs\) and \((s-4)(r+4)=rs\)
2. These give \(2r-s=4\) and \(s-r=4\)
3. So \(r=8\), \(s=12\) and the number of students is \(96\)

Teacher's Note:
1. Form two equations from the two arrangements; the rs terms cancel.
2. Check: 16 x 6 = 96 and 8 x 12 = 96.

Answer: (a) 628 and 372

Explanation:
1. \(x+y=1000\) and \(x^2-y^2=256000\)
2. \((x-y)(x+y)=256000\), so \(x-y=256\)
3. \(x=628\), \(y=372\)

Teacher's Note:
1. Use the identity x^2 - y^2 = (x + y)(x - y).
2. Divide to get the difference directly.

Answer: (d) 50 km/h and 30 km/h

Explanation:
1. Same direction: \(8(x-y)=160\), so \(x-y=20\)
2. Opposite directions: \(2(x+y)=160\), so \(x+y=80\)
3. \(x=50\), \(y=30\)

Teacher's Note:
1. Relative speed is the difference for the same direction and the sum for opposite directions.
2. Distance = speed x time.

Answer: (c) 19 m and 10 m

Explanation:
1. \((x-5)(y+3)=xy-8\) gives \(3x-5y=7\)
2. \((x+3)(y+2)=xy+74\) gives \(2x+3y=68\)
3. Solving: \(x=19\), \(y=10\)

Teacher's Note:
1. Expand both products; the xy terms cancel.
2. Check: 14 x 13 = 182 = 190 - 8.

Answer: (b) Rs. 250 and Rs. 8

Explanation:
1. \(x+110y=1130\) and \(x+200y=1850\)
2. Subtracting: \(90y=720\), so \(y=8\)
3. \(x=1130-880=250\)

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

Answer: (a) 40 km/h and 30 km/h

Explanation:
1. Same direction: \(7(x-y)=70\), so \(x-y=10\)
2. Opposite: \(x+y=70\)
3. \(x=40\), \(y=30\)

Teacher's Note:
1. Use relative speed in both situations.
2. The faster car starts from A.

Answer: (c) No, because the left side becomes 7

Explanation:
1. Put \(x=12\): \(\dfrac{12}3+\dfrac{12}4=4+3=7\)
2. \(7\ne8\), so \(x=12\) is not a solution (the solution is \(x=\dfrac{96}{7}\))

Teacher's Note:
1. A value is a solution only if it makes both sides equal.
2. Always substitute, never assume.

Answer: (b) Yes, both sides equal \(-11\)

Explanation:
1. LHS \(=(4(-4)+7)-2=-9-2=-11\)
2. RHS \(=3(-4)+1=-11\)
3. Both sides are equal, so \(x=-4\) is a solution

Teacher's Note:
1. Substitute carefully with brackets when the value is negative.
2. Compare the two numbers obtained.

Answer: (a) No

Explanation:
1. At \(x=\dfrac12\): \(\dfrac{6.5}4-\dfrac{-0.5}2=1.625+0.25=1.875\)
2. \(1.875\ne5\), so \(x=\dfrac12\) is not a solution

Teacher's Note:
1. Evaluate each fraction separately before subtracting.
2. Notice the sign change when subtracting a negative value.

Answer: (d) Yes, both sides equal 3

Explanation:
1. LHS \(=6-4+1=3\)
2. RHS \(=9-6=3\)
3. Both sides equal 3, so \(x=3\) is a solution

Teacher's Note:
1. Check both sides separately.
2. Equal values mean the number is a solution.

Answer: (c) \(x=3\)

Explanation:
1. Bring the terms with x together: \(\dfrac6x-\dfrac3x=12-11\)
2. \(\dfrac3x=1\), so \(x=3\)

Teacher's Note:
1. Collect the fractions on one side and the numbers on the other.
2. Check the answer by substituting it in both original equations.

Answer: (b) \(2\)

Explanation:
1. \(2x=16\), so \(x=8\)
2. \(x^2+x-70=64+8-70=2\)

Teacher's Note:
1. Find x first and then substitute into the expression.
2. Take care to square 8 correctly.