CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 16

Answer: (a) consistent with unique solution.

Explanation:
1. The two lines are not parallel, so they cut at exactly one point.
2. One intersection point means a consistent pair with a unique solution.

Teacher's Note:
1. Parallel lines never meet. Lines that cross meet once.
2. Look for the number of common points.

Answer: (c) \(3x + y = 900,\ 5x + 3y = 2100\)

Explanation:
1. 3 chairs and 1 table: \(3x + y = 900\).
2. 5 chairs and 3 tables: \(5x + 3y = 2100\).

Teacher's Note:
1. Write one equation for each sentence.
2. The number of chairs goes with x, the number of tables with y.

Answer: (b) 54

Explanation:
1. Let the digits be x and y, so the number is \(10x + y\).
2. \(\frac{5}{6}(10x + y) = 10y + x\) gives \(4x = 5y\).
3. With \(x - y = 1\): \(x = 5\), \(y = 4\). The number is 54.

Teacher's Note:
1. Turn the fraction statement into an equation.
2. Check: \(\frac{5}{6} \times 54 = 45\).

Answer: (c) 48

Explanation:
1. Let the tens digit be y and the units digit be x. Then \(x = 2y\).
2. \(10y + x + 36 = 10x + y\), so \(x - y = 4\).
3. Solving: \(x = 8\), \(y = 4\). The number is 48.

Teacher's Note:
1. Use the two conditions to write two equations.
2. Check: 48 + 36 = 84, the reverse of 48.

Answer: (c) 7 : 1

Explanation:
1. Cross-multiply: \(12x + 16y = 9x + 18y\), so \(3x = 2y\), i.e. \(x = \frac{2y}{3}\).
2. Then \(\dfrac{3x + 5y}{3x - y} = \dfrac{2y + 5y}{2y - y} = \dfrac{7y}{y} = 7\), so the ratio is 7 : 1.

Teacher's Note:
1. Cross-multiply first, then substitute x in terms of y.
2. The ratio of the two expressions is what is asked.

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

Explanation:
1. \(3^{x+y} = 3^4\), so \(x + y = 4\).
2. \(81^{x-y} = 3\) means \(3^{4(x-y)} = 3\), so \(x - y = \frac{1}{4}\).
3. Solving: \(x = \frac{17}{8} = 2\frac{1}{8}\) and \(y = \frac{15}{8} = 1\frac{7}{8}\).

Teacher's Note:
1. Write both powers of 3 as the same base.
2. Compare the exponents.

Answer: (d) 15

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

Teacher's Note:
1. Turn each condition into an equation in x and y.
2. Check: (15 + 1)/(3 + 1) = 4.

Answer: (a) all real values except 10

Explanation:
1. \(\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}\) and \(\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}\).
2. Parallel needs \(\frac{c_1}{c_2} \ne \frac{1}{2}\), i.e. \(\frac{-5}{-p} \ne \frac{1}{2}\), so \(p \ne 10\).

Teacher's Note:
1. Parallel lines have equal ratios of a and b, but unequal ratio of c.
2. Solve \(\frac{5}{p} = \frac{1}{2}\) and exclude it.

Answer: (b) 11

Explanation:
1. The numbers are \(10x + y\) and \(10y + x\). Their sum is \(11(x + y)\).
2. For this to be a perfect square, \(x + y = 11\).

Teacher's Note:
1. Add the two numbers and factorise 11 out.
2. 11 must be the missing factor to make a square.

Answer: (d) 10

Explanation:
1. Inconsistent means \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\).
2. \(\frac{1}{5} = \frac{2}{k} \ne \frac{-3}{7}\), so \(k = 10\).

Teacher's Note:
1. Use the ratio of the coefficients of x and y.
2. Check that the constant ratio is different.

Answer: (d) 2

Explanation:
1. \(\frac{1}{2} = \frac{1}{k} \ne \frac{-4}{-3}\) is needed for no solution.
2. \(\frac{1}{2} = \frac{1}{k}\) gives \(k = 2\). Check: \(\frac{-4}{-3} \ne \frac{1}{2}\).

Teacher's Note:
1. Equate the ratios of x and y coefficients.
2. Then check that the constant ratio is different.

Answer: (a) \(k \ne -3\)

Explanation:
1. \(\frac{a_1}{a_2} = \frac{2k}{6}\) and \(\frac{b_1}{b_2} = \frac{5}{-5} = -1\).
2. Unique solution needs \(\frac{2k}{6} \ne -1\), so \(k \ne -3\).

Teacher's Note:
1. A unique solution needs the ratios of x-coefficients and y-coefficients to differ.
2. Solve the single condition for k.

Answer: (d) no solution

Explanation:
1. \(\frac{a_1}{a_2} = \frac{1}{-3}\) and \(\frac{b_1}{b_2} = \frac{2}{-6} = \frac{1}{-3}\).
2. \(\frac{c_1}{c_2} = \frac{5}{1}\), which is different, so the lines are parallel and there is no solution.

Teacher's Note:
1. Compare the three ratios.
2. If the first two are equal but the third is not, there is no solution.

Answer: (c) intersecting or coincident

Explanation:
1. A consistent pair has either \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\) (intersecting), or all three ratios equal (coincident).
2. So the lines are intersecting or coincident.

Teacher's Note:
1. Consistent means the lines have at least one common point.
2. Parallel lines have none, so they are excluded.

Answer: (d) no solution

Explanation:
1. Both lines are parallel to the x-axis.
2. Parallel lines never meet, so there is no solution.

Teacher's Note:
1. Draw the two horizontal lines in your mind.
2. They are at different heights, so they never meet.

Answer: (c) 2

Explanation:
1. Coincident lines need \(\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16}\).
2. \(\frac{3}{6} = \frac{1}{2}\), and \(\frac{-1}{-k} = \frac{1}{k} = \frac{1}{2}\), so \(k = 2\).

Teacher's Note:
1. Coincident lines have all three ratios equal.
2. Use the x and y ratios first, then check the constant.

Answer: (d) intersecting at \((a, b)\)

Explanation:
1. \(x = a\) is a vertical line and \(y = b\) is a horizontal line.
2. They meet at the point \((a, b)\).

Teacher's Note:
1. Vertical and horizontal lines are perpendicular.
2. Read the point of meeting as (x, y).

Answer: (c) \(\frac{15}{4}\)

Explanation:
1. \(\frac{3}{2} = \frac{2k}{5} \ne \frac{-2}{1}\) is required for parallel lines.
2. \(\frac{3}{2} = \frac{2k}{5}\), so \(2k = \frac{15}{2}\) and \(k = \frac{15}{4}\).

Teacher's Note:
1. Parallel lines have equal ratios of x and y coefficients.
2. Solve the ratio equation for k.

Answer: (d) no value

Explanation:
1. Infinitely many solutions need \(\frac{c}{6} = \frac{-1}{-2} = \frac{2}{3}\).
2. \(\frac{c}{6} = \frac{1}{2}\) gives \(c = 3\), but \(\frac{2}{3} \ne \frac{1}{2}\).
3. So no value of c gives infinitely many solutions.

Teacher's Note:
1. For infinitely many solutions, all three ratios must be equal.
2. The ratio of the constants does not match, so no c works.

Answer: (d) \(10x - 14y = -4\)

Explanation:
1. Multiply \(-5x + 7y = 2\) by \(-2\): \(10x - 14y = -4\).
2. This is the same line, so the pair is dependent.

Teacher's Note:
1. Dependent equations are multiples of each other.
2. Multiply the given equation by -2.

Answer: (c) 3 and 1

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

Teacher's Note:
1. Add the equations to remove y.
2. Check: 3 - 1 = 2 and 3 + 1 = 4.

Answer: (d) 25 and 25

Explanation:
1. Let x be the number of Re 1 coins and y the number of Rs 2 coins. Then \(x + y = 50\).
2. \(x + 2y = 75\). Subtracting, \(y = 25\), so \(x = 25\).

Teacher's Note:
1. Write the coin count and the money as two equations.
2. Check: 25 + 2 × 25 = 75.

Answer: (c) 6 and 36

Explanation:
1. Let the father's age be x and the son's age be y. Then \(x = 6y\).
2. After 4 years: \(x + 4 = 4(y + 4)\), so \(x - 4y = 12\).
3. Solving: \(y = 6\) and \(x = 36\).

Teacher's Note:
1. Write the present-age condition and the future-age condition.
2. Check: after 4 years 40 = 4 × 10.

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

Explanation:
1. \(\frac{9}{18} = \frac{3}{6} = \frac{12}{24} = \frac{1}{2}\), so the two equations are multiples of each other.
2. The reason gives the condition for infinitely many solutions, and it explains the assertion.

Teacher's Note:
1. Divide the coefficients of the second equation by a common factor.
2. A reason that gives the exact condition explains the answer.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. For \(k = 2\): \(\frac{1}{2} = \frac{1}{2} \ne \frac{-4}{-3}\), so there is no solution. Assertion (A) is true.
2. Reason (R) is true, but it talks about consistency in general and does not explain why k = 2 gives no solution.

Teacher's Note:
1. Check the ratios when k = 2.
2. A true reason need not explain the assertion.