NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 3: Pair of Linear Equations in Two Variables

Exercise 3.3

Write both in standard form.

\(\lambda x + y - \lambda^2 = 0\)

\(x + \lambda y - 1 = 0\)

Here \(a_1 = \lambda,\ b_1 = 1,\ c_1 = -\lambda^2\).

And \(a_2 = 1,\ b_2 = \lambda,\ c_2 = -1\).

Unique solution if the determinant is non-zero.

\(\Delta = a_1 b_2 - a_2 b_1 = \lambda^2 - 1\).

So, unique solution when \(\lambda^2 - 1 \neq 0\).

That is \(\lambda \neq \pm 1\).

Parallel (no solution) when

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}.\)

Compute: \(\dfrac{a_1}{a_2} = \lambda\), \(\dfrac{b_1}{b_2} = \dfrac{1}{\lambda}\).

Equality gives \(\lambda = \dfrac{1}{\lambda}\Rightarrow \lambda^2=1\).

At \(\lambda = -1\), the constants ratio is

\(\dfrac{c_1}{c_2} = \dfrac{-\lambda^2}{-1} = \lambda^2 = 1\).

The first two ratios are \(-1\) but the third is \(1\).

Hence, lines are parallel ⇒ no solution.

Coincident (infinitely many) when all three ratios are equal.

At \(\lambda = 1\), we have \(\dfrac{a_1}{a_2} = 1\), \(\dfrac{b_1}{b_2} = 1\),

and \(\dfrac{c_1}{c_2} = 1\).

So the lines coincide ⇒ infinitely many solutions.

Standard form:

\(kx + 3y - (k - 3) = 0\)

\(12x + ky - k = 0\)

For no solution: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}\).

First two ratios equal when

\(\dfrac{k}{12} = \dfrac{3}{k}\Rightarrow k^2 = 36\Rightarrow k = \pm 6\).

Check constants ratio \(\dfrac{c_1}{c_2} = \dfrac{-(k-3)}{-k} = \dfrac{k-3}{k}\).

For \(k = 6\): all three ratios are \(\dfrac{1}{2}\) ⇒ coincident, not “no solution”.

For \(k = -6\): the first two ratios are \(-\dfrac{1}{2}\) but \(\dfrac{c_1}{c_2} = \dfrac{3}{2}\).

Hence unequal ⇒ parallel distinct lines ⇒ no solution.

Write coefficients:

\(a_1 = 1,\ b_1 = 2,\ c_1 = -1\).

\(a_2 = a-b,\ b_2 = a+b,\ c_2 = -(a+b-2)\).

For infinitely many solutions,

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.\)

From the first two ratios:

\(\dfrac{1}{a-b} = \dfrac{2}{a+b}\Rightarrow a+b = 2(a-b)\).

So, \(-a + 3b = 0\Rightarrow a = 3b\).

Match with constants ratio:

\(\dfrac{1}{a-b} = \dfrac{-1}{-a-b+2} = \dfrac{1}{a+b-2}\).

Thus \(a-b = a+b-2\Rightarrow -b = b - 2\Rightarrow b = 1\).

Then \(a = 3b = 3\).

(i).

Slopes are equal when

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\Rightarrow \dfrac{3}{6} = \dfrac{-1}{-2} = \dfrac{1}{2}.\)

This holds for all \(p\).

To avoid coincidence, keep constants ratio different:

\(\dfrac{c_1}{c_2} = \dfrac{-5}{-p} = \dfrac{5}{p} \neq \dfrac{1}{2}\Rightarrow p \neq 10\).

(ii).

No solution when

\(\dfrac{-1}{p} = \dfrac{p}{-1} \ne \dfrac{-1}{-1}\).

The first equality gives \(\dfrac{-1}{p} = -p\Rightarrow p^2 = 1\Rightarrow p=\pm 1\).

For \(p=1\): coefficient ratios are \(-1\) and constants ratio is \(1\) ⇒ parallel.

For \(p=-1\): all three ratios are \(1\) ⇒ coincident.

Hence, \(p=1\) gives no solution.

(iii).

Unique point when slopes are different.

Check equality of ratios:

\(\dfrac{-3}{2p} \ne \dfrac{5}{-3} = -\dfrac{5}{3}.\)

Equality would give

\(\dfrac{-3}{2p} = -\dfrac{5}{3}\Rightarrow 3 = \dfrac{10p}{3}\Rightarrow p = \dfrac{9}{10}.\)

So for \(p \ne \dfrac{9}{10}\) the intersection is unique.

(iv).

Require \(\dfrac{2}{p} \ne \dfrac{3}{-6} = -\dfrac{1}{2}.\)

Equality occurs when \(\dfrac{2}{p} = -\dfrac{1}{2}\Rightarrow p = -4\).

Hence unique solution for all \(p \ne -4\).

(v).

Write second in standard form:

\(2px + (p+q)y - 28 = 0\).

Match ratios with first line \(2x + 3y - 7 = 0\).

\(\dfrac{2}{2p} = \dfrac{3}{p+q} = \dfrac{-7}{-28} = \dfrac{1}{4}.\)

From the first equality: \(\dfrac{1}{p} = \dfrac{1}{4}\Rightarrow p = 4\).

Then \(\dfrac{3}{p+q} = \dfrac{1}{4}\Rightarrow p + q = 12\Rightarrow q = 8\).

Step 1: Write the first line equation:

\(x - 3y = 2\)

Step 2: Multiply the whole equation by \(-2\) so that the \(x\)-term looks like the second equation:

\(-2) \times (x - 3y) = (-2) \times 2\)

\(-2x + 6y = -4\)

Step 3: Now compare with the second line:

\(-2x + 6y = 5\)

Step 4: The left-hand sides are the same (\(-2x + 6y\)).

But the right-hand sides are different: one is \(-4\) and the other is \(5\).

Step 5: This means both lines have the same slope, so they are parallel.

Since the constants are not equal, they are not the same line — they are distinct.

Final Answer: Parallel and distinct lines never meet. So the paths do not cross each other.

We are told that the solution is \(x = -1\) and \(y = 3\).

Step 1: Write the general form of a linear equation in two variables:

\(a x + b y = c\)

Step 2: Substitute the values \(x = -1\) and \(y = 3\).

\(a(-1) + b(3) = c\)

\(-a + 3b = c\)

So any equation of the form \(a x + b y = -a + 3b\) will be satisfied by the point \((-1,3)\).

Step 3: To form a pair of equations, we need two different equations. Let us take:

Equation 1: \(x + y = 2\)

Equation 2: \(2x - y = -5\)

Step 4: Check if \(x = -1, y = 3\) satisfies both.

For Equation 1: \((-1) + 3 = 2\) ✔

For Equation 2: \(2(-1) - 3 = -2 - 3 = -5\) ✔

Since both are true, the pair \(x + y = 2\) and \(2x - y = -5\) has the solution \((-1,3)\).

Step 5: How many such pairs can we write?

There are infinitely many choices because we can pick any two different equations of the form \(a x + b y = -a + 3b\), as long as the two equations are not multiples of each other (so that the lines are not parallel).

Step 1: Write the two equations clearly.

Equation (1): \(2x + y = 23\)

Equation (2): \(4x - y = 19\)

Step 2: Add the two equations to remove \(y\).

\((2x + y) + (4x - y) = 23 + 19\)

\(2x + 4x + y - y = 42\)

\(6x = 42\)

So, \(x = 7\).

Step 3: Put the value of \(x\) into Equation (1).

Equation (1): \(2x + y = 23\)

\(2(7) + y = 23\)

\(14 + y = 23\)

So, \(y = 23 - 14 = 9\).

Step 4: Now calculate \(5y - 2x\).

\(5y - 2x = 5(9) - 2(7)\)

\(= 45 - 14\)

\(= 31\)

Step 5: Calculate \(\dfrac{y}{x} - 2\).

\(\dfrac{y}{x} - 2 = \dfrac{9}{7} - 2\)

Write 2 as \(\dfrac{14}{7}\).

So, \(\dfrac{9}{7} - \dfrac{14}{7} = \dfrac{-5}{7}\).

Final Answer:

\(5y - 2x = 31\)

\(\dfrac{y}{x} - 2 = -\dfrac{5}{7}\)

In a rectangle, opposite sides are equal in length.

So we can write two equations:

Top = Bottom → \(x + 3y = 13\)

Left = Right → \(3x + y = 7\)

Step 1: Start with the second equation.

\(3x + y = 7\)

Subtract \(3x\) from both sides:

\(y = 7 - 3x\)

Step 2: Put this value of \(y\) into the first equation.

The first equation is:

\(x + 3y = 13\)

Replace \(y\) with \(7 - 3x\):

\(x + 3(7 - 3x) = 13\)

Step 3: Expand the brackets:

\(x + 21 - 9x = 13\)

Step 4: Combine like terms:

\(-8x + 21 = 13\)

Step 5: Subtract 21 from both sides:

\(-8x = -8\)

Step 6: Divide both sides by \(-8\):

\(x = 1\)

Step 7: Now find \(y\) using \(y = 7 - 3x\):

\(y = 7 - 3(1)\)

\(y = 7 - 3 = 4\)

So the solution is:

\(x = 1\) and \(y = 4\).

(i)

We are given two equations:

1) \(x + y = 3.3\)

2) \(\dfrac{0.6}{3x - 2y} = -1\)

From (2):

\(\dfrac{0.6}{3x - 2y} = -1\)

This means: \(3x - 2y = -0.6\).

So the system is:

\(x + y = 3.3\)

\(3x - 2y = -0.6\)

Multiply the first by 2:

\(2x + 2y = 6.6\)

Now add with the second:

\(2x + 2y + 3x - 2y = 6.6 - 0.6\)

\(5x = 6.0\)

So, \(x = 1.2\).

Put in \(x + y = 3.3\):

\(1.2 + y = 3.3\)

So, \(y = 2.1\).

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(ii)

Equations:

\(\dfrac{x}{3} + \dfrac{y}{4} = 4\)

\(\dfrac{5x}{6} - \dfrac{y}{8} = 4\)

Clear fractions by multiplying:

First ×12: \(4x + 3y = 48\)

Second ×24: \(20x - 3y = 96\)

Add them:

\(4x + 20x + 3y - 3y = 48 + 96\)

\(24x = 144\)

So, \(x = 6\).

Put in \(4x + 3y = 48\):

\(24 + 3y = 48\)

\(3y = 24\)

\(y = 8\).

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(iii)

Equations:

\(4x + \dfrac{6}{y} = 15\)

\(6x - \dfrac{8}{y} = 14\)

Let \(t = \dfrac{1}{y}\).

Then:

\(4x + 6t = 15\)

\(6x - 8t = 14\)

Multiply first by 3:

\(12x + 18t = 45\)

Multiply second by 2:

\(12x - 16t = 28\)

Subtract:

\(34t = 17\)

\(t = \dfrac{1}{2}\)

So, \(y = 2\).

Put in first: \(4x + 6(1/2) = 15\)

\(4x + 3 = 15\)

\(4x = 12\)

\(x = 3\).

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(iv)

Equations:

\(\dfrac{1}{2x} - \dfrac{1}{y} = -1\)

\(\dfrac{1}{x} + \dfrac{1}{2y} = 8\)

Let \(u = \dfrac{1}{x}\), \(v = \dfrac{1}{y}\).

So equations become:

\(\dfrac{1}{2}u - v = -1\)

\(u + \dfrac{1}{2}v = 8\)

Multiply both by 2:

\(u - 2v = -2\)

\(2u + v = 16\)

From first: \(u = -2 + 2v\).

Substitute in second:

\(2(-2 + 2v) + v = 16\)

\(-4 + 4v + v = 16\)

\(5v = 20\)

\(v = 4\)

So, \(y = 1/v = 1/4\).

Then \(u = -2 + 2(4) = 6\)

So, \(x = 1/u = 1/6\).

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(v)

Equations:

\(43x + 67y = -24\)

\(67x + 43y = 24\)

Add them:

\(110x + 110y = 0\)

\(x + y = 0\)

So, \(y = -x\).

Put in first:

\(43x + 67(-x) = -24\)

\(-24x = -24\)

\(x = 1\)

So, \(y = -1\).

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(vi)

Equations:

\(\dfrac{x}{a} + \dfrac{y}{b} = a + b\)

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

Let \(U = \dfrac{x}{a}\), \(V = \dfrac{y}{b}\).

Then:

\(U + V = a + b\)

\(\dfrac{U}{a} + \dfrac{V}{b} = 2\)

Multiply the second by \(ab\):

\(Ub + Va = 2ab\)

But from first: \(U + V = a + b\).

Check consistency: solution is \(U = a, V = b\).

So, \(x = a^2, y = b^2\).

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(vii)

Equation 1:

\(\dfrac{2xy}{x + y} = \dfrac{3}{2}\)

Cross multiply:

\(4xy = 3(x + y)\)

Rearrange:

\(y(4x - 3) = 3x\)

So, \(y = \dfrac{3x}{4x - 3}\).

Equation 2:

\(\dfrac{xy}{2x - y} = -\dfrac{3}{10}\)

Substitute y:

\(\dfrac{x(3x/(4x - 3))}{2x - (3x/(4x - 3))} = -3/10\)

Simplify (step by step):

\(\dfrac{3x^2}{8x^2 - 18x + 9} = -3/10\)

Cancel 3:

\(\dfrac{x^2}{8x^2 - 18x + 9} = -1/10\)

Cross multiply:

\(10x^2 = -(8x^2 - 18x + 9)\)

\(10x^2 = -8x^2 + 18x - 9\)

\(18x^2 - 18x + 9 = 0\)

\(2x^2 - 2x + 1 = 0\) (divided by 9)

Check discriminant: \(D = (-2)^2 - 4*2*1 = 4 - 8 = -4\).

Oops, mistake in earlier simplification.

Let’s go back carefully:

From earlier, simpler substitution works:

\(\dfrac{3x}{8x - 9} = -3/10\)

Cancel 3:

\(\dfrac{x}{8x - 9} = -1/10\)

Cross multiply: \(10x = -(8x - 9)\)

\(10x = -8x + 9\)

\(18x = 9\)

\(x = 1/2\).

Find y:

\(y = \dfrac{3(1/2)}{4(1/2) - 3} = \dfrac{1.5}{2 - 3} = -3/2\).

Step 1: Write the equations clearly.

Equation (1): \(\dfrac{x}{10} + \dfrac{y}{5} - 1 = 0\)

Equation (2): \(\dfrac{x}{8} + \dfrac{y}{6} = 15\)

Step 2: Remove fractions in Equation (1).

Multiply the whole equation by 10:

\(x + 2y - 10 = 0\)

So, \(x = 10 - 2y\).

Step 3: Remove fractions in Equation (2).

The denominators are 8 and 6. The LCM is 24. Multiply the whole equation by 24:

\(3x + 4y = 360\)

Step 4: Substitute the value of \(x\) from Step 2.

Substitute \(x = 10 - 2y\) into \(3x + 4y = 360\):

\(3(10 - 2y) + 4y = 360\)

Step 5: Simplify.

\(30 - 6y + 4y = 360\)

\(30 - 2y = 360\)

Step 6: Solve for \(y\).

\(-2y = 360 - 30\)

\(-2y = 330\)

\(y = -165\)

Step 7: Solve for \(x\).

From Step 2, \(x = 10 - 2y\).

Put \(y = -165\):

\(x = 10 - 2(-165)\)

\(x = 10 + 330 = 340\)

Step 8: Use the given relation \(y = \lambda x + 5\).

Substitute \(x = 340\) and \(y = -165\):

\(-165 = 340\lambda + 5\)

Step 9: Solve for \(\lambda\).

\(340\lambda = -165 - 5\)

\(340\lambda = -170\)

\(\lambda = -\dfrac{170}{340}\)

\(\lambda = -\dfrac{1}{2}\)

(i).

First equation: \(3x + y + 4 = 0\).

Move terms: \(y = -3x - 4\).

This is a straight line with slope \(-3\).

Second equation: \(6x - 2y + 4 = 0\).

Rearrange: \(-2y = -6x - 4\).

Divide by \(-2\): \(y = 3x + 2\).

This line has slope \(3\).

Since slopes are different, the two lines will intersect at one point.

To find the point, set the two expressions for \(y\) equal:

\(-3x - 4 = 3x + 2\).

Bring terms together: \(-3x - 3x = 2 + 4\).

\(-6x = 6\).

So, \(x = -1\).

Put \(x = -1\) in \(y = -3x - 4\):

\(y = -3(-1) - 4 = 3 - 4 = -1\).

Solution is \((x, y) = (-1, -1)\).

Since they intersect, the system is consistent and has a unique solution.

(ii).

First equation: \(x - 2y = 6\).

Second equation: \(3x - 6y = 0\).

Notice: if we multiply the first equation by 3, we get:

\(3x - 6y = 18\).

But the second equation is \(3x - 6y = 0\).

Left-hand sides are the same, but right-hand sides are different (18 vs 0).

This means the two lines are parallel and never meet.

So, the system is inconsistent (no solution).

(iii).

First equation: \(x + y = 3\).

Second equation: \(3x + 3y = 9\).

Divide the second equation by 3:

\(x + y = 3\).

This is exactly the same as the first equation.

So both equations represent the same line.

That means there are infinitely many solutions (all points on the line \(x + y = 3\)).

The system is consistent and dependent.

Step 1: Find where each line cuts the y-axis.

On the y-axis, \(x = 0\).

For the first line: \(2x + y = 4\).

Put \(x = 0\):

\(2(0) + y = 4 \Rightarrow y = 4\).

So the point is \((0, 4)\).

For the second line: \(2x - y = 4\).

Put \(x = 0\):

\(2(0) - y = 4 \Rightarrow -y = 4 \Rightarrow y = -4\).

So the point is \((0, -4)\).

Step 2: Find where the two lines meet (their intersection).

We have:

\(2x + y = 4\) … (i)

\(2x - y = 4\) … (ii)

Add (i) and (ii):

\((2x + y) + (2x - y) = 4 + 4\)

\(4x = 8\)

So, \(x = 2\).

Put \(x = 2\) in (i):

\(2(2) + y = 4\)

\(4 + y = 4\)

\(y = 0\).

So the point is \((2, 0)\).

Step 3: Identify the triangle.

The three vertices are \((0, 4)\), \((0, -4)\), and \((2, 0)\).

Step 4: Find the area.

The base is on the y-axis between \((0, 4)\) and \((0, -4)\).

Length of base = \(4 - (-4) = 8\).

The height is the distance from the y-axis to the point \((2, 0)\).

This distance is just the x-coordinate, which is \(2\).

Area of triangle = \(\tfrac{1}{2} \times \text{base} \times \text{height}\)

= \(\tfrac{1}{2} \times 8 \times 2\)

= \(8\) square units.

Step 1: We are given two equations:

\(x + y = 2\)

\(2x - y = 1\)

Step 2: From the first equation:

\(x + y = 2\)

So, \(y = 2 - x\).

Step 3: Put this value of \(y\) into the second equation:

\(2x - y = 1\)

\(2x - (2 - x) = 1\)

Step 4: Simplify:

\(2x - 2 + x = 1\)

\(3x - 2 = 1\)

Step 5: Add 2 to both sides:

\(3x = 3\)

Step 6: Divide by 3:

\(x = 1\)

Step 7: Now, put \(x = 1\) into \(y = 2 - x\):

\(y = 2 - 1 = 1\)

So, the solution point is \((1,1)\).

Step 8: A line through this point can be written in slope form:

\(y - 1 = m(x - 1)\)

Here, \(m\) is the slope and can be any real number.

Final Result: Infinitely many lines can pass through the point \((1,1)\).

Step 1: Apply the Factor Theorem

If \(x + 1\) is a factor, then putting \(x = -1\) in the polynomial should give 0.

Step 2: Substitute \(x = -1\)

Polynomial: \(2x^3 + ax^2 + 2bx + 1\)

\(= 2(-1)^3 + a(-1)^2 + 2b(-1) + 1\)

Step 3: Simplify each term

\((-1)^3 = -1\), so \(2(-1)^3 = -2\).

\((-1)^2 = 1\), so \(a(-1)^2 = a\).

\(2b(-1) = -2b\).

And \(+1\) stays the same.

Step 4: Combine all terms

\(-2 + a - 2b + 1 = 0\)

\(a - 2b - 1 = 0\)

So, \(a = 2b + 1\). (Equation 1)

Step 5: Use the second condition

We are also given: \(2a - 3b = 4\).

Step 6: Substitute \(a = 2b + 1\)

\(2(2b + 1) - 3b = 4\)

\(4b + 2 - 3b = 4\)

\(b + 2 = 4\)

\(b = 2\)

Step 7: Find \(a\)

From Equation (1): \(a = 2b + 1\).

\(a = 2(2) + 1 = 4 + 1 = 5\).

Final Answer: \(a = 5, b = 2\).

Step 1: In any triangle, the sum of the three angles is always \(180^\circ\).

So,

\(x + y + 40 = 180\)

\(x + y = 180 - 40\)

\(x + y = 140\)

Step 2: We are also told that the difference between \(x\) and \(y\) is \(30^\circ\).

This means:

\(|x - y| = 30\)

Step 3: Now we have two equations:

1) \(x + y = 140\)

2) \(|x - y| = 30\)

Step 4: Let us assume \(x > y\). Then \(x - y = 30\).

Step 5: Add the two equations:

\((x + y) + (x - y) = 140 + 30\)

\(2x = 170\)

\(x = 85\)

Step 6: Put \(x = 85\) in equation (1):

\(85 + y = 140\)

\(y = 55\)

Step 7: If we had assumed \(y > x\), then we would get \(y = 85\) and \(x = 55\).

Final Answer: The two angles are \(85^\circ\) and \(55^\circ\).

Step 1: Assume present ages.

Let Salim's present age be \(S\) years.

Let his daughter's present age be \(D\) years.

Step 2: Write the first condition.

Two years ago, Salim was thrice his daughter's age.

That means:

\(S - 2 = 3(D - 2)\)

Simplify:

\(S - 2 = 3D - 6\)

\(S = 3D - 4\) … (1)

Step 3: Write the second condition.

Six years later, Salim will be four years older than twice her age.

That means:

\(S + 6 = 2(D + 6) + 4\)

Simplify:

\(S + 6 = 2D + 12 + 4\)

\(S + 6 = 2D + 16\)

\(S = 2D + 10\) … (2)

Step 4: Solve the two equations.

From (1): \(S = 3D - 4\)

From (2): \(S = 2D + 10\)

Equating them:

\(3D - 4 = 2D + 10\)

\(3D - 2D = 10 + 4\)

\(D = 14\)

Step 5: Find Salim's age.

Put \(D = 14\) in (1):

\(S = 3(14) - 4\)

\(S = 42 - 4\)

\(S = 38\)

Final Answer:

Salim is 38 years old and his daughter is 14 years old.

Step 1: Let the sum of the children's present ages be \(C\).

Step 2: Let the father's present age be \(F\).

Step 3: According to the question, the father's age is twice the sum of the children's ages.

So, \(F = 2C\).

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Step 4: After 20 years:

The father's age will be: \(F + 20\).

The children's ages together will be: \(C + 20 + 20 = C + 40\).

(Because there are two children, each gets 20 years added.)

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Step 5: At that time, father's age will equal the children's total age.

So, \(F + 20 = C + 40\).

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Step 6: Now substitute \(F = 2C\) into the equation:

\(2C + 20 = C + 40\).

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Step 7: Solve step by step:

\(2C + 20 = C + 40\)

Subtract \(C\) from both sides: \(C + 20 = 40\)

Subtract 20 from both sides: \(C = 20\)

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Step 8: Father's present age:

\(F = 2C = 2 \times 20 = 40\).

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Final Answer: The father is 40 years old.

Step 1: Suppose the two numbers are in the form of \(5k\) and \(6k\).

Step 2: According to the question, if we subtract 8 from each number, the new ratio becomes \(4:5\).

This means:

\(\dfrac{5k - 8}{6k - 8} = \dfrac{4}{5}\)

Step 3: Now we cross multiply:

\(5 \times (5k - 8) = 4 \times (6k - 8)\)

Step 4: Multiply both sides:

\(25k - 40 = 24k - 32\)

Step 5: Bring like terms together:

\(25k - 24k = -32 + 40\)

Step 6: Simplify:

\(k = 8\)

Step 7: Put the value of \(k\) back into the numbers:

First number = \(5k = 5 \times 8 = 40\)

Second number = \(6k = 6 \times 8 = 48\)

Final Answer: The two numbers are \(40\) and \(48\).

Step 1: Let the original number of students in Hall A be \(a\).

Let the original number of students in Hall B be \(b\).

First condition:

If 10 students go from A to B, then

Students in A = \(a - 10\)

Students in B = \(b + 10\)

According to the question, they become equal:

\(a - 10 = b + 10\)

Rearranging:

\(a - b = 20\)   … (1)

Second condition:

If 20 students go from B to A, then

Students in A = \(a + 20\)

Students in B = \(b - 20\)

According to the question, A becomes double of B:

\(a + 20 = 2(b - 20)\)

Expand the right side:

\(a + 20 = 2b - 40\)

Rearranging:

\(a - 2b = -60\)   … (2)

Step 2: Solve the equations.

From (1): \(a = b + 20\)

Substitute into (2):

\((b + 20) - 2b = -60\)

\(-b + 20 = -60\)

\(-b = -80\)

\(b = 80\)

Now, put \(b = 80\) into (1):

\(a - 80 = 20\)

\(a = 100\)

Final Answer:

Hall A has 100 students and Hall B has 80 students.

Step 1: Suppose the fixed charge for the first two days is \(F\).

Step 2: Suppose the extra charge per day (after the first two days) is \(r\).

For Latika:

She stayed for 6 days.

The first 2 days are included in the fixed charge \(F\).

So extra days = \(6 - 2 = 4\).

Extra charge = \(4 \times r = 4r\).

Total money paid by Latika = \(F + 4r = 22\). … (Equation 1)

For Anand:

He stayed for 4 days.

The first 2 days are included in the fixed charge \(F\).

So extra days = \(4 - 2 = 2\).

Extra charge = \(2 \times r = 2r\).

Total money paid by Anand = \(F + 2r = 16\). … (Equation 2)

Now solve the equations:

From Equation (1): \(F + 4r = 22\).

From Equation (2): \(F + 2r = 16\).

Subtract Equation (2) from Equation (1):

(F + 4r) – (F + 2r) = 22 – 16

\(F - F + 4r - 2r = 6\)

\(2r = 6\)

\(r = 3\).

Find F:

Put \(r = 3\) in Equation (2):

\(F + 2(3) = 16\)

\(F + 6 = 16\)

\(F = 10\).

Final Answer:

Fixed charge = Rs 10, and additional daily charge = Rs 3.

Step 1: Let the number of correct answers be \(c\).

Step 2: Then the number of wrong answers will be the rest, that is \(120 - c\).

Step 3: For every correct answer she gets +1 mark.

So, marks from correct answers = \(c\).

Step 4: For every wrong answer she loses half a mark (\(\tfrac{1}{2}\)).

So, marks lost from wrong answers = \(\tfrac{1}{2} \times (120 - c)\).

Step 5: Total score = (marks from correct answers) – (marks lost from wrong answers).

That means:

\[ \text{Score} = c - \tfrac{1}{2}(120 - c) \]

Step 6: We know her score is 90, so:

\[ c - \tfrac{1}{2}(120 - c) = 90 \]

Step 7: Expand the brackets:

\[ c - 60 + \tfrac{1}{2}c = 90 \]

Step 8: Combine like terms:

\[ 1.5c - 60 = 90 \]

Step 9: Add 60 on both sides:

\[ 1.5c = 150 \]

Step 10: Divide both sides by 1.5:

\[ c = 100 \]

Therefore, Jayanti answered 100 questions correctly.

Step 1: In a cyclic quadrilateral, the sum of opposite angles is \(180^\circ\).

Step 2: Take opposite angles \(A\) and \(C\).

\(\angle A = 6x + 10\)

\(\angle C = x + y\)

So,

\((6x + 10) + (x + y) = 180\)

\(7x + y + 10 = 180\)

\(7x + y = 170\) … (Equation 1)

Step 3: Now take the other pair of opposite angles, \(B\) and \(D\).

\(\angle B = 5x\)

\(\angle D = 3y - 10\)

So,

\(5x + (3y - 10) = 180\)

\(5x + 3y - 10 = 180\)

\(5x + 3y = 190\) … (Equation 2)

Step 4: Solve the two equations:

Equation (1): \(7x + y = 170\)

Equation (2): \(5x + 3y = 190\)

From Equation (1):

\(y = 170 - 7x\)

Substitute this value of \(y\) into Equation (2):

\(5x + 3(170 - 7x) = 190\)

\(5x + 510 - 21x = 190\)

\(-16x + 510 = 190\)

\(-16x = 190 - 510\)

\(-16x = -320\)

\(x = 20\)

Step 5: Put \(x = 20\) in Equation (1):

\(7(20) + y = 170\)

\(140 + y = 170\)

\(y = 30\)

Step 6: Now calculate each angle:

\(\angle A = 6x + 10 = 6(20) + 10 = 130^\circ\)

\(\angle B = 5x = 5(20) = 100^\circ\)

\(\angle C = x + y = 20 + 30 = 50^\circ\)

\(\angle D = 3y - 10 = 3(30) - 10 = 80^\circ\)

Final Answer: \(x = 20\), \(y = 30\). Angles are: \(A = 130^\circ, B = 100^\circ, C = 50^\circ, D = 80^\circ\).