Class 10 Maths Sample Paper Set I (NCERT Exemplar Solutions)

Given:

The rational number is $\frac{47}{2^{3} \cdot 5^{2}}$.

To Find:

The number of decimal places after which the decimal expansion terminates.

Solution:

According to the Fundamental Theorem of Arithmetic, for a rational number in the form $\frac{p}{q}$ (where $p$ and $q$ are co-primes), the decimal expansion terminates if the prime factorisation of the denominator $q$ is of the form $2^m \cdot 5^n$, where $m, n$ are non-negative integers.

The number of decimal places after which the expansion terminates is equal to the maximum of $m$ and $n$.

In the given number $\frac{47}{2^{3} \cdot 5^{2}}$:

Since the maximum value between $3$ and $2$ is $3$, the decimal expansion will terminate after 3 decimal places.

Verification (Alternate Solution):

To convert the denominator into a power of $10$, we multiply both the numerator and the denominator by $5^1$:

Now, substituting the value back into the equation:

We can clearly see that there are 3 digits after the decimal point.

Conclusion:

The decimal expansion terminates after 3 decimal places. Therefore, the correct option is (C).

Given:

Euclid’s division lemma states that for two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$.

To Find:

The condition on the remainder $r$ in the statement of Euclid's division lemma.

Solution:

Euclid's division lemma (also known as the Euclidean Algorithm) is a fundamental theorem in number theory. It precisely defines the relationship between the dividend ($a$), the divisor ($b$), the quotient ($q$), and the remainder ($r$) when one positive integer is divided by another.

The statement of Euclid's division lemma is:

For any two positive integers $a$ and $b$, there exist unique integers $q$ (quotient) and $r$ (remainder) such that:

The crucial condition on the remainder $r$ is that it must be non-negative and strictly less than the divisor $b$. This ensures the uniqueness of $q$ and $r$.

The condition is expressed as:

This means the remainder $r$ can be zero (when $a$ is perfectly divisible by $b$) or any positive integer less than $b$.

Conclusion:

The condition on the remainder $r$ in Euclid’s division lemma is $0 \leq r < b$.

The correct option is (D) $0 \leq r < b$.

Given:

The polynomial is $p(x) = (x - 2)^{2} + 4$.

To Find:

The number of real zeroes the polynomial $p(x)$ can have.

Solution:

A zero of a polynomial $p(x)$ is the value of $x$ for which $p(x) = 0$.

We know that for any real number $x$, the square of a quantity, i.e., $(x - 2)^{2}$, is always non-negative (greater than or equal to zero).

Since a non-negative number cannot be equal to a negative number ($-4$), there is no real value of $x$ that satisfies equation (i).

Therefore, the polynomial has zero real zeroes.

Alternate Solution (Using Discriminant):

Let us expand the polynomial:

$p(x) = (x - 2)^{2} + 4$

$p(x) = (x^{2} - 4x + 4) + 4$

Comparing this with the standard quadratic form $ax^{2} + bx + c$, we get:

$a = 1, \ b = -4, \ c = 8$

The Discriminant ($D$) is given by $b^{2} - 4ac$:

$D = (-4)^{2} - 4(1)(8)$

$D = 16 - 32$

Since the Discriminant ($D < 0$) is negative, the quadratic polynomial has no real roots (zeroes).

Conclusion:

The number of real zeroes is 0. Thus, the correct option is (C).

Given:

A pair of linear equations:

To Find:

The condition for the pair of linear equations to be inconsistent.

Solution:

A system of linear equations is said to be inconsistent if it has no solution. Graphically, this corresponds to two parallel lines that do not intersect.

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, the consistency and the nature of solutions are determined by comparing the ratios of the coefficients:

1. Unique Solution (Consistent): If the lines intersect at exactly one point, there is a unique solution. The condition for this is:

2. Infinitely Many Solutions (Consistent): If the lines are coincident (one lies exactly on top of the other), there are infinitely many solutions. The condition for this is:

3. No Solution (Inconsistent): If the lines are parallel and distinct (they never intersect), there is no solution. The condition for this is:

The question asks for the condition when the system is inconsistent, which means there is no solution.

Comparing the condition for no solution with the given options, we find that option (C) matches the condition for an inconsistent system.

Conclusion:

A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is inconsistent if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$.

The correct option is (C) $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$.

Given:

The given quadratic equation is:

$x^2 + kx + 9 = 0$

The equation is stated to have real roots.

To Find:

The smallest value of $k$ that satisfies the condition for real roots.

Solution:

A quadratic equation of the form $ax^2 + bx + c = 0$ has real roots if its discriminant $D$ is greater than or equal to zero ($D \geq 0$).

Comparing the given equation $x^2 + kx + 9 = 0$ with the standard form $ax^2 + bx + c = 0$, we have:

$a = 1$

$b = k$

$c = 9$

The formula for the discriminant is:

$D = b^2 - 4ac$

For real roots, the condition is:

Substituting the values of $a$, $b$, and $c$ in the inequality (i):

$k^2 - 4(1)(9) \geq 0$

$k^2 - 36 \geq 0$

$k^2 \geq 36$

To solve for $k$, we take the square root on both sides:

$\sqrt{k^2} \geq \sqrt{36}$

$|k| \geq 6$

This absolute value inequality represents two possible ranges for $k$:

Now, let us examine the given options based on the condition derived in (ii):

(A) $-6$: This satisfies $k \leq -6$. It is a valid value for real roots.

(B) $6$: This satisfies $k \geq 6$. It is a valid value for real roots.

(C) $36$: This satisfies $k \geq 6$. It is a valid value for real roots.

(D) $-3$: This does not satisfy $k \leq -6$ or $k \geq 6$ (since $(-3)^2 = 9$, which is less than $36$).

The question asks for the smallest value of $k$. Comparing the valid options $-6$, $6$, and $36$, the smallest numerical value is $-6$.

Alternate Method (Checking Options):

We can substitute the values from the options into the discriminant formula $D = k^2 - 36$ and check for $D \geq 0$.

1. For $k = -6$: $D = (-6)^2 - 36 = 36 - 36 = 0$ (Roots are real and equal).

2. For $k = 6$: $D = (6)^2 - 36 = 36 - 36 = 0$ (Roots are real and equal).

3. For $k = 36$: $D = (36)^2 - 36 = 1296 - 36 = 1260$ (Roots are real and distinct).

4. For $k = -3$: $D = (-3)^2 - 36 = 9 - 36 = -27$ (Roots are imaginary).

Among the values that result in real roots ($-6, 6, 36$), the smallest value is $-6$.

Final Answer:

The smallest value of $k$ for which the equation has real roots is -6.

Hence, the correct option is (A).

Given:

Coordinates of point $P(x_1, y_1) = (4, -3)$

Coordinates of point $Q(x_2, y_2) = (-1, 7)$

Point $R$ lies on the line segment $PQ$ such that:

To Find:

The abscissa (x-coordinate) of point $R$.

Solution:

Since point $R$ lies on the line segment $PQ$, the total length $PQ$ is the sum of $PR$ and $RQ$.

From the given ratio in equation (i), we can say that if the total length $PQ$ is $5$ units, then $PR$ is $3$ units.

To find the length of $RQ$ in terms of the ratio:

$RQ = PQ - PR$

$RQ = 5 - 3 = 2 \text{ units}$

Therefore, point $R$ divides the line segment $PQ$ internally in the ratio $m : n$, where:

$m : n = PR : RQ = 3 : 2$

By using the Section Formula, the coordinates of a point $R(x, y)$ dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m : n$ are given by:

$x = \frac{mx_2 + nx_1}{m + n}$

Here, we only need to find the abscissa ($x$):

$m = 3$

$n = 2$

$x_1 = 4$

$x_2 = -1$

Substituting these values into the formula:

$x = \frac{3(-1) + 2(4)}{3 + 2}$

$x = \frac{-3 + 8}{5}$

$x = \frac{5}{5}$

$x = 1$

Final Answer:

The abscissa of point $R$ is 1.

Hence, the correct option is (D).

Given:

PA and PB are tangents drawn from an external point P to a circle with centre O. OAPB is a quadrilateral formed by joining the points O, A, P, and B.

To Determine:

The type of quadrilateral OAPB.

Solution:

We know the following properties related to tangents drawn from an external point to a circle:

1. The radius through the point of contact is perpendicular to the tangent.

Therefore, the radius OA is perpendicular to the tangent PA, and the radius OB is perpendicular to the tangent PB.

Now consider the quadrilateral OAPB. The sum of the interior angles of any quadrilateral is $360^\circ$.

So, in quadrilateral OAPB:

Substitute the values of $\angle OAP$ and $\angle OBP$:

We have found that the sum of the opposite angles $\angle AOB$ and $\angle APB$ is $180^\circ$.

We also know that the sum of the other pair of opposite angles is $\angle OAP + \angle OBP = 90^\circ + 90^\circ = 180^\circ$.

A quadrilateral is called a cyclic quadrilateral if the sum of each pair of opposite angles is $180^\circ$.

Since both pairs of opposite angles ($\angle OAP$ and $\angle OBP$; $\angle AOB$ and $\angle APB$) sum up to $180^\circ$, the quadrilateral OAPB is a cyclic quadrilateral.

Let's consider the other options:

• Square: A square requires all angles to be $90^\circ$ and all sides equal (OA=AP=PB=BO). While $\angle OAP = \angle OBP = 90^\circ$, $\angle AOB$ and $\angle APB$ are not necessarily $90^\circ$. Also, OA (radius) is not necessarily equal to AP (tangent length).
• Rhombus: A rhombus requires all sides to be equal (OA=AP=PB=BO). Again, OA is not necessarily equal to AP.
• Parallelogram: A parallelogram requires opposite angles to be equal and opposite sides parallel. While $\angle OAP = \angle OBP = 90^\circ$ (so opposite angles are equal), $\angle AOB$ is not necessarily equal to $\angle APB$.

The property that the sum of opposite angles is $180^\circ$ always holds for OAPB when PA and PB are tangents from P. This confirms that OAPB is a cyclic quadrilateral.

Conclusion:

The quadrilateral OAPB must be a cyclic quadrilateral.

The correct option is (C) cyclic quadrilateral.

Given:

$\cot 2\theta = \frac{1}{\sqrt{3}}$ and $2\theta \leq 90^\circ$.

To Find:

The value of $\sin 3\theta$.

Solution:

We are given that $\cot 2\theta = \frac{1}{\sqrt{3}}$.

We know the standard trigonometric values for common angles. The cotangent of $60^\circ$ is $\frac{1}{\sqrt{3}}$.

Comparing the given equation with the standard value, we have:

Since the angle $2\theta$ is in the range $2\theta \leq 90^\circ$, and the cotangent value is positive, $2\theta$ must be an angle in the first quadrant.

In the first quadrant, if $\cot A = \cot B$, then $A = B$.

Therefore, we can equate the angles:

Now, we can find the value of $\theta$:

We need to find the value of $\sin 3\theta$. Substitute the value of $\theta$ we found:

We know the value of $\sin 90^\circ$:

So, the value of $\sin 3\theta$ is 1.

Conclusion:

The value of $\sin 3\theta$ is 1.

The correct option is (B) 1.

Given:

Side of the square ($s$) = $4\text{ cm}$

Radius of each quadrant at the corners ($r$) = $1\text{ cm}$

Diameter of the central circle = $2\text{ cm}$

To Find:

The area of the remaining (shaded) portion of the square.

Solution:

To find the area of the remaining portion, we need to subtract the area of the four quadrants and the area of the central circle from the total area of the square.

Step 1: Calculate the area of the square

The formula for the area of a square is $\text{side} \times \text{side}$.

Step 2: Calculate the area of the four quadrants

A quadrant is one-fourth of a circle. Therefore, four quadrants of the same radius together form a complete circle.

$\text{Area of 4 quadrants} = 4 \times \left( \frac{1}{4} \pi r^2 \right) = \pi r^2$

Substituting $r = 1\text{ cm}$:

Step 3: Calculate the area of the central circle

The radius of the central circle is also $1\text{ cm}$.

$\text{Area of the central circle} = \pi r^2$

Step 4: Calculate the area of the remaining portion

$\text{Remaining Area} = \text{Area of Square} - (\text{Area of 4 quadrants} $$ + \text{Area of central circle})$

Using values from equations (i), (ii), and (iii):

$\text{Remaining Area} = 16 - (\pi + \pi)$

$\text{Remaining Area} = (16 - 2\pi)\text{ cm}^2$

Final Answer:

The area of the remaining (shaded) portion is $(16 - 2\pi)\text{ cm}^2$.

Hence, the correct option is (A).

Given:

The word provided is ‘MATHEMATICS’.

A letter is chosen at random from the English alphabets.

To Find:

The probability that the chosen letter is one of the letters present in the word ‘MATHEMATICS’.

Solution:

Total number of letters in the English alphabet is 26.

Now, let us identify the unique letters in the word ‘MATHEMATICS’:

The letters are: $M, \ A, \ T, \ H, \ E, \ M, \ A, \ T, \ I, \ C, \ S$.

The unique letters (distinct outcomes) are: M, A, T, H, E, I, C, S.

Counting the unique letters:

1. M

2. A

3. T

4. H

5. E

6. I

7. C

8. S

The formula for the probability of an event $E$ is given by:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(E) = \frac{n(E)}{n(S)}$

Substituting the values from equations (i) and (ii):

$P(E) = \frac{8}{26}$

Simplifying the fraction by dividing both numerator and denominator by 2:

$P(E) = \frac{\cancel{8}^{4}}{\cancel{26}_{13}}$

$P(E) = \frac{4}{13}$

Final Answer:

The probability that a chosen letter is a letter of the word ‘MATHEMATICS’ is $\frac{4}{13}$.

Hence, the correct option is (D).

To Find:

Whether there exists any natural number $n$ for which the expression $4^n$ ends with the digit 0.

Solution:

For a number to end with the digit 0, it must be divisible by 10.

The prime factors of 10 are 2 and 5. Therefore, any number ending with the digit 0 must have both 2 and 5 as its prime factors.

Let us examine the prime factorization of the given number $4^n$:

$4^n = (2 \times 2)^n$

From equation (i), it is evident that the only prime factor of $4^n$ is 2.

The Fundamental Theorem of Arithmetic states that the prime factorization of every composite number is unique, apart from the order in which the prime factors occur.

Since the prime factorization of $4^n$ contains only the prime 2 and does not contain the prime 5, $4^n$ is not divisible by 5.

Because $4^n$ is not divisible by 5, it cannot be divisible by 10.

Therefore, there is no natural number $n$ for which $4^n$ ends with the digit 0.

Final Answer:

No, there is no natural number $n$ for which $4^n$ ends with the digit 0, because its prime factorization does not contain the factor 5.

Question 12

Given:

The Arithmetic Progression (AP) is: $5, \ 17, \ 29, \ 41, \ \dots$

To Find:

The term of the AP which is $120$ more than its $15^{th}$ term, without using the $n^{th}$ term formula directly.

Solution:

First, we find the common difference ($d$) of the given AP:

$d = \text{Second term} - \text{First term}$

In an AP, each term is obtained by adding the common difference $d$ to the preceding term. Therefore, the difference between any two terms, say the $n^{th}$ term and the $m^{th}$ term, is always a multiple of the common difference $d$.

Specifically, the difference between the $n^{th}$ term and the $15^{th}$ term is given by:

According to the question, this difference is $120$. So,

$(n - 15) \times 12 = 120$

Dividing both sides by 12:

$(n - 15) = \frac{120}{12}$

$(n - 15) = 10$

Adding 15 to both sides:

$n = 10 + 15$

$n = 25$

This means the $25^{th}$ term is $10$ common differences ahead of the $15^{th}$ term.

Justification: Since each step in an AP increases the value by $d = 12$, to increase the value by $120$, we need $\frac{120}{12} = 10$ more steps. Adding $10$ steps to the $15^{th}$ term brings us to the $(15 + 10) = 25^{th}$ term.

Final Answer: The 25^th term of the AP will be 120 more than its 15^th term.

OR Question

Given:

The Arithmetic Progression (AP) is: $3, \ 7, \ 11, \ \dots$

Target value = $144$

To Find:

Whether $144$ is a term of the given AP.

Solution:

For the given AP:

First term ($a$) = $3$

Common difference ($d$) = $7 - 3 = 4$

Let $144$ be the $n^{th}$ term of the AP.

The formula for the $n^{th}$ term ($a_n$) of an AP is:

$a_n = a + (n - 1)d$

Substituting the known values:

$144 = 3 + (n - 1)4$

Subtracting 3 from both sides:

$144 - 3 = (n - 1)4$

$141 = (n - 1)4$

Dividing by 4:

$(n - 1) = \frac{141}{4}$

$(n - 1) = 35.25$

$n = 35.25 + 1$

$n = 36.25$

Justification: The position of a term ($n$) in an AP must always be a natural number (i.e., $1, 2, 3, \dots$). Since the value of $n$ calculated here is $36.25$, which is not a whole number or natural number, $144$ cannot be a term of the given progression.

Final Answer: No, 144 is not a term of the AP: 3, 7, 11, ...

Given:

The coordinates of the vertices of the triangle are:

$P(x_1, y_1) = (3, 4)$

$Q(x_2, y_2) = (3, -4)$

$R(x_3, y_3) = (-3, 4)$

Stated area of $\Delta PQR = 24$ sq. units.

To Find:

Verify whether the area of $\Delta PQR$ is indeed $24$ sq. units and justify the answer.

Solution:

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Substituting the given coordinates into the formula:

$\text{Area} = \frac{1}{2} |3(-4 - 4) + 3(4 - 4) + (-3)(4 - (-4))|$

$\text{Area} = \frac{1}{2} |3(-8) + 3(0) - 3(4 + 4)|$

$\text{Area} = \frac{1}{2} |-24 + 0 - 3(8)|$

$\text{Area} = \frac{1}{2} |-24 - 24|$

$\text{Area} = \frac{1}{2} |-48|$

Justification: Since the calculated area matches the stated area, the given statement is correct.

Alternate Solution:

We can also justify this by observing the positions of the points on a Cartesian plane.

1. Points $P(3, 4)$ and $Q(3, -4)$ have the same x-coordinate ($x=3$). Therefore, $PQ$ is a vertical line segment.

2. Points $P(3, 4)$ and $R(-3, 4)$ have the same y-coordinate ($y=4$). Therefore, $PR$ is a horizontal line segment.

3. Since $PQ$ is vertical and $PR$ is horizontal, they are perpendicular to each other. This means $\Delta PQR$ is a right-angled triangle at $P$.

$\text{Area of } \Delta PQR = \frac{1}{2} \times \text{Base} \times \text{Height}$

$\text{Area} = \frac{1}{2} \times 6 \times 8$

$\text{Area} = 3 \times 8 = 24$ sq. units.

Final Answer:

Yes, the area of $\Delta PQR$ is 24 sq. units. The calculation through the area formula and the geometric property of the right-angled triangle both confirm this result.

Given:

Length of the line segment = $10$ units.

Coordinates of one end point $P(x_1, y_1) = (2, -3)$.

Coordinates of the other end point $Q(x_2, y_2) = (10, y)$.

To Find:

The possible values of the ordinate ($y$) of point $Q$.

Solution:

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is calculated using the Distance Formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting the given values into the formula:

$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$

To eliminate the square root, we square both sides of the equation:

$(10)^2 = (8)^2 + (y + 3)^2$

$100 = 64 + (y + 3)^2$

Subtracting 64 from both sides:

$100 - 64 = (y + 3)^2$

$36 = (y + 3)^2$

Taking the square root on both sides:

$\pm \sqrt{36} = y + 3$

This gives us two distinct cases for the value of $y$:

Case 1: Taking the positive root

$y + 3 = 6$

$y = 6 - 3$

Case 2: Taking the negative root

$y + 3 = -6$

$y = -6 - 3$

Justification:

The distance formula leads to a quadratic equation in terms of $y$. Since a quadratic equation generally has two solutions, we obtain two possible values for the ordinate. Geometrically, this means there are two points on the vertical line $x = 10$ that are exactly $10$ units away from the point $(2, -3)$, one located above the level of the first point and one located below it.

Final Answer:

The ordinate of the other end of the line segment is indeed either 3 or -9.

Given:

The trigonometric expression is:

$\frac{3}{\text{cosec} \theta}$

To Find:

The maximum value of the given expression.

Solution:

We know the reciprocal relationship between the sine and cosecant functions:

By using this identity, the given expression can be rewritten as:

$\frac{3}{\text{cosec} \theta} = 3 \times \frac{1}{\text{cosec} \theta}$

Now, let us observe the values of $\sin \theta$ for angles between $0^\circ$ and $90^\circ$ as provided in the standard trigonometric table:

For $\theta = 0^\circ, \ \sin 0^\circ = 0$

For $\theta = 30^\circ, \ \sin 30^\circ = \frac{1}{2} = 0.5$

For $\theta = 45^\circ, \ \sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$

For $\theta = 60^\circ, \ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$

From the above values, we can see that as the angle $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\sin \theta$ also increases from 0 to 1.

Therefore, the maximum value of $\sin \theta$ for $0^\circ \leq \theta \leq 90^\circ$ is 1.

Using this in equation (i):

$\text{Maximum value of } 3 \sin \theta = 3 \times (\text{Maximum value of } \sin \theta)$

$\text{Maximum value} = 3 \times 1$

$\text{Maximum value} = 3$

Final Answer:

The maximum value of $\frac{3}{\text{cosec} \theta}$ is 3.

Question 16 (First Part)

Given:

The quadratic polynomial is $p(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}$.

(Note: Based on the standard NCERT Exemplar Question 16, the middle term is $+ 5x$. We will proceed with this corrected polynomial.)

To Find:

1. The zeroes of the polynomial $p(x)$.

2. Verify the relationship between the zeroes and the coefficients.

Solution:

To find the zeroes, we set $p(x) = 0$:

$4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0$

We use the method of splitting the middle term. We need two numbers whose sum is $5$ and whose product is $(4\sqrt{3}) \times (-2\sqrt{3}) = -8 \times 3 = -24$.

The numbers are $8$ and $-3$ (since $8 - 3 = 5$ and $8 \times -3 = -24$).

$4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0$

$4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0$

$(\sqrt{3}x + 2)(4x - \sqrt{3}) = 0$

This gives two cases:

Case 1: $\sqrt{3}x + 2 = 0 \implies x = -\frac{2}{\sqrt{3}}$

Case 2: $4x - \sqrt{3} = 0 \implies x = \frac{\sqrt{3}}{4}$

Verification:

Comparing $4\sqrt{3}x^2 + 5x - 2\sqrt{3}$ with $ax^2 + bx + c$, we have:

$a = 4\sqrt{3}, \ b = 5, \ c = -2\sqrt{3}$

1. Sum of zeroes:

$\alpha + \beta = -\frac{2}{\sqrt{3}} + \frac{\sqrt{3}}{4} = \frac{-8 + 3}{4\sqrt{3}} = -\frac{5}{4\sqrt{3}}$

2. Product of zeroes:

$\alpha \beta = \left( -\frac{2}{\sqrt{3}} \right) \left( \frac{\sqrt{3}}{4} \right) = -\frac{2}{4} = -\frac{1}{2}$

The relationship between zeroes and coefficients is verified.

OR Question

Given:

Dividend $f(x) = x^3 - 5x^2 + 6x - 4$

Quotient $q(x) = x - 3$

Remainder $r(x) = -3x + 5$

To Find:

The divisor polynomial $g(x)$.

Solution:

According to the Division Algorithm for Polynomials:

Substituting the given values:

$x^3 - 5x^2 + 6x - 4 = g(x) \cdot (x - 3) + (-3x + 5)$

Subtracting the remainder from the dividend:

$g(x) \cdot (x - 3) = (x^3 - 5x^2 + 6x - 4) - (-3x + 5)$

$g(x) \cdot (x - 3) = x^3 - 5x^2 + 6x - 4 + 3x - 5$

$g(x) \cdot (x - 3) = x^3 - 5x^2 + 9x - 9$

To find $g(x)$, we divide $(x^3 - 5x^2 + 9x - 9)$ by $(x - 3)$ using Long Division:

$\begin{array}{r} x^2 - 2x + 3 \phantom{x^3-5x^2)} \\ x-3{\overline{\smash{\big)}\,x^3-5x^2+9x-9\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-3x^2)\phantom{-b--)}} \\ -2x^2+9x\phantom{)} \\ \underline{-~\phantom{()}(-2x^2+6x)} \\ 3x-9\phantom{)} \\ \underline{-~\phantom{()}(3x-9)} \\ 0\phantom{)} \end{array}$

Final Answer:

The polynomial $g(x)$ is $x^2 - 2x + 3$.

Given:

The pair of linear equations are:

To Find:

The solution of the given system of equations graphically.

Solution:

To solve the equations graphically, we first find at least two or three points (coordinates) for each equation.

Step 1: Points for Equation (i) $5x - y = 5$

We can rewrite the equation as $y = 5x - 5$.

1. If $x = 0$, then $y = 5(0) - 5 = -5$.

2. If $x = 1$, then $y = 5(1) - 5 = 0$.

3. If $x = 2$, then $y = 5(2) - 5 = 5$.

Step 2: Points for Equation (ii) $3x - y = 3$

We can rewrite the equation as $y = 3x - 3$.

1. If $x = 0$, then $y = 3(0) - 3 = -3$.

2. If $x = 1$, then $y = 3(1) - 3 = 0$.

3. If $x = 2$, then $y = 3(2) - 3 = 3$.

Step 3: Plotting the Graph

Plot the points from both tables on a graph paper and join them to form two straight lines.

From the graph, we observe that the two lines intersect at a specific point.

The coordinates of the point of intersection are $x = 1$ and $y = 0$.

Final Answer:

The solution of the given pair of linear equations is $x = 1$ and $y = 0$.

Question 18

Given:

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = 4n - n^2$

To Find:

The $n^{th}$ term ($a_n$) and the $10^{th}$ term ($a_{10}$).

Solution:

We know that the $n^{th}$ term of an AP can be found using the relationship between the $n^{th}$ term and the sum of terms:

First, let us find the expression for $S_{n-1}$ by replacing $n$ with $(n-1)$ in the given formula:

$S_{n-1} = 4(n-1) - (n-1)^2$

$S_{n-1} = 4n - 4 - (n^2 - 2n + 1)$

$S_{n-1} = 4n - 4 - n^2 + 2n - 1$

Now, substitute the expressions for $S_n$ and $S_{n-1}$ into equation (i):

$a_n = (4n - n^2) - (-n^2 + 6n - 5)$

$a_n = 4n - n^2 + n^2 - 6n + 5$

$a_n = 4n - 6n + 5$

$a_n = 5 - 2n$

This is the expression for the $n^{th}$ term.

To find the $10^{th}$ term, we substitute $n = 10$ in the expression for $a_n$:

$a_{10} = 5 - 2(10)$

$a_{10} = 5 - 20$

$a_{10} = -15$

Final Answer:

The $n^{th}$ term is $5 - 2n$ and the $10^{th}$ term is $-15$.

OR Question

Given:

The Arithmetic Progression (AP) is: $9, \ 17, \ 25, \ \dots$

Sum required ($S_n$) = $636$

To Find:

The number of terms ($n$) that must be taken to get the sum $636$.

Solution:

From the given AP:

First term ($a$) = $9$

Common difference ($d$) = $17 - 9 = 8$

The formula for the sum of the first $n$ terms is:

$S_n = \frac{n}{2} [2a + (n-1)d]$

Substituting the values of $S_n$, $a$, and $d$:

$636 = \frac{n}{2} [2(9) + (n-1)8]$

$636 = \frac{n}{2} [18 + 8n - 8]$

$636 = \frac{n}{2} [10 + 8n]$

Taking 2 common from the bracket:

$636 = \frac{n}{2} \times 2 [5 + 4n]$

$636 = n(5 + 4n)$

$636 = 5n + 4n^2$

Rearranging into a standard quadratic equation form:

We solve this quadratic equation using the Quadratic Formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 4$, $b = 5$, and $c = -636$.

$D = b^2 - 4ac$

$D = (5)^2 - 4(4)(-636)$

$D = 25 + 10176$

$D = 10201$

Now, finding $n$:

$n = \frac{-5 \pm \sqrt{10201}}{2(4)}$

$n = \frac{-5 \pm 101}{8}$

This gives two possible values for $n$:

1. $n = \frac{-5 + 101}{8} = \frac{96}{8} = 12$

2. $n = \frac{-5 - 101}{8} = \frac{-106}{8} = -13.25$

Since the number of terms $n$ must be a natural number, we reject the negative and fractional value.

Final Answer:

12 terms of the AP must be taken to give a sum of 636.

Given:

The vertices of a parallelogram taken in order are:

$A(x_1, y_1) = (1, 2)$

$B(x_2, y_2) = (4, y)$

$C(x_3, y_3) = (x, 6)$

$D(x_4, y_4) = (3, 5)$

To Find:

The values of $x$ and $y$.

Solution:

In a parallelogram, the diagonals bisect each other. This means that the midpoint of the diagonal $AC$ is the same as the midpoint of the diagonal $BD$.

The Midpoint Formula for a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Step 1: Find the midpoint of diagonal $AC$

Using vertices $A(1, 2)$ and $C(x, 6)$:

$\text{Midpoint of } AC = \left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right)$

Step 2: Find the midpoint of diagonal $BD$

Using vertices $B(4, y)$ and $D(3, 5)$:

$\text{Midpoint of } BD = \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right)$

Step 3: Equate the coordinates

Since the diagonals bisect each other, the midpoints from (i) and (ii) must be equal.

Comparing the x-coordinates:

$\frac{1 + x}{2} = \frac{7}{2}$

$1 + x = 7$

Comparing the y-coordinates:

$4 = \frac{y + 5}{2}$

$8 = y + 5$

$y = 8 - 5$

Final Answer:

The value of $x$ is 6 and the value of $y$ is 3.

Given:

In $\Delta ABC$, $AD$ is the median to the side $BC$.

In $\Delta PQR$, $PM$ is the median to the side $QR$.

It is given that:

To Prove:

$\Delta ABC \sim \Delta PQR$

Image:

Proof:

Since $AD$ is the median of $\Delta ABC$, $D$ is the midpoint of $BC$.

Similarly, since $PM$ is the median of $\Delta PQR$, $M$ is the midpoint of $QR$.

Now, substituting these values into the given ratio from equation (i):

$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM}$

By canceling the common factor 2 in the middle term:

In $\Delta ABD$ and $\Delta PQM$, all three corresponding sides are proportional as shown in equation (ii). Therefore, by SSS Similarity Criterion:

$\Delta ABD \sim \Delta PQM$

Since corresponding angles of similar triangles are equal, we have:

Now, consider the larger triangles $\Delta ABC$ and $\Delta PQR$:

1. From the given equation (i), we have:

$\frac{AB}{PQ} = \frac{BC}{QR}$

2. From equation (iii), we have:

$\angle B = \angle Q$

Since two pairs of sides are proportional and the included angles are equal, by the SAS Similarity Criterion:

$\Delta ABC \sim \Delta PQR$

Hence Proved.

Given:

A triangle $ABC$ circumscribes a circle with center $O$ and radius $r = 4\text{ cm}$.

The circle touches the side $BC$ at point $D$ such that $BD = 8\text{ cm}$ and $DC = 7\text{ cm}$.

To Find:

The lengths of the sides AB and AC.

Construction:

1. Let the points of contact of the circle with sides $AC$ and $AB$ be $E$ and $F$ respectively.

2. Join the center $O$ to the points of contact $D, E, F$ and to the vertices $A, B, C$.

Solution:

We know that the lengths of tangents drawn from an external point to a circle are equal.

Let the length of the tangents from point $A$ be $x$.

Now, the lengths of the sides of the triangle are:

$a = BC = 8 + 7 = 15\text{ cm}$

$b = AC = x + 7\text{ cm}$

$c = AB = x + 8\text{ cm}$

The semi-perimeter ($s$) of $\Delta ABC$ is:

$s = \frac{a + b + c}{2} = \frac{15 + (x + 7) + (x + 8)}{2}$

Step 1: Area of $\Delta ABC$ using Heron's Formula

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$

$\text{Area} = \sqrt{(x + 15)(x + 15 - 15)(x + 15 - (x + 7))(x + 15 - (x + 8))}$

$\text{Area} = \sqrt{(x + 15)(x)(8)(7)}$

Step 2: Area of $\Delta ABC$ using In-radius

The total area of the triangle is the sum of the areas of triangles $OBC$, $OAC$, and $OAB$.

$\text{Area} = \text{Area}(\Delta OBC) + \text{Area}(\Delta OAC) + \text{Area}(\Delta OAB)$

$\text{Area} = \frac{1}{2}(BC \times r) + \frac{1}{2}(AC \times r) + \frac{1}{2}(AB \times r)$

$\text{Area} = \frac{1}{2} \times r \times (a + b + c) = r \times s$

Substituting $r = 4$ and $s = x + 15$:

Step 3: Equating the two area expressions

From (iii) and (iv):

$\sqrt{56x(x + 15)} = 4(x + 15)$

Squaring both sides:

$56x(x + 15) = 16(x + 15)^2$

Dividing both sides by $8(x + 15)$ (since $x + 15 \neq 0$):

$7x = 2(x + 15)$

$7x = 2x + 30$

$5x = 30$

Step 4: Finding the sides

$AB = x + 8 = 6 + 8 = 14\text{ cm}$

$AC = x + 7 = 6 + 7 = 13\text{ cm}$

Final Answer:

The sides of the triangle are AB = 14 cm and AC = 13 cm.

Given:

1. An isosceles triangle with base = $6\text{ cm}$ and altitude = $5\text{ cm}$.

2. Scale factor for the new triangle = $\frac{7}{5}$.

To Construct:

A triangle $\Delta A'BC'$ similar to the isosceles triangle $\Delta ABC$ such that its sides are $\frac{7}{5}$ of the corresponding sides of $\Delta ABC$.

Steps of Construction:

1. Draw a line segment $BC = 6\text{ cm}$.

2. Draw the perpendicular bisector of $BC$ and let it intersect $BC$ at point $M$.

3. Mark a point $A$ on the perpendicular bisector such that $MA = 5\text{ cm}$.

4. Join $AB$ and $AC$. Now, $\Delta ABC$ is the required isosceles triangle.

5. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.

6. Locate 7 points $B_1, B_2, B_3, B_4, B_5, B_6,$ and $B_7$ on $BX$ such that:

7. Join $B_5$ (the $5^{th}$ point, as the denominator is 5) to $C$.

8. Draw a line through $B_7$ (the $7^{th}$ point, as the numerator is 7) parallel to $B_5C$, intersecting the extended line segment $BC$ at $C'$.

9. Draw a line through $C'$ parallel to $CA$, intersecting the extended line segment $BA$ at $A'$.

10. $\Delta A'BC'$ is the required triangle.

Image:

Justification:

By construction, $C'A' \parallel CA$. Therefore, by the AA Similarity Criterion:

$\Delta ABC \sim \Delta A'BC'$

Also, since $B_7C' \parallel B_5C$, by the Basic Proportionality Theorem in $\Delta BB_7C'$:

$\frac{BC'}{BC} = \frac{BB_7}{BB_5} = \frac{7}{5}$

Since the triangles are similar, the ratio of all corresponding sides will be the same:

This justifies that the sides of the new triangle are $\frac{7}{5}$ of the corresponding sides of the original triangle.

Question 23

To Prove:

$\frac{\cos\theta - \sin\theta + 1}{\sin\theta + \cos\theta - 1} = \frac{1}{\text{cosec}\theta - \cot\theta}$

Proof:

Let us consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\cos\theta - \sin\theta + 1}{\sin\theta + \cos\theta - 1}$

To convert this into the terms of $\text{cosec}\theta$ and $\cot\theta$, we divide both the numerator and the denominator by $\sin\theta$:

$LHS = \frac{\frac{\cos\theta}{\sin\theta} - \frac{\sin\theta}{\sin\theta} + \frac{1}{\sin\theta}}{\frac{\sin\theta}{\sin\theta} + \frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta}}$

Rearranging the terms in the numerator:

$LHS = \frac{(\text{cosec}\theta + \cot\theta) - 1}{1 - \text{cosec}\theta + \cot\theta}$

We know the trigonometric identity:

Substituting the value of $1$ from equation (i) into the numerator:

$LHS = \frac{(\text{cosec}\theta + \cot\theta) - (\text{cosec}^2\theta - \cot^2\theta)}{1 - \text{cosec}\theta + \cot\theta}$

Using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$:

$LHS = \frac{(\text{cosec}\theta + \cot\theta) - [(\text{cosec}\theta - \cot\theta)(\text{cosec}\theta + \cot\theta)]}{1 - \text{cosec}\theta + \cot\theta}$

Taking $(\text{cosec}\theta + \cot\theta)$ as common from the numerator:

$LHS = \frac{(\text{cosec}\theta + \cot\theta) [1 - (\text{cosec}\theta - \cot\theta)]}{1 - \text{cosec}\theta + \cot\theta}$

$LHS = \frac{(\text{cosec}\theta + \cot\theta) (1 - \text{cosec}\theta + \cot\theta)}{1 - \text{cosec}\theta + \cot\theta}$

Canceling the common term $(1 - \text{cosec}\theta + \cot\theta)$:

Now, let us simplify the Right Hand Side (RHS):

$RHS = \frac{1}{\text{cosec}\theta - \cot\theta}$

To rationalize the denominator, multiply and divide by $(\text{cosec}\theta + \cot\theta)$:

$RHS = \frac{1}{\text{cosec}\theta - \cot\theta} \times \frac{\text{cosec}\theta + \cot\theta}{\text{cosec}\theta + \cot\theta}$

$RHS = \frac{\text{cosec}\theta + \cot\theta}{\text{cosec}^2\theta - \cot^2\theta}$

Using the identity $\text{cosec}^2\theta - \cot^2\theta = 1$:

From equations (ii) and (iii), we have:

LHS = RHS

Hence Proved.

OR Question

To Evaluate:

$\left( \frac{3\cos 43^\circ}{\sin 47^\circ} \right)^2 - \frac{\cos 37^\circ \text{cosec} 53^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ}$

Solution:

We will use the Trigonometric Ratios of Complementary Angles: $\sin(90^\circ - \theta) = \cos\theta$, $\cos(90^\circ - \theta) = \sin\theta$, and $\tan(90^\circ - \theta) = \cot\theta$.

Step 1: Evaluate the first term

$\sin 47^\circ = \sin(90^\circ - 43^\circ) = \cos 43^\circ$

Therefore, $\left( \frac{3\cos 43^\circ}{\sin 47^\circ} \right)^2 = \left( \frac{3\cos 43^\circ}{\cos 43^\circ} \right)^2 = (3)^2 = 9$

Step 2: Evaluate the numerator of the second term

$\text{cosec} 53^\circ = \text{cosec}(90^\circ - 37^\circ) = \sec 37^\circ$

Numerator = $\cos 37^\circ \text{cosec} 53^\circ = \cos 37^\circ \times \sec 37^\circ$

Step 3: Evaluate the denominator of the second term

The denominator is: $\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ$

$\tan 85^\circ = \tan(90^\circ - 5^\circ) = \cot 5^\circ$

$\tan 65^\circ = \tan(90^\circ - 25^\circ) = \cot 25^\circ$

$\tan 45^\circ = 1$

Substituting these in the denominator:

$\text{Denominator} = (\tan 5^\circ \cot 5^\circ) \times (\tan 25^\circ \cot 25^\circ) \times 1$

Step 4: Calculate the final value

Value $= 9 - \frac{1}{1}$

Value $= 9 - 1$

Value $= 8$

Final Answer:

The value of the expression is 8.

Given:

Triangle $ABC$ is right-angled at $A$.

Side $AB = 6$ cm

Side $AC = 8$ cm

Three semicircles are drawn with $AB$, $AC$, and $BC$ as their respective diameters.

To Find:

The area of the shaded region.

Solution:

Step 1: Calculate the length of the hypotenuse $BC$

In right-angled triangle $ABC$, by Pythagoras Theorem:

$BC^2 = AB^2 + AC^2$

$BC^2 = 6^2 + 8^2$

$BC^2 = 36 + 64 = 100$

Step 2: Calculate the area of the triangle $ABC$

$\text{Area of } \Delta ABC = \frac{1}{2} \times \text{Base} \times \text{Height}$

$\text{Area of } \Delta ABC = \frac{1}{2} \times 6 \times 8$

Step 3: Calculate the areas of the semicircles

Let $A_1$, $A_2$, and $A_3$ be the areas of the semicircles on $AB$, $AC$, and $BC$ respectively.

Step 4: Calculate the area of the shaded region

From the figure, the area of the shaded region is the sum of the areas of the two smaller semicircles and the triangle, minus the area of the largest semicircle.

$\text{Area of Shaded Region} = (A_1 + A_2 + \text{Area of } \Delta ABC) - A_3$

Substituting the values from equations (ii), (iii), (iv), and (v):

$\text{Area of Shaded Region} = (4.5\pi + 8\pi + 24) - 12.5\pi$

$\text{Area of Shaded Region} = (12.5\pi + 24) - 12.5\pi$

$\text{Area of Shaded Region} = 24 \text{ cm}^2$

Final Answer:

The area of the shaded region is 24 cm².

Given:

Probability of getting a white ball, $P(\text{White}) = \frac{3}{10}$

Probability of getting a black ball, $P(\text{Black}) = \frac{2}{5}$

Number of black balls in the bag = $20$

To Find:

1. The probability of getting a red ball, $P(\text{Red})$.

2. The total number of balls in the bag.

Solution:

Step 1: Finding the probability of getting a red ball

Since the bag contains only white, black, and red balls, these are the only possible outcomes. The sum of the probabilities of all mutually exclusive and exhaustive events is always 1.

Substituting the given probabilities into equation (i):

$\frac{3}{10} + \frac{2}{5} + P(\text{Red}) = 1$

To add the fractions, we take the L.C.M. of the denominators (10 and 5), which is 10:

$\frac{3}{10} + \frac{4}{10} + P(\text{Red}) = 1$

$\frac{7}{10} + P(\text{Red}) = 1$

$P(\text{Red}) = 1 - \frac{7}{10}$

$P(\text{Red}) = \frac{10 - 7}{10}$

Step 2: Finding the total number of balls in the bag

Let the total number of balls in the bag be $N$.

We know that the probability of an event is given by:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

For black balls:

Substituting the known values into equation (iii):

$\frac{2}{5} = \frac{20}{N}$

By cross-multiplication:

$2 \times N = 20 \times 5$

$2N = 100$

$N = \frac{\cancel{100}^{50}}{\cancel{2}_{1}}$

Final Answer:

The probability of getting a red ball is $\frac{3}{10}$ and the total number of balls in the bag is 50.

Question 26

Given:

Total amount spent = $\textsf{₹} 300$

Reduction in the price of each book = $\textsf{₹} 5$

Number of extra books that can be bought = $5$

To Find:

The original list price of the book.

Solution:

Let the original list price of the book be $\textsf{₹} x$.

Then, the number of books originally bought for $\textsf{₹} 300$ is:

When the price is reduced by $\textsf{₹} 5$, the new price of the book becomes $\textsf{₹} (x - 5)$.

The number of books that can be bought at the new price is:

According to the question, the difference between the new number of books and the original number of books is $5$.

$\frac{300}{x - 5} - \frac{300}{x} = 5$

Taking $300$ common and solving the fraction:

$300 \left[ \frac{x - (x - 5)}{x(x - 5)} \right] = 5$

$300 \left[ \frac{5}{x^2 - 5x} \right] = 5$

Dividing both sides by $5$:

$\frac{300}{x^2 - 5x} = 1$

$x^2 - 5x = 300$

We solve this quadratic equation by splitting the middle term. We need two numbers whose sum is $-5$ and product is $-300$. These numbers are $-20$ and $15$.

$x^2 - 20x + 15x - 300 = 0$

$x(x - 20) + 15(x - 20) = 0$

$(x - 20)(x + 15) = 0$

This gives two possible values for $x$:

$x - 20 = 0 \implies x = 20$

$x + 15 = 0 \implies x = -15$

Since the price of a book cannot be negative, we reject $x = -15$.

Final Answer:

The original list price of the book is $\textsf{₹} 20$.

OR Question

Given:

Sum of the present ages of two friends = $20$ years.

Product of their ages $4$ years ago = $48$.

To Find:

Whether this situation is possible, and if so, determine their present ages.

Solution:

Let the present age of the first friend be $x$ years.

Then, the present age of the second friend is $(20 - x)$ years.

Four years ago:

Age of the first friend = $(x - 4)$ years.

Age of the second friend = $(20 - x - 4) = (16 - x)$ years.

According to the given condition, the product of their ages was $48$:

$(x - 4)(16 - x) = 48$

$16x - x^2 - 64 + 4x = 48$

$-x^2 + 20x - 64 = 48$

$-x^2 + 20x - 64 - 48 = 0$

$-x^2 + 20x - 112 = 0$

Multiplying the whole equation by $-1$:

To check if the situation is possible, we need to find the discriminant $D$ of the quadratic equation $ax^2 + bx + c = 0$.

Here, $a = 1, b = -20, c = 112$.

$D = b^2 - 4ac$

$D = (-20)^2 - 4(1)(112)$

$D = 400 - 448$

Since the discriminant $D < 0$, the quadratic equation has no real roots.

This means there are no real values of $x$ that satisfy the given conditions.

Final Answer:

No, this situation is not possible because the discriminant of the resulting quadratic equation is negative ($D < 0$).

Question 27 (Part I)

Given:

A circle with center $O$ and a point $P$ outside the circle. $PQ$ and $PR$ are two tangents drawn from point $P$ to the circle at points $Q$ and $R$ respectively.

To Prove:

The lengths of the tangents are equal, i.e., $PQ = PR$.

Construction:

Join $OQ, OR,$ and $OP$.

Proof:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

In $\Delta OQP$ and $\Delta ORP$:

Therefore, by RHS (Right-angle Hypotenuse Side) congruence criterion:

$\Delta OQP \cong \Delta ORP$

By CPCT (Corresponding Parts of Congruent Triangles):

$PQ = PR$

Hence Proved.

Question 27 (Part II - Application)

Given:

A quadrilateral $ABCD$ circumscribes a circle. The sides $AB, BC, CD,$ and $DA$ touch the circle at points $P, Q, R,$ and $S$ respectively.

To Prove:

$AB + CD = AD + BC$

Proof:

Using the theorem proved above, the lengths of tangents from an external point to a circle are equal.

Adding equations (iii), (iv), (v), and (vi):

$AP + BP + CR + DR = AS + BQ + CQ + DS$

Rearranging the terms:

$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$

From the figure, we can see that:

$AP + BP = AB$

$CR + DR = CD$

$AS + DS = AD$

$BQ + CQ = BC$

Substituting these values:

$AB + CD = AD + BC$

Hence Proved.

OR Question (Part I - Theorem)

Given:

Two triangles $ABC$ and $PQR$ such that $\Delta ABC \sim \Delta PQR$.

To Prove:

$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$

Construction:

Draw altitudes $AM \perp BC$ and $PN \perp QR$.

Proof:

Since $\Delta ABC \sim \Delta PQR$, their corresponding sides are in the same ratio and corresponding angles are equal.

In $\Delta ABM$ and $\Delta PQN$:

$\angle B = \angle Q$ (Given similar triangles)

$\angle M = \angle N = 90^\circ$ (Construction)

So, $\Delta ABM \sim \Delta PQN$ by AA Similarity Criterion.

Comparing (i) and (ii):

Now, $\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \frac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times QR \times PN} = \frac{BC}{QR} \times \frac{AM}{PN}$

Substituting the ratio from (iii):

$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \frac{BC}{QR} \times \frac{BC}{QR} = \frac{BC^2}{QR^2}$

Hence Proved.

OR Question (Part II - Application)

Given:

$\Delta ABC$ is an isosceles triangle, right-angled at $B$.

$\Delta ABE$ and $\Delta ACD$ are equilateral triangles constructed on sides $AB$ and $AC$.

Solution:

In isosceles right-angled $\Delta ABC$, let $AB = BC = a$.

By Pythagoras Theorem:

$AC^2 = AB^2 + BC^2$

$AC^2 = a^2 + a^2 = 2a^2$

Since $\Delta ABE$ and $\Delta ACD$ are equilateral triangles, they are similar (all angles are $60^\circ$).

Using the theorem proved above:

$\frac{\text{Area}(\Delta ABE)}{\text{Area}(\Delta ACD)} = \left( \frac{AB}{AC} \right)^2$

Substituting the values of $AB$ and $AC$:

$\frac{\text{Area}(\Delta ABE)}{\text{Area}(\Delta ACD)} = \left( \frac{a}{a\sqrt{2}} \right)^2$

$\frac{\text{Area}(\Delta ABE)}{\text{Area}(\Delta ACD)} = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$

Final Answer:

The ratio of the areas of $\Delta ABE$ and $\Delta ACD$ is 1 : 2.

Given:

Height of the building ($CD$) = $50\text{ m}$

Angle of depression of the top of the building ($C$) from the top of the tower ($A$) = $30^\circ$

Angle of depression of the bottom of the building ($D$) from the top of the tower ($A$) = $60^\circ$

To Find:

1. The height of the tower ($AB$).

2. The horizontal distance between the building and the tower ($BD$).

Construction:

Let $AB$ be the tower of height $H$ and $CD$ be the building of height $50\text{ m}$.

Draw $CE \perp AB$ such that $E$ lies on $AB$.

Therefore, $CE = BD = x$ (horizontal distance) and $BE = CD = 50\text{ m}$.

The height $AE = AB - BE = H - 50$.

By alternate interior angles, $\angle ACE = 30^\circ$ and $\angle ADB = 60^\circ$.

Solution:

In right-angled triangle $\Delta ABD$:

$\tan 60^\circ = \frac{AB}{BD}$

$\sqrt{3} = \frac{H}{x}$

In right-angled triangle $\Delta ACE$:

$\tan 30^\circ = \frac{AE}{CE}$

$\frac{1}{\sqrt{3}} = \frac{H - 50}{x}$

Equating the values of $x$ from equations (i) and (ii):

$\frac{H}{\sqrt{3}} = \sqrt{3}(H - 50)$

$H = 3(H - 50)$

$H = 3H - 150$

$150 = 3H - H$

$2H = 150$

Now, substitute the value of $H$ in equation (i) to find the horizontal distance $x$:

$x = \frac{75}{\sqrt{3}}$

To rationalize the denominator, multiply and divide by $\sqrt{3}$:

$x = \frac{75 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$x = \frac{75\sqrt{3}}{3}$

Using the value $\sqrt{3} \approx 1.732$:

$x = 25 \times 1.732$

$x = 43.3\text{ m}$

Final Answer:

The height of the tower is 75 m and the horizontal distance between the building and the tower is $25\sqrt{3}$ m (or approximately 43.3 m).

Given:

For the cylindrical well:

$\text{Depth or Height } (h) = 14\text{ m}$

For the embankment (circular ring):

$\text{Width of the embankment} = 4\text{ m}$

To Find:

The height of the embankment ($H$).

Solution:

The earth taken out from the well is used to form the embankment. Therefore, the volume of the earth dug out is equal to the volume of the embankment formed.

Step 1: Calculate the volume of earth dug out

The well is in the shape of a cylinder. The volume of a cylinder is given by $\pi r^2 h$.

$\text{Volume of earth} = \pi \times (1.5)^2 \times 14$

Step 2: Calculate the volume of the embankment

The embankment is in the shape of a hollow cylinder (circular ring) with height $H$. Its volume is given by the area of its base multiplied by its height.

$\text{Volume of embankment} = \text{Area of ring} \times H$

$\text{Volume of embankment} = \pi (r_2^2 - r_1^2) \times H$

Substituting values from equation (ii):

$\text{Volume of embankment} = \pi (5.5^2 - 1.5^2) \times H$

Using the identity $a^2 - b^2 = (a - b)(a + b)$:

$\text{Volume of embankment} = \pi [(5.5 - 1.5)(5.5 + 1.5)] \times H$

$\text{Volume of embankment} = \pi [4 \times 7] \times H$

Step 3: Equating the volumes

According to the problem:

Substituting values from (iii) and (iv) into (v):

$28\pi H = 2.25 \times 14 \times \pi$

Dividing both sides by $\pi$ and $14$:

$2H = 2.25$

$H = \frac{2.25}{2}$

$H = 1.125\text{ m}$

Final Answer:

The height of the embankment is 1.125 m.

Given:

The distribution of ages of patients is as follows:

To Find:

1. The Mode of the data.

2. The Mean of the data.

Solution:

1. Calculation of Mean

To calculate the mean, we first find the class mark ($x_i$) for each class interval using the formula $x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$.

The formula for Mean ($\bar{x}$) is:

$\bar{x} = \frac{\sum\limits f_i x_i}{\sum\limits f_i}$

Substituting the values:

$\bar{x} = \frac{2830}{80}$

$\bar{x} = \frac{\cancel{2830}}{\cancel{80}}$

$\bar{x} = 35.375$

Therefore, the Mean age of the patients is 35.38 years (approx).

2. Calculation of Mode

From the given data, the maximum frequency is 23.

The class corresponding to the maximum frequency is 35 - 45. Therefore, the Modal Class is 35 - 45.

The formula for Mode is:

$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$

Substituting the values:

$\text{Mode} = 35 + \left( \frac{23 - 21}{2(23) - 21 - 14} \right) \times 10$

$\text{Mode} = 35 + \left( \frac{2}{46 - 35} \right) \times 10$

$\text{Mode} = 35 + \left( \frac{2}{11} \right) \times 10$

$\text{Mode} = 35 + \frac{20}{11}$

$\text{Mode} = 35 + 1.818...$

$\text{Mode} = 36.818...$

Therefore, the Mode age of the patients is 36.8 years (approx).

Conclusion:

The Mode of the data is 36.8 years and the Mean of the data is 35.38 years.

Interpretation: The maximum number of patients admitted to the hospital are of the age 36.8 years (approx), while on average, the age of a patient admitted is 35.38 years.