Step 1: Understanding the Concept:

The Least Common Multiple (LCM) of two numbers is the smallest positive integer that is perfectly divisible by both numbers.

If one number is a multiple of the other, the larger number is the LCM.


Step 2: Key Formula or Approach:

We check the divisibility of the larger number by the smaller number.

If \( a = k \times b \), then \( LCM(a, b) = a \).


Step 3: Detailed Explanation:

Given numbers are 960 and 240.

Let's divide 960 by 240:
\[ \frac{960}{240} = \frac{96}{24} = 4 \]

Since 960 is exactly divisible by 240 (multiplied by 4), 960 is a multiple of 240.

By the property of multiples, the LCM of 960 and 240 is 960.


Step 4: Final Answer:

The LCM of 960 and 240 is 960.
Quick Tip: For any two numbers where the larger is a multiple of the smaller, the larger is the LCM and the smaller is the HCF.

Step 1: Understanding the Concept:

A prime number is a natural number greater than 1 with exactly two factors (1 and itself).

A composite number is a natural number greater than 1 with more than two factors.


Step 2: Detailed Explanation:

The number 1 only has one factor, which is 1 itself.

Since it does not have exactly two distinct factors, it is not prime.

Since it does not have more than two factors, it is not composite.

By definition, 1 is excluded from the categories of both prime and composite numbers.


Step 3: Final Answer:

The natural number 1 is neither prime nor composite.
Quick Tip: 2 is the smallest prime number and the only even prime number. 4 is the smallest composite number.

Step 1: Understanding the Concept:

The last digit (unit digit) of a number raised to a power depends on the cyclicity of that digit.


Step 2: Detailed Explanation:

Let's calculate the first few powers of 5:
\[ 5^1 = 5 \]
\[ 5^2 = 25 \]
\[ 5^3 = 125 \]
\[ 5^4 = 625 \]

Observing the pattern, the unit digit is always 5.

Mathematically, the product of any number ending in 5 with 5 will always result in a number ending in 5.


Step 3: Final Answer:

For any natural number \( n \), \( 5^n \) ends with the digit 5.
Quick Tip: Numbers ending in 0, 1, 5, or 6 always have the same digit at the units place for any natural power \( n \).

Step 1: Understanding the Concept:

The zeroes of a polynomial function \( f(x) \) correspond to the points where the graph intersects the x-axis.


Step 2: Detailed Explanation:

By observing the provided graph, we look for the points where the curve crosses or touches the horizontal axis (x-axis).

The curve intersects the x-axis at:

1. One point on the negative x-axis side.

2. Two distinct points on the positive x-axis side.

Counting these intersection points, we get a total of 3 distinct points.


Step 3: Final Answer:

The number of distinct zeroes is 3.
Quick Tip: Do not count intersections with the y-axis. Only the points on the x-axis represent the zeroes of the function \( y = f(x) \).

Step 1: Understanding the Concept:

For a quadratic polynomial \( ax^2 + bx + c \), the sum of zeroes \( \alpha + \beta = -b/a \) and the product of zeroes \( \alpha\beta = c/a \).


Step 2: Key Formula or Approach:

Given: \( f(x) = px^2 - 2x + 3p \).

Here, \( a = p \), \( b = -2 \), and \( c = 3p \).


Step 3: Detailed Explanation:

Sum of zeroes \( \alpha + \beta = -\frac{-2}{p} = \frac{2}{p} \).

Product of zeroes \( \alpha\beta = \frac{3p}{p} = 3 \).

According to the question, \( \alpha + \beta = \alpha\beta \):
\[ \frac{2}{p} = 3 \]
\[ 2 = 3p \]
\[ p = \frac{2}{3} \]


Step 4: Final Answer:

The value of \( p \) is \( \frac{2}{3} \).
Quick Tip: When solving for \( p \), always ensure \( p \neq 0 \) since it is the leading coefficient of a quadratic equation.

Step 1: Understanding the Concept:

A system of equations is 'consistent' if it has at least one solution. It is 'dependent' if it has infinitely many solutions (the two lines coincide).


Step 2: Detailed Explanation:

The three conditions for linear equations are:

1. Unique Solution: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (Consistent and Independent).

2. No Solution: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) (Inconsistent).

3. Infinitely Many Solutions: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) (Consistent and Dependent).

The question explicitly asks for the consistent and dependent case.


Step 3: Final Answer:

The condition is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Quick Tip: Dependent lines are just the same line written differently. For example, \( x+y=2 \) and \( 2x+2y=4 \).

Step 1: Understanding the Concept:

An Arithmetic Progression (A.P.) is a sequence where the difference between consecutive terms is constant.


Step 2: Detailed Explanation:

Let's check the common difference \( d \) for each:

(A) \( 2.5 - 2 = 0.5 \); \( 3 - 2.5 = 0.5 \). It is an A.P.

(B) \( -3.2 - (-1.2) = -2 \); \( -5.2 - (-3.2) = -2 \). It is an A.P.

(C) \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2} \dots \). Difference is \( \sqrt{2} \). It is an A.P.

(D) \( 1, 9, 25, 49 \). Differences are: \( 9 - 1 = 8 \) and \( 25 - 9 = 16 \).

Since \( 8 \neq 16 \), the common difference is not constant.


Step 3: Final Answer:

The sequence \( 1^2, 3^2, 5^2, 7^2, \dots \) is not an A.P.
Quick Tip: Always simplify radical terms like \( \sqrt{8} \) to \( 2\sqrt{2} \) before checking the common difference.

Step 1: Understanding the Concept:

By AA (Angle-Angle) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.


Step 2: Detailed Explanation:

Given: \( \angle A = \angle Q \) and \( \angle B = \angle R \).

Therefore, \( \triangle ABC \sim \triangle QRP \). (Order: \( A \rightarrow Q \), \( B \rightarrow R \), \( C \rightarrow P \)).

In similar triangles, corresponding sides are in the same ratio:
\[ \frac{AB}{QR} = \frac{BC}{RP} = \frac{AC}{QP} \]

From the ratio \( \frac{AB}{QR} = \frac{AC}{QP} \), we can write:
\[ \frac{AB}{AC} = \frac{QR}{QP} \]


Step 3: Final Answer:
\( AB : AC = QR : QP \).
Quick Tip: The order of vertices in a similarity statement is crucial. Match the equal angles to set up the correct ratio.

Step 1: Understanding the Concept:

The distance of any point \( (x, y) \) from the x-axis is given by the magnitude of its y-coordinate, i.e., \( |y| \).


Step 2: Detailed Explanation:

The given point is \( A(4a, 3a) \).

The y-coordinate of this point is \( 3a \).

The distance from the x-axis is \( |3a| \).

Assuming \( a \) is positive as per standard distance problems, the value is \( 3a \).


Step 3: Final Answer:

The distance is \( 3a \).
Quick Tip: Distance from x-axis = y-coordinate value. Distance from y-axis = x-coordinate value.

Step 1: Understanding the Concept:

In a right-angled triangle, \( \cos A = \frac{Base}{Hypotenuse} \) and \( \tan A = \frac{Perpendicular}{Base} \).


Step 2: Key Formula or Approach:

Use Pythagoras Theorem: \( P^2 + B^2 = H^2 \).


Step 3: Detailed Explanation:

Let \( Base = 4k \) and \( Hypotenuse = 5k \).
\[ Perpendicular^2 = (5k)^2 - (4k)^2 \]
\[ P^2 = 25k^2 - 16k^2 = 9k^2 \implies P = 3k \]

Now, \( \tan A = \frac{P}{B} = \frac{3k}{4k} = \frac{3}{4} \).


Step 4: Final Answer:

The value is \( \frac{3}{4} \).
Quick Tip: Memorizing common Pythagorean triplets like (3, 4, 5) helps solve these problems instantly.

Step 1: Understanding the Concept:

Solve for angle A and substitute its value into the required trigonometric expression.


Step 2: Detailed Explanation:
\( 2 \sin A = 1 \implies \sin A = \frac{1}{2} \).

We know \( \sin 30^\circ = \frac{1}{2} \), so \( A = 30^\circ \).

Evaluate: \( \tan 30^\circ + \cot 30^\circ \).
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \quad and \quad \cot 30^\circ = \sqrt{3} \]
\[ \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]


Step 3: Final Answer:

The value is \( \frac{4}{\sqrt{3}} \).
Quick Tip: \( \tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta} \). This identity is also useful for faster calculations.

Step 1: Understanding the Concept:

This is an application of basic trigonometry using the tangent ratio in a right-angled triangle.


Step 2: Detailed Explanation:

Let \( h \) be the height and \( d \) be the distance (60 m).
\( \tan 45^\circ = \frac{Height}{Distance} \).

We know \( \tan 45^\circ = 1 \).
\[ 1 = \frac{h}{60} \implies h = 60 m \]


Step 3: Final Answer:

The height of the tower is 60 m.
Quick Tip: Whenever the angle of elevation is \( 45^\circ \), the height is always equal to the distance from the base.

Step 1: Understanding the Concept:

Radius is perpendicular to the tangent. Sum of angles in a triangle is \( 180^\circ \).


Step 2: Detailed Explanation:

In \( \triangle OAB \), \( OA = OB \) (radii).

So, \( \angle OBA = \angle OAB = 15^\circ \).
\( \angle AOB = 180^\circ - (15^\circ + 15^\circ) = 150^\circ \).

In quadrilateral OAPB, \( \angle OAP = \angle OBP = 90^\circ \).

Sum of angles \( \angle APB + \angle AOB = 180^\circ \) (supplementary).
\[ \angle APB = 180^\circ - 150^\circ = 30^\circ \]


Step 3: Final Answer:
\( \angle APB = 30^\circ \).
Quick Tip: The angle between tangents is always double the angle between the chord and the tangent (\( \angle APB = 2 \times \angle OAB \)).

Step 1: Understanding the Concept:

The angle between the tangents from an external point is supplementary to the angle subtended by the radii at the centre.


Step 2: Detailed Explanation:

In quadrilateral OAPB:
\( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

The sum of angles in a quadrilateral is \( 360^\circ \).
\[ \angle APB + \angle AOB = 180^\circ \]
\[ \angle APB + 130^\circ = 180^\circ \]
\[ \angle APB = 50^\circ \]


Step 3: Final Answer:
\( \angle APB = 50^\circ \).
Quick Tip: Just subtract the central angle from 180 to find the angle between tangents.

Step 1: Understanding the Concept:

Area of Segment = Area of Sector - Area of Triangle.


Step 2: Detailed Explanation:

Central angle \( \theta = 60^\circ \).

Area of sector = \( \frac{60}{360} \pi r^2 = \frac{\pi r^2}{6} \).

Triangle with \( 60^\circ \) and two equal radii is an equilateral triangle.

Area of equilateral triangle = \( \frac{\sqrt{3}}{4} r^2 \).

Area of segment = \( \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 \).


Step 3: Final Answer:

The area is \( \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2 \).
Quick Tip: For \( \theta = 60^\circ \), the triangle is equilateral. For \( \theta = 90^\circ \), the triangle area is \( \frac{1}{2}r^2 \).

Step 1: Understanding the Concept:

Volume of material in a hollow hemisphere = Outer Volume - Inner Volume.


Step 2: Detailed Explanation:

Outer Radius \( R = 6 cm \).

Thickness = 1 cm.

Inner Radius \( r = 6 - 1 = 5 cm \).

Volume = \( \frac{2}{3} \pi R^3 - \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (6^3 - 5^3) \).
\[ V = \frac{2}{3} \pi (216 - 125) = \frac{2}{3} \pi (91) = \frac{182}{3} \pi \]


Step 3: Final Answer:

The volume is \( \frac{182}{3} \pi cm^3 \).
Quick Tip: Always calculate the inner radius first by subtracting the thickness from the outer radius.

Step 1: Understanding the Concept:

Use the empirical relationship: \( Mode = 3 Median - 2 Mean \).


Step 2: Detailed Explanation:

Given Mean = 43 and Median = 43.4.
\[ Mode = 3(43.4) - 2(43) \]
\[ Mode = 130.2 - 86 = 44.2 \]


Step 3: Final Answer:

The mode is 44.2.
Quick Tip: Remember the 3-2-1 rule: 3 Median - 2 Mean = 1 Mode.

Step 1: Understanding the Concept:

Count total composite numbers and divide by total numbers (25).


Step 2: Detailed Explanation:

Primes between 1-25: 2, 3, 5, 7, 11, 13, 17, 19, 23 (Total = 9).

The number '1' is neither prime nor composite.

Total non-composite numbers = \( 9 + 1 = 10 \).

Total composite numbers = \( 25 - 10 = 15 \).

Probability = \( \frac{15}{25} \).


Step 3: Final Answer:

The probability is \( \frac{15}{25} \).
Quick Tip: Always subtract both prime count and '1' from the total to get the composite count.

Step 1: Understanding the Concept:

Total Surface Area of cuboid = \( 2(lb + bh + hl) \).


Step 2: Detailed Explanation:

Joined cubes dimensions: \( L = 8, B = 4, H = 4 \).

Area = \( 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32+16+32) = 160 \). (A is True).

Reason R states area is \( (lb + bh + hl) \). It lacks the factor of 2. (R is False).


Step 3: Final Answer:

Assertion is true, Reason is false.
Quick Tip: Watch out for partial formulas in the Reason section of Assertion-Reason questions.

Step 1: Understanding the Concept:

Mean = Sum / Count.


Step 2: Detailed Explanation:

Sum = \( \frac{n(n+1)}{2} \). (Reason R is True).

Mean = \( \frac{n(n+1)}{2} \times \frac{1}{n} = \frac{n+1}{2} \).

Assertion A says \( \frac{n-1}{2} \). (Assertion A is False).


Step 3: Final Answer:

(A) is false, (R) is true.
Quick Tip: The mean of an A.P. is simply the average of the first and last terms: \( (1 + n)/2 \).

Step 1: Understanding the Concept:

For a quadratic polynomial \( ax^2 + bx + c \), the sum of zeroes is \( \alpha + \beta = -b/a \) and the product of zeroes is \( \alpha\beta = c/a \).


Step 2: Key Formula or Approach:

From the given polynomial \( px^2 + qx + r \):

Sum of zeroes \( \alpha + \beta = -\frac{q}{p} \)

Product of zeroes \( \alpha\beta = \frac{r}{p} \)


Step 3: Detailed Explanation:

We need to evaluate the expression:
\[ E = \alpha^3\beta + \beta^3\alpha \]

Taking common factor \( \alpha\beta \):
\[ E = \alpha\beta(\alpha^2 + \beta^2) \]

Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\[ E = \alpha\beta [(\alpha + \beta)^2 - 2\alpha\beta] \]

Substituting the values of sum and product:
\[ E = \left(\frac{r}{p}\right) \left[ \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) \right] \]
\[ E = \left(\frac{r}{p}\right) \left[ \frac{q^2}{p^2} - \frac{2r}{p} \right] \]

Taking LCM inside the bracket:
\[ E = \left(\frac{r}{p}\right) \left[ \frac{q^2 - 2pr}{p^2} \right] \]
\[ E = \frac{r(q^2 - 2pr)}{p^3} \]


Step 4: Final Answer:

The value is \( \frac{r(q^2 - 2pr)}{p^3} \).
Quick Tip: Express any symmetric function of roots in terms of \((\alpha + \beta)\) and \((\alpha\beta)\) to solve polynomial relation problems easily.

Step 1: Understanding the Concept:

In similar triangles, the ratios of the lengths of corresponding sides are equal.


Step 2: Key Formula or Approach:

Given \( \triangle AHK \sim \triangle ABC \), the corresponding sides are:
\[ \frac{AH}{AB} = \frac{HK}{BC} = \frac{AK}{AC} \]


Step 3: Detailed Explanation:

From the given data: \( AK = 10 cm \), \( BC = 3.5 cm \), \( HK = 7 cm \).

We use the ratio:
\[ \frac{HK}{BC} = \frac{AK}{AC} \]

Substituting the values:
\[ \frac{7}{3.5} = \frac{10}{AC} \]

Since \( 3.5 \times 2 = 7 \):
\[ 2 = \frac{10}{AC} \]
\[ AC = \frac{10}{2} = 5 cm \]


Step 4: Final Answer:

The length of AC is 5 cm.
Quick Tip: Identify the pair of sides where both values are known to find the scale factor of similarity first.

Step 1: Understanding the Concept:

By the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.


Step 2: Detailed Explanation:

Since \( XY \parallel QR \), by BPT:
\[ \frac{PX}{XQ} = \frac{PY}{YR} \]

Adding 1 to both sides leads to the corollary:
\[ \frac{PX + XQ}{XQ} = \frac{PY + YR}{YR} \implies \frac{PQ}{XQ} = \frac{PR}{YR} \]

Given \( \frac{PQ}{XQ} = \frac{7}{3} \) and \( PR = 6.3 cm \):
\[ \frac{7}{3} = \frac{6.3}{YR} \]
\[ YR = \frac{6.3 \times 3}{7} \]
\[ YR = 0.9 \times 3 = 2.7 cm \]


Step 3: Final Answer:

The length of YR is 2.7 cm.
Quick Tip: If the full length of a side is given, use the full-side to partial-side ratio directly to save time.

Step 1: Understanding the Concept:

In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal AC is the same as the midpoint of diagonal BD.


Step 2: Key Formula or Approach:

Midpoint formula: \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).


Step 3: Detailed Explanation:

Midpoint of AC:
\[ Midpoint_{AC} = \left( \frac{4 + 4}{2}, \frac{5 + 3}{2} \right) = (4, 4) \]

Midpoint of BD:
\[ Midpoint_{BD} = \left( \frac{m + 1}{2}, \frac{6 + n}{2} \right) \]

Since the midpoints must coincide:

1. \( \frac{m + 1}{2} = 4 \implies m + 1 = 8 \implies m = 7 \)

2. \( \frac{6 + n}{2} = 4 \implies 6 + n = 8 \implies n = 2 \)

Wait, looking at the coordinates again: A(4,5), B(m,6), C(4,3), D(1,n).

Midpoint of AC: \( x = (4+4)/2 = 4 \), \( y = (5+3)/2 = 4 \). Correct.

Midpoint of BD: \( x = (m+1)/2 = 4 \implies m = 7 \).
\( y = (6+n)/2 = 4 \implies n = 2 \).


Step 4: Final Answer:

The values are \( m = 7 \) and \( n = 2 \).
Quick Tip: Using the midpoint of diagonals is the standard and fastest way to find missing coordinates of a parallelogram.

Step 1: Understanding the Concept:

We use the algebraic identity \( (a + b)^2 = a^2 + b^2 + 2ab \).


Step 2: Detailed Explanation:

Given: \( \tan \theta + \frac{1}{\tan \theta} = 2 \).

Squaring both sides:
\[ \left( \tan \theta + \frac{1}{\tan \theta} \right)^2 = (2)^2 \]
\[ \tan^2 \theta + \left( \frac{1}{\tan \theta} \right)^2 + 2(\tan \theta) \left( \frac{1}{\tan \theta} \right) = 4 \]

Since \( \tan \theta \times \frac{1}{\tan \theta} = 1 \):
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} + 2 = 4 \]
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 4 - 2 = 2 \]


Step 3: Final Answer:

The value is 2.
Quick Tip: If \( x + \frac{1}{x} = 2 \), then \( x^n + \frac{1}{x^n} \) is always 2 for any natural number \( n \).

Step 1: Understanding the Concept:

To eliminate the square root, rationalize the expression inside by multiplying by the conjugate of the denominator.


Step 2: Detailed Explanation:

LHS = \( \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \)

Multiply numerator and denominator by \( (1 - \sin \theta) \) inside the root:
\[ LHS = \sqrt{\frac{(1 - \sin \theta)(1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}} \]
\[ LHS = \sqrt{\frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}} \]

Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \):
\[ LHS = \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}} = \frac{1 - \sin \theta}{\cos \theta} \]

Splitting the terms:
\[ LHS = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \sec \theta - \tan \theta = RHS \]


Step 3: Final Answer:

LHS = RHS. Hence Proved.
Quick Tip: Rationalizing helps simplify complex roots in trigonometry quickly.

Step 1: Understanding the Concept:

A quadrilateral is cyclic if the sum of either pair of opposite angles is \( 180^\circ \).


Step 2: Detailed Explanation:

In quadrilateral PQOR:

1. \( \angle OQP = 90^\circ \) (Radius OQ is perpendicular to tangent PQ).

2. \( \angle ORP = 90^\circ \) (Radius OR is perpendicular to tangent PR).

Sum of this pair of opposite angles:
\[ \angle OQP + \angle ORP = 90^\circ + 90^\circ = 180^\circ \]

In any quadrilateral, the sum of all angles is \( 360^\circ \):
\[ \angle QOR + \angle QPR = 360^\circ - 180^\circ = 180^\circ \]

Since opposite angles are supplementary, PQOR is a cyclic quadrilateral.


Step 3: Final Answer:

The quadrilateral PQOR is cyclic. Hence Proved.
Quick Tip: The angle between two tangents from an external point is always supplementary to the central angle subtended by the radii at points of contact.

Step 1: Understanding the Concept:

We use the method of contradiction by assuming \( \sqrt{5} \) is rational.


Step 2: Detailed Explanation:

Suppose \( \sqrt{5} \) is rational. Then \( \sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \).

Squaring both sides: \( 5 = \frac{a^2}{b^2} \implies a^2 = 5b^2 \).

This implies \( a^2 \) is divisible by 5, which means \( a \) is also divisible by 5.

Let \( a = 5c \) for some integer \( c \).

Substitute: \( (5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2 \).

This implies \( b^2 \) is divisible by 5, which means \( b \) is also divisible by 5.

Thus, \( a \) and \( b \) have a common factor 5, contradicting that they are co-prime.

Our assumption is wrong; therefore, \( \sqrt{5} \) is irrational.


Step 3: Final Answer:
\( \sqrt{5} \) is irrational. Hence Proved.
Quick Tip: If a prime \( p \) divides \( a^2 \), then \( p \) must divide \( a \). This is the core logic of irrationality proofs.

Step 1: Understanding the Concept:

Trisection divides a line segment into three equal parts. Two points, P and Q, divide AB in the ratios 1:2 and 2:1 respectively.


Step 2: Detailed Explanation:

Point P divides AB in ratio \( 1:2 \):
\( x = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-5}{3} \).
\( y = \frac{1(-2) + 2(4)}{1 + 2} = \frac{6}{3} = 2 \).

So, \( P = (-5/3, 2) \).

Point Q divides AB in ratio \( 2:1 \):
\( x = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-7}{3} \).
\( y = \frac{2(-2) + 1(4)}{2 + 1} = \frac{0}{3} = 0 \).

So, \( Q = (-7/3, 0) \).


Step 3: Final Answer:

The trisection points are \( (-5/3, 2) \) and \( (-7/3, 0) \).
Quick Tip: The trisection points are also the midpoints of segments formed by other trisection points. For example, Q is the midpoint of PB.

Step 1: Understanding the Concept:

We use fundamental trigonometric identities such as \( \sec^2 \theta - 1 = \tan^2 \theta \) and \( cosec^2 \theta - 1 = \cot^2 \theta \) to simplify the expressions.

Converting complex terms into sine and cosine usually simplifies the verification of identities.


Step 2: Key Formula or Approach:

Identity 1: \( \sec^2 \theta - 1 = \tan^2 \theta \).

Identity 2: \( cosec^2 \theta - 1 = \cot^2 \theta \).

Identity 3: \( \sec \theta = \frac{1}{\cos \theta} \), \( cosec \theta = \frac{1}{\sin \theta} \), \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).


Step 3: Detailed Explanation:

LHS \( = \frac{\sec^3 \theta}{\tan^2 \theta} + \frac{cosec^3 \theta}{\cot^2 \theta} \).

Converting to sine and cosine:
\[ LHS = \frac{1/\cos^3 \theta}{\sin^2 \theta / \cos^2 \theta} + \frac{1/\sin^3 \theta}{\cos^2 \theta / \sin^2 \theta} \]

Simplifying each term:
\[ LHS = \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta} \]

Taking the LCM of the denominators, which is \( \sin^2 \theta \cos^2 \theta \):
\[ LHS = \frac{\cos \theta + \sin \theta}{\sin^2 \theta \cos^2 \theta} \]

Now, let's look at the RHS:
\[ RHS = \sec \theta \cdot cosec \theta (\sec \theta + cosec \theta) \]
\[ RHS = \frac{1}{\cos \theta \sin \theta} \left( \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \right) \]
\[ RHS = \frac{1}{\cos \theta \sin \theta} \left( \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \right) \]
\[ RHS = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta} \]

Since LHS \( = \) RHS, the identity is proved.


Step 4: Final Answer:

LHS \( = \) RHS. Hence Proved.
Quick Tip: When you see higher powers or fractions, simplify denominators using identities first, then convert the whole expression to \(\sin\) and \(\cos\).

Step 1: Understanding the Concept:

The goal is to substitute the given values of \( p \) and \( q \) into the expression and use the identity \( \sec^2 \alpha - \tan^2 \alpha = 1 \).


Step 2: Key Formula or Approach:

Trigonometric Identity: \( \sec^2 \alpha - \tan^2 \alpha = 1 \).


Step 3: Detailed Explanation:

Start with the term \( (p^2 - q^2) \):
\[ p^2 - q^2 = \left( \frac{\sec \alpha}{cosec \beta} \right)^2 - \left( \frac{\tan \alpha}{cosec \beta} \right)^2 \]
\[ p^2 - q^2 = \frac{\sec^2 \alpha - \tan^2 \alpha}{cosec^2 \beta} \]

Using the identity \( \sec^2 \alpha - \tan^2 \alpha = 1 \):
\[ p^2 - q^2 = \frac{1}{cosec^2 \beta} \]

Now, substitute this into the LHS of the equation to be proved:
\[ LHS = (p^2 - q^2) \sec^2 \alpha \]
\[ LHS = \frac{1}{cosec^2 \beta} \cdot \sec^2 \alpha \]
\[ LHS = \frac{\sec^2 \alpha}{cosec^2 \beta} \]

Since \( p = \frac{\sec \alpha}{cosec \beta} \), then \( p^2 = \frac{\sec^2 \alpha}{cosec^2 \beta} \).

Thus, LHS \( = p^2 \), which is the RHS.


Step 4: Final Answer:

LHS \( = \) RHS. Hence Proved.
Quick Tip: Isolating the common denominator in substitution problems often makes the algebra much cleaner.

Step 1: Understanding the Concept:

We use the properties of a circle and congruent triangles. The radius is perpendicular to the tangent at the point of contact.


Step 2: Detailed Explanation:

Given: A circle with centre O and two tangents PQ and PR drawn from an external point P.

To Prove: \( PQ = PR \).

Construction: Join OQ, OR and OP.

Proof:

In \( \triangle OQP \) and \( \triangle ORP \):

1. \( OQ = OR \) (Radii of the same circle).

2. \( \angle OQP = \angle ORP = 90^\circ \) (The tangent at any point of a circle is perpendicular to the radius through the point of contact).

3. \( OP = OP \) (Common side).

Therefore, \( \triangle OQP \cong \triangle ORP \) by the RHS (Right angle-Hypotenuse-Side) congruence rule.

By CPCT (Corresponding Parts of Congruent Triangles):
\[ PQ = PR \]


Step 3: Final Answer:

Lengths of tangents from an external point are equal. Hence Proved.
Quick Tip: A neat diagram showing the circle, tangents, and radii is essential for full marks in geometry proofs.

Step 1: Understanding the Concept:

We use the fact that tangents from an external point are equal and properties of isosceles triangles.


Step 2: Detailed Explanation:

Let \( \angle PTQ = \theta \).

Since \( TP = TQ \) (tangents from an external point), \( \triangle TPQ \) is an isosceles triangle.

Therefore, \( \angle TPQ = \angle TQP \).

In \( \triangle TPQ \):
\[ \angle PTQ + \angle TPQ + \angle TQP = 180^\circ \]
\[ \theta + 2\angle TPQ = 180^\circ \]
\[ \angle TPQ = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2} \]

We know that the radius OP is perpendicular to the tangent TP at point P.
\[ \angle OPT = 90^\circ \]

From the figure, \( \angle OPT = \angle OPQ + \angle TPQ \).
\[ 90^\circ = \angle OPQ + \left( 90^\circ - \frac{\theta}{2} \right) \]
\[ \angle OPQ = \frac{\theta}{2} \]
\[ \theta = 2\angle OPQ \]
\[ \angle PTQ = 2\angle OPQ \]


Step 3: Final Answer:
\( \angle PTQ = 2 \angle OPQ \). Hence Proved.
Quick Tip: Remember that \( \triangle TPQ \) is isosceles and the radius is perpendicular to the tangent to link the angles.

Step 1: Understanding the Concept:

The area of a sector depends on the central angle and the total area of the circle.


Step 2: Key Formula or Approach:

Area of sector \( = \frac{\theta}{360} \times \pi r^2 \).

Area of major sector \( = Total Area - Area of minor sector \).


Step 3: Detailed Explanation:

Given: \( r = 42 cm \), \( \theta = 30^\circ \).

1. Area of minor sector:
\[ Area = \frac{30}{360} \times \frac{22}{7} \times 42 \times 42 \]
\[ Area = \frac{1}{12} \times 22 \times 6 \times 42 \]
\[ Area = \frac{1}{2} \times 22 \times 42 = 11 \times 42 = 462 cm^2 \]

2. Area of major sector:

The central angle of the major sector is \( 360^\circ - 30^\circ = 330^\circ \).
\[ Area of major sector = \frac{330}{360} \times \frac{22}{7} \times 42 \times 42 \]
\[ Area of major sector = \frac{11}{12} \times 22 \times 6 \times 42 \]
\[ Area of major sector = 11 \times 11 \times 42 = 121 \times 42 = 5082 cm^2 \]

Alternatively: Total Area \( = \frac{22}{7} \times 42 \times 42 = 5544 cm^2 \).

Major sector area \( = 5544 - 462 = 5082 cm^2 \).


Step 4: Final Answer:

Minor Sector Area \( = 462 cm^2 \); Major Sector Area \( = 5082 cm^2 \).
Quick Tip: Using \( 360^\circ - \theta \) for the major sector is often faster than calculating the full circle area and subtracting.

Step 1: Understanding the Concept:

When two dice are thrown, each die has 6 possible outcomes.

The total number of possible outcomes for the pair is the product of individual outcomes.


Step 2: Key Formula or Approach:

Total outcomes \( (n(S)) = 6 \times 6 = 36 \).

Probability of an event \( P(E) = \frac{Number of favorable outcomes}{Total number of outcomes} \).


Step 3: Detailed Explanation:

Let \( E \) be the event that the sum of the numbers on the two dice is 5.

The possible pairs \( (x, y) \) such that \( x + y = 5 \) are:

1. \( (1, 4) \)

2. \( (2, 3) \)

3. \( (3, 2) \)

4. \( (4, 1) \)

The number of favorable outcomes \( n(E) = 4 \).

Applying the probability formula:
\[ P(E) = \frac{4}{36} \]

Simplifying the fraction:
\[ P(E) = \frac{1}{9} \]


Step 4: Final Answer:

The probability that the sum is 5 is \( \frac{1}{9} \).
Quick Tip: For a sum \( S \) in two dice, the number of favorable outcomes is \( (S-1) \) if \( S \leq 7 \), and \( (13-S) \) if \( S > 7 \). Here \( 5-1 = 4 \).

Step 1: Understanding the Concept:

The difference refers to the absolute value of the difference between the two numbers shown on the dice.


Step 2: Detailed Explanation:

Total possible outcomes \( n(S) = 36 \).

Let \( F \) be the event that the difference is 3. We list the pairs \( (x, y) \) such that \( |x - y| = 3 \):

1. \( (1, 4) \) and \( (4, 1) \)

2. \( (2, 5) \) and \( (5, 2) \)

3. \( (3, 6) \) and \( (6, 3) \)

Counting these pairs, the number of favorable outcomes \( n(F) = 6 \).

Calculating the probability:
\[ P(F) = \frac{6}{36} \]

Simplifying the fraction:
\[ P(F) = \frac{1}{6} \]


Step 3: Final Answer:

The probability that the difference is 3 is \( \frac{1}{6} \).
Quick Tip: Always remember to count both orders \( (x, y) \) and \( (y, x) \) unless the numbers are the same (which isn't possible for a non-zero difference).

Step 1: Understanding the Concept:

We need to form two equations based on the cost of items and find their intersection point on a graph.


Step 2: Key Formula or Approach:

Let the price of one pencil be \( x \) and the price of one chocolate be \( y \).

According to Aarush's purchase: \( 2x + 3y = 11 \) (Equation 1)

According to Tanish's purchase: \( x + 2y = 7 \) (Equation 2)


Step 3: Detailed Explanation:

To plot these graphically, find at least two points for each line:

For \( 2x + 3y = 11 \):

If \( x = 1 \), then \( 2(1) + 3y = 11 \implies 3y = 9 \implies y = 3 \). Point: \( (1, 3) \).

If \( x = 4 \), then \( 2(4) + 3y = 11 \implies 3y = 3 \implies y = 1 \). Point: \( (4, 1) \).


For \( x + 2y = 7 \):

If \( y = 3 \), then \( x + 2(3) = 7 \implies x = 1 \). Point: \( (1, 3) \).

If \( y = 2 \), then \( x + 2(2) = 7 \implies x = 3 \). Point: \( (3, 2) \).


When plotted on a graph, the two lines intersect at the point \( (1, 3) \).

This intersection point represents the solution where \( x = 1 \) and \( y = 3 \).


Step 4: Final Answer:

The price of 1 pencil is Rs 1 and the price of 1 chocolate is Rs 3.
Quick Tip: While solving graphically, choose values for \( x \) that result in whole numbers for \( y \) to make plotting more accurate and easier.

Step 1: Understanding the Concept:

Total budget remains the same. If the number of days increases, the average daily expenditure must decrease.


Step 2: Key Formula or Approach:

Let the original duration of the tour be \( x \) days.

Daily expense = \( \frac{Total Budget}{Duration} \).


Step 3: Detailed Explanation:

Original daily expense = \( \frac{4200}{x} \).

New duration = \( x + 3 \) days.

New daily expense = \( \frac{4200}{x + 3} \).

According to the problem, the difference between original and new daily expenses is Rs 70.
\[ \frac{4200}{x} - \frac{4200}{x + 3} = 70 \]

Divide the entire equation by 70:
\[ \frac{60}{x} - \frac{60}{x + 3} = 1 \]

Multiply by \( x(x + 3) \) to clear denominators:
\[ 60(x + 3) - 60x = x(x + 3) \]
\[ 60x + 180 - 60x = x^2 + 3x \]
\[ 180 = x^2 + 3x \implies x^2 + 3x - 180 = 0 \]

Factoring the quadratic equation:
\[ x^2 + 15x - 12x - 180 = 0 \]
\[ x(x + 15) - 12(x + 15) = 0 \]
\[ (x - 12)(x + 15) = 0 \]

So, \( x = 12 \) or \( x = -15 \).

Since duration cannot be negative, we take \( x = 12 \).


Step 4: Final Answer:

The original duration of the tour was 12 days.
Quick Tip: In time-expenditure problems, the equation always follows the structure: \( Higher Rate - Lower Rate = Difference in Rate \).

Step 1: Understanding the Concept:

We use the area formula for a triangle and then apply the Pythagoras theorem to find the third side.


Step 2: Key Formula or Approach:

Area = \( \frac{1}{2} \times Base \times Altitude \).

Pythagoras Theorem: \( Base^2 + Altitude^2 = Hypotenuse^2 \).


Step 3: Detailed Explanation:

Let the altitude be \( x \) cm.

Then the base is \( (x + 10) \) cm.

Area \( = \frac{1}{2} \times (x + 10) \times x = 600 \).
\[ x(x + 10) = 1200 \implies x^2 + 10x - 1200 = 0 \]

Factoring the quadratic:
\[ x^2 + 40x - 30x - 1200 = 0 \]
\[ x(x + 40) - 30(x + 40) = 0 \]
\[ (x - 30)(x + 40) = 0 \]

Thus, \( x = 30 \) (ignoring \( x = -40 \) as length is positive).

Altitude = 30 cm.

Base = \( 30 + 10 = 40 cm \).

To find the hypotenuse (\( h \)):
\[ h^2 = 30^2 + 40^2 = 900 + 1600 = 2500 \]

\[ h = \sqrt{2500 = 50 \text{ cm \).


Step 4: Final Answer:

The dimensions of the triangle are 30 cm, 40 cm, and 50 cm.
Quick Tip: Notice that the dimensions (30, 40, 50) are just 10 times the standard Pythagorean triplet (3, 4, 5).

Step 1: Understanding the Concept:

This is the Basic Proportionality Theorem (Thales Theorem). We prove it using the areas of triangles with common heights.


Step 2: Detailed Explanation:

Given: In \( \triangle ABC \), a line \( DE \parallel BC \) intersects \( AB \) at \( D \) and \( AC \) at \( E \).

To Prove: \( \frac{AD}{DB} = \frac{AE}{EC} \).

Construction: Join \( BE \) and \( CD \). Draw \( DM \perp AC \) and \( EN \perp AB \).

Proof:

Area(\( \triangle ADE \)) = \( \frac{1}{2} \times base AD \times height EN \).

Area(\( \triangle BDE \)) = \( \frac{1}{2} \times base DB \times height EN \).

Therefore, \( \frac{Area(\triangle ADE)}{Area(\triangle BDE)} = \frac{AD}{DB} \) (1)

Similarly:

Area(\( \triangle ADE \)) = \( \frac{1}{2} \times base AE \times height DM \).

Area(\( \triangle CED \)) = \( \frac{1}{2} \times base EC \times height DM \).

Therefore, \( \frac{Area(\triangle ADE)}{Area(\triangle CED)} = \frac{AE}{EC} \) (2)

Now, \( \triangle BDE \) and \( \triangle CED \) are on the same base \( DE \) and between the same parallels \( DE \parallel BC \).

So, Area(\( \triangle BDE \)) = Area(\( \triangle CED \)) (3)

From (1), (2), and (3), we get:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]


Step 3: Final Answer:

The ratio of the segments is equal. Hence Proved.
Quick Tip: While proving theorems, always provide a neat diagram and clearly label "Given", "To Prove", and "Construction".

Step 1: Understanding the Concept:

This problem involves similar triangles. The lamp post, the girl, and their shadows form triangles with the same angle of elevation from the ground.


Step 2: Detailed Explanation:

Let AB be the lamp post and CD be the girl. Let AE be the ground level and \( x \) be the shadow length DE.

Height of lamp post (\( AB \)) = 3.6 m.

Height of girl (\( CD \)) = 90 cm = 0.9 m.

Speed of girl = 1.2 m/s. Time = 4 s.

Distance walked by girl (\( BD \)) = \( Speed \times Time = 1.2 \times 4 = 4.8 m \).

In \( \triangle ABE \) and \( \triangle CDE \):
\( \angle B = \angle D = 90^\circ \)
\( \angle E = \angle E \) (Common)

So, \( \triangle ABE \sim \triangle CDE \) by AA similarity.

The ratio of corresponding sides:
\[ \frac{AB}{CD} = \frac{BE}{DE} \]
\[ \frac{3.6}{0.9} = \frac{4.8 + x}{x} \]
\[ 4 = \frac{4.8 + x}{x} \]
\[ 4x = 4.8 + x \]
\[ 3x = 4.8 \implies x = 1.6 m \].


Step 3: Final Answer:

The length of her shadow after 4 seconds is 1.6 m.
Quick Tip: Always ensure all units are consistent (convert cm to m) before substituting values into similarity ratios.

Step 1: Understanding the Concept:

We use the grouped data formulas for Mode and Median.


Step 2: Key Formula or Approach:

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

Median \( = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \).


Step 3: Detailed Explanation:

Part 1: Calculating Mode

Highest frequency is 33 in the class interval 35 - 40.
\( l = 35 \), \( f_1 = 33 \), \( f_0 = 21 \), \( f_2 = 11 \), \( h = 5 \).
\[ Mode = 35 + \left( \frac{33 - 21}{66 - 21 - 11} \right) \times 5 \]
\[ Mode = 35 + \left( \frac{12}{34} \right) \times 5 = 35 + 1.764... \approx 36.76 \]

Part 2: Calculating Median

Total Frequency \( N = 100 \). \( N/2 = 50 \).

Cumulative Frequency (CF) table:

15-20: 2

20-25: 6

25-30: 24

30-35: 45

35-40: 78 (Median class)

For class 35-40: \( l = 35 \), \( cf (previous) = 45 \), \( f = 33 \), \( h = 5 \).
\[ Median = 35 + \left( \frac{50 - 45}{33} \right) \times 5 \]
\[ Median = 35 + \frac{25}{33} = 35 + 0.757... \approx 35.76 \]


Step 4: Final Answer:

The modal age is approx 36.76 years and the median age is approx 35.76 years.
Quick Tip: Median class is the first class whose cumulative frequency is greater than or equal to \( N/2 \).

Step 1: Understanding the Concept:

The monthly instalments form an Arithmetic Progression (A.P.) because there is a constant increase in the payment every month.


Step 2: Key Formula or Approach:

The \( n^{th} \) term of an A.P. is given by the formula:
\[ a_n = a + (n - 1)d \]

Where:
\( a = \) first term (Rs 1,000)
\( d = \) common difference (Rs 100)
\( n = 30 \) for the \( 30^{th} \) instalment.


Step 3: Detailed Explanation:

Substitute the values into the formula:
\[ a_{30} = 1000 + (30 - 1) \times 100 \]
\[ a_{30} = 1000 + 29 \times 100 \]
\[ a_{30} = 1000 + 2900 \]
\[ a_{30} = 3900 \]


Step 4: Final Answer:

The amount paid in the \( 30^{th} \) instalment is Rs 3,900.
Quick Tip: Always identify 'a' and 'd' correctly from the word problem. The "starting amount" is 'a' and the "monthly increase" is 'd'.

Step 1: Understanding the Concept:

The last instalment refers to the \( 40^{th} \) term of the A.P. sequence.


Step 2: Detailed Explanation:

Here, \( n = 40 \), \( a = 1000 \), and \( d = 100 \).

Using the formula \( a_n = a + (n - 1)d \):
\[ a_{40} = 1000 + (40 - 1) \times 100 \]
\[ a_{40} = 1000 + 39 \times 100 \]
\[ a_{40} = 1000 + 3900 \]
\[ a_{40} = 4900 \]


Step 3: Final Answer:

The amount paid in the last (\( 40^{th} \)) instalment is Rs 4,900.
Quick Tip: The "last instalment" is simply the term corresponding to the total number of periods given in the problem.

Step 1: Understanding the Concept:

To find the remaining amount, we first calculate the sum of the first 30 instalments and subtract it from the total loan amount.


Step 2: Key Formula or Approach:

Sum of first \( n \) terms of an A.P.:
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \]


Step 3: Detailed Explanation:

Total loan = Rs 1,18,000.

Sum of first 30 instalments (\( S_{30} \)):
\[ S_{30} = \frac{30}{2} [2(1000) + (30 - 1) \times 100] \]
\[ S_{30} = 15 [2000 + 2900] \]
\[ S_{30} = 15 \times 4900 = 73,500 \]

Amount remaining = Total loan \( - S_{30} \)
\[ Remaining = 1,18,000 - 73,500 = 44,500 \]


Step 4: Final Answer:

The amount left to pay after the \( 30^{th} \) instalment is Rs 44,500.
Quick Tip: When asked for "remaining amount", always calculate the sum paid so far using the \( S_n \) formula, not just the \( n^{th} \) term.

Step 1: Understanding the Concept:

Ratio is the comparison of two quantities by division. We need the values of the \( 10^{th} \) and \( 40^{th} \) instalments.


Step 2: Detailed Explanation:

Calculate \( 10^{th} \) instalment (\( a_{10} \)):
\[ a_{10} = 1000 + (10 - 1) \times 100 = 1000 + 900 = 1,900 \]

Calculate last (\( 40^{th} \)) instalment (\( a_{40} \)):

From part (ii), \( a_{40} = 4,900 \).

Ratio = \( a_{10} : a_{40} \)
\[ Ratio = \frac{1900}{4900} = \frac{19}{49} \]


Step 3: Final Answer:

The ratio is 19 : 49.
Quick Tip: Always simplify the ratio by canceling out common factors (like zeros) to reach the simplest integer form.

Step 1: Understanding the Concept:

Angle of depression is equal to the angle of elevation due to alternate interior angles. We use trigonometric ratios in the right-angled triangle.


Step 2: Key Formula or Approach:

In \( \triangle ABC \) (where A is the top and BC is the base), \( \tan \theta = \frac{Perpendicular}{Base} \).


Step 3: Detailed Explanation:

Let the height of the building be \( h \).

At \( 60^\circ \) depression, the car is at point C, 25 m from the base B.

In right \( \triangle ABC \):
\[ \tan 60^\circ = \frac{AB}{BC} \]
\[ \sqrt{3} = \frac{h}{25} \]
\[ h = 25\sqrt{3} m \]


Step 4: Final Answer:

The height of the building is \( 25\sqrt{3} \) m.
Quick Tip: Draw a clear diagram. The angle closer to the base of the building is always the larger angle (60\(^\circ\) in this case).

Step 1: Understanding the Concept:

We need to find the distance CD between the car's initial position (D) at \( 30^\circ \) and its second position (C) at \( 60^\circ \).


Step 2: Detailed Explanation:

In right \( \triangle ABD \):
\[ \tan 30^\circ = \frac{AB}{BD} \]
\[ \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD} \]
\[ BD = 25\sqrt{3} \times \sqrt{3} = 25 \times 3 = 75 m \]

Distance between positions (\( CD \)) = \( BD - BC \)
\[ CD = 75 - 25 = 50 m \]


Step 3: Final Answer:

The distance between the two positions of the car is 50 m.
Quick Tip: Distance CD is found by calculating the full horizontal distance from the base and subtracting the known partial distance.

Step 1: Understanding the Concept:

First, find the uniform speed of the car using the distance it covered in 6 seconds. Then find the total time for the total distance.


Step 2: Detailed Explanation:

Distance \( CD = 50 \) m was covered in 6 seconds.

Speed \( v = \frac{Distance}{Time} = \frac{50}{6} = \frac{25}{3} m/s \).

The total distance from starting point D to foot B is \( BD = 75 \) m.

Total time taken = \( \frac{Total distance BD}{Speed} \)
\[ Time = \frac{75}{25/3} = \frac{75 \times 3}{25} = 3 \times 3 = 9 seconds \]


Step 3: Final Answer:

The total time taken is 9 seconds.
Quick Tip: Alternatively, since distance 50m takes 6s, 25m takes 3s. Total time = 6 + 3 = 9 seconds.

Step 1: Understanding the Concept:

The distance of the observer (at A) from the car (at C) is the length of the hypotenuse AC in \( \triangle ABC \).


Step 2: Key Formula or Approach:

Use \( \cos \theta = \frac{Base}{Hypotenuse} \).


Step 3: Detailed Explanation:

In right \( \triangle ABC \):
\[ \cos 60^\circ = \frac{BC}{AC} \]
\[ \frac{1}{2} = \frac{25}{AC} \]
\[ AC = 25 \times 2 = 50 m \]


Step 4: Final Answer:

The distance of the observer from the car is 50 m.
Quick Tip: You can also use Pythagoras theorem \( h^2 + 25^2 = AC^2 \) but using cosine is much faster.

Step 1: Understanding the Concept:

The diagonal of a cube connects two opposite corners through the center of the cube.


Step 2: Key Formula or Approach:

Diagonal of a cube with side \( a \) is given by:
\[ Diagonal = a\sqrt{3} \]


Step 3: Detailed Explanation:

Given edge length \( a = 6 \) cm.

Substituting in the formula:
\[ Diagonal = 6\sqrt{3} cm \]


Step 4: Final Answer:

The length of the diagonal is \( 6\sqrt{3} \) cm.
Quick Tip: Do not confuse the face diagonal (\( a\sqrt{2} \)) with the space diagonal (\( a\sqrt{3} \)).

Step 1: Understanding the Concept:

Volume represents the space occupied by the cube.


Step 2: Key Formula or Approach:

Volume of cube \( = a^3 \).


Step 3: Detailed Explanation:

Given edge \( a = 7 \) cm.
\[ Volume = 7 \times 7 \times 7 = 343 cm^3 \]


Step 4: Final Answer:

The volume of the cube is 343 \( cm^3 \).
Quick Tip: Always include the correct unit of measurement. Volume is measured in cubic units (\( cm^3 \)).

Step 1: Understanding the Concept:

Curved Surface Area (CSA) refers to the area of the outer curved part of the hemisphere.


Step 2: Key Formula or Approach:

CSA of hemisphere \( = 2\pi r^2 \).


Step 3: Detailed Explanation:

Given radius \( r = 7 \) cm. Use \( \pi = \frac{22}{7} \).
\[ CSA = 2 \times \frac{22}{7} \times 7 \times 7 \]
\[ CSA = 2 \times 22 \times 7 \]
\[ CSA = 44 \times 7 = 308 cm^2 \]


Step 4: Final Answer:

The curved surface area is 308 \( cm^2 \).
Quick Tip: Total Surface Area would be \( 3\pi r^2 \), but the question specifically asks for Curved Surface Area.

Step 1: Understanding the Concept:

When two cubes are joined end-to-end, only the length changes while breadth and height remain the same.


Step 2: Key Formula or Approach:

Surface Area of cuboid \( = 2(lb + bh + hl) \).


Step 3: Detailed Explanation:

Dimensions of the new cuboid:

Length (\( l \)) = \( 4 + 4 = 8 \) cm.

Breadth (\( b \)) = 4 cm.

Height (\( h \)) = 4 cm.
\[ Surface Area = 2(8 \times 4 + 4 \times 4 + 4 \times 8) \]
\[ Surface Area = 2(32 + 16 + 32) \]
\[ Surface Area = 2(80) = 160 cm^2 \]


Step 4: Final Answer:

The surface area of the resulting cuboid is 160 \( cm^2 \).
Quick Tip: The area decreases because two faces (one from each cube) are now hidden in the joint.