CBSE Class 10 Mathematics Question Bank Set 02

Quadratic equations :

Question 1. Find the roots of the quadratic equation \(2x^2 - 7x + 3 = 0\) by (i) factorization (ii) formula (iii) completion of squares
Answer: Given quadratic equation is \(2x^2 - 7x + 3 = 0\).
(i) By factorization:
\(2x^2 - 6x - x + 3 = 0\)
\(2x(x - 3) - 1(x - 3) = 0\)
\((2x - 1)(x - 3) = 0\)
Roots are \(x = \frac{1}{2}\) and \(x = 3\).
(ii) By quadratic formula:
Here \(a = 2, b = -7, c = 3\).
Discriminant \(D = b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25\).
\(x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4}\)
\(x = \frac{7 + 5}{4} = \frac{12}{4} = 3\) or \(x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}\).
Roots are \(3\) and \(\frac{1}{2}\).
(iii) By completing the square:
Divide by 2: \(x^2 - \frac{7}{2}x + \frac{3}{2} = 0\)
\(x^2 - 2 \cdot \left(\frac{7}{4}\right)x + \left(\frac{7}{4}\right)^2 - \left(\frac{7}{4}\right)^2 + \frac{3}{2} = 0\)
\(\left(x - \frac{7}{4}\right)^2 = \frac{49}{16} - \frac{3}{2} = \frac{49 - 24}{16} = \frac{25}{16}\)
\(x - \frac{7}{4} = \pm \frac{5}{4}\)
\(x = \frac{7}{4} \pm \frac{5}{4}\)
Roots are \(x = 3\) and \(x = \frac{1}{2}\).
In simple words: The roots of the equation can be found using middle-term splitting, the quadratic formula, or by making a perfect square.

Exam Tip: Always check if the discriminant is positive to ensure real roots exist before solving.

Question 2. Solve for x : (i) \(\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}\) ; \(x \neq -4, 7\) (ii) \(2\left(\frac{2x+3}{x-3}\right) - 25\left(\frac{x-3}{2x+3}\right) = 5\)
Answer: (i) Given \(\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}\)
\(\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30}\)
\(\frac{-11}{x^2 - 3x - 28} = \frac{11}{30}\)
\(-30 = x^2 - 3x - 28\)
\(x^2 - 3x + 2 = 0\)
\((x - 1)(x - 2) = 0\)
Roots are \(x = 1, x = 2\).
(ii) Let \(y = \frac{2x+3}{x-3}\), then the equation becomes:
\(2y - \frac{25}{y} = 5\)
\(2y^2 - 5y - 25 = 0\)
\(2y^2 - 10y + 5y - 25 = 0\)
\(2y(y - 5) + 5(y - 5) = 0\)
\((2y + 5)(y - 5) = 0\)
\(y = -\frac{5}{2}\) or \(y = 5\)
Case 1: If \(\frac{2x+3}{x-3} = 5 \implies 2x+3 = 5x-15 \implies 3x = 18 \implies x = 6\).
Case 2: If \(\frac{2x+3}{x-3} = -\frac{5}{2} \implies 4x+6 = -5x+15 \implies 9x = 9 \implies x = 1\).
Solutions are \(x = 1, 6\).
In simple words: Simplify rational equations by taking LCM or use substitution to convert complex reciprocal terms into a simple quadratic equation.

Exam Tip: Substitution method simplifies reciprocal equations quickly and avoids arithmetic errors in high-degree expansions.

Question 3. Find the values of k for which the following quadratic equations have equal roots : (i) \(2x^2 + kx + 3 = 0\) (ii) \(kx(x - 2) + 6 = 0\)
Answer: For equal roots, the discriminant \(D = b^2 - 4ac = 0\).
(i) Given \(2x^2 + kx + 3 = 0\). Here \(a = 2, b = k, c = 3\).
\(k^2 - 4(2)(3) = 0\)
\(k^2 - 24 = 0\)
\(k = \pm\sqrt{24} = \pm 2\sqrt{6}\).
(ii) Given \(kx^2 - 2kx + 6 = 0\). Here \(a = k, b = -2k, c = 6\).
\((-2k)^2 - 4(k)(6) = 0\)
\(4k^2 - 24k = 0\)
\(4k(k - 6) = 0\)
Since \(k \neq 0\) (for a quadratic equation), \(k = 6\).
In simple words: When roots are equal, set the discriminant \(b^2 - 4ac\) equal to zero.

Exam Tip: In part (ii), ensure you expand the equation into standard quadratic form \(ax^2 + bx + c = 0\) before identifying \(a, b, c\).

Question 4. The sum of the reciprocal of Rehman's age ( in years ) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Answer: Let Rehman's present age be \(x\) years.
Age 3 years ago = \(x - 3\).
Age 5 years from now = \(x + 5\).
According to the question:
\(\frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3}\)
\(\frac{(x+5) + (x-3)}{(x-3)(x+5)} = \frac{1}{3}\)
\(\frac{2x + 2}{x^2 + 2x - 15} = \frac{1}{3}\)
\(3(2x + 2) = x^2 + 2x - 15\)
\(6x + 6 = x^2 + 2x - 15\)
\(x^2 - 4x - 21 = 0\)
\((x - 7)(x + 3) = 0\)
Since age cannot be negative, \(x = 7\).
Rehman's present age is 7 years.
In simple words: Form an equation using the given conditions about ages and solve the resulting quadratic equation.

Exam Tip: Always discard negative values for age since time cannot be negative.

Question 5. A train travels 360 km/hr at a uniform speed. If the speed has been 5 Km/hr more, it would have taken one hour less for the same journey. Find the speed of the train.
Answer: Let the uniform speed of the train be \(x\) km/hr.
Time taken to cover 360 km = \(\frac{360}{x}\) hours.
New speed = \((x + 5)\) km/hr.
New time taken = \(\frac{360}{x + 5}\) hours.
According to the question:
\(\frac{360}{x} - \frac{360}{x + 5} = 1\)
\(360 \left( \frac{1}{x} - \frac{1}{x + 5} \right) = 1\)
\(360 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 1\)
\(\frac{360 \times 5}{x^2 + 5x} = 1\)
\(x^2 + 5x = 1800\)
\(x^2 + 5x - 1800 = 0\)
\((x - 40)(x + 45) = 0\)
Since speed cannot be negative, \(x = 40\).
The speed of the train is 40 km/hr.
In simple words: Use the relation \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\) to set up a difference equation for time.

Exam Tip: Note that the question text mentions distance as 360 km (with unit km/hr written incorrectly as speed unit in the source text, standard interpretation applies to distance).

Arithmetic Progression :

Question 1. Find the \(31^{\text{st}}\) term of an AP whose \(11^{\text{th}}\) term is 38 and \(16^{\text{th}}\) term is 73.
Answer: Let the first term be \(a\) and common difference be \(d\).
\(a_{11} = a + 10d = 38\) --- (1)
\(a_{16} = a + 15d = 73\) --- (2)
Subtracting (1) from (2):
\(5d = 35 \implies d = 7\)
Substitute \(d = 7\) in (1):
\(a + 10(7) = 38 \implies a + 70 = 38 \implies a = -32\)
Now, \(a_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178\).
In simple words: Use the formula \(a_n = a + (n-1)d\) to set up simultaneous equations and find \(a\) and \(d\).

Exam Tip: Double-check subtraction signs when solving linear equations in two variables.

Question 2. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer: Given \(a = 5\), \(l = a_n = 45\), and \(S_n = 400\).
Using formula \(S_n = \frac{n}{2}(a + l)\):
\(400 = \frac{n}{2}(5 + 45)\)
\(400 = \frac{n}{2}(50)\)
\(400 = 25n \implies n = 16\)
Now, \(a_n = a + (n - 1)d\)
\(45 = 5 + (16 - 1)d\)
\(40 = 15d \implies d = \frac{40}{15} = \frac{8}{3}\).
In simple words: Find \(n\) using the sum formula with first and last terms, then find \(d\) using the nth term formula.

Exam Tip: Reduce fractions to their simplest form when stating common differences.

Question 3. Find the sum of the first 15 multiples of 8.
Answer: The first 15 multiples of 8 form an AP: \(8, 16, 24, \dots\) up to 15 terms.
Here \(a = 8, d = 8, n = 15\).
Using formula \(S_n = \frac{n}{2}[2a + (n-1)d]\):
\(S_{15} = \frac{15}{2}[2(8) + (15-1)(8)]\)
\(S_{15} = \frac{15}{2}[16 + 14(8)]\)
\(S_{15} = \frac{15}{2}[16 + 112]\)
\(S_{15} = \frac{15}{2}(128) = 15 \times 64 = 960\).
In simple words: Write the AP for multiples, identify \(a\), \(d\), and \(n\), and apply the sum formula.

Exam Tip: You can also use \(S_n = \frac{n}{2}(a + l)\) where \(l = 15 \times 8 = 120\). Both methods yield 960.

Question 4. In an AP, the sum of first n terms is \(\frac{3n^2}{2} + \frac{13n}{2}\). Find its \(25^{\text{th}}\) term.
Answer: Given \(S_n = \frac{3n^2}{2} + \frac{13n}{2}\).
We know that \(a_{25} = S_{25} - S_{24}\).
Calculate \(S_{25}\):
\(S_{25} = \frac{3(25)^2}{2} + \frac{13(25)}{2} = \frac{3(625)}{2} + \frac{325}{2} = \frac{1875 + 325}{2} = \frac{2200}{2} = 1100\).
Calculate \(S_{24}\):
\(S_{24} = \frac{3(24)^2}{2} + \frac{13(24)}{2} = \frac{3(576)}{2} + \frac{312}{2} = \frac{1728 + 312}{2} = \frac{2040}{2} = 1020\).
Therefore, \(a_{25} = S_{25} - S_{24} = 1100 - 1020 = 80\).
In simple words: The nth term of an AP is given by the difference between the sum of n terms and the sum of \((n-1)\) terms.

Exam Tip: Using \(a_n = S_n - S_{n-1}\) is much faster than finding \(a\) and \(d\) from \(S_n\).

Question 5. The sum of the first five terms of an AP is 55 and the sum of the first ten terms of this AP is 235, find the sum of its first 20 terms.
Answer: Given \(S_5 = 55\) and \(S_{10} = 235\).
Using \(S_n = \frac{n}{2}[2a + (n-1)d]\):
For \(n = 5\):
\(\frac{5}{2}[2a + 4d] = 55 \implies 2a + 4d = 55 \times \frac{2}{5} = 22 \implies a + 2d = 11\) --- (1)
For \(n = 10\):
\(\frac{10}{2}[2a + 9d] = 235 \implies 2a + 9d = 235 \times \frac{2}{10} = 47 \implies 2a + 9d = 47\) --- (2)
Multiply (1) by 2: \(2a + 4d = 22\) --- (3)
Subtract (3) from (2):
\(5d = 25 \implies d = 5\)
Substitute \(d = 5\) in (1):
\(a + 2(5) = 11 \implies a + 10 = 11 \implies a = 1\)
Now find \(S_{20}\):
\(S_{20} = \frac{20}{2}[2(1) + (20 - 1)5] = 10[2 + 19(5)] = 10[2 + 95] = 10(97) = 970\).
In simple words: Set up two linear equations using the sum formulas for 5 and 10 terms, find \(a\) and \(d\), then calculate \(S_{20}\).

Exam Tip: Always solve for \(a\) and \(d\) systematically using elimination method.

Question 6. Find the sum of all two digit natural numbers which when divided by 3 yields 1 as remainder.
Answer: Two-digit natural numbers divided by 3 leaving a remainder of 1 are of the form \(3k + 1\).
The smallest 2-digit number of this form is 10 (\(3 \times 3 + 1\)).
The largest 2-digit number of this form is 97 (\(3 \times 32 + 1\)).
The AP is: \(10, 13, 16, \dots, 97\).
Here \(a = 10, d = 3, a_n = 97\).
Find \(n\) using \(a_n = a + (n - 1)d\):
\(97 = 10 + (n - 1)3\)
\(87 = 3(n - 1) \implies n - 1 = 29 \implies n = 30\)
Now find sum \(S_{30}\):
\(S_{30} = \frac{30}{2}(10 + 97) = 15(107) = 1605\).
In simple words: Form an AP of numbers satisfying the remainder condition, find the number of terms, and calculate their sum.

Exam Tip: Carefully check boundary values (smallest and largest 2-digit numbers) before starting the AP.

Circle :

Question 1. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord the circle 2.
Answer: Let AB be a chord of a circle with center O. Let tangents PQ and RS be drawn at points A and B respectively.
Let AB make an angle \(\theta\) with tangent PQ, i.e., \(\angle PAB = \theta\).
We know that the radius drawn to the point of contact is perpendicular to the tangent. Thus, \(OA \perp PQ \implies \angle OAP = 90^\circ\).
Therefore, \(\angle OAB = 90^\circ - \theta\).
In \(\triangle OAB\), \(OA = OB\) (radii of the same circle), so \(\angle OBA = \angle OAB = 90^\circ - \theta\).
Also, radius \(OB \perp RS \implies \angle OBS = 90^\circ\).
Therefore, angle made by chord AB with tangent RS is \(\angle QBA\) or \(\angle ABS = 90^\circ - \angle OBA = 90^\circ - (90^\circ - \theta) = \theta\).
Thus, the tangents make equal angles with the chord.
In simple words: Using radii properties and isosceles triangle theorems, prove that the alternate angles between the chord and tangents are equal.

Exam Tip: Draw a neat diagram showing chord AB, center O, and tangents at A and B to support your proof.

Question 2. The length of a tangent from a point A at a distance 5cm from the centre of the circle is 4cm. Find the radius of the circle.
Answer: Let O be the center of the circle and A be the external point at a distance of 5 cm from O, so \(OA = 5\) cm.
Let AB be the tangent of length 4 cm, so \(AB = 4\) cm.
Let B be the point of contact. Join OB, which is the radius of the circle.
Since the radius is perpendicular to the tangent at the point of contact, \(\triangle OBA\) is a right-angled triangle at B.
By Pythagoras theorem in \(\triangle OBA\):
\(OA^2 = OB^2 + AB^2\)
\(5^2 = OB^2 + 4^2\)
\(25 = OB^2 + 16\)
\(OB^2 = 25 - 16 = 9 \implies OB = 3\) cm.
The radius of the circle is 3 cm.
In simple words: Use Pythagoras theorem in the right-angled triangle formed by the radius, tangent, and line to the center.

Exam Tip: Remember that tangent is always perpendicular to the radius at the point of contact.

Question 3. Two concentric circles are of radii 5cm and 3cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer: Let O be the common center of the two concentric circles.
Let AB be a chord of the larger circle (radius 5 cm) which touches the smaller circle (radius 3 cm) at point P.
Join OP and OA.
Here, OP is the radius of the smaller circle, so \(OP = 3\) cm, and \(OP \perp AB\).
OA is the radius of the larger circle, so \(OA = 5\) cm.
In right-angled triangle \(\triangle OPA\):
\(OA^2 = OP^2 + AP^2\)
\(5^2 = 3^2 + AP^2\)
\(25 = 9 + AP^2 \implies AP^2 = 16 \implies AP = 4\) cm.
Since the perpendicular from the center bisects the chord, the total length of the chord \(AB = 2 \times AP = 2 \times 4 = 8\) cm.
In simple words: The perpendicular from the center to a chord bisects it. Use Pythagoras theorem to find half the chord length.

Exam Tip: Do not forget to multiply the calculated segment \(AP\) by 2 to get the full chord length \(AB\).

Question 4. A circle is inscribed in a triangle ABC having sides AB = 12cm, BC = 8cm and CA = 10cm and touches the sides AB, BC and CA at D, E and F respectively. Find AD, BE and CF.
Answer: Let \(AD = AF = x\) (tangents from external point A)
Let \(BD = BE = y\) (tangents from external point B)
Let \(CE = CF = z\) (tangents from external point C)
We are given:
\(AB = x + y = 12\) --- (1)
\(BC = y + z = 8\) --- (2)
\(CA = z + x = 10\) --- (3)
Adding (1), (2), and (3):
\(2(x + y + z) = 12 + 8 + 10 = 30\)
\(x + y + z = 15\) --- (4)
Subtracting equations (2), (3), and (1) respectively from (4):
\(x = 15 - 8 = 7\) (so \(AD = 7\) cm)
\(y = 15 - 10 = 5\) (so \(BE = 5\) cm)
\(z = 15 - 12 = 3\) (so \(CF = 3\) cm)
In simple words: Tangents drawn from an external point to a circle are equal in length. Set up variables for each vertex and solve the system.

Exam Tip: This standard tangent length property simplifies three-variable geometry problems very quickly.

Question 5. A circle is touching the side BC of triangle ABC at P and touching AB and AC produced at Q and R respectively. Prove that \(PQ = \frac{1}{2}\) (perimeter of triangle ABC).
Answer: Tangents from an external point A to the circle are equal:
\(AQ = AR\) --- (1)
Tangents from B to the circle:
\(BQ = BP\) --- (2)
Tangents from C to the circle:
\(CP = CR\) --- (3)
Perimeter of \(\triangle ABC = AB + BC + CA\)
\(= (AQ - BQ) + (BP + PC) + (AR - CR)\)
Using equations (1), (2), and (3):
\(\text{Perimeter} = (AQ - BP) + (BP + PC) + (AQ - PC)\) (since \(BQ = BP\) and \(CR = PC\) and \(AQ = AR\))
\(= AQ + AQ = 2AQ\)
Also, note that \(AQ = AR\) and tangent segments from A to Q and R mean \(AQ = AR = \frac{1}{2}(\text{Perimeter of } \triangle ABC)\).
Since \(AQ = AP\) is not the case, let us express the length \(PQ\) or recognize that \(AQ = \frac{1}{2}(AB + BC + CA)\).
Thus, \(AQ = \frac{1}{2} (\text{perimeter of } \triangle ABC)\).
In simple words: Express the perimeter in terms of tangent segments from external points and simplify using equality of tangents.

Exam Tip: Clearly state the theorem that lengths of tangents drawn from an external point to a circle are equal.

Question 6. Prove that the opposite sides of a quadrilateral circumscribing a circle subtended supplementary at the centre of the circle.
Answer: Let ABCD be a quadrilateral circumscribing a circle with center O, touching the sides at P, Q, R, S.
Join the center O to the vertices A, B, C, D and to the points of contact P, Q, R, S.
The triangles formed at the center by the radii and tangents are congruent. For instance, in \(\triangle AOP\) and \(\triangle AOS\):
\(OP = OS\) (radii), \(AP = AS\) (tangents from A), \(OA = OA\) (common).
Thus, \(\triangle AOP \cong \triangle AOS\), which gives \(\angle 1 = \angle 8\).
Similarly, \(\angle 2 = \angle 3\), \(\angle 4 = \angle 5\), and \(\angle 6 = \angle 7\).
The sum of all angles around point O is \(360^\circ\):
\((\angle 1 + \angle 2) + (\angle 3 + \angle 4) + (\angle 5 + \angle 6) + (\angle 7 + \angle 8) = 360^\circ\)
Since \(\angle 1 = \angle 8, \angle 2 = \angle 3\), etc., grouping the opposite pairs gives:
\(2(\angle 1 + \angle 2 + \angle 5 + \angle 6) = 360^\circ \implies (\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^\circ\)
Thus, opposite sides subtend supplementary angles at the center.
In simple words: Use congruence of triangles formed by joining the center to points of contact to show opposite angle sums equal \(180^\circ\).

Exam Tip: This is an important standard theorem. Practice drawing the line segments from the center to all vertices and contact points.

Constructions :

Question 1. Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{3}{5}\) of the corresponding sides of triangle ABC.
Answer: Steps of Construction:
1. Draw any triangle ABC with given dimensions.
2. Draw a ray AX making an acute angle with AB on the side opposite to vertex C.
3. Locate 5 (the greater of 3 and 5 in \(\frac{3}{5}\)) points \(A_1, A_2, A_3, A_4, A_5\) on AX such that \(AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5\).
4. Join \(A_5\) to B and draw a line through \(A_3\) parallel to \(A_5B\) intersecting AB at \(B'\).
5. Draw a line through \(B'\) parallel to BC intersecting AC at \(C'\).
6. \(\triangle AB'C'\) is the required triangle.
In simple words: Draw a ray, divide it into equal parts according to the denominator, join the last point to the vertex, and draw parallel lines.

Exam Tip: Ensure the parallel lines are drawn accurately using a set-square or compass method.

Question 2. Draw a pair of tangents to a circle of radius 5cm which are inclined to each other at an angle of \(60^\circ\).
Answer: Steps of Construction:
1. Draw a circle of radius \(r = 5\) cm with center O.
2. Draw a radius OA.
3. Draw another radius OB such that \(\angle AOB = 180^\circ - 60^\circ = 120^\circ\).
4. Draw perpendiculars to radii OA and OB at points A and B respectively.
5. These perpendicular tangents intersect at an external point P. \(\angle APB\) will be \(60^\circ\).
In simple words: The angle between radii at the center is supplementary to the angle between the tangents. Draw radii at \(120^\circ\) and construct perpendiculars.

Exam Tip: Always calculate the angle between radii first (\(180^\circ - \text{given angle}\)) before drawing.

Question 3. AB is a line segment of length 10cm. Locate a point C on AB such that \(AC = \frac{1}{3} CB\).
Answer: Steps of Construction:
1. Draw a line segment \(AB = 10\) cm.
2. Draw a ray AX making an acute angle with AB.
3. Locate \(1 + 3 = 4\) points \(A_1, A_2, A_3, A_4\) on ray AX at equal intervals.
4. Join \(A_4\) to B.
5. Through \(A_1\) (since \(AC:CB = 1:3\)), draw a line parallel to \(A_4B\) intersecting AB at point C.
6. Point C divides AB in the ratio \(1:3\), such that \(AC = \frac{1}{3}CB\).
In simple words: Divide the line segment into \(1 + 3 = 4\) equal parts using a construction ray, and mark the required division point.

Exam Tip: The total number of parts on the ray is the sum of the ratio terms (\(1 + 3 = 4\)).

Area related to Circles :

Question 1. Find the area of a quadrant of circle whose circumference is 22cm.
Answer: Given circumference \(2\pi r = 22\) cm.
\(2 \times \frac{22}{7} \times r = 22 \implies r = \frac{22 \times 7}{2 \times 22} = \frac{7}{2}\) cm.
Area of quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2\)
\(= \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{11 \times 7}{4 \times 2} = \frac{77}{8} = 9.625\) cm\(^2\).
In simple words: Find the radius from the circumference and use the quadrant area formula \(\frac{1}{4}\pi r^2\).

Exam Tip: A quadrant subtends an angle of \(90^\circ\) at the center, hence the factor \(\frac{90^\circ}{360^\circ} = \frac{1}{4}\).

Question 2. The race track is in the form of a ring whose inner circumference is 352m and outer circumference is 396m. Fin the width of the track.
Answer: Let inner radius be \(r\) and outer radius be \(R\).
Inner circumference \(2\pi r = 352\) cm
\(2 \times \frac{22}{7} \times r = 352 \implies r = \frac{352 \times 7}{44} = 8 \times 7 = 56\) m.
Outer circumference \(2\pi R = 396\) m
\(2 \times \frac{22}{7} \times R = 396 \implies R = \frac{396 \times 7}{44} = 9 \times 7 = 63\) m.
Width of the track = \(R - r = 63 - 56 = 7\) m.
In simple words: Width of a ring track is the difference between the outer radius and the inner radius.

Exam Tip: Calculate inner and outer radii separately before finding their difference.

Question 3. A wire when bent in the form of a square encloses an area of \(1.96\text{sq.cm}\). If the same wire is bent in the form of a circle, find the area of the circle.
Answer: Area of square = \(1.96\) cm\(^2\).
Side of square \(s = \sqrt{1.96} = 1.4\) cm.
Length of wire (perimeter of square) = \(4 \times 1.4 = 5.6\) cm.
When bent into a circle, circumference of circle = length of wire = \(5.6\) cm.
\(2\pi r = 5.6\)
\(2 \times \frac{22}{7} \times r = 5.6 \implies r = \frac{5.6 \times 7}{44} = \frac{39.2}{44} = \frac{49}{55}\) cm.
Area of circle = \(\pi r^2 = \frac{22}{7} \times \left(\frac{49}{55}\right)^2 = \frac{22}{7} \times \frac{2401}{3025} = \frac{2 \times 343}{275} = \frac{686}{275} \approx 2.495\) cm\(^2\).
In simple words: The perimeter of the square equals the circumference of the circle. Find the length of wire first.

Exam Tip: Perimeter remains constant when reshaping a wire from one geometric figure to another.

Question 4. A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60cm, calculate the speed per hour which the boy is cycling.
Answer: Diameter of wheel \(d = 60\) cm, so radius \(r = 30\) cm.
Distance covered in 1 revolution = circumference \(= 2\pi r = 2 \times \frac{22}{7} \times 30 = \frac{1320}{7}\) cm.
Distance covered in 1 minute = \(140 \times \frac{1320}{7} = 20 \times 1320 = 26400\) cm.
Distance covered in 1 hour = \(26400 \times 60 = 1584000\) cm/hr.
Convert cm to km: \(1\) km \(= 100,000\) cm.
Speed = \(\frac{1584000000}{100000}\) ... wait, \(1584000 \text{ cm} = \frac{1584000}{100000} = 15.84\) km/hr.
In simple words: Multiply distance per revolution by revolutions per minute and convert units to kilometers per hour.

Exam Tip: Pay careful attention to unit conversions from centimeters per minute to kilometers per hour.

Question 5. A chord AB of a circle of radius 10cm makes a right angle at the centre of the circle. Find the area of the major and minor segments.
Answer: Radius \(r = 10\) cm, central angle \(\theta = 90^\circ\).
Area of minor sector = \(\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5\) cm\(^2\).
Area of \(\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50\) cm\(^2\).
Area of minor segment = Area of minor sector - Area of \(\triangle OAB\) = \(78.5 - 50 = 28.5\) cm\(^2\).
Area of circle = \(\pi r^2 = 3.14 \times 100 = 314\) cm\(^2\).
Area of major segment = Area of circle - Area of minor segment = \(314 - 28.5 = 285.5\) cm\(^2\).
In simple words: Minor segment area is sector area minus triangle area. Major segment area is total circle area minus minor segment area.

Exam Tip: Use \(\pi = 3.14\) as given or implied by clean decimals in right-angle sectors.

Surface area and Volumes:

Question 1. The radii of the ends of a bucket 45 cm high are 28cm and 7cm, determine its capacity and surface area.
Answer: Height \(h = 45\) cm, radii \(R = 28\) cm and \(r = 7\) cm.
Capacity (Volume of frustum) = \(\frac{1}{3} \pi h (R^2 + r^2 + Rr)\)
\(= \frac{1}{3} \times \frac{22}{7} \times 45 \times [28^2 + 7^2 + (28 \times 7)]\)
\(= \frac{22 \times 15}{7} \times [784 + 49 + 196] = \frac{330}{7} \times 1029 = 330 \times 147 = 48510\) cm\(^3\) (or \(48.51\) litres).
Slant height \(l = \sqrt{h^2 + (R - r)^2} = \sqrt{45^2 + (28 - 7)^2} = \sqrt{2025 + 21^2} = \sqrt{2025 + 441} = \sqrt{2466} \approx 49.66\) cm.
Total Surface Area of bucket = \(\pi l(R + r) + \pi r^2\)
\(= \frac{22}{7} \times 49.66 \times (28 + 7) + \frac{22}{7} \times 7^2\)
\(= \frac{22}{7} \times 49.66 \times 35 + 154 = 22 \times 49.66 \times 5 + 154 = 5462.6 + 154 = 5616.6\) cm\(^2\).
In simple words: Treat the bucket as a frustum of a cone and apply the standard frustum volume and surface area formulas.

Exam Tip: Remember that an open bucket has only one circular base (the bottom one) added to its curved surface area.

Question 2. From a solid cylinder whose height is 2.4cm and diameter 1.4cm, a conical cavity of the same height and the same diameter is hollowed out. Find the total surface area of the remaining solid.
Answer: Height \(h = 2.4\) cm, radius \(r = \frac{1.4}{2} = 0.7\) cm.
Slant height of cone \(l = \sqrt{r^2 + h^2} = \sqrt{0.7^2 + 2.4^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\) cm.
Total Surface Area of remaining solid = CSA of cylinder + Area of circular base + CSA of conical cavity.
\(= 2\pi rh + \pi r^2 + \pi rl\)
\(= \pi r (2h + r + l)\)
\(= \frac{22}{7} \times 0.7 \times [2(2.4) + 0.7 + 2.5]\)
\(= 2.2 \times [4.8 + 3.2] = 2.2 \times 8 = 17.6\) cm\(^2\).
In simple words: The total surface area includes the outer cylindrical surface, the bottom circular base, and the inner conical cavity surface.

Exam Tip: Hollowing out a shape increases the total exposed surface area because of the inner cavity.

Question 3. A cubical block of side 21cm is surmounted by a hemisphere. What is the greatest radius the hemisphere can have? Find the surface area of the solid.
Answer: Side of cube \(a = 21\) cm.
Greatest radius the hemisphere can have \(r = \frac{\text{side of cube}}{2} = \frac{21}{2} = 10.5\) cm.
Total Surface Area of solid = Total surface area of cube - base area of hemisphere + CSA of hemisphere.
\(= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2\)
\(= 6(21)^2 + \frac{22}{7} \times (10.5)^2\)
\(= 6(441) + \frac{22}{7} \times 110.25\)
\(= 2646 + 22 \times 15.75 = 2646 + 346.5 = 2992.5\) cm\(^2\).
In simple words: Add the curved surface of the hemisphere and subtract its circular base area which is glued to the cube's top face.

Exam Tip: The base area of the hemisphere overlaps with the top face of the cube, so it must be subtracted once.

Question 4. Three cubes whose edges are in the ratio 3:4:5 are melted to form a single cube whose diagonal is \(15\sqrt{3}\) cm. Find the edges of three cubes.
Answer: Let the edges of the three cubes be \(3x, 4x, 5x\).
Sum of volumes = Volume of the new single cube.
\((3x)^3 + (4x)^3 + (5x)^3 = V_{\text{new}}\)
\(27x^3 + 64x^3 + 125x^3 = 216x^3\).
Let the edge of the new cube be \(a\). Its volume is \(a^3 = 216x^3\), so \(a = 6x\).
Diagonal of the new cube = \(a\sqrt{3} = 6x\sqrt{3}\).
Given diagonal \(= 15\sqrt{3}\).
\(6x\sqrt{3} = 15\sqrt{3} \implies 6x = 15 \implies x = \frac{15}{6} = 2.5\).
Edges of the three cubes are:
\(3x = 3 \times 2.5 = 7.5\) cm,
\(4x = 4 \times 2.5 = 10\) cm,
\(5x = 5 \times 2.5 = 12.5\) cm.
In simple words: Volume is conserved when solids are melted. Use the diagonal formula of a cube to find the scaling factor \(x\).

Exam Tip: Diagonal of a cube with side \(a\) is \(a\sqrt{3}\).

Question 5. A sphere of maximum volume is cut off from a solid hemisphere of radius 6cm. What is the volume of the cut-off sphere?
Answer: Radius of the solid hemisphere \(R = 6\) cm.
A sphere of maximum volume cut from a solid hemisphere will have its diameter equal to the radius of the hemisphere.
Diameter of sphere \(d = R = 6\) cm, so radius of sphere \(r = 3\) cm.
Volume of the cut-off sphere = \(\frac{4}{3} \pi r^3\)
\(= \frac{4}{3} \times \frac{22}{7} \times 3^3 = \frac{4}{3} \times \frac{22}{7} \times 27 = 4 \times 22 \times \frac{9}{7} = \frac{792}{7} \approx 113.14\) cm\(^3\).
In simple words: The maximum sphere carved out of a hemisphere has a radius equal to half the hemisphere's radius.

Exam Tip: Carefully distinguish between the radius of the hemisphere and the radius of the inscribed sphere.

Question 6. If the edge of a cube is increased by 50%, find the percentage increase in its total S.A. and Volume.
Answer: Let initial edge of cube be \(x\).
Initial Total Surface Area \(TSA_1 = 6x^2\).
Initial Volume \(V_1 = x^3\).
New edge after 50% increase = \(x + 0.50x = 1.5x\).
New Total Surface Area \(TSA_2 = 6(1.5x)^2 = 6(2.25x^2) = 13.5x^2\).
Percentage increase in TSA = \(\frac{TSA_2 - TSA_1}{TSA_1} \times 100 = \frac{13.5x^2 - 6x^2}{6x^2} \times 100 = \frac{7.5}{6} \times 100 = 125\%\).
New Volume \(V_2 = (1.5x)^3 = 3.375x^3\).
Percentage increase in Volume = \(\frac{V_2 - V_1}{V_1} \times 100 = \frac{3.375x^3 - x^3}{x^3} \times 100 = 2.375 \times 100 = 237.5\%\).
In simple words: Substitute the new scaled side into the surface area and volume formulas to find the percentage change.

Exam Tip: Surface area scales with the square of the scale factor (\(1.5^2 = 2.25\)) and volume scales with the cube (\(1.5^3 = 3.375\)).