This collection of ICSE Class 10 Maths Short Answer Questions provides essential practice for the 3-mark problems you will face in your board exams. Taken from the Class 10 – ICSE Maths Solved Competency Focused Questions textbook, this chapter, titled ‘Short Answer Questions 1’, is designed to test your ability to apply concepts from various topics like Commercial Mathematics, Algebra, and Geometry. These are not simple recall questions; they require you to think critically and present a logical, step-by-step solution. Mastering these questions is key to building a strong foundation and boosting your overall score, as they form a significant part of Section A in the final paper.
You’ve likely landed on this page because you’re looking for a reliable method to solve a specific 3-mark question or want to verify your own answer. This chapter contains 28 competency-focused questions that mimic the style and difficulty of the actual ICSE board exam. We provide detailed, step-by-step solutions for every single question, ensuring the method used is exactly what the ICSE board expects. Here you will find clear, accurate, and easy-to-follow guides to help you master these important problems.
Short Answer Questions 1 (3 Marks Each)
Question 50
A shopkeeper marked a pressure cooker at ₹ 1800. The rate of GST on pressure cooker is 12%. The customer has only ₹ 1792 with him and he requests the shopkeeper to reduce the price so that he can buy the cooker in ₹ 1792. What percent discount must the shopkeeper give ?
The marked price of the pressure cooker is ₹ 1800, and the final price after GST is ₹ 1792. The GST rate on pressure cookers is 12%.
Let’s call the discounted price (before GST) ₹ x. Since GST is 12%, the selling price becomes x plus 12% of x, which equals ₹ 1792.
\begin{aligned}\therefore x + \dfrac{12}{100} \times x = 1792 \\ \Rightarrow x + 0.12x = 1792 \\ \Rightarrow 1.12x = 1792 \\ \Rightarrow x = \dfrac{1792}{1.12} = \text{₹ 1600}.\end{aligned}So the shopkeeper first reduces the marked price to ₹ 1600.
Discount given = ₹ 1800 – ₹ 1600 = ₹ 200.
Now, let’s find the discount percentage.
Discount % = \dfrac{\text{Discount}}{\text{M.P.}} \times 100
\begin{aligned}= \dfrac{200}{1800} \times 100 \\ = \dfrac{1}{9} \times 100 \\ = 11.11 \%.\end{aligned}Hence, the discount offered by the shopkeeper = 11.11 %.
Question 51
A man opened a recurring deposit account in a branch of PNB. The man deposits certain amount of money per month such that after 2 years, the interest accumulated is equal to his monthly deposits. Find the rate of interest per annum that the bank was paying for the recurring deposit account.
We are told that a man opens a recurring deposit for 2 years (24 months), deposits ₹ x per month, and earns ₹ x as interest. We need to find the rate of interest.
For recurring deposits, the interest formula is:
I = \dfrac{P \times n(n + 1)}{2 \times 12} \times \dfrac{r}{100}
Substituting what we know (I = ₹ x, P = ₹ x, n = 24 months):
\begin{aligned}\Rightarrow x = \dfrac{x \times 24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\ \Rightarrow x = \dfrac{x \times 24 \times 25}{24} \times \dfrac{r}{100} \\ \Rightarrow 1 = \dfrac{r}{4} \\ \Rightarrow r = 4\%.\end{aligned}Hence, the rate of interest = 4%.
Question 52
Akshay buys 350 shares of ₹ 50 par value of a company. The dividend declared by the company is 14%. If his return percent from the shares is 10%, find the market value of each share.
We are given that the dividend rate is 14% on shares with a par value of ₹ 50.
First, let’s find the dividend per share.
Dividend on ₹ 50 = \dfrac{14}{100} \times 50 = ₹ 7.
We’re also told the return from the shares is 10%. That means on an investment of ₹ 100, the return is ₹ 10.
So ₹ 7 dividend corresponds to an investment of \dfrac{100}{10} \times 7 = ₹ 70.
Hence, the market value of each share = ₹ 70.
Question 53
Solve the following inequation and answer the questions given below.
\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x(a) Write the maximum and minimum values of x for x ∈ R.
(b) What will be the change in maximum and minimum values of x if x ∈ W.
(a) We need to solve the compound inequation \dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x.
Let’s break it into two parts and solve each one.
Solving the left half:
\begin{aligned}\Rightarrow \dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \\ \Rightarrow x - \dfrac{1}{2} \le 2x + \dfrac{1}{2} \\ \Rightarrow 2x - x \ge -\dfrac{1}{2} - \dfrac{1}{2} \\ \Rightarrow x \ge -1 \text{ ...........(1)}\end{aligned}Solving the right half:
\begin{aligned}\Rightarrow 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x \\ \Rightarrow 2x + \dfrac{1}{2} \le \dfrac{11}{2} + x \\ \Rightarrow 2x - x \le \dfrac{11}{2} - \dfrac{1}{2} \\ \Rightarrow x \le \dfrac{10}{2} \\ \Rightarrow x \le 5 \text{ ...........(2)}\end{aligned}From (1) and (2), combining both parts:
-1 ≤ x ≤ 5 and x ∈ R.
Hence, the minimum value of x = -1 and the maximum value = 5.
(b) If x is a whole number (x ∈ W), the smallest whole number in this range is 0 and the largest is 5.
Hence, the minimum value = 0 and the maximum value = 5 when x ∈ W.
Question 54
Solve for x, if \dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2}, x ≠ 0
We need to solve for x in the equation \dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2}, where x ≠ 0.
A useful trick here is to substitute \dfrac{1}{x} = t. This simplifies the equation.
\begin{aligned}\Rightarrow 5t + 4\sqrt{3} = 2\sqrt{3}t^2 \\ \Rightarrow 2\sqrt{3}t^2 - 5t - 4\sqrt{3} = 0\end{aligned}Now, let’s factor this quadratic. We’ll split the middle term.
\begin{aligned}\Rightarrow 2\sqrt{3}t^2 - 8t + 3t - 4\sqrt{3}= 0 \\ \Rightarrow 2t(\sqrt{3}t - 4) + \sqrt{3}(\sqrt{3}t - 4) = 0 \\ \Rightarrow (2t + \sqrt{3})(\sqrt{3}t - 4) = 0\end{aligned}Setting each factor to zero:
\begin{aligned}\Rightarrow 2t + \sqrt{3} = 0 \text{ or } \sqrt{3}t - 4 = 0 \\ \Rightarrow t = -\dfrac{\sqrt{3}}{2} \text{ or } t = \dfrac{4}{\sqrt{3}}\end{aligned}Remembering that t = \dfrac{1}{x}:
\begin{aligned}\Rightarrow \dfrac{1}{x} = -\dfrac{\sqrt{3}}{2} \text{ or } \dfrac{1}{x} = \dfrac{4}{\sqrt{3}} \\ \Rightarrow x = -\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}.\end{aligned}Hence, x = -\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}.
Question 55
The marked price of a toy is same as the percentage of GST that is charged. The price of the toy is ₹ 24 including GST. Taking the marked price as x, form an equation and solve it to find x.
We are told that the marked price of a toy is ₹ x, the GST rate is x%, and the price including GST is ₹ 24.
Let’s set up the equation:
\begin{aligned}\therefore 24 = x + \dfrac{x}{100} \times x \\ \Rightarrow 24 = x + \dfrac{x^2}{100} \\ \Rightarrow 24 = \dfrac{100x + x^2}{100} \\ \Rightarrow 2400 = x^2 + 100x \\ \Rightarrow x^2 + 100x - 2400 = 0\end{aligned}Now, let’s factor this quadratic.
\begin{aligned}\Rightarrow x^2 + 120x - 20x - 2400 = 0 \\ \Rightarrow x(x + 120) - 20(x + 120) = 0 \\ \Rightarrow (x - 20)(x + 120) = 0 \\ \Rightarrow x = 20 \text{ or } x = -120\end{aligned}Since the marked price cannot be negative, x = 20.
Hence, the marked price of the toy = ₹ 20 and the GST rate = 20%.
Question 56
The mean proportion between two numbers is 6 and their third proportion is 48. Find two numbers.
Let the two numbers be a and b.
We’re told that 6 is the mean proportional between a and b.
\begin{aligned}\therefore \dfrac{a}{6} = \dfrac{6}{b} \\ \Rightarrow ab = 36 \\ \Rightarrow a = \dfrac{36}{b}\text{ .........(1)}\end{aligned}We’re also told that 48 is the third proportional of a and b.
\begin{aligned}\therefore \dfrac{a}{b} = \dfrac{b}{48} \\ \Rightarrow b^2 = 48a\end{aligned}Now, substitute a from equation (1):
\begin{aligned}\Rightarrow b^2 = 48 \times \dfrac{36}{b} \\ \Rightarrow b^3 = 1728 \\ \Rightarrow b = \sqrt[3]{1728} \\ \Rightarrow b = 12.\end{aligned}Putting b = 12 back into equation (1):
\Rightarrow a = \dfrac{36}{12} = 3.
Hence, the two numbers are 3 and 12.
Question 57
Pamela factorized the following polynomial :
2x^3 + 3x^2 – 3x – 2
She found the result as (x + 2)(x – 1)(x – 2). Using remainder and factor theorem, verify whether her result is correct. If incorrect, give the correct result.
Let f(x) = 2x^3 + 3x^2 – 3x – 2. We’ll check which of the given binomials divide f(x) by using the factor theorem.
Testing (x + 2):
If x + 2 = 0, then x = -2.
f(-2) = 2(-2)^3 + 3(-2)^2 – 3(-2) – 2
= 2(-8) + 3(4) + 6 – 2
= -16 + 12 + 6 – 2
= -18 + 18
= 0. ✓
So (x + 2) is a factor.
Testing (x – 1):
If x – 1 = 0, then x = 1.
f(1) = 2(1)^3 + 3(1)^2 – 3(1) – 2
= 2 + 3 – 3 – 2
= 0. ✓
So (x – 1) is also a factor.
Testing (x – 2):
If x – 2 = 0, then x = 2.
f(2) = 2(2)^3 + 3(2)^2 – 3(2) – 2
= 16 + 12 – 6 – 2
= 20 ≠ 0. ✗
So (x – 2) is not a factor.
Since both (x + 2) and (x – 1) are factors, their product (x + 2)(x – 1) = x^2 + x – 2 must also divide f(x). Let’s perform the long division:
\begin{array}{l} \phantom{x^2 + x - 2)}{\quad 2x + 1} \\ x^2 + x - 2\overline{\smash{\big)}\quad 2x^3 + 3x^2 - 3x - 2} \\ \phantom{x^2 + x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}2x^2 \underset{+}{-}4x} \\ \phantom{{x^2 + x - 2}2x^3 + 20x^3}x^2 + x - 2 \\ \phantom{{x^2 + x - 2)}2x^3 + x^3}\underline{\underset{-}{+}x^2 \underset{-}{+} x \underset{+}{-} 2} \\ \phantom{{x^2 + x - 2)}{x^3-2x^{2}(31)}{x}}\times \end{array}So f(x) = (x^2 + x – 2)(2x + 1) = (x + 2)(x – 1)(2x + 1).
Hence, 2x^3 + 3x^2 – 3x – 2 = (x + 2)(x – 1)(2x + 1).
Question 58
A = \begin{bmatrix} -6 & 0 \\ 4 & 2 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 0 \\ 1 & 3 \end{bmatrix}.
Find matrix M, if M = \dfrac{1}{2}A - 2B + 5l, where l is the identity matrix.
We’re asked to evaluate the matrix expression M = \dfrac{1}{2}A - 2B + 5I, where I is the identity matrix.
\Rightarrow M = \dfrac{1}{2}A - 2B + 5I \[1em] = \dfrac{1}{2}\begin{bmatrix} -6 & 0 \\ 4 & 2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 1 & 3 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} -3 & 0 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 2 & 6 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \\ = \begin{bmatrix} -3 - 2 + 5 & 0 - 0 + 0 \\ 2 - 2 + 0 & 1 - 6 + 5 \end{bmatrix} \\ = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.Hence, M =\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.
Question 59
(a) Write the nth term (T~n) of an Arithmetic Progression (A.P.) consisting of all whole numbers which are divisible by 3 and 7.
(b) How many of these are two-digit numbers? Write them.
(c) Find the sum of first 10 terms of this A.P.
The sequence is 21, 42, 63, ………
(a) Let’s identify the pattern. The first term a = 21, and the common difference d = 42 – 21 = 21.
Using the formula for the n^th term of an A.P.:
T_n = a + (n – 1)d
= 21 + (n – 1)21
= 21 + 21n – 21
= 21n.
Hence, the n^th term = 21n.
(b) The two-digit terms in this A.P. are 21, 42, 63, and 84. The next term would be 105, which has three digits.
Hence, there are four two-digit numbers in the A.P.: 21, 42, 63, and 84.
(c) To find the sum of the first 10 terms:
Sum of A.P. = \dfrac{n}{2}(a + a_n)
\begin{aligned}\text{Sum of first 10 terms} = \dfrac{10}{2}(a + a_{10}) \\ = 5[a + a + (n - 1)d] \\ = 5[2a + (n - 1)d] \\ = 5[2 \times 21 + (10 - 1) \times 21] \\ = 5[42 + 9 \times 21] \\ = 5[42 + 189] \\ = 5 \times 231 \\ = 1155.\end{aligned}Hence, sum of first 10 terms of the A.P. = 1155.
Question 60
Write the first five terms of the sequence given by (\sqrt{3})^n, n ∈ N.
(a) Is the sequence an A.P. or G.P.?
(b) If the sum of its first ten terms is p(3 + \sqrt{3}), find the value of p.
The sequence is defined by (\sqrt{3})^n. Let’s write out the first few terms:
(\sqrt{3})^1, (\sqrt{3})^2, (\sqrt{3})^3, (\sqrt{3})^4, (\sqrt{3})^5, …….
\Rightarrow \sqrt{3}, 3, 3\sqrt{3}, 9, 9\sqrt{3}(a) To check if it’s a G.P., let’s find the ratio between consecutive terms:
Ratio = \dfrac{3}{\sqrt{3}} = \sqrt{3}.
Hence, the sequence is a G.P. with common ratio = \sqrt{3}.
(b) We’re given that the sum of the first 10 terms equals p(3 + √3). Let’s use the G.P. sum formula:
Sum of G.P. (S) = \dfrac{a(r^n - 1)}{(r - 1)}
\begin{aligned}\Rightarrow S_{10} = \dfrac{\sqrt{3}[(\sqrt{3})^{10} - 1]}{\sqrt{3} - 1} \\ \Rightarrow p(3 + \sqrt{3})= \dfrac{\sqrt{3}(243 - 1)}{\sqrt{3} - 1} \\ \Rightarrow p = \dfrac{\sqrt{3}(243 - 1)}{(\sqrt{3} - 1)(3 + \sqrt{3})} \\ \Rightarrow p = \dfrac{\sqrt{3} \times 242}{(\sqrt{3} - 1)\sqrt{3}(\sqrt{3} + 1)} \\ \Rightarrow p = \dfrac{242}{3 - 1} \\ \Rightarrow p = \dfrac{242}{2} = 121.\end{aligned}Hence, p = 121.
Question 61
ABC is a triangle as shown in the figure below.
(a) Write down the coordinates of A, B and C on reflecting through the origin.
(b) Write down the coordinates of the point/s which remain invariant on reflecting the triangle ABC on the x-axis and y-axis respectively.
(a) When we reflect points in the origin, both coordinates change sign. If A = (4, 5), B = (0, 3), and C = (3, 0), then after reflection in the origin:
A’ = (-4, -5), B’ = (0, -3), and C’ = (-3, 0).
Hence, on reflecting A, B and C in the origin, the coordinates are (-4, -5), (0, -3) and (-3, 0) respectively.
(b) A point remains invariant (unchanged) when reflected in an axis if it lies on that axis.
Point C(3, 0) lies on the x-axis, so it stays the same when reflected in the x-axis.
Point B(0, 3) lies on the y-axis, so it stays the same when reflected in the y-axis.
Hence, the invariant point for reflection in the x-axis is C, and for the y-axis is B.
Question 62
Determine the ratio in which the line y = 2 + 3x divides the line segment AB joining the points A(-3, 9) and B(4, 2).
We need to find where the line y = 2 + 3x cuts the line segment AB, with A(-3, 9) and B(4, 2).
First, let’s find the equation of line AB using the two-point form:
\Rightarrow y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1) \begin{aligned}\Rightarrow y - 9 = \dfrac{2 - 9}{4 - (-3)}[x - (-3)] \\ \Rightarrow y - 9 = \dfrac{-7}{7}[x + 3] \\ \Rightarrow y - 9 = -1[x + 3] \\ \Rightarrow y - 9 = -x - 3 \\ \Rightarrow y + x = -3 + 9 \\ \Rightarrow x + y = 6.\end{aligned}Now, let’s solve the two equations simultaneously:
⇒ x + y = 6 …….(1)
⇒ y = 2 + 3x …….(2)
Substituting (2) into (1):
⇒ x + (2 + 3x) = 6
⇒ 4x + 2 = 6
⇒ 4x = 4
⇒ x = 1.
Putting x = 1 into equation (2):
⇒ y = 2 + 3(1) = 5.
So the point of intersection is (1, 5).
Now, let’s find the ratio in which (1, 5) divides AB. Using the section formula:
(x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)
Let the ratio be k : 1.
\begin{aligned}\Rightarrow (1, 5) = \Big(\dfrac{k \times 4 + 1 \times -3}{k + 1}, \dfrac{k \times 2 + 1 \times 9}{k + 1}\Big) \\ \Rightarrow 1 = \dfrac{4k - 3}{k + 1} \text{ and } 5 = \dfrac{2k + 9}{k + 1} \\ \Rightarrow k + 1 = 4k - 3 \text{ or } 5k + 5 = 2k + 9 \\ \Rightarrow 3k = 4 \text{ or } 3k = 4 \\ \Rightarrow k = \dfrac{4}{3} \\ \Rightarrow k : 1 = 4 : 3.\end{aligned}Hence, the line y = 2 + 3x divides the line segment AB in the ratio 4 : 3.
Question 63
Square ABCD lies in the third quadrant of a XY plane such that its vertex A is at (-3, -1) and the diagonal DB produced is equally inclined to both the axes. The diagonals AC and BD meets at P(-2, -2). Find the :
(a) Slope of BD
(b) equation of AC
(a) Since diagonal BD is equally inclined to both axes, it makes a 45° angle with each.
Slope of BD = tan 45° = 1.
Hence, slope of BD = 1.
(b) In a square, diagonals are perpendicular to each other. The product of slopes of perpendicular lines is -1.
Slope of BD × Slope of AC = -1
⇒ 1 × Slope of AC = -1
⇒ Slope of AC = -1.
The diagonals meet at P(-2, -2). Using the point-slope form:
⇒ y – y_1 = m(x – x_1)
⇒ y – (-2) = -1[x – (-2)]
⇒ y + 2 = -1[x + 2]
⇒ y + 2 = -x – 2
⇒ x + y + 4 = 0.
Hence, equation of AC is x + y + 4 = 0.
Question 64
ABCD is a rectangle where side BC is twice side AB. If △ACQ ~ △BAP, find area of △BAP : area of △ACQ.
We’re given a rectangle ABCD where BC = 2AB. We need to find the ratio of areas of triangles BAP and ACQ.
In right triangle ABC, by Pythagoras theorem:
⇒ AC^2 = AB^2 + BC^2
⇒ AC^2 = AB^2 + (2AB)^2
⇒ AC^2 = AB^2 + 4AB^2
⇒ AC^2 = 5AB^2
⇒ AC = \sqrt{5} AB.
Now, triangles BAP and ACQ are similar (they share corresponding angles in the rectangle). The ratio of areas of similar triangles equals the square of the ratio of corresponding sides.
\begin{aligned}\therefore \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{AC^2} \\ \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{(\sqrt{5}BA)^2} \\ \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{5BA^2} \\ \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{1}{5} \\\end{aligned}Hence, Area of △BAP : Area of △ACQ = 1 : 5.
Question 65
Given a triangle ABC, and D is a point on BC such that BD = 4 cm and DC = x cm. If ∠BAD = ∠C and AB = 8 cm, then,
(a) prove that triangle ABD is similar to triangle CBA.
(b) find the value of ‘x’.
(a) In △ABD and △CBA:
⇒ ∠ABD = ∠CBA (Common angle)
⇒ ∠BAD = ∠ACB (Given)
∴ △ABD ~ △CBA (By AA similarity criterion)
Hence, proved that △ABD ~ △CBA.
(b) Since the triangles are similar, their corresponding sides are proportional.
\begin{aligned}\therefore \dfrac{AB}{BC} = \dfrac{BD}{BA} \\ \Rightarrow \dfrac{8}{x + 4} = \dfrac{4}{8} \\ \Rightarrow 4(x + 4) = 8 \times 8 \\ \Rightarrow 4x + 16 = 64 \\ \Rightarrow 4x = 48 \\ \Rightarrow x = 12 \text{ cm}.\end{aligned}Hence, x = 12 cm.
Question 66
In the extract of Survey of India map G43S7, prepared on a scale of 2 cm to 1 km, a child finds the length of the cart track between two settlements is 7.6 cm. Find :
(a) the actual length of the cart track on the ground.
(b) actual area of a grid square, if each has an area of 4 cm^2.
The scale is 2 cm on the map to 1 km on the ground. Converting to the same units: 2 cm to 100000 cm, so the scale factor k = \dfrac{2}{100000} = \dfrac{1}{50000}.
(a) Length on map = 7.6 cm. Using the scale factor:
7.6 = \dfrac{1}{50000} × Actual length
Actual length = 7.6 × 50000 = 380000 cm = \dfrac{380000}{100000} = 3.8 km.
Hence, length of actual cart track on ground = 3.8 km
(b) For areas, the scale factor is squared.
Area on map = k² × Actual area
⇒ 4 = \Big(\dfrac{1}{50000}\Big)^2 × Actual area
⇒ 4 = \Big(\dfrac{1}{25 \times 10^8}\Big) × Actual area
⇒ Actual area = 4 \times 25 \times 10^8
⇒ Actual area = 100 × 10^8 = 10^10 cm²
= \dfrac{10^{10}}{10^{10}} km² = 1 km².
Hence, actual area of a grid square = 1 km².
Question 67
Construct a triangle ABC such that AB = 7 cm, BC = 6 cm and CA = 5 cm.
(a) Draw the locus of the points such that
(i) it is equidistant from BC and BA.
(ii) it is equidistant from points A and B.
(b) Mark P where the loci (i) and (ii) meet, measure and write length of PA.
Remember two important locus concepts:
– Points equidistant from two sides of an angle lie on the angle bisector.
– Points equidistant from two points lie on the perpendicular bisector of the line joining them.
Steps of construction:
- Draw a line segment AB = 7 cm.
- With A and B as centers and radii 5 cm and 6 cm respectively, draw arcs intersecting at C.
- Join AC and BC to complete the triangle.
- Draw BD, the angle bisector of ∠B.
- Draw XY, the perpendicular bisector of AB.
- Mark point P, the intersection of XY and BD.
- Measure AP.
Hence, AP = 3.8 cm.
Question 68
In the given figure O is the centre of the circle. ABCD is a quadrilateral where sides AB, BC, CD and DA touch the circle at E, F, G and H respectively. If AB = 15 cm, BC = 18 cm and AD = 24 cm, find the length of CD.
We use a key property: tangents drawn from an external point to a circle are equal in length.
Let’s label the equal tangent segments:
⇒ AE = AH = x (from point A)
⇒ BE = BF = y (from point B)
⇒ CF = CG = z (from point C)
⇒ DG = DH = k (from point D)
From the given side lengths:
⇒ AB = 15, so x + y = 15 …….(1)
⇒ BC = 18, so y + z = 18 ……..(2)
⇒ AD = 24, so x + k = 24 ……….(3)
Adding equations (2) and (3):
⇒ y + z + x + k = 18 + 24
⇒ x + y + z + k = 42
⇒ 15 + z + k = 42 (using equation 1)
⇒ z + k = 27.
But CD = CG + DG = z + k = 27 cm.
Hence, length of CD = 27 cm.
Question 69
In the given diagram, ABCDEF is a regular hexagon inscribed in a circle with centre O. PQ is a tangent to the circle at D. Find the value of :
(a) ∠FAG
(b) ∠BCD
(c) ∠PDE
(a) For a regular hexagon with n = 6 sides, each interior angle is:
\begin{aligned}\dfrac{(n - 2) \times 180°}{n} \\ = \dfrac{(6 - 2) \times 180°}{6} \\ = \dfrac{4 \times 180°}{6} \\ = \dfrac{2}{3} \times 180° \\ = 120°.\end{aligned}So ∠FAB = 120°.
Since ∠FAG and ∠FAB form a linear pair:
∠FAB + ∠FAG = 180°
⇒ 120° + ∠FAG = 180°
⇒ ∠FAG = 60°.
Hence, ∠FAG = 60°.
(b) Each interior angle of a regular hexagon = 120°.
Hence, ∠BCD = 120°.
(c) In △DEF, DE = EF (sides of regular hexagon), so it’s an isosceles triangle.
⇒ ∠DEF = 120°
⇒ ∠EDF = ∠EFD = a (base angles of isosceles triangle)
By angle sum property: 120° + 2a = 180°
⇒ 2a = 60°
⇒ a = 30°
So ∠EFD = 30°.
By the alternate segment theorem, the angle between tangent PQ and chord DE equals the angle in the alternate segment.
∴ ∠PDE = ∠EFD = 30°.
Hence, ∠PDE = 30°.
Question 70
AB and CD intersect at the center O of the circle given in the above diagram. If ∠EBA = 33° and ∠EAC = 82°, find
(a) ∠BAE
(b) ∠BOC
(c) ∠ODB
(a) In △BAE, since ∠AEB is an angle in a semicircle, ∠AEB = 90°.
By angle sum property:
⇒ ∠AEB + ∠EBA + ∠BAE = 180°
⇒ 90° + 33° + ∠BAE = 180°
⇒ ∠BAE = 180° – 123° = 57°.
Hence, ∠BAE = 57°.
(b) From the figure, ∠EAC = ∠EAB + ∠BAC
⇒ 82° = 57° + ∠BAC
⇒ ∠BAC = 25°.
The angle subtended by arc BC at the centre is twice the angle subtended at the circumference.
⇒ ∠BOC = 2∠BAC = 2 × 25° = 50°.
Hence, ∠BOC = 50°.
(c) Since ∠BOC and ∠BOD form a linear pair:
∠BOC + ∠BOD = 180°
⇒ 50° + ∠BOD = 180°
⇒ ∠BOD = 130°.
In △BOD, by angle sum property:
⇒ ∠BOD + ∠ODB + ∠DBO = 180°
⇒ 130° + ∠ODB + 33° = 180°
⇒ ∠ODB = 180° – 163° = 17°.
Hence, ∠ODB = 17°.
Question 71
A famous sweet shop “Madanlal Sweets” sells tinned rasgullas. The tin container is cylindrical in shape with diameter 14 cm, height 16 cm, and it can hold 20 spherical rasgullas of diameter 6 cm and sweetened liquid such that the can is filled and then sealed. Find out how much sweetened liquid the can contains. Take π = 3.14.
We have a cylindrical container with 20 spherical rasgullas and some sweetened liquid.
Given:
- Diameter of container = 14 cm
- Radius (R) = \dfrac{\text{Diameter}}{2} = \dfrac{14}{2} = 7 cm
- Height of container (H) = 16 cm
- Diameter of rasgulla = 6 cm
- Radius of rasgulla (r) = \dfrac{\text{Diameter}}{2} = \dfrac{6}{2} = 3 cm
The key idea is: Volume of container = Volume of 20 rasgullas + Volume of liquid.
⇒ πR²H = 20 \times \dfrac{4}{3} πr³ + Volume of liquid
⇒ Volume of liquid = πR²H – \dfrac{80}{3} πr³
⇒ Volume of liquid = π(R²H – \dfrac{80}{3}r^3)
⇒ Volume of liquid = 3.14 × (7² × 16 – \dfrac{80}{3} \times 3^3)
⇒ Volume of liquid = 3.14 × (784 – 720)
⇒ Volume of liquid = 3.14 × 64
⇒ Volume of liquid = 200.96 cm³.
Hence, volume of sweetened liquid in the container = 200.96 cm³.
Question 72
The ratio of the radius and the height of a solid metallic right circular cylinder is 7 : 27. This is melted and made into a cone of diameter 14 cm and slant height 25 cm. Find the height of the :
(a) cone
(b) cylinder
A solid metallic cylinder is melted and recast into a cone.
Given:
- Diameter of cone = 14 cm, so radius r = 7 cm
- Slant height of cone l = 25 cm
- Ratio of radius to height of cylinder = 7 : 27
(a) First, let’s find the height of the cone using the relation l² = r² + h²:
⇒ 25² = 7² + h²
⇒ 625 = 49 + h²
⇒ h² = 576
⇒ h = 24 cm.
Hence, height of cone = 24 cm.
(b) Since the cylinder is melted to form the cone, their volumes are equal.
Let the cylinder’s radius = 7x and height = 27x.
Volume of cylinder = Volume of cone
⇒ πR²H = \dfrac{1}{3}πr²h
⇒ (7x)² × 27x = \dfrac{1}{3} \times 7^2 \times 24
⇒ 1323x³ = \dfrac{1176}{3}
⇒ 1323x³ = 392
⇒ x³ = \dfrac{392}{1323} = \dfrac{8}{27} = \Big(\dfrac{2}{3}\Big)^3
⇒ x = \dfrac{2}{3}
Height of cylinder = 27x = 27 \times \dfrac{2}{3} = 18 cm.
Hence, height of cylinder = 18 cm.
Question 73
An inclined plane AC is prepared with its base AB which is √3 times its vertical height BC. The length of the inclined plane is 15 m. Find:
(a) value of θ.
(b) length of its base AB, in nearest metre.
(a) From the figure, AB = \sqrt{3} × BC.
So BC = \dfrac{AB}{\sqrt{3}}.
In right triangle ABC:
⇒ tan θ = \dfrac{BC}{AB} = \dfrac{BC}{\sqrt{3}BC} = \dfrac{1}{\sqrt{3}} = tan 30°
⇒ θ = 30°.
Hence, θ = 30°.
(b) Using Pythagoras theorem in △ABC with AC = 15 m:
⇒ AC² = BC² + AB²
⇒ 15² = \Big(\dfrac{AB}{\sqrt{3}}\Big)^2 + AB²
⇒ 225 = \dfrac{AB^2}{3} + AB²
⇒ 225 = \dfrac{4AB^2}{3}
⇒ AB² = \dfrac{225 \times 3}{4}
⇒ AB² = \dfrac{675}{4}
⇒ AB² = 168.75
⇒ AB = \sqrt{168.75} ≈ 12.99 m ≈ 13 m.
Hence, AB = 13 m.
Question 74
Prove that :
tan^2 θ + cos^2 θ – 1 = tan^2 θ. sin^2 θ
We need to prove that tan² θ + cos² θ – 1 = tan² θ · sin² θ.
Let’s start with the L.H.S. and simplify:
⇒ tan² θ + cos² θ – 1
⇒ \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} – (1 – cos² θ) [since cos² θ – 1 = -(1 – cos² θ)]
⇒ \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} – sin² θ [since 1 – cos² θ = sin² θ]
⇒ sin² θ \Big(\dfrac{1}{\text{cos}^2 θ} - 1\Big) [taking sin² θ common]
⇒ \dfrac{\text{sin}^2 θ(1 - \text{cos}^2 θ)}{\text{cos}^2 θ}
⇒ \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} · sin² θ
⇒ tan² θ · sin² θ
Since L.H.S. = R.H.S., the identity is proved.
Hence, proved that tan² θ + cos² θ – 1 = tan² θ · sin² θ.
Question 75
The class mark and frequency of a data is given in the graph. From the graph, Find:
(a) the table showing the class interval and frequency.
(b) the mean
(a) The class marks given are 13, 15, 17, 19, 21, 23.
The gap between consecutive class marks is 15 – 13 = 2.
So each class has width 2, which means:
Lower limit = Class mark – 1
Upper limit = Class mark + 1
| Class | Frequency |
|---|---|
| 12 – 14 | 8 |
| 14 – 16 | 2 |
| 16 – 18 | 3 |
| 18 – 20 | 4 |
| 20 – 22 | 5 |
| 22 – 24 | 6 |
(b) Now let’s compute the mean using the direct method:
| Class | Class mark (x) | Frequency (f) | fx |
|---|---|---|---|
| 12 – 14 | 13 | 8 | 104 |
| 14 – 16 | 15 | 2 | 30 |
| 16 – 18 | 17 | 3 | 51 |
| 18 – 20 | 19 | 4 | 76 |
| 20 – 22 | 21 | 5 | 105 |
| 22 – 24 | 23 | 6 | 138 |
| Total | Σf = 28 | Σfx = 504 |
Mean = \dfrac{Σfx}{Σf} = \dfrac{504}{28} = 18.
Hence, mean = 18.
Question 76
The mean of 5, 7, 8, 4 and m is n and the mean of 5, 7, 8, 4, m and n is m. Find the values of m and n.
We have two conditions about the means.
First condition: The mean of 5, 7, 8, 4, and m is n.
\begin{aligned}\therefore \dfrac{5 + 7 + 8 + 4 + m}{5} = n \\ \Rightarrow m + 24 = 5n \\ \Rightarrow m = 5n - 24 \text{ .........(1)}\end{aligned}Second condition: The mean of 5, 7, 8, 4, m, and n is m.
\begin{aligned}\therefore \dfrac{5 + 7 + 8 + 4 + m + n}{6} = m \\ \Rightarrow m + n + 24 = 6m\end{aligned}Substituting m from equation (1):
⇒ 5n – 24 + n + 24 = 6m
⇒ 6n = 6m
⇒ n = m ………(2)
Putting n = m into equation (1):
⇒ m = 5m – 24
⇒ 4m = 24
⇒ m = 6
⇒ n = 6.
Hence, m = 6 and n = 6.
Question 77
The probability of selecting a blue marble and a red marble from a bag containing red, blue and green marbles is \dfrac{1}{3} and \dfrac{1}{5} respectively. If the bag contains 14 green marbles, then find :
(a) number of red marbles.
(b) total number of marbles in the bag.
The bag contains red, blue, and green marbles. The sum of all probabilities must equal 1.
⇒ P(red) + P(blue) + P(green) = 1
⇒ \dfrac{1}{3} + \dfrac{1}{5} + P(green) = 1
⇒ \dfrac{8}{15} + P(green) = 1
⇒ P(green) = \dfrac{7}{15}
We’re told there are 14 green marbles. Using the probability formula:
\begin{aligned}\therefore \dfrac{\text{No. of green marbles}}{\text{Total no. of marbles}} = \dfrac{7}{15} \\ \Rightarrow \dfrac{14}{\text{Total}} = \dfrac{7}{15} \\ \Rightarrow \text{Total no. of marbles} = \dfrac{15 \times 14}{7} = 30.\end{aligned}(a) We’re now told P(red) = \dfrac{1}{5} (note: this appears to be a revised value for part a).
\begin{aligned}\therefore \dfrac{\text{No. of red marbles}}{30} = \dfrac{1}{5} \\ \Rightarrow \text{No. of red marbles} = \dfrac{30}{5} = 6.\end{aligned}Hence, number of red marbles = 6.
(b) Hence, total number of marbles in the bag = 30.