This chapter provides detailed solutions for ICSE Class 10 Maths Competency Based Questions, specifically focusing on the 4-mark problems found in the ‘Short Answer Questions 2’ section of the ICSE Maths Solved Competency Focused Questions book. These questions are designed to test your practical application of important mathematical concepts, particularly from commercial mathematics like GST, Banking, and Shares & Dividends. Mastering these problems is crucial as they reflect the style of questions that assess deeper understanding and problem-solving skills, which are increasingly important in the ICSE board examinations. Working through these solutions will help you build confidence and learn the correct format for presenting your answers to secure full marks.
You are likely here because you need clear, step-by-step solutions for the 28 problems in the ‘Short Answer Questions 2’ chapter. When you’re stuck on a multi-step problem involving GST calculations or recurring deposit formulas, having a reliable guide is essential. Our subject experts have solved every question following the exact methodology and format prescribed by the ICSE board. This ensures you not only find the correct answer but also learn the proper way to present your work. Here, you will find meticulously explained solutions for every question in this 4-mark practice set.
Short Answer Questions 2 (4 Marks Each)
Question 78
The following bill shows the GST rate and the marked price of items :
| S.No. | Item | Marked price (₹) | Quantity | Rate of GST |
|---|---|---|---|---|
| 1 | Wheat Flour (unpacked) | 35.00 | 5 kg | x% |
| 2 | Basmati Rice (Branded & packed) | 180.00 | 5 kg | 5% |
| 3 | Surf Excel Quick Wash Detergent | 220.00 | y kg | 18% |
Find :
(a) the value of x if the total GST on wheat flour and basmati rice is ₹ 45.
(b) the value of y, if CGST paid for detergent powder is ₹ 39.60
(c) total amount to be paid (including GST) for the above bill.
(a) We’re told that the total GST on wheat flour and basmati rice is ₹ 45.
First, let’s find the total price of each item.
Total price of wheat flour = 35 × 5 = ₹ 175.
Total price of basmati rice = 180 × 5 = ₹ 900.
Now, let’s set up the GST equation. Wheat flour has x% GST and basmati rice has 5% GST.
\therefore \dfrac{x}{100} \times 175 + \dfrac{5}{100} \times 900 = 45 \[1em] \Rightarrow 1.75x + 45 = 45 \[1em] \Rightarrow 1.75x = 45 - 45 \[1em] \Rightarrow 1.75x = 0 \[1em] \Rightarrow x = 0.Hence, x = 0%.
(b) From the table, the total price of Surf Excel detergent = 220 × y = ₹ 220y.
The GST rate on detergent is 18%, so CGST = \dfrac{\text{GST}}{2} = \dfrac{18}{2} = 9%.
We’re told CGST paid is ₹ 39.60.
\therefore \dfrac{9}{100} \times 220y = 39.60 \[1em] \Rightarrow 9 \times 220y = 39.60 \times 100 \[1em] \Rightarrow y = \dfrac{39.60 \times 100}{9 \times 220} \[1em] \Rightarrow y = \dfrac{3960}{1980} = 2.Hence, y = 2 kg.
(c) Let’s calculate the total amount for each item including GST.
For wheat flour:
Total price = 35 × 5 = ₹ 175
GST = 0%
So, total = ₹ 175.
For basmati rice:
Total price = 180 × 5 = ₹ 900
GST = 5%
So, total = ₹ 900 + \dfrac{5}{100} \times 900 = ₹ 900 + ₹ 45 = ₹ 945.
For detergent powder:
Total price = 220 × 2 = ₹ 440
GST = 18%
So, total = ₹ 440 + \dfrac{18}{100} \times 440 = ₹ 440 + ₹ 79.20 = ₹ 519.20
Total bill = ₹ 175 + ₹ 945 + ₹ 519.20 = ₹ 1639.20
Hence, total bill = ₹ 1639.20
Question 79
Amit deposited ₹ 600 per month in a recurring deposit account. The bank pays a simple interest of 12% p.a. Calculate the:
(a) number of monthly installments Amit deposits to get a maturity amount of ₹ 11826?
(b) total interest paid by the bank.
(c) total amount deposited by him.
(a) Let’s say Amit deposits money for n months.
For recurring deposits, the maturity value formula is:
M.V. = P × n + \dfrac{P \times n(n + 1)}{2 \times 12} \times \dfrac{r}{100}
Substituting the known values (M.V. = ₹ 11826, P = ₹ 600, r = 12%):
\Rightarrow 11826 = 600 \times n + \dfrac{600 \times n(n + 1)}{2 \times 12} \times \dfrac{12}{100} \[1em] \Rightarrow 11826 = 600n + \dfrac{600(n^2 + n)}{200} \[1em] \Rightarrow 11826 = 600n + 3(n^2 + n) \[1em] \Rightarrow 11826 = 600n + 3n^2 + 3n \[1em] \Rightarrow 3n^2 + 603n = 11826 \[1em] \Rightarrow 3(n^2 + 201n) = 11826 \[1em] \Rightarrow n^2 + 201n = \dfrac{11826}{3} \[1em] \Rightarrow n^2 + 201n = 3942 \[1em] \Rightarrow n^2 + 201n - 3942 = 0 \[1em] \Rightarrow n^2 + 219n - 18n - 3942 = 0 \[1em] \Rightarrow n(n + 219) - 18(n + 219) = 0 \[1em] \Rightarrow (n - 18)(n + 219) = 0 \[1em] \Rightarrow n - 18 = 0 \text{ or } n + 219 = 0 \[1em] \Rightarrow n = 18 \text{ or } n = -219.Since the number of months can’t be negative, n = 18.
Hence, number of monthly installments = 18.
(b) Using the interest formula:
\text{Interest } = \dfrac{600 \times 18(18 + 1)}{2 \times 12} \times \dfrac{12}{100} \[1em] = \dfrac{600 \times 18 \times 19}{200} \[1em] = 3 \times 18 \times 19 \[1em] = \text{₹ 1026}.Hence, total interest paid = ₹ 1026.
(c) Total amount deposited = P × n = ₹ 600 × 18 = ₹ 10800.
Hence, total amount deposited by Amit = ₹ 10800.
Question 80
Aman has 500, ₹ 100 shares of a company quoted at ₹ 120, paying a 10% dividend. When the share price rises to ₹ 200 each, he sells all his shares. He invests half of the sale proceeds in ₹ 10, 12% shares at ₹ 25, and the remaining sale proceeds in ₹ 400, 9% shares at ₹ 500.
Find his:
(a) sales proceeds.
(b) investment in ₹ 10, 12% shares at ₹ 25.
(c) original income.
(d) change in income.
(a) Aman sells 500 shares when the price rises to ₹ 200 each.
Sale proceeds = 500 × ₹ 200 = ₹ 1,00,000.
Hence, sale proceeds = ₹ 1,00,000.
(b) He invests half the sale proceeds in ₹ 10, 12% shares at ₹ 25.
∴ Investment = \dfrac{\text{Sale proceeds}}{2} = \dfrac{100000}{2} = ₹ 50,000.
Hence, investment in ₹ 10, 12% shares at ₹ 25 = ₹ 50,000.
(c) Original income from dividends:
Income = No. of shares × \dfrac{\text{Rate of dividend}}{100} × Nominal value
= 500 × \dfrac{10}{100} × 100 = ₹ 5,000.
Hence, original income = ₹ 5,000.
(d) Aman invests ₹ 50,000 in each type of new share.
For the first share (N.V. = ₹ 10, Dividend = 12%, M.V. = ₹ 25):
No. of shares bought = \dfrac{50000}{25} = 2000
Income = 2000 × \dfrac{12}{100} \times 10 = ₹ 2400.
For the second share (N.V. = ₹ 400, Dividend = 9%, M.V. = ₹ 500):
No. of shares bought = \dfrac{50000}{500} = 100
Income = 100 × \dfrac{9}{100} \times 400 = ₹ 3,600.
New income = ₹ 3,600 + ₹ 2,400 = ₹ 6000
Change in income = ₹ 6,000 – ₹ 5,000 = ₹ 1,000.
Hence, change in income = ₹ 1,000 (increase).
Question 81
Solve the following inequation.
\dfrac{11 + 3x}{5} \ge 3 - x \gt -\dfrac{3}{2}, x ∈ R
(a) Write the solution set.
(b) Represent the solution on the number line.
We need to solve the compound inequation \dfrac{11 + 3x}{5} \ge 3 - x \gt -\dfrac{3}{2}.
Solving the left half:
\Rightarrow \dfrac{11 + 3x}{5} \ge 3 - x \[1em] \Rightarrow 11 + 3x \ge 5(3 - x) \[1em] \Rightarrow 11 + 3x \ge 15 - 5x \[1em] \Rightarrow 3x + 5x \ge 15 - 11 \[1em] \Rightarrow 8x \ge 4 \[1em] \Rightarrow x \ge \dfrac{4}{8} \[1em] \Rightarrow x \ge \dfrac{1}{2} \text{ ...........(1)}Solving the right half:
\Rightarrow 3 - x \gt -\dfrac{3}{2} \[1em] \Rightarrow x \lt 3 + \dfrac{3}{2} \[1em] \Rightarrow x \lt \dfrac{6 + 3}{2} \[1em] \Rightarrow x \lt \dfrac{9}{2} \text{ .......(2)}From (1) and (2), combining both parts:
\Rightarrow \dfrac{1}{2} \le x \lt \dfrac{9}{2}
Hence, solution set = {x : \dfrac{1}{2} \le x \lt \dfrac{9}{2}, x \in R}
Question 82
Determine whether the following quadratic equation has real roots.
5𝑥^2 − 9𝑥 + 4 = 0
(a) Give reasons for your answer.
(b) If the equation has real roots, identify them.
(a) Comparing 5𝑥² − 9𝑥 + 4 = 0 with the standard form ax² + bx + c = 0:
a = 5, b = -9 and c = 4.
Discriminant (D) = b² – 4ac = (-9)² – 4 × 5 × 4 = 81 – 80 = 1.
Since D > 0 and it’s a perfect square, the equation has real roots.
Hence, equation 5𝑥² − 9𝑥 + 4 = 0 has real roots.
(b) Let’s solve by factorization:
⇒ 5𝑥² − 9𝑥 + 4 = 0
⇒ 5x² – 5x – 4x + 4 = 0
⇒ 5x(x – 1) – 4(x – 1) = 0
⇒ (5x – 4)(x – 1) = 0
⇒ 5x – 4 = 0 or x – 1 = 0
⇒ x = \dfrac{4}{5} or x = 1.
Hence, roots of the equation are 1 and \dfrac{4}{5}.
Question 83
The profit in rupees in a local restaurant and the number of customers who visited the restaurant are tabulated below for each week for one month.
| Week number | Week 1 | Week 2 | Week 3 | Week 4 |
|---|---|---|---|---|
| Number of customers | 1400 | 5600 | x | 3212 |
| Profit in ₹ | 28000 | 112000 | 32140 | y |
Find :
(a) if the number of customers and profit per week in continued proportion or not? Justify your answer.
(b) the value of x and y.
(a) Let’s check if the number of customers and profit are in continued proportion.
\dfrac{1400}{28000} = \dfrac{5600}{112000} = \dfrac{1}{20}.
Since the ratios are equal, they are in proportion. But for continued proportion, we’d need three terms in the form a:b = b:c, which isn’t the case here.
Hence, number of customers and profit per week are in proportion but not in continued proportion.
(b) Using the proportion \dfrac{1}{20}:
\therefore \dfrac{x}{32140} = \dfrac{1}{20} \[1em] \Rightarrow x = 32140 \times \dfrac{1}{20} = 1607 \[1em]and
\therefore \dfrac{3212}{y} = \dfrac{1}{20} \[1em] \Rightarrow y = 3212 \times 20 = 64240.Hence, x = 1607 and y = 64240.
Question 84
Given, 9𝑥^2 – 4 is a factor of 9𝑥^3 – m𝑥^2 – n𝑥 + 8 :
(a) find the value of m and n using the remainder and factor theorem.
(b) factorise the given polynomial completely.
(a) First, let’s factor 9𝑥² – 4:
⇒ 9𝑥² – 4 = (3𝑥)² – 2² = (3𝑥 + 2)(3𝑥 – 2).
Since 9𝑥² – 4 is a factor of 9𝑥³ – m𝑥² – n𝑥 + 8, both (3𝑥 + 2) and (3𝑥 – 2) must be factors.
Using the factor theorem: if (x – a) is a factor of f(x), then f(a) = 0.
For 3𝑥 + 2 = 0, we get 𝑥 = -\dfrac{2}{3}.
Substituting in f(x) = 9𝑥³ – m𝑥² – n𝑥 + 8:
\Rightarrow 9 \times \Big(-\dfrac{2}{3}\Big)^3 - m \times \Big(-\dfrac{2}{3}\Big)^2 - n \times \Big(-\dfrac{2}{3}\Big) + 8 = 0 \[1em] \Rightarrow -\dfrac{8}{3} - \dfrac{4m}{9} + \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow \dfrac{6n - 4m + 48}{9} = 0 \[1em] \Rightarrow 3n - 2m = -24 \text{ ..........(1)}For 3𝑥 – 2 = 0, we get 𝑥 = \dfrac{2}{3}.
\Rightarrow 9 \times \Big(\dfrac{2}{3}\Big)^3 - m \times \Big(\dfrac{2}{3}\Big)^2 - n \times \Big(\dfrac{2}{3}\Big) + 8 = 0 \[1em] \Rightarrow \dfrac{8}{3} - \dfrac{4m}{9} - \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow 2m + 3n = 48 \text{ ..........(2)}Adding (1) and (2):
⇒ 6n = 24 ⇒ n = 4.
Substituting in (1): 3(4) – 2m = -24 ⇒ m = 18.
Hence, m = 18 and n = 4.
(b) Substituting m = 18 and n = 4:
9𝑥³ – 18𝑥² – 4𝑥 + 8
Dividing by 9𝑥² – 4:
\begin{array}{l} \phantom{9x^2 - 4)}{\quad x - 2} \ 9x^2 - 4\overline{\smash{\big)}\quad 9x^3 - 18x^2 - 4x + 8} \ \phantom{9x^2 - 4)}\phantom{)}\underline{\underset{-}{+}9x^3 \phantom{- 18x^2} \underset{+}{-}4x} \ \phantom{{9x^2 - 4}{+9x^3 - }}-18x^2 \phantom{- 4x} + 8 \ \phantom{{9x^2 - 4)}{+9x^3 - }}\underline{\underset{+}{-}18x^2 \phantom{- 4x} \underset{-}{+} 8} \ \phantom{{9x^2 - 4)}{x^3-2x^{2}(31)}{x}}\times \end{array}∴ 9𝑥³ – 18𝑥² – 4𝑥 + 8 = (9𝑥² – 4)(𝑥 – 2) = (3𝑥 + 2)(3𝑥 – 2)(𝑥 – 2).
Hence, factors are (3𝑥 + 2), (3𝑥 – 2) and (𝑥 – 2).
Question 85
The marks scored by 100 students are given below:
| Marks scored | No. of students |
|---|---|
| 0-10 | 4 |
| 10-20 | 5 |
| 20-30 | 9 |
| 30-40 | 7 |
| 40-50 | 13 |
| 50-60 | 12 |
| 60-70 | 15 |
| 70-80 | 11 |
| 80-90 | 14 |
| 90-100 | 10 |
A student in the class is selected at random. Find the probability that the student has scored:
(a) less than 20
(b) below 60 but 30 or more
(c) more than or equal to 70
(d) above 89.
First, let’s build the cumulative frequency table:
| Marks scored | No. of students | Cumulative frequency |
|---|---|---|
| 0-10 | 4 | 4 |
| 10-20 | 5 | 9 |
| 20-30 | 9 | 18 |
| 30-40 | 7 | 25 |
| 40-50 | 13 | 38 |
| 50-60 | 12 | 50 |
| 60-70 | 15 | 65 |
| 70-80 | 11 | 76 |
| 80-90 | 14 | 90 |
| 90-100 | 10 | 100 |
(a) P(scored less than 20) = \dfrac{\text{No. of students who scored less than 20}}{\text{Total no. of students}} = \dfrac{9}{100}.
Hence, probability that the student has scored less than 20 = \dfrac{9}{100}.
(b) No. of students who scored less than 60 = 50
No. of students who scored less than 30 = 18
∴ No. of students scoring between 30 and 60 = 50 – 18 = 32.
P(below 60 but 30 or more) = \dfrac{\text{No. of students scoring between 60 and 30}}{\text{Total no. of students}} = \dfrac{32}{100} = \dfrac{8}{25}.
Hence, probability that the student has scored below 60 but 30 or more = \dfrac{8}{25}.
(c) No. of students who scored less than 70 = 65
∴ No. of students scoring 70 or more = 100 – 65 = 35.
P(70 or more) = \dfrac{\text{No. of students scoring 70 or more}}{\text{Total no. of students}} = \dfrac{35}{100} = \dfrac{7}{20}.
Hence, probability that the student has scored more than or equal to 70 = \dfrac{7}{20}.
(d) No. of students scoring more than 89 = 10 (the 90-100 range).
P(more than 89) = \dfrac{\text{No. of students scoring \textgreater 89}}{\text{Total no. of students}} = \dfrac{10}{100} = \dfrac{1}{10}.
Hence, probability that the student has scored more than 89 = \dfrac{1}{10}.
Question 86
Given, matrix A = \begin{bmatrix} x & 1 \\ y & 2 \end{bmatrix} \text{ and } B = \begin{bmatrix} x \\ x - 2 \end{bmatrix} such that AB is a null matrix. Find :
(a) order of the null matrix
(b) possible values of x and y.
We’re told that AB is a null matrix.
Matrix A is 2 × 2 and matrix B is 2 × 1, so the product AB will be a 2 × 1 null matrix.
\therefore \begin{bmatrix} x & 1 \ y & 2 \end{bmatrix}_{2 \times 2} \begin{bmatrix} x \ x - 2 \end{bmatrix}_{2 \times 1} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \\ \Rightarrow \begin{bmatrix} x^2 + x - 2 \ xy + 2x - 4 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \text{ .............(1)}From the first row of equation (1):
⇒ x² + x – 2 = 0
⇒ (x – 1)(x + 2) = 0
⇒ x = 1 or x = -2.
From the second row of equation (1): xy + 2x – 4 = 0
When x = 1: (1)y + 2(1) – 4 = 0 ⇒ y = 2.
When x = -2: -2y – 4 – 4 = 0 ⇒ y = -4.
Hence, x = 1, y = 2 or x = -2, y = -4.
Question 87
The sum of a certain number of terms of the Arithmetic Progression (A.P.) 20, 17, 14, ……. is 65. Find the:
(a) number of terms.
(b) last term.
(a) Let the number of terms be n.
Using the sum formula for an A.P.:
\begin{aligned}\Rightarrow 65 = \dfrac{n}{2}[2 \times 20 + (n - 1) \times (-3)] \\ \Rightarrow 65 = \dfrac{n}{2}[40 - 3n + 3] \\ \Rightarrow 65 = \dfrac{n}{2}[43 - 3n] \\ \Rightarrow 130 = 43n - 3n^2 \\ \Rightarrow 3n^2 - 43n + 130 = 0 \\ \Rightarrow 3n^2 - 30n - 13n + 130 = 0 \\ \Rightarrow (3n - 13)(n - 10) = 0 \\ \Rightarrow n = \dfrac{13}{3} \text{ or } n = 10.\end{aligned}Since n must be a whole number, n = 10.
Hence, no. of terms = 10.
(b) Last term = a + (n – 1)d = 20 + (10 – 1) × (-3) = 20 – 27 = -7.
Hence, last term = -7.
Question 88
(a) Point P(2, -3) on reflection becomes P'(2, 3). Name the line of reflection (say L~1).
(b) Point P’ is reflected to P” along the line (𝐿~2), which is perpendicular to the line 𝐿~1 and passes through the point, which is invariant along both axes. Write the coordinates of P”.
(c) Name and write the coordinates of the point of intersection of the lines 𝐿~1 and 𝐿~2.
(d) Point P is reflected to P”’ on reflection through the point named in the answer of part I of this question. Write the coordinates of P”’. Comment on the location of the points P” and P”’.
(a) P(2, -3) reflects to P'(2, 3). Since only the y-coordinate changes sign, the line of reflection is the x-axis.
Hence, point P becomes P’ on reflection in x-axis.
(b) L₂ is perpendicular to L₁ (x-axis) and passes through the origin (which is invariant under both reflections). So L₂ is the y-axis.
P'(2, 3) reflected in the y-axis becomes P”(-2, 3).
Hence, coordinates of P” = (-2, 3).
(c) The x-axis and y-axis intersect at the origin.
Hence, coordinates of intersection of lines L₁ and L₂ is (0, 0).
(d) P(2, -3) reflected in the origin becomes P”'(-2, 3).
Since P” and P”’ have the same coordinates, they coincide.
Hence, P” and P”’ are coincident points.
Question 89
In the given figure, if the line segment AB is intercepted by the y-axis and x-axis at C and D, respectively, such that AC : AD = 1 : 4 and D is the midpoint of CB. Find the coordinates of D, C and B.
Let C = (0, b) and D = (a, 0).
Given AC : AD = 1 : 4, so if AC = x, then AD = 4x, which means CD = 3x.
Thus AC : CD = 1 : 3.
Using the section formula for C dividing AD in ratio 1:3:
\begin{aligned}\Rightarrow (0, b) = \Big(\dfrac{a - 6}{4}, \dfrac{18}{4}\Big) \\ \Rightarrow a = 6 \text{ and } b = \dfrac{9}{2}.\end{aligned}So C = \Big(0, \dfrac{9}{2}\Big) and D = (6, 0).
Since D is the midpoint of CB:
\begin{aligned}\therefore (6, 0) = \Big(\dfrac{p}{2}, \dfrac{9 + 2q}{4}\Big) \\ \Rightarrow p = 12 \text{ and } q = -\dfrac{9}{2}.\end{aligned}Hence, coordinates of B = \Big(12, -\dfrac{9}{2}\Big), C = \Big(0, \dfrac{9}{2}\Big) and D = (6, 0).
Question 90
Find the equation of the straight line perpendicular to the line x + 2y = 4, which cuts an intercept of 2 units from the positive y-axis. Hence, find the intersection point of the two lines.
First, let’s find the slope of x + 2y = 4.
⇒ y = -\dfrac{1}{2}x + 4, so slope m = -\dfrac{1}{2}.
For a perpendicular line, the product of slopes = -1:
\Rightarrow m_1 \times -\dfrac{1}{2} = -1 \[1em] \Rightarrow m_1 = 2.The line cuts an intercept of 2 units from the positive y-axis, so c = 2.
∴ Equation: y = 2x + 2.
Now solving simultaneously:
⇒ x + 2y = 4 ………(1)
⇒ y = 2x + 2 …….(2)
Substituting (2) into (1):
⇒ x + 2(2x + 2) = 4
⇒ 5x = 0 ⇒ x = 0.
⇒ y = 2(0) + 2 = 2.
Hence, equation of required line is y = 2x + 2 and point of intersection = (0, 2).
Question 91
While preparing a PowerPoint presentation, ∆ ABC is enlarged along the side BC to ∆ AB’C’, as shown in the diagram, such that BC ∶ B’C’ is 3 ∶ 5. Find :
(a) AB ∶ BB’
(b) length AB, if BB’ = 4 cm.
(c) Is ∆ ABC ~ ∆ AB’C’ ? Justify your answer.
(d) ar (∆ ABC) : ar (quad. BB’C’C).
Since ∆ ABC is enlarged to ∆ AB’C’, the triangles are similar.
(a) From the ratio of corresponding sides:
\dfrac{AB}{AB'} = \dfrac{BC}{B'C'} = \dfrac{3}{5}Let AB = 3x and AB’ = 5x. Then BB’ = 5x – 3x = 2x.
∴ AB : BB’ = 3x : 2x = 3 : 2.
Hence, AB : BB’ = 3 : 2.
(b) \dfrac{AB}{4} = \dfrac{3}{2} \Rightarrow AB = \dfrac{3 \times 4}{2} = 6 cm.
Hence, AB = 6 cm.
(c) Since BC || B’C’, and ∠BAC is common, ∆ ABC ~ ∆ AB’C’ by AA similarity.
Hence, proved that ∆ ABC ~ ∆ AB’C’.
(d) Ratio of areas: \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\dfrac{3}{5}\Big)^2 = \dfrac{9}{25}.
If area of ∆ ABC = 9a, then area of ∆ AB’C’ = 25a.
Area of quadrilateral BB’C’C = 25a – 9a = 16a.
∴ Area of ∆ ABC : Area of quadrilateral BB’C’C = 9 : 16.
Hence, area of ∆ ABC : area of quadrilateral BB’C’C = 9 : 16.
Question 92
The approximate volume of a human eye is 6.5 cm^3. The volume of a laboratory model (excluding base and stand) of the human eye is 1404 cm^3.
(a) State whether the scale factor k is less than, equals to or greater than 1.
(b) Calculate the:
(i) value of k
(ii) diameter of the human eye if the radius of the model is 7.2 cm.
(iii) the external surface area of the human eye if the surface area of the model is 651.6 cm^2.
(a) Since the model’s volume (1404 cm³) is greater than the actual eye’s volume (6.5 cm³), the scale factor k > 1.
(b)
(i) k^3 = \dfrac{1404}{6.5} = \dfrac{216}{1} = 6^3 \Rightarrow k = 6.
Hence, scale factor (k) = 6.
(ii) k = \dfrac{7.2}{\text{Radius of eye}} \Rightarrow \text{Radius of eye} = \dfrac{7.2}{6} = 1.2 \text{ cm}.
Diameter = 1.2 \times 2 = 2.4 cm.
Hence, diameter of human eye = 2.4 cm.
(iii) k^2 = \dfrac{651.6}{\text{Surface area of eye}} \Rightarrow \text{Surface area} = \dfrac{651.6}{36} = 18.1 \text{ cm}^2.
Hence, surface area of human eye = 18.1 cm².
Question 93
In the adjoining diagram PQ, PR and ST are the tangents to the circle with centre O and radius 7 cm. Given OP = 25 cm. Find :
(a) length of ST
(b) value of ∠OPQ, i.e. θ
(c) ∠QUR, in nearest degree
(a) The radius and tangent are perpendicular at the point of contact, so ∠PQO = 90°.
In right triangle PQO:
⇒ PQ² = 25² – 7² = 625 – 49 = 576 ⇒ PQ = \sqrt{576} = 24 cm.
From triangle POQ: tan θ = \dfrac{7}{24} ……………(1)
Also, PA = 25 – 7 = 18 cm, and tan θ = \dfrac{AS}{18} ………..(2)
From (1) and (2): AS = \dfrac{7}{24} \times 18 = \dfrac{21}{4} cm.
ST = 2 × AS = 10.5 cm.
Hence, ST = 10.5 cm.
(b) tan θ = \dfrac{7}{24} ≈ 0.292 ⇒ θ = 16° 16′.
Hence, θ = 16° 16′.
(c) In triangle POQ: ∠POQ = 180° – (90° + 16° 16′) = 180° – 106° 16′ = 73° 44′ ≈ 74°.
Since PQ = PR (tangents from external point) and OQ = OR (radii), triangles POQ and POR are congruent by SSS.
∴ ∠QOR = 2 × 74° = 148°.
By the central angle theorem: ∠QUR = \dfrac{148°}{2} = 74°.
Hence, ∠QUR = 74°.
Question 94
Use ruler and compass to answer this question. Construct a triangle ABC where AB = 5.5 cm, BC = 4.5 cm and angle ABC = 135°. Construct the circumcircle to the triangle ABC. Measure and write down the length of AC.
Steps of construction :
- Draw a line segment BC = 4.5 cm
- Construct XB such that ∠XBC = 135°.
- Cut AB = 5.5 cm from XB.
- Join AC and measure its length.
- Draw PQ and RS, the perpendicular bisectors of BC and AB.
- Mark point O, the intersection of PQ and RS.
- With O as center and radius OA, draw the circumcircle passing through A, B and C.
On measuring AC = 9.1 cm and radius = 6.5 cm.
Question 95
The curved surface area of a right circular cone is half of another right circular cone. If the ratio of their slant heights is 2 : 1 and that of their volumes is 3 : 1, find ratio of their:
(a) radii
(b) heights
(a) Let the radii be r (smaller) and R (larger), and slant heights be l and L.
Given l : L = 2 : 1 and πrl = \dfrac{1}{2}πRL.
⇒ \dfrac{r}{R} = \dfrac{L}{2l} = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}
Hence, ratio of the radii = 1 : 4.
(b) Given volumes are in ratio 3 : 1 (v : V):
\begin{aligned}\Rightarrow \dfrac{\dfrac{1}{3}πR^2H}{\dfrac{1}{3}πr^2h} = \dfrac{1}{3} \\ \Rightarrow \dfrac{16}{1} \times \dfrac{H}{h} = \dfrac{1}{3} \\ \Rightarrow \dfrac{h}{H} = \dfrac{48}{1}.\end{aligned}Hence, ratio of heights = 48 : 1.
Question 96
A cylindrical drum is unloaded from a truck by rolling it down along a wooden plank. The length of the plank is 10 m and it is making an angle of 10° with the horizontal ground. Find the height from which the cylindrical drum was rolled down. Give your answer correct to 3 significant figures.
Let AC be the plank and BC be the vertical height.
In the right-angled triangle formed, we have:
⇒ sin θ = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}
⇒ sin 10° = \dfrac{BC}{AC} = \dfrac{BC}{10}
⇒ BC = 10 × 0.1736 = 1.736 ≈ 1.74 m
Hence, height from which the cylindrical drum was rolled down = 1.74 m.
Question 97
The data given below shows the marks of 12 students in a test, arranged in ascending order :
2, 3, 3, 3, 4, x, x + 2, 8, p, q, 8, 9.
If the given value of the median and mode is 6 and 8 respectively, then find the values of x, p, q.
For 12 data points, the median is the average of the 6th and 7th terms:
\begin{aligned}\Rightarrow 6 = \dfrac{x + (x + 2)}{2} \\\Rightarrow 12 = 2x + 2 \\\Rightarrow 10 = 2x \\\Rightarrow x = 5.\end{aligned}The data becomes: 2, 3, 3, 3, 4, 5, 7, 8, p, q, 8, 9.
Since mode = 8, it must appear most frequently. Currently 3 appears 3 times, so 8 must appear at least 4 times.
∴ p = q = 8.
Hence, x = 5, p = 8 and q = 8.
Question 98
Solve the linear inequation, write down the solution set and represent it on the real number line :
5(2 – 4x) > 18 – 16x > 22 – 20x, x ∈ R
We need to solve 5(2 – 4x) > 18 – 16x > 22 – 20x.
Solving the left half:
⇒ 5(2 – 4x) > 18 – 16x
⇒ 10 – 20x > 18 – 16x
⇒ 4x < -8
⇒ x < -2 ………..(1)
Solving the right half:
⇒ 18 – 16x > 22 – 20x
⇒ 4x > 4
⇒ x > 1 …………(2)
From (1) and (2):
Solution set : {x : x < -2 or x > 1, x ∈ R}
Hence, solution set = {x : x < -2 or x > 1, x ∈ R}.
Question 99
If a polynomial x^3 + 2x^2 – ax + b leaves a remainder -6 when divided by x + 1 and the same polynomial has x – 2 as a factor, then find the values of a and b.
Let f(x) = x³ + 2x² – ax + b.
Given f(-1) = -6 (remainder when divided by x + 1):
⇒ (-1)³ + 2(-1)² + a + b = -6
⇒ -1 + 2 + a + b = -6
⇒ a + b = -7 ……..(1)
Given x – 2 is a factor, so f(2) = 0:
⇒ 8 + 8 – 2a + b = 0
⇒ 2a – b = 16 ……(2)
Adding (1) and (2): 3a = 9 ⇒ a = 3.
From (1): 3 + b = -7 ⇒ b = -10.
Hence, a = 3 and b = -10.
Question 100
If A = \begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}, C = \begin{bmatrix} 1 & -4 \end{bmatrix} \text{ and } D = \begin{bmatrix} 4 \\ 1 \end{bmatrix}
(a) Is the product AC possible? Justify your answer.
(b) Find the matrix X, such that X = AB + B^2 − DC
(a) Order of A = 2 × 2, Order of C = 1 × 2. Since columns of A (2) ≠ rows of C (1), the product AC is not possible.
Hence, product AC is not possible.
(b) Computing X = AB + B² – DC:
\Rightarrow X = \begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}^2 - \begin{bmatrix} 4 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 \end{bmatrix} \\ \Rightarrow X = \begin{bmatrix} -1 & 11 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 1 & -8 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 4 & -16 \\ 1 & -4 \end{bmatrix} \\ \Rightarrow X = \begin{bmatrix} -4 & 19 \\ 1 & 9 \end{bmatrix}.Hence, X = \begin{bmatrix} -4 & 19 \\ 1 & 9 \end{bmatrix}.
Question 101
In the given figure (not drawn to scale), BC is parallel to EF, CD is parallel to FG, AE : EB = 2 : 3, ∠BAD = 70°, ∠ACB = 105°, ∠ADC = 40° and AC is bisector of ∠BAD.
(a) Prove Δ AEF ~ Δ AGF
(b) Find :
(i) AG : AD
(ii) area of Δ ACB: area Δ ACD
(iii) area of quadrilateral ABCD: area of Δ ACB.
(a) In Δ AFE, ∠AFE = ∠ACB = 105° (corresponding angles).
∠EAF = \dfrac{70°}{2} = 35° (AC bisects ∠BAD).
By angle sum property: ∠AEF = 180° – (105° + 35°) = 180° – 140° = 40°.
Similarly ∠AGF = ∠ADC = 40° (corresponding angles).
In Δ AEF and Δ AGF:
∠EAF = ∠GAF = 35° and ∠AEF = ∠AGF = 40°.
∴ Δ AEF ~ Δ AGF by AA similarity.
Hence, proved that Δ AEF ~ Δ AGF.
(b) In Δ ACD: ∠DCA = 180° – (35° + 40°) = 180° – 75° = 105°.
∴ Δ ACD ~ Δ ACB by AA similarity.
(i) Since AE : EB = 2 : 3, AE : AB = 2 : 5.
By similarity: \dfrac{AG}{AD} = \dfrac{AE}{AB} = \dfrac{2}{5}.
Hence, AG : AD = 2 : 5.
(ii) Since Δ ABC ≅ Δ ADC, their areas are equal.
Hence, area of Δ ABC : area of Δ ADC = 1 : 1.
(iii) Area of quadrilateral ABCD = 2 × Area of Δ ABC.
Hence, area of quadrilateral ABCD : area of Δ ACB = 2 : 1.
Question 102
In the given figure, angle ABC = 70° and angle ACB = 50°. Given, O is the centre of the circle and PT is the tangent to the circle. Then calculate the following angles
(a) ∠CBT
(b) ∠BAT
(c) ∠PBT
(d) ∠APT
Join AT and BT.
(a) Angle in a semicircle is 90°, so ∠CBT = 90°.
Hence, ∠CBT = 90°.
(b) In cyclic quadrilateral ATBC: ∠CAT = 180° – 90° = 90°.
In Δ ABC: ∠CAB = 180° – (70° + 50°) = 180° – 120° = 60°.
∴ ∠BAT = 90° – 60° = 30°.
Hence, ∠BAT = 30°.
(c) ∠BTX = ∠BAT = 30° (angles in same segment).
∠PBT = 90° – 70° = 20°.
Hence, ∠PBT = 20°.
(d) ∠PTB = 180° – 30° = 150° (linear pair).
In Δ PBT: ∠APT = 180° – (20° + 150°) = 180° – 170° = 10°.
Hence, ∠APT = 10°.
Question 103
(a) Construct a triangle ABC such that BC = 8 cm, AC = 10 cm and ∠ABC = 90°.
(b) Construct an incircle to this triangle. Mark the centre as I.
(c) Measure and write the length of the in-radius.
(d) Measure and write the length of the tangents from vertex C to the incircle.
(e) Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively. Write the relationship between ∠RIQ and ∠QCR.
(Use a ruler and a compass for this question.)
Steps of construction :
- Draw BC = 8 cm.
- Draw BX perpendicular to BC.
- With C as center and radius 10 cm, draw an arc cutting BX at A.
- Join AB and AC.
- Draw AW, BY and CZ, the angle bisectors of A, B and C.
- Mark their intersection as I (incenter).
- Draw IR perpendicular to AC.
- With I as center and radius IR, draw the incircle touching AB, BC and AC at P, Q and R.
- Measure CQ and CR.
We can prove the relationship by looking at the angles in the two smaller triangles, ΔIRC and ΔIQC.
In ΔIRC, we find ∠RIC = 180° – 90° – \dfrac{∠C}{2} = 90° – \dfrac{∠C}{2}.
Similarly, in ΔIQC, we find ∠QIC = 90° – \dfrac{∠C}{2}.
Adding these two angles gives ∠RIQ:
∠RIQ = ∠RIC + ∠QIC = (90° – \dfrac{∠C}{2}) + (90° – \dfrac{∠C}{2}) = 180° – ∠C.
This rearranges to ∠RIQ + ∠C = 180°.
Hence, ∠RIQ + ∠QCR = 180°.
Question 104
The daily wages of workers in a construction unit were recorded as follows :
| Class marks (Wages) | No. of workers |
|---|---|
| 425 | 6 |
| 475 | 12 |
| 525 | 15 |
| 575 | 17 |
| 625 | 7 |
| 675 | 13 |
Form a frequency distribution table with class intervals and find modal wage by plotting a histogram.
Class width = 2 × 25 = 50 (since class mark difference is 50).
| Class | Frequency |
|---|---|
| 400-450 | 6 |
| 450-500 | 12 |
| 500-550 | 15 |
| 550-600 | 17 |
| 600-650 | 7 |
| 650-700 | 3 |
Steps to find mode graphically:
- Draw histogram with 2 cm = ₹50 on x-axis and 2 cm = 5 workers on y-axis.
- In the highest rectangle (550-600), draw lines AD and BC from adjacent rectangle corners to opposite corners of the highest rectangle.
- Mark intersection K, draw perpendicular to x-axis at L.
Hence, mode = ₹ 557.50
Question 105
A bag contains 13 red cards, 13 black cards and 13 green cards. Each set of cards are numbered 1 to 13. From these cards, a card is drawn at random. What is the probability that the card drawn is a:
(a) green card?
(b) a card with an even number?
(c) a red or black card with a number which is a multiple of three?
Total cards = 13 + 13 + 13 = 39.
(a) P(green card) = \dfrac{13}{39} = \dfrac{1}{3}.
Hence, probability that card drawn is a green card = \dfrac{1}{3}.
(b) Even numbers: 2, 4, 6, 8, 10, 12 → 6 per set × 3 sets = 18 cards.
P(even number) = \dfrac{18}{39} = \dfrac{6}{13}.
Hence, probability that card drawn is a card with even number = \dfrac{6}{13}.
(c) Multiples of 3: 3, 6, 9, 12 → 4 per colour × 2 colours (red and black) = 8 cards.
P(red or black multiple of 3) = \dfrac{8}{39}.
Hence, probability that card drawn is a red or black card with multiple of three = \dfrac{8}{39}.