Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2015 is the best way to boost your preparation for the board exams.
ICSE Class 10 Maths Question Paper 2015 Solved
Time Allowed: 2 1/2 hours
Max. Marks: 80
Answers to this paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers.
Attempt all questions from Section A and any four questions from Section B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of questions are given in brackets [ ]
Mathematical tables are provided.
Section-A (40 Marks)
(Attempt all questions from this section)
Question 1.
(a) A shopkeeper bought an article for ₹ 3,450. He marks the price of the article 16% above the cost price. The rate of sales tax charged on the article is 10%. Find the:
(i) marked price of the article.
(ii) price paid by a customer who buys the article. [3]
(b) Solve the following inequation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R
Represent the solution on a real number line. [3]
(c) Without using trigonometric tables evaluate:
\(\frac{\sin 65^{\circ}}{\cos 25^{\circ}}\) – \(\frac{\cos 32^{\circ}}{\sin 58^{\circ}}\) – sin 28°. sec 62° + cosec2 30°. [4]
Solution:
(a) Cost price of article for shopkeeper = ₹ 3450
Marked price of article = 16% above the c.p.of shopkeeper
= 16% of ₹ 3450 + ₹ 3450
= ₹(552 + 3450)
= ₹ 4002.
Now, Marked price = ₹ 4002
Rate of sales tax charged = 10%
Price paid by the customer to buy the article
= ₹(4002 + \(\frac{10}{100} \times 4002\))
= ₹ (4002 + 400.20)
= ₹ 4402.20
(b) Given 13x – 5 < 15x + 4 < 7x + 12, x ∈ R
⇒ 13x – 5 < 15x + 4
and 15x + 4 < 7x + 12
⇒ 13x – 5 + (-15x + 5) < 15x + 4 + (-15x + 5)
and 15x + 4 + (-7x – 4) < 7x + 12 + (-7x – 4)
⇒ 13x – 15x < 4 + 5
and 15x – 7x < 12 – 4
-2x < 9
and 8x < 8 x > \(-\frac{9}{2}\)
and x < 1
\(-\frac{9}{2}\) < x < 1, x ∈ R
Hence, the required solution set on the number line is :
(c)
Question 2.
(a) If A = \(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
9 & 16 \\
0 & -y
\end{array}\right]\), find x and y when A2 = B. [3]
(b) The present population of a town is 2,00,000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of the two years. [3]
(c) Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. [4]
Solution:
(a)
On comparing corresponding elements of equal matrices, we obtain
4x = 16 ⇒ x = 4 and
1 = -y ⇒ y = -1
Hence, x = 4 and y = -1
(b) Present population (P0) of the town = 2,00,000
Rate of increase of population in first year = R1 = 10%
Rate of increase of population in second year = R2 = 15%
Population after 2 years = P2
= P0\(\left(1+\frac{\mathrm{R}_1}{100}\right)\left(1+\frac{\mathrm{R}_2}{100}\right)\)
= 2,00,000\(\left(1+\frac{10}{100}\right)\left(1-\frac{15}{100}\right)\)
= 2,00,000 × \(\frac{11}{10}\) × \(\frac{23}{20}\)
= 2,53,000
Thus, the population of the town at the end of 2 years is 2,53,000.
(c) Consider the parallelogram ABCD with coordinates of the vertices are A (3, 6), B (5, 10), C(3, 2) and D(x, y)
Since in a parallelogram diagonals bisect each other.
Let in the parallelogram ABCD, diagonals AC and BD bisect each other at O.
Now, coordinates of O, using diagonal AC and mid-point formula
O\(\left(\frac{3+3}{2}, \frac{6+2}{2}\right)\) = O(3, 4)
By using diagonal BD and mid-point formula and coordinates of O, we have
\(\frac{x+5}{2}\) = 3
⇒ x + 5 = 6
⇒ x = 1
and \(\frac{y+10}{2}\) = 4
⇒ y = -2
Thus, the coordinates of D(x, y) are D (1, -2).
(ii) Length of diagonal BD = |BD|
= \(\sqrt{(1-5)^2+(-2-10)^2}\) [using distance formula]
= \(\sqrt{(-4)^2+(-12)^2}\)
= \(\sqrt{16+144}\) = \(\sqrt{160}\) = 4\(\sqrt{10}\) units
(iii) Equation of side AB is:
y – 6 = \(\frac{10-6}{5-3}\)(x – 3)
⇒ y – 6 = \(\frac{4}{2}\)(x – 3)
⇒ y – 6 = 2(x – 3)
⇒ 2x – 6 – y + 6 = 0
⇒ 2x – y = 0
Question 3.
(a) In the given figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semi circles are drawn with AD and BC as diameters. Find the area of the shaded region. (Take π = \(\frac{22}{7}\)) [3]
(b) The marks obtained by 30 students in a class assessment of 5 marks is given below:
| Marks | No. of students |
| 0 | 1 |
| 1 | 3 |
| 2 | 6 |
| 3 | 10 |
| 4 | 5 |
| 5 | 5 |
Calculate the mean, median and mode of the above distribution. [3]
(c) In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the value of x, y and z. [4]
Solution:
(a) We know that the diagonals of a square are equal and bisect each other at right angles.
Since in square ABCD, we have
AB = BC = CD = DA = 21 cm
and OA = OB = OC = OD
We obtain 4 congruent right triangles i.e.,
ΔAOB, ΔBOC, ΔCOD and ΔDOA of equal areas
Hence, shaded area inside the square = \(\frac{1}{2}\)(area of square)
= \(\frac{1}{2}\) × 21 × 21 = 220.5 cm2
Shaded area outside the square
ABCD = Area of two semicircles with AD and BC as diameters
Radius (r) = \(\frac{\mathrm{AD}}{2}\) = \(\frac{21}{2}\) cm.
Area of semicircle = \(\frac{\pi r^2}{2}\) = \(\frac{22 \times 21 \times 21}{7 \times 2 \times 2 \times 2}\)
= 173.25 cm2
Area of two semicircles = 2 × 173.25 = 346,5 cm2
Hence, area of shaded region = shaded area inside the square + shaded area outside the square = (220.5 + 346.5) cm2 = 567 cm2
(b)
Marks is taken as ‘xi‘ and number of students as ‘fi‘
Median: The given variates (xi) are already in ascending order. We construct the cumulative frequency table. Here, n (total number of students) = 30, which is an even number.
[∵ all observations from 11th to 20th correspond to xi = 3]
Mode: Clearly 3 occurs 10 times which is maximum.
Hence, mode 3.
(c) Clearly, O is the centre of the circle and SP is a tangent to the circle.
∴ OS ⊥ SP
[radius at the point of contact is perpendicular to the tangent]
⇒ ∠OSP = ∠OSR = ∠TSR = 90°
Using angles sum property of a triangle, in ΔTSR, we have
∠TSR + ∠SRT + ∠RTS = 180°
⇒ 90° + 65° + x = 180°
⇒ x = 25°
We know that angle subtended at the centre is twice the angle subtended at any point on remaining part of the circle by an arc
⇒ y = 2x
⇒ y = 2 × 25° = 50°
Now, using angles sum property of Δ in ΔOPS, we have
∠OSP + ∠SPO + ∠POS = 180°
90° + z + 50° = 180°
⇒ z = 40°
Thus, x = 25°, y = 50° and z = 40°
Question 4.
(a) Katrina opened a recurring deposit account with a National Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly instalment is ₹ 1,000, find the:
(i) interest earned in 2 years.
(ii) matured value. [3]
(b) Find the value of ‘k’ for which x = 3 is a solution of the quadratic equation, (k + 2)x2 – kx + 6 = 0.
Thus find the other root of the equation.[3]
(c) Construct a regular hexagon of side 5 cm. Construct a circle circumscribing the hexagon. All traces of construction must be clearly shown. [4]
Solution:
(a)
Monthly instalment = ₹ 1000
Rate = 6% per annum
Time = 2 years
⇒ n = 24
Interest earned in 2 years
= ₹ [1000 × \(\frac{24(24+1)}{2}\) × \(\frac{1}{12}\) × \(\frac{6}{200}\)]
= ₹(60 × 25) = ₹ 1500
Maturity amount = ₹ (1000 × 24 + 1500)
= ₹ (24000 + 1500)
= ₹ 25500
(b) Since x = 3 is a solution of the quadratic equation.
(k + 2)x2 – kx + 6 = 0
⇒ (k + 2)(3)2 – k(3) + 6 = 0
⇒ (k + 2)9 – 3k + 6 = 0
⇒ 9k + 18 – 3k + 6 = 0
⇒ 6k + 24 = 0
⇒ 6k = – 24
⇒ k = -4.
Now, we have the quadratic equation
(-4 + 2) x2 – (-4)x + 6 = 0
⇒ -2x2 + 4x + 6 = 0
or x2 – 2x – 3 = 0
⇒ x2 – 3x + x – 3 = 0
⇒ x (x – 3) + 1 (x – 3) = 0
⇒ (x + 1) (x – 3) = 0
⇒ x + 1 = 0; x – 3 = 0
⇒ x = – 1 ; x = 3
Hence, x = -1 is other root of the given equation.
(c) Steps of Construction :
(i) Draw a circle of radius 5 cm with centre O.
(ii) Since \(\frac{360^{\circ}}{6}\) = 60°, draw radii OA and OB, such that ∠AOB = 60°.
(iii) Cut off arcs BC, CD, DE, EF and each equal to arc AB on given circle.
(iv) Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.
Section-B (40 Marks)
(Attempt any four questions from this Section)
Question 5.
(a) Use a graph paper for this question taking 1 cm = 1 unit along both the x and y years:
(i) Plot the points A (0, 5), B (2, 5), C (5, 2), D (5, -2), E (2, -5) and F (0, -5).
(ii) Reflect the points B, C D and E on the y-axis and name them respectively as B’, C’, D’ and E’.
(iii) Write the coordinates of B’, C’, D’ and E’.
(iv) Name the figure formed by B C D E D D’ C’ B’.
(v) Name a line of symmetry for the figure formed. [5]
(b) Virat opened has Savings Bank account in a bank on 16th April 2010. His pass book shows the following entries.
Calculate the interest Viral earned at the end of 31st July, 2010 at 4% per annum interest. What sum of money will he receive if he closes the account on 1st August, 2010? [5]
Solution:
(a)
(i) and (ii)
(iii) B'(-2, 5), C'(-5,2), D'(-5, -2) and E'(-2, – 5)
(iv) Clearly, an octagon is formd by jouning BCDEE’D’C’B’.
(b) Minimum balance for the month of
April = ₹ 0
May = ₹ 4650
June = ₹ 4750
July = ₹ 8950
Total = ₹ 18350
Now, I = ₹ \(\left(\frac{18350 \times 4 \times 1}{12 \times 100}\right)\)
= ₹ 61.17
Hence, interest earned at the end of 31<supst July is ₹ 61.17
Money received on 1st Aug 2010
= ₹ (8950 + 61.17) = ₹ 9011.17
Question 6.
(a) If a, b, c are in continued proportion, prove that
(a + b + c) (a – b + c) = a2 + b2 + c2. [3]
(b) In the given figure ABC is a triangle and BC is parallel to the y-axis. AB and AC intersects the y-axis at P and Q respectively.
(i) Write the coordinates of A.
(ii) Find the length of AB and AC.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC.
(c) Calculate the mean of the following distribution:
| Class Interval | Frequency |
| 0-10 | 8 |
| 10-20 | 5 |
| 20-30 | 12 |
| 30-40 | 35 |
| 40-50 | 24 |
| 50-60 | 16 |
Solution:
(a) Given a, b, c are in continued proportion
∴ \(\frac{a}{b}\) = \(\frac{b}{c}\) = k(say).
⇒ b = ck
Also, a = bk ⇒ a = (ck)k = ck2
Now, (a + b + c)(a – b + c) = (ck2 + ck + c)(ck2 – ck + c)
= c2(k2 + k + 1)(k2 – k + 1)
= c2(k2 + 1 + k)(k2 + 1 – k) = c2 . [(k2 + 1)2 – (k)2]
= c2k4 + c2 + c2k2 = (ck2) + (ck)2 + c2
Hence, (a + b + c)(a – b + c) = a2 + b2 + c2
(b) (i) The coordinates of A are A (4, 0).
(ii) The length of AB = |AB|
= \(\sqrt{(-2-4)^2+(3-0)^2}\)
[using distance formula]
= \(\sqrt{36+9}\) = \(\sqrt{45}\) = 3latex]\sqrt{5}[/latex] units
The length of AC = |AC|
= \(\sqrt{(-2-4)^2+(-4-0)^2}\)
[using distance formula]
= \(\sqrt{36+9}\) = \(\sqrt{45}\) = 3\(\sqrt{5}\) units
The length of AC = |AC|
= \(\sqrt{(-2-4)^2+(-4-0)^2}\)
= \(\sqrt{36+16}\) = \(\sqrt{52}\) = 2\(\sqrt{13}\) units
(iii) Let Q(0, y) divides AC in the ratio k : 1
∴ \(\frac{-2 k+4}{k+1}\) = 0
⇒ -2k + 4 = 0
⇒ -2k = -4
⇒ k = 2
Thus required ratio is 2 : 1
And y = \(\frac{2 \times(-4)+1 \times 0}{2+1}\) = \(\frac{-8+0}{3}\)
y = \(\frac{-8}{3}\)
Hence, the coordinates of Q are Q(0, \(\frac{-8}{3}\))
(iv) Equation of line AC
y – 0 = \(\frac{-4-0}{-2-4}\)(x – 4)
⇒ y = \(\frac{-4}{-6}\)(x – 4)
⇒ y = \(\frac{2}{3}\)(x – 4)
⇒ 2x – 3y – 8 = 0
(c)
Question 7.
(a) Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed. [3]
(b) Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3. [3]
(c) Prove that \(\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}\) = cos θ + sin θ. [4]
Solution:
Radius of smaller sphere = 2 cm
Volume of sphere melted = \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \pi \times 8\)
= \(\frac{32}{3} \pi \mathrm{cm}^3\)
Radius of bigger sphere = 4 cm
Volume of sphere melted = \(\frac{4}{3} \times \pi \times 64\)
= \(\frac{256}{3} \pi \mathrm{cm}^3\)
Volume of both spheres melted = \(\frac{32 \pi}{3}+\frac{256 \pi}{3}\)
= \(\frac{288}{3} \pi \mathrm{cm}^3\)
We know that volume of the cone formed = Volume of spheres melted
Now, height of the cone formed = 8 cm
∴ \(\frac{1}{3} \pi\) × r2 × 8 = \(\frac{288}{3} \pi\)
⇒ 8r2 = 288
⇒ r2 = 36
⇒ r = 6
Hence, the radius of the cone so formed is 6 cm.
(b) Let p(x) = ax3 + 3x2 – 9 and q(x) = 2x3 + 4x + a
By remainder theorem, when p(x) is divided by x + 3,
remainder = f(-3), and
When q(x) is divided by x + 3. the remainder = q (-3)
Since the polynomials p(x) and q (x) when divided by (x + 3) leave the same remainder.
∴ p(-3) = q(-3)
⇒ a(-3)3 + 3(-3)2 – 9 = 2(-3)3 + 4(-3) + a
⇒ -27a + 27 – 9 = – 54 – 12 + a
⇒ -28a = -84
⇒ a = 3
(c)
\(\frac{\sin \theta}{1-\cot \theta}\) + \(\frac{\cos \theta}{1-\tan \theta}\) = cos θ + sin θ
Question 8.
(a) AB and CD are two chords of a circle intersecting at R Prove that AP × PB = CP × PD. [3]
(b) A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball
(iii) is neither a green ball nor a white ball. [3]
(c) Rohit invested ₹ 9,600 on ₹ 100 shares at ₹ 20 premium paying 8% dividend. Rohit sold the shares when the price rose to ₹ 160. He invested the proceeds (excluding dividend) in 10% ₹ 50 shares at ₹40. Find the:
(i) original number of shares.
(ii) sale proceeds.
(iii) new number of shares.
(iv) change in the two dividends. [4]
Solution:
(a) Given : Two chords AB and CD of a circle intersecting at P.
To prove : AP × PB = CP × PD
Construction : Join AC and BD
Proof: In ΔACP and ΔDBP, we have
∠APC = ∠BPD [vert. opp. ∠s)
∠CAP = ∠PDB
[angles in same segment]
ΔACP \(\sim\) ΔDBP [A.A. similarity rule]
\(\frac{\mathrm{AP}}{\mathrm{PD}}\) = \(\frac{\mathrm{CP}}{\mathrm{PB}}\)
[corresponding sides of similar triangles are proportional]
⇒AP × PB = CP × PD
(b) There are 5 white baIls. 6 red balls and 9 green balls
∴ Total number of balls = 5 +6+ 920
(i) Total number of possible outcomes = 20
Number of green balls = 9
Number of favourable outcomes 9
∴ Required probability = \(\frac{11}{20}\)
(ii) Number of white and red ball = 6 + 5 = 11
Number of favourable outcomes = 11
∴ Required probability = \(\frac{11}{20}\)
(iii) Number of balls which is neither a green ball nor a white ball = red balls = 6
Number of favourable outcomes = 6
Required probability = \(\frac{6}{20}\) = \(\frac{3}{10}\)
(c) Face (nominal) value of 1 share = ₹ 1oo
Market value of I share = (100 + 20)
= ₹ 120
Money invested on shares = ₹ 9600
Required number of shares purchased = ₹ \(\frac{9600}{120}\) = 80
Selling price of 1 share = ₹ 160
Money received = ₹ 80 × 160 = ₹ 12800
Face value of 1 share = ₹ 50
Market value of ₹ 50’s share = ₹ 40
Money invested on shares = ₹ 12800
Total number of shares purchased = \(\frac{12800}{40}\) = 320
Dividend received on one share of ₹ 100
= ₹ 8 [8% of ₹ 100]
∴ Dividend received on 80 shares of ₹ 100 = ₹ 8 × 80
= ₹ 640
Dividend received on one share of ₹ 50 = 10% of ₹ 50
= ₹ 5
Dividend received on 320 shares of ₹ 50 = ₹ 5 × 320
= ₹ 1600
Change in the two dividends = ₹ (1600 – 640) = ₹ 960
Question 9.
(a) The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30° and 24° respectively.
Find the height of the two towers. Give your answer correct to 3 significant figures. [4]
(b) The weight of 50 workers is given below:
| Weight(in kg) | No. of workers |
| 50-60 | 4 |
| 60-70 | 7 |
| 70-80 | 11 |
| 80-90 | 14 |
| 90-100 | 6 |
| 100-110 | 5 |
| 110-120 | 3 |
Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use a graph to estimate the following:
(i) the upper and lower quartiles.
(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight. [6]
Solution:
(a) Here, AB be the first tower and CD be the second tower.
In ΔAEC, we have
\(\frac{\mathrm{AE}}{\mathrm{EC}}\) = tan 30°
⇒ \(\frac{\mathrm{AE}}{120}\) = \(\frac{1}{\sqrt{3}}\) [∵ EC = BD = 120 m (given)]
⇒ AE = \(\frac{120}{\sqrt{3}}\) = \(\frac{40 \sqrt{3} \times \sqrt{3}}{\sqrt{3}}\) = 40\(\sqrt{3}\)
= 69.28 m
Let h be the height of second tower, then CD = h m
AB = AE + EB = 69.28 + h
Now, in ri ΔCDB = tan 24°
⇒ \(\frac{h}{120}\) = 0.4452
h = 120 × 0.4452
= 53.424 m
Thus, height of the first tower
= (53.424 + 69.28) m = 122.704 m
(b) To draw an ogive
Since the given frequency distribution is continuous.
So, cumulative frequency table for the given data is:
| Weight(in kg) | No. of workers | Cumulative frequency |
| 50-60 | 4 | 4 |
| 60-70 | 7 | 11(4 + 7) |
| 70-80 | 11 | 22(11 + 11) |
| 80-90 | 14 | 36(22 + 14) |
| 90-100 | 6 | 42(36 + 6) |
| 100-110 | 5 | 47(42 + 5) |
| 110 – 120 | 3 | 50(47 + 3) |
| Total | 50 |
Since the x-axis starts at 50, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 50.
Take 2 cm = 10 kg on x-axis and Take 2 cm
= 5 workers on y-axis
Plot the points (50, 0), (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47), (120, 50)
Join these points by using free hand drawing. The required ogive is drawn on the graph paper.
Here n = number of workers = 50
(i) To find upper quartile:
Let A be the point on y-axis representing frequency
= \(\frac{3 n}{4}\) = \(\frac{3 \times 50}{4}\) = 37.5
Through A, draw a horizontal line to meet the ogive at B. Through B draw a vertical line to meet the x-axis at C. The abscissa of the point C represents 92.5 kg.
∴ The upper quartile = 92.5 kg
To find the lower quartile:
Let D be the point on y-axis representing frequency
= \(\frac{n}{4}\) = \(\frac{50}{4}\) = 12.5
Through D, draw a horizontal line to meet the ogive at E. Through E draw a vertical line to meet the x-axis at F. The abscissa of the point F represents 72 kg
∴ The lower quartile is 72 kg
(ii) On the graph point G represents 95 kg. Through G, draw a vertical line to meet the ogive at H. Through H, draw a horizontal line to meet y-axis at I. The ordinate of the point I represents 40 workers ony-axis
∴ The number of workers who are 95 kg and above = Total number of workers – number of workers of
weight less than 95 kg = 50 – 40 = 10
Question 10.
(a) A wholesaler buys a TV from the manufacturer for ₹ 25,000. He marks the price of the TV 20% above his cost price and sells it to a retailer at a 10% discount on the marked price. If the rate of VAT is 8%, find the:
(i) marked price.
(ii) retailer’s cost price inclusive of tax.
(iii) VAT paid by the wholesaler. [3]
(b) If A = \(\left[\begin{array}{ll}
3 & 7 \\
2 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
0 & 2 \\
5 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & -5 \\
-4 & 6
\end{array}\right]\). Find AB – 5C.
(c) ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
(i) ΔADE ~ ΔACB.
(ii) If aC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of ΔADE: area of quadrialteral BCED. [4]
Solution:
(a) For the wholesaler:
Cost price of the TV = ₹ 25000
Tax paid = 8% of ₹ 25000
= ₹ 2000
Now, price marked by the wholesaler is 20% above cost price
= ₹ (25000 + 20% of 25000)
= ₹ (25000 + 5000) = ₹ 30000
Discount = 10% of marked price
= \(\left(\frac{10 \times 30000}{100}\right)\) = ₹ 3000
Selling price of the TV = ₹ (30000 – 30000)
= ₹ 27000
For the retailor, cost price of the TV = ₹ 27000
VAT = 8% of ₹ 27000
= ₹ \(\frac{27000 \times 8}{100}\) = ₹ 2160
Cost price of the TV for retailer, inclusive of tax = ₹ (27000 + 2160) = ₹ 29160
VAT paid by the wholesaler = Tax recovered on sale – Tax paid on purchase = ₹ (2160 – 2000) = ₹ 160
(b)
(c)
In ΔADE and ΔACB
∠AED = ∠ABC = 90° ….. (given)
∠A = ∠A (common)
ΔADE \(\sim\) ΔACB
(A.A. axiom of similarity)
(ii) In right ΔABC, we have
AC = 13 cm, BC = 5 cm
We know, area of rt.
ΔABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 5 × 12 = 30 cm2
Since ΔADE \(\sim\) ΔACB
(iii) For area of ΔADE : area of quad. BCED
Area of quad. BCED
= area of ΔABC – area of ΔADE
= 30 – \(\frac{10}{3}\) = \(\frac{80}{3}\) cm2
⇒ Area of ΔADE : Area of quad. BCED = 1 : 8
Question 11.
(a) Sum of two natural numbers is 8 and the difference of their reciprocal is \(\frac{2}{15}\). Find the numbers. [3]
(b) Given \(\frac{x^3+12 x}{6 x^2+8}\) = \(\frac{y^3+27 y}{9 y^2+27}\). Using componendo and dividendo find x:y. [3]
(c) Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC. [4]
Solution:
(a) Consider two natural numbers x and y such that x > y.
Now, \(\frac{1}{x}\) < \(\frac{1}{y}\)
By using the statement of question, we have
x + y = 8 …. (i)
\(\frac{1}{y}\) – \(\frac{1}{x}\) = \(\frac{2}{15}\) …(ii)
From (i) x = 8 – y … (iii)
From (ii) and (iii), we have
\(\frac{1}{y}\) – \(\frac{1}{8-y}\) = \(\frac{2}{15}\) ⇒ \(\frac{8-y-y}{y(8-y)}\) = \(\frac{2}{15}\)
⇒ \(\frac{8-2 y}{8 y-y^2}\) = \(\frac{2}{15}\)
⇒ 120 – 30y = 16y – 2y2
⇒ 2y2 – 46y + 120 = 0
⇒ y2 – 23y + 60 = 0
⇒ y2 – 20y – 3y + 60 = 0
⇒ y(y – 20) – 3 (y – 20) = 0
⇒ (y – 3)(y – 20) = 0
Now, y – 3 = 0 ⇒ y = 3
or y – 20 = 0 ⇒ y = 20
Putting y = 3 in (iii), we have x = 5
Putting y = 20 in (iii), we have x = -12
Since x = -12 is not a natural number, neglecting this value of y
Hence, the numbers are 3 and 5.
(b)
\(\frac{x^3+12 x}{6 x^2+8}\) = \(\frac{y^3+27 y}{9 y^2+27}\)
Apply componendo and dividendo
Again, applying componendo and dividendo, we have
\(\frac{x+2+x-2}{x+2-x+2}\) = \(\frac{y+3+y-3}{y+3-y+3}\)
⇒ \(\frac{2 x}{4}\) = \(\frac{2 y}{6}\) ⇒ \(\frac{x}{2}\) = \(\frac{y}{3}\)
⇒ \(\frac{x}{y}\) = \(\frac{2}{3}\) ⇒ x : y = 2 : 3
(c) Steps of construction :
Constract the triangle ABC with AB = 5.5 cm
∠BAC = 105° and AC = 6 cm
(i) Points which are equidistant from BA and BC lies on the bisector of ∠ABC. Construct bisector of ∠ABC.
(ii) Points equidistant from B and C lies on the perpendicular bisector of BC. Draw perpendicular bisector of BC. The required point P is the point of intersection of the bisector of ∠ABC and the perpendicular bisector of BC.
(iii) Required length of PC = 4.8 cm.
