Other Solved Mathematics Board Papers


MATHEMATICS (ICSE – Class X Board Paper 2019)

Two and Half HourAnswers 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 form Section A and any four questions from Section BAll 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 the loss of marks.

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables are provided.


SECTION A [40 Marks]

(Answer all questions from this Section.)


Question 1:

(a) Solve the following in-equation and write down the solution set:

11x - 4 < 15x + 4 \leq 13x + 14 , x \in W

(b) A man invests Rs. 4500 in shares of a company which is paying 7.5\% dividend. If Rs. 100 shares are available at a discount of 10\% . Find:

(i) Number of shares he purchases.

(ii) His annual income.

(c) In a class of 40 students, marks obtained by the students in a class test (out of 10 ) are given below: Calculate the following for the given distribution:

Marks 1 2 3 4 5 6 7 8 9 10
Number of Students 1 2 3 3 6 10 5 4 3 3

(i) Median

(ii) Mode

Answer:

(a) Given 11x - 4 < 15x + 4 \leq 13x + 14 , x \in W

Case 1: 11x - 4 < 15x + 4

\Rightarrow -4x < 8   \Rightarrow -x < 2 \ or \  x > -2

Case 2: 15x + 4 \leq 13x + 14

\Rightarrow 2x \leq 10   \Rightarrow x \leq 5

\therefore x \in [ -1, 0, 1, 2, 3, 4, 5]

(b)  Rs. 100 shares at a discount of 10\% will cost = 100 - 10 = 90 Rs.

i) Therefore Number of shares = \frac{4500}{90} = 50

ii) His annual income at 7.5\% dividend = \frac{7.5}{100} \times 50 \times 100 = 337.5 Rs.

(c)

Marks (x) No. of Students Cumulative Frequency
1 1 1
2 2 3
3 3 6
4 3 9
5 6 15
6 10 25
7 5 30
8 4 34
9 3 37
10 3 40
Total 40

i) Total number of Students = 40 which is even

Median = \frac{1}{2} \Bigg[ \Big( \frac{n}{2} \Big) \ term + \Big( \frac{n}{2} \Big)^{th} \ term \Bigg]

= \frac{1}{2} \Bigg[ \Big( \frac{40}{2} \Big) \ term + \Big( \frac{40}{2} \Big)^{th} \ term \Bigg]

= \frac{1}{2} [ 20^{th} term + 21^{st} term ]

= \Big(  \frac{41}{2} \Big)^{th} term

= 20.5^{th} term

Which is between 15 and 25 . Therefore Median = 6

ii) Mode frequency of 6 is the highest. Therefore Mode = 3

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Question 2: 

(a) Using the factor theorem, show that (x - 2) is a factor of x^3 + x^2 - 4x - 4 . Hence factorize the polynomial completely.

(b) Prove that:  ( \mathrm{cosec} \theta - \sin \theta )(\sec \theta - \cos \theta)(\tan \theta+\cot \theta)=1

(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the:

(i) first term

(ii) common difference

(iii) sum of the first 20 terms.

Answer:

(a)  f(x) = x^3 + x^2 - 4x - 4

Let x-2 = 0 \Rightarrow x = 2

\therefore f(2) = (2)^3 + (2)^2 - 4(2) - 4

\Rightarrow f(2) = 8 + 4 - 8 - 4 = 0

f(2) = 0

Therefore (x-2) is a factor of f(x)

x-2 ) \overline{x^3 + x^2 - 4x - 4} ( x^2 + 3x + 2 \\ \hspace*{0.5cm} (-)  \underline{x^3 - 2x^2 \hspace*{1.8cm}} \\ \hspace*{2.1cm} {3x^2 - 4x - 4} \\ \hspace*{1.5cm} (-) \underline{3x^2 - 6x \hspace*{0.8cm}} \\ \hspace*{3.1cm}2x-4 \\ \hspace*{2.5cm} (-) \underline {2x-4} \\ \hspace*{3.7cm}0

Hence x^3 + x^2 - 4x - 4 = (x-2)(x^2 + 3x + 2)

\Rightarrow x^3 + x^2 - 4x - 4 = (x-2)(x+2)(x+1)

(b)   To prove: ( \mathrm{cosec} \theta - \sin \theta )(\sec \theta - \cos \theta)(\tan \theta+\cot \theta)=1

LHS = ( \mathrm{cosec} \theta - \sin \theta )(\sec \theta - \cos \theta)(\tan \theta+\cot \theta)

= \Big( \frac{1}{\sin \theta} - \sin \theta \Big) \Big( \frac{1}{\cos \theta} - \cos \theta\Big) \Big( \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \Big)

= \Big( \frac{1 - \sin^2 \theta}{\sin \theta} \Big) . \Big( \frac{1 - \cos^2 \theta}{\cos \theta} \Big) . \Big( \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta . \cos \theta} \Big)

= \Big( \frac{\cos^2 \theta}{\sin \theta} \Big) . \Big( \frac{\sin^2 \theta}{\cos \theta} \Big) . \Big( \frac{1}{\sin \theta . \cos \theta} \Big)

= 1 = RHS. Hence proved.

(c)  Let the first term of the sequence is a and the common difference is d .

a_4 = a + 3d = 8    … … … … … i)

a_6 = a + 5d = 14    … … … … … ii)

Solving i) and ii) we get d = 3 , and a = -1

Therefore

i) First term (a) = -1

ii) Common difference (d) = 3

iii) Sum of the first 20 terms = S_n = \frac{n}{2} \Big[ 2a + (n-1)d \Big]

= \frac{20}{2} \Big[ -2 + (20-1)d \Big] = 10 \times 55 = 550

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Question 3: 

(a) Simplify: \sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}

(b) M and N are two points on the X axis and Y axis respectively. P (3, 2) divides the line segment MN in the ratio 2 : 3 . Find:

(i) the coordinates of M and N

(ii) slope of the line MN .

(c) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the:

(i) radius of the cylinder

(ii) curved surface area of the cylinder     [Take \pi = 3.1 ]

Answer:

(a)  Given \sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}

\Rightarrow \begin{bmatrix} \sin^2 A & - \sin A \cos A \\ \sin A \cos A & \sin^2 A \end{bmatrix} +   \begin{bmatrix} \cos^2 A & \cos A \sin A \\ - \cos A \sin A & \cos^2 A \end{bmatrix}

\Rightarrow \begin{bmatrix} \sin^2 A + \cos^2 A & - \sin A \cos A + \cos A \sin A \\ \sin A \cos A - \cos A \sin A & \sin^2 A + \cos^2 A \end{bmatrix}

\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

(b) Let coordinates of M is (x_1, 0) and N is (0, y_2) . Point P divides MN in 2 : 3 ratio

\therefore 3 = \frac{3x_1+ 2 \times 0}{3+2}  \Rightarrow 3x_1 = 15 \Rightarrow x_1 = 5

and 2 = \frac{3 \times 0 + 2 \times y_2}{3+2} \Rightarrow 3y_2 = 10 \Rightarrow y_2 = 5

Therefore i) The coordinates of M is (5, 0) and N(0, 5)

ii) Slope of line MN = \frac{0-5}{5-0} = - 1

(c)  Let the radius of the sphere is r_1 and radius of cylinder is r_2 and the height of the cylinder is h .

Therefore  Volume of sphere = Volume of cylinder

\Rightarrow \frac{4}{3} \pi {r_1}^3 = \pi {r_2}^2 h

\Rightarrow \frac{4}{3} ( {6}^3) =  {r_2}^2 (32)

\Rightarrow {r_2}^2 = \frac{12 \times 6}{8} = 9

\Rightarrow r_2 = 3

Therefore Radius of the cylinder is 3 cm

Curved surface area = 2 \pi r h = 2 \times 3.1 \times 3 \times 32 = 595.2 \ cm^2

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Question 4:

(a) The following numbers, K + 3, K + 2, 3K - 7 and 2K - 3 are in proportion. Find K .

(b) Solve for x the quadratic equation x^2 - 4x - 8 = 0 . Give your answer correct to three significant figures

(c) Use ruler and compass only for answering this question.

Draw a circle of radius 4 cm. Mark the center as O . Mark a point P outside the circle at a distance of 7 cm from the center. Construct two tangents to the circle from the external point P .

Measure and write down the length of any one tangent.

Answer:

(a)  Given K + 3, K + 2, 3K - 7 and 2K - 3 are in proportion.

\Rightarrow \frac{K+3}{K+2} = \frac{3K-7}{2K-3}

\Rightarrow (K+3)(2K-3) = (3K-7)(K+2)

\Rightarrow 2K^2 + 6K - 3K - 9 = 3K^2 - 7K + 6K - 14

\Rightarrow 2K^2 + 3K - 9 = 3K^2 - K  - 14

\Rightarrow K^2 - 4K-5 = 0

\Rightarrow K^2 - 5K + K -5 = 0

\Rightarrow K(K-5) + (K-5) = 0

\Rightarrow (K-5)(K+1) = 0

\Rightarrow K = 5 \ or \  K = -1

(b)  Given x^2 - 4x - 8 = 0

Comparing the above equation by ax^2 + bx + c = 0

a =1, b = -4, c = -8

x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}

x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-8)} }{2(1)}

x = \frac{4 \pm \sqrt{16+32} }{2}

x = \frac{ 4 \pm \sqrt{48} }{2}

x = \frac{4 \pm 6.928}{2} 2019-10-01_8-30-07

x = 5.464 or x = 1.464

(c)  i. Draw a line segment OP = 7 cm

ii. With center O and radius 4 cm, draw a circle.

iii. Draw the mid point of OP .

iv. With center M and diameter OP , draw a circle which intersect the circle at T and S .

v. Joint PT and PS

PT and PS are the required tangent on measuring the length of PT = PS = 5.74 cm

SECTION B (40 Marks)
Attempt any four questions from this Section

Question 5:

(a) There are 25 discs numbered 1 to 25 . They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. Find the probability that the number on the disc is:

(i) an odd number

(ii) divisible by 2 and 3 both.

(iii) a number less than 16 .

(b) Rekha opened a recurring deposit account for 20 months. The rate of interest is 9\% per annum and Rekha receives Rs. 441 as interest at the time of maturity. Find the amount Rekha deposited each month.

(c) Use a graph sheet for this question. Take 1 cm = 1 unit along both x and y axis.

(i) Plot the following points: A(0,5), B(3,0), C(1,0) and D(1,-5)

(ii) Reflect the points B, C and D on the y axis and name them as B', C' and D' respectively.

(iii) Write down the coordinates of B', C' and D' .

(iv) Join the points A, B, C, D, D', C', B', A in order and give a name to the closed figure ABCDD'C'B' .

Answer:

(a) Total number of cases = 25

(i) an odd number \{ 1,3,5,7,9,11,13,15,17,19,21,23,25 \}

Therefore Probability of an odd number = \frac{13}{25} 

ii) Divisible by 2 and 3 both \{ 6,12,18,24 \}

Probability of the number divisible by 2 and 3 both = \frac{4}{25} 

iii) Probability a number less than 16 = \frac{15}{25} = \frac{3}{5} 

(b)  Let the monthly installment i.e P = Rs. x

Since n = 20 months and r = 9\%

Therefore Interest = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

= \frac{x \times 20(20+1)}{2 \times 12} \times \frac{9}{100}

= \frac{x \times 20 \times 21 }{2 \times 12} \times \frac{9}{100}

= \frac{x \times 5 \times 7 }{2 } \times \frac{9}{100}

Interest  = 1.575x Rs.

\Rightarrow 441 = 1.575x \Rightarrow x = 280 Rs.

(c)  Please refer to the graph below for answers. the shape of the figure is “Arrow Head”.

2019-10-02_7-01-29

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Question 6:2019-09-26_10-09-34

(a) In the given figure, \angle PQR = \angle PST = 90^o, PQ = 5 cm and PS = 2 cm.

(i) Prove that \triangle PQR \sim \triangle PST .

(ii) Find Area of \triangle PQR : Area of quadrilateral SRQT .

(b) The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2 , find:

(i) 'n' the number of terms of the G.P.

(ii) Sum of the n terms.2019-09-26_9-44-18

(c) A hemispherical and a conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows:

The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm. Give your answer correct to the nearest whole number. Take \pi = \frac{22}{7} .

Answer:

(a)  

i)    To prove \triangle PQR \sim \triangle PST

Consider \triangle PQR and \triangle PST

\angle PQR = \angle PST = 90^o      (Given)

\angle P is common

\therefore \triangle PQR \sim \triangle PST    (By AA criterion)

ii)   \frac{ Ar \triangle PQR}{ Ar \triangle PST } = \frac{ \frac{1}{2} \times PQ \times QR}{\frac{1}{2} \times PS \times ST} = \frac{5}{2} \times \frac{5}{2} = \frac{25}{4}                       \Big[ \because \frac{PQ}{PS} = \frac{QR}{ST} \Big]

Taking the reciprocals on both sides

\frac{ Ar \triangle PST }{ Ar \triangle PQR} = \frac{4}{25}

Now deducting both sides by 1

1 - \frac{ Ar \triangle PST }{ Ar \triangle PQR} = 1 - \frac{4}{25}

\Rightarrow \frac{ Ar \triangle PQR - Ar \triangle PST }{ Ar \triangle PQR} = \frac{25 - 4}{25}

\Rightarrow \frac{ Ar \ of \ Quadrilateral \ QRST }{ Ar \triangle PQR} = \frac{21}{25}

(b) Given first term (a) = 3 and last term (l) = 96

i)    r = 2

We know l = a r^{n-1}

\Rightarrow 96 = 3 \times 2^{n-1}

\Rightarrow 32 = 2^{n-1}

\Rightarrow 2^5 = 2^{n-1}

\therefore n = 6

ii) Sum of the n terms = (S_n) = \frac{a r^{n-1}}{r-1} = \frac{(3) 2^{6-1}}{2-1} = 3 \times 63 = 189

(c) Required volume = Volume of cylinder - Volume of hemisphere - Volume of cone

Volume of cone = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (3)^2 \times 3 = 9 \pi \ cm^3

Volume of hemisphere = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \pi \times 3^3 = 18 \pi \ cm^3

Volume of cylinder = \pi r^2 h = \pi \times 3^3 \times 7 = 63 \pi \ cm^3

Therefore required volume = 63 \pi - 18\pi - 9 \pi = 36 \pi = 36 \times \frac{22}{7} = 113.14 \ cm^3

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Question 7:

(a) In the given figure AC is a tangent to the circle with center O . If \angle ADB = 55^o , find x and y . Give reasons for your answers.

2019-09-26_9-37-34

(b) The model of a building is constructed with the scale factor 1 : 30 .

(i) If the height of the model is 80 cm, find the actual height of the building in meters.

(ii) If the actual volume of a tank at the top of the building is 27 \ m^3 , find the volume of the tank on the top of the model.

(c) Given \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} , where M is a matrix and I is unit matrix of order 2x^2 .

(i) State the order of matrix M .

(ii) Find the matrix M .

Answer:2019-09-26_9-37-34

(a) In \triangle ABD, \angle BAD = 90^o

\therefore \angle ABD+ \angle BAD+ \angle ADB = 180^o   [Sum of angles in a triangle]

\therefore \angle ABD + 90^o + 55^o= 180^o

\therefore \angle ABD= 35^o

Also \angle AOE = 2\angle ABE = 2\angle ABD

\Rightarrow y = 70^o

In \triangle AOC,

\angle OAC + x + y = 180^o

\therefore 90^o + x + 70^o=180^o

\therefore 90^o + x + 70^o=180^o

\Rightarrow x = 20^o

(b)

i) If the scale factor is 1 : 30 , then actual height will be 30 times the height of the model.

The height of the model is 80 cm

Therefore actual height = 80 cm \times 30 = 2400 cm =24 \ m

ii) Now, actual volume of a tank will be (30)^3 times the volume of a tank in the model.

Therefore Volume of tank in model \times (30)^3 = Actual volume of a tank

Therefore volume of tank = \frac{27}{30^3} \ m^3 = 0.001 \ m^3 = 1000 \ cm^3

(c) Given

i) \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} M = 6I

\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} M = 6 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} M = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

Therefore M has the order of 2 \times 2

ii) Let us assume M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

\therefore \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

\Rightarrow \begin{bmatrix} 4a+2c & 4b+2d \\ -a + c & -b+d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

\Rightarrow -a+c = 0 \Rightarrow a = c

Similarly, 4b+2d = 0 \Rightarrow d = - 2b

Now, 4a + 2c = 6 \Rightarrow  4a + 2a = 6 \Rightarrow 6a = 6 \Rightarrow a = 1 

also, -b + d = 6  \Rightarrow -b + (-2b) = 6  \Rightarrow -3b =6  \Rightarrow b= -2

\therefore a =1, b =-  2, c =1, d = 4

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Question 8:

(a) The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52 . Find the first term and the common difference.

(b) The vertices of a \triangle ABC are A(3, 8), B(-1, 2) and C(6, -6) . Find:

(i) Slope of BC .

(ii) Equation of a line perpendicular to BC and passing through A .

(c) Using ruler and a compass only construct a semi-circle with diameter BC = 7 cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C . Complete the cyclic quadrilateral ABCD , such that D is equidistant from AB and BC . Measure \angle ADC and write it down.

Answer:

(a)  Let the three times terms of an A.P. be (a - d ),a, (a + d)

Sum: 42=  (a - d)+ a + (a + d) \Rightarrow 42 = 3a \Rightarrow a = 14

Also (a -d) (a + d) = 52

\Rightarrow a^2 - d^2 = 52

d^2 = a^2 - 52 = 196 - 52 = 144

\Rightarrow d = \pm 12

When d = 12, a = 14-  12 = 2

When d = - 12, a = 14 + 12 = 26

Hence the two APs are 2, 14, 26, \cdots and 26, 14, 2, \cdots

(b) Given A(3, 8), B(-1, 2), C(6, -6)

i) Slope  of BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6-2}{6-(-1)} = - \frac{8}{7}

ii) Slope of line perpendicular to BC = \frac{-1}{-\frac{8}{7}} = \frac{7}{8}

Therefore required line is y - y_1 = m(x-x_1)

\Rightarrow y - 8 = \frac{7}{8} (x-3)

\Rightarrow 8y - 64 = 7x - 21

\Rightarrow 7x - 8y  + 43 = 0

(c)2019-10-02_12-14-11.png

i. Draw a line segment BC = 7 cm

ii. Taking mid point of BC as center O , draw a semi-circle with radius = 3.5 cm

iii. Now, the semicircle circumscribes the \triangle ABC

iv. Draw angle bisector of \angle ABC and make it intersect the semi-circle at D .

v. Measure the angle \angle DBC which comes out to be 22.5^o

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Question 9:

(a) The data on the number of patients attending a hospital in a month are given below. Find the average (mean) number of patients attending the hospital in a month by using the shortcut method. Take the assumed mean as 45 . Give your answer correct to 2 decimal places.

Number of Patients 10-20 20-30 30-40 40-50 50-60 60-70
Number of Days 5 2 7 9 2 5

(b) Using properties of proportion solve for x , given \frac{\sqrt{5x} + \sqrt{2x-6}}{\sqrt{5x} + \sqrt{2x-6}} = 4

(c) Sachin invests Rs. 8500 in 10\% , Rs. 100 shares at Rs. 170 . He sells the shares when the price of each share rises by Rs. 30 . He invests the proceeds in 12\% Rs. 100 shares at Rs. 125 . Find:

(i) the sale proceeds.

(ii) the number of Rs. 125 shares he buys.

(iii) the change in his annual income.

Answer:

(a)

Number of Patients Number of Days (f_i) Mid Value (x_i) Assumed Mean A = 45   \therefore d = x_i - A f_i \times d_i
10-20 5 15 -30 -150
20-30 2 25 -20 -40
30-40 7 35 -10 -70
40-50 9 45 0 0
50-60 2 55 10 20
60-70 5 65 20 100
Total 30 -140

Mean = A + \frac{ \Sigma f_id_i  }{f_i} = 45 + \frac{-140}{30} = 45 - \frac{14}{3} = \frac{121}{3} = 40.33

(b) Given \frac{\sqrt{5x} + \sqrt{2x-6}}{\sqrt{5x} + \sqrt{2x-6}} = 4

Using componendo and dividendo on both sides

\frac{(\sqrt{5x} + \sqrt{2x-6})+ (\sqrt{5x} + \sqrt{2x-6})}{\sqrt{5x} + \sqrt{2x-6}) - (\sqrt{5x} + \sqrt{2x-6})} = \frac{4+1}{4-1}

\frac{\sqrt{5x}}{\sqrt{2x-6}} = \frac{5}{3}

On squaring both sides

\frac{5x}{2x-6} = \frac{25}{9}

\Rightarrow 45x = 50x -150

\Rightarrow 150 =  5x

\Rightarrow x = 30

(c)

i) Market value of shares = 170 Rs.

Therefore the number of shares bought = \frac{8500}{170} = 50

Total face value of 50 shares = 50 \times 100 = 5000 Rs.

Income form these shares = \frac{10}{100} \times 5000 = 500 Rs.

Selling price of the shares = 170 + 30 = 200 Rs.

Sale proceed = 50 \times 200 = 10000 Rs.

ii) Market value of the shares bought = 125 Rs.

Therefore number of Rs. 125 shares bought = \frac{10000}{125} = 80

Total face value of 80 shares = 80 \times 100 = 8000 Rs.

iii) Income from Rs. 125 shares = \frac{12}{100} \times 8000 = 960 Rs.

Therefore increase in income = 960 - 500 = 460 Rs.

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Question 10: 

(a) Use graph paper for this question. The marks obtained by 120 students in an English test are given below:

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Number of Students 5 6 16 22 26 18 11 6 4 3

Draw the ogive and hence, estimate:

(i) the median marks.

(ii) the number of students who did not pass the test if the pass percentage was 50 .

(iii) the upper quartile marks.

(b) A man observes the angle of elevation of the top of the tower to be 45^o . He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to 60^o . Find the height of the tower correct to 2 significant figures.

Answer:

(a)

Class Interval Frequency Cumulative Frequency
0-10 5 5
10-20 9 14
20-30 16 30
30-40 22 52
40-50 26 78
50-60 18 96
60-70 11 107
70-80 6 113
80-90 4 117
90-100 3 120

2019-10-02_18-30-32

Here n = 120 , therefore \frac{n}{2} = 60

i) Median = 43

ii) Number of students who failed = 78

iii) The upper Quartile marks = \frac{3}{4} n^{th} term = 90^{th} term

(b) Let the height of the tower be h m.

In \triangle ADC, \tan 45^o = \frac{h}{20+x}

\Rightarrow 1 = \frac{h}{20+x}

\Rightarrow h = 20 + x    … … … … … i)

In \triangle BDC, \tan 60^o = \frac{h}{x}

\sqrt{3} = \frac{h}{x}

\Rightarrow x = \frac{h}{\sqrt{3}}    … … … … … ii)

Using i) and ii), we get

h = 20 + \frac{h}{\sqrt{3}}

\Rightarrow h - \frac{h}{\sqrt{3}} = 20 

\Rightarrow h \Big(  \frac{\sqrt{3} -1 }{\sqrt{3}} \Big) = 20 

\Rightarrow h = 47.32  m

Therefore Height of the tower = 47.32 m

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Question 11:

(a) Using the Remainder Theorem find the remainders obtained when x^3 + (kx + 8)x + k is divided by x + 1 and x - 2 . Hence find k if the sum of the two remainders is 1 .

(b) The product of two consecutive natural numbers which are multiples of 3 is equal to 810 . Find the two numbers.

2019-09-26_9-21-18(c) In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC \parallel AE . If \angle BAC = 50^o , find giving reasons:

(i) \angle ACB

(ii) \angle EDC

(iii) \angle BEC

Hence prove that BE is also a diameter.

Answer:

(a) Remainder theorem:

Dividend = Divisors \times Quotient + Remainder

 \therefore Let f (x) = x^3 + (kx + 8) x + k = x^3 + kx^2 + 8x + k

Dividing f (x) by x + 1 gives remainder as R_1

 \therefore f(-1) = R_1

Also f(2) = R_2

 \therefore f(-1) = (-1)^3 + k(-1)^2 + 8(-1) + k = -1 + k - 8 + k = 2k-9 = R_1

 f(2) = (2)^3 + k (2)^2 + 8 \times 2 + k = 8 + 4k + 16 + k = 5k + 24  = R_2

Sum of remainder = R_1 + R_2 = 1

 \therefore 2k-9 + 5k + 24 = 1

 \Rightarrow 7k + 15 = 1

 \Rightarrow k = -2

(b)    Let the numbers be 3x, 3(x+1)

Therefore (3x) \times 3(x+1) = 810

9(x^2 + x) = 810

x^2 + x = 90

x^2 + x - 90 = 0

x^2 + 10x - 9 x - 90 = 0

x(x+10) - 9 (x+10) = 0

(x+10)(x-9) = 0

x = 9 or x = -10

But x \neq -10 , because 3x must be natural number.2019-10-02_19-34-22.png

Therefore x = 9

Therefore the numbers are 27, 30

(c)  Let \angle ACB = x, \angle EDC = y and \angle BEC = z

We know \angle ABC = 90^o      Angle in a semi-circle

i) In \triangle ABC, \angle ABC + \angle BAC + \angle ACB = 180^o

\Rightarrow 90^o + 50^o + \angle ACB = 180^o

\Rightarrow \angle ACB = 40^o= x

ii) We know \angle EAC = \angle ACB   

Alternate interior angles since AC is transversal to parallel  lines  AE and BC

\therefore \angle EAC  = 40^o

Also \angle EAC + \angle EDC = 180^o

\therefore 40^o + \angle EDC = 180^o

\Rightarrow \angle EDC = 140^o = y

iii)  \angle EBC + \angle EDC = 180^o

\Rightarrow \angle EBC + 140^o = 180^o

\Rightarrow \angle EBC = 40^o

Therefore in \triangle EBC

\angle BEC + \angle ECB + \angle EBC =180^o

\angle BEC + 90^o + 40^o =180^o

\angle BEC = 50^o

Also in \triangle EAB

\angle EAB = \angle EAC + \angle BAC = 40^o + 50^o

\angle EAB = 90^o

We know, if an angle of a triangle in a circle is 90^o then, the hypotenuse must be the diameter of the circle. Hence, BE is the diameter of the circle.