Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2018)

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) Find the value of $\displaystyle x \text{ and } y$ if

$\displaystyle 2 \begin{bmatrix} x & y \\ 9 & y-5 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix} \hspace{6.0cm} [3]$

(b) Sonia had a recurring deposit in a bank and deposited Rs. 600 per month for 2.5 years. If the rate of interest was 10% per annum, find the maturity value of this account.                                                                                                             [3]

(c) Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from this bag. Find a probability of getting a card which is:

(i) a prime number

(ii) a number divisible by 4

(iii) a number that is multiple of 6

(iv) an odd number                                                                                              [4]

$\displaystyle \text{(a) } 2 \begin{bmatrix} x & y \\ 9 & y-5 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x & 2y \\ 18 & 2y-10 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x+6 & 2y-7 \\ 22 & 2y-5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$

$\displaystyle \Rightarrow 2x+6=10 \Rightarrow x = 2$

and $\displaystyle 27-5 = 15 \Rightarrow y = 10$

Hence, $\displaystyle x =2 \text{ and } y = 10$

(b) Monthly investments $\displaystyle (x) = 600 \text{ Rs.}$

$\displaystyle n = 30 \text{ months} , \ \ \ r = 10\%$

$\displaystyle I = \frac{n(n+1)}{2} \times x \times \frac{1}{12} \times \frac{r}{100}$

$\displaystyle = \frac{30(31)}{2} \times 600 \times \frac{1}{12} \times \frac{10}{100}$

$\displaystyle = 2325 \text{ Rs.}$

Maturity value $\displaystyle = x \times n + I = 600 \times 30 + 2325 = 20325 \text{ Rs.}$

(c) Cards: $\displaystyle 2, 4, 6, 8, 10, 12, 14, 16, 18 \text{ and } 20$

$\displaystyle n(S) = 10$

(i) Prime numbers: $\displaystyle 2$

$\displaystyle n(E) = 1$

$\displaystyle \text{Probability (a prime number) } = \frac{n(E)}{n(S)} = \frac{1}{10}$

(ii) Numbers divisible by $\displaystyle 4: 4, 8, 12, 16, 20$

$\displaystyle n(E) = 5$

$\displaystyle \text{Probability (a prime number) } = \frac{n(E)}{n(S)} = \frac{5}{10} = 0.5$

(iii) Numbers multiple of $\displaystyle 6: 6, 12, 18$

$\displaystyle n(E) = 3$

$\displaystyle \text{Probability (a prime number) } = \frac{n(E)}{n(S)} = \frac{3}{10} = 0.5$

(iv) Odd numbers: None

$\displaystyle n(E) = 0$

$\displaystyle \text{Probability (a prime number) } = \frac{n(E)}{n(S)} = \frac{0}{10} = 0$

$\displaystyle \\$

Question 2:

(a) The circumference of the base of a cylindrical vessel is $\displaystyle 132 \text{ cm }$ and its height is $\displaystyle 25 \text{ cm }$. Find the

$\displaystyle \text{(ii) volume of the cylinder (use } \pi = \frac{22}{7} ) \hspace{6.5cm} [3]$

(b) If $\displaystyle (k-3), (2k+1) \text{ and } (4k+3)$ are three consecutive terms of an A.P., find the value of $\displaystyle k$.                                                                                                                              [3]

(c) $\displaystyle PQRS$ is a cyclic quadrilateral. Given that $\displaystyle \angle QPS = 73^{\circ}, \angle PQS = 55^{\circ} \text{ and } \angle PSR = 82^{\circ}$, calculate

(i) $\displaystyle \angle QRS$

(ii) $\displaystyle \angle RQS$

(iii) $\displaystyle \angle PRQ$                                                                                                                 [4]

(a) Circumference $\displaystyle = 132 \text{ cm }$

Height $\displaystyle = 25 \text{ cm }$

Circumference $\displaystyle = 2\pi r = 132$

$\displaystyle r = \frac{132 \times 7}{2 \times 22} = 21 \text{ cm }$

Volume $\displaystyle = \pi r ^2 h$

$\displaystyle = \frac{22}{7} \times 21 \times 21 \times 25$

$\displaystyle = 34650 \text{ cm}^3$

(b) $\displaystyle (k-3), (2k+1) \text{ and } (4k+3)$ are in AP

$\displaystyle \Rightarrow (2k+1) - (k-3) = (4k+3) -(2k+1)$

$\displaystyle \Rightarrow k+4 = 2k+2$

$\displaystyle \Rightarrow k = 2$

Therefore the terms are $\displaystyle -1, 5, 11$ and the common difference is $\displaystyle 6$

(c) Join $\displaystyle PR$

(i) $\displaystyle \angle SPQ + \angle QRS = 180^{\circ}$

$\displaystyle \Rightarrow \angle QRS = 180^{\circ} - 73^{\circ} = 107^{\circ}$ (opposite angles of a cyclic quadrilateral)

(ii) $\displaystyle \angle PSR + \angle PQR = 180^{\circ}$

$\displaystyle \Rightarrow 82^{\circ}+55^{\circ}+\angle SQR = 180^{\circ}$

$\displaystyle \Rightarrow \angle SQR = 180^{\circ} -82^{\circ}-55^{\circ} = 43^{\circ}$

(iii) $\displaystyle \angle PSR = 55^{\circ} = \angle PQS$ (angles in the same segment)

Therefore $\displaystyle \angle SRQ = 180^{\circ} - 73^{\circ} = 107^{\circ}$

$\displaystyle \angle PRQ = 105^{\circ} - 55^{\circ} = 52^{\circ}$

$\displaystyle \\$

Question 3:

(a) If $\displaystyle (x+2$) and $\displaystyle (x+3)$ are factors of $\displaystyle x^3+ax+b$, find the values of $\displaystyle a \text{ and } b$.     [3]

(b) Prove that $\displaystyle \sqrt{\sec^2 \theta + \mathrm{cosec}^2 \theta} = \tan \theta + \cot \theta$                                                              [3]

(c) Using a graph paper draw a histogram for the given distribution showing the number of runs scored by 50 batsman. Estimate the mode of the data.     [4]

 Runs Scored 3000-4000 4000-5000 5000-6000 6000-7000 7000-8000 8000-9000 9000-10000 No. of Batsman 4 18 9 6 7 2 4

(a) $\displaystyle f(x) = x^3+ax+b$

Factors: $\displaystyle x+2 = 0 \Rightarrow x = -2$

$\displaystyle x+3 = 0 \Rightarrow x = -3$

$\displaystyle f(-2) = (-2)^3+a(-2)+b =0$

$\displaystyle \Rightarrow 2a-b = -8 \text{ ... ... ... ... ... (i)}$

$\displaystyle f(-3) = (-3)^3+a(-3)+b =0$

$\displaystyle \Rightarrow -3a+b = 27 \text{ ... ... ... ... ... (ii)}$

Solving (i) and (ii)

$\displaystyle 2a-b=-8 \\ \underline{-3a+b=27} \\ -a = 19$

$\displaystyle \Rightarrow a = -19$

Substituting in (ii)

$\displaystyle 2(-19)-b = -8$

$\displaystyle \Rightarrow b = -30$

(b) $\displaystyle \sqrt{\sec^2 \theta + \mathrm{cosec}^2 \theta} = \tan \theta + \cot \theta$

$\displaystyle LHS = \sqrt{\sec^2 \theta + \mathrm{cosec}^2 \theta}$

$\displaystyle = \sqrt{1+\tan^2 \theta + 1 + \cot^2 \theta}$

$\displaystyle = \sqrt{2+\tan^2 \theta + \cot^2 \theta}$

$\displaystyle = \sqrt{2 \tan \theta \cot \theta+\tan^2 \theta + \cot^2 \theta}$

$\displaystyle =\sqrt{(\tan \theta + \cot \theta)^2}$

$\displaystyle = \tan \theta + \cot \theta$

$\displaystyle = \text{ RHS }$. Hence proved.

(c)

Mode $\displaystyle = 4600$

$\displaystyle \\$

Question 4:

(a) Solve the following inequation, write down the solution set and represent it on a real number line.

$\displaystyle -2+10x \leq 13x + 10 \leq 24 + 10x,\ \ \ x \in Z \hspace{7.0cm} [3]$

(b) If the straight lines $\displaystyle 3x-5y=7 \text{ and } 4x +ay+9 =0$ are perpendicular to one another, find the value of $\displaystyle a$.                                                                                              [3]

(c) Solve $\displaystyle x^2 + 7x = 7$ and give your answer correct to two decimal places.         [4]

(a) $\displaystyle -2+10x \leq 13x + 10 \leq 24 + 10x, x \in Z$

First equation: $\displaystyle -2+10x \leq 13x + 10$

$\displaystyle \Rightarrow -12 \leq 3x$

$\displaystyle \Rightarrow -4 \leq x$

Second equation: $\displaystyle 13x + 10 \leq 24 + 10x$

$\displaystyle \Rightarrow 3x < 14$

$\displaystyle \Rightarrow x < \frac{14}{3}$

$\displaystyle \therefore -4 \leq x < \frac{14}{3}$

Hence the solution set $\displaystyle = \{x \in Z: -4, -3, -2, -1, 0, 1, 2, 3, 4 \}$

(b) Two equations are:

$\displaystyle 3x-5y=7 \Rightarrow y = \frac{3}{5} x - \frac{7}{5} \Rightarrow m_1 = \frac{3}{5}$

$\displaystyle 4x+ay+9=0 \Rightarrow y = \frac{-4}{a} x - \frac{1}{a} \Rightarrow m_2 = \frac{-4}{a}$

Since the two lines are perpendicular to each other

$\displaystyle m_1 \times m_2 = -1$

$\displaystyle \frac{3}{5} \times \frac{-4}{a} = -1 \Rightarrow a = \frac{12}{5}$

(c) $\displaystyle x^2+7x=7$

$\displaystyle x^2+7x-7=0$

$\displaystyle \Rightarrow a = 1, b = 7, c = -7$

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

$\displaystyle = \frac{-7 \pm \sqrt{7^2+4.1.17}}{2.1}$

$\displaystyle = \frac{-7 \pm \sqrt{77}}{2}$

$\displaystyle = \frac{-7 \pm 8.77}{2}$

$\displaystyle \text{Therefore } x = \frac{-7 + 8.77}{2.1}, \frac{-7 - 8.77}{2}$

$\displaystyle \text{or } x = 0.885 \approx 0.89, -7.885 \approx 7.89$

$\displaystyle \\$

SECTION B [40 Marks]

(Attempt any four questions from this Section.)

Question 5:

(a) The $\displaystyle 4^{th}$ term of a G.P is 16 and the $\displaystyle 7^{th}$ term is 128. Find the first term and the common ratio of the series.                                                                                           [3]

(b) A man invests Rs. 22500 in Rs. 50 shares available at 10% discount. If the dividend paid by the company is 12%, calculate:

(i) The number of shares purchased

(iii) The rate of return he gets on his investment. Give your answer correct to the nearest whole number.                                                                              [3]

(c) Use graph paper for this question (Take 2 cm = 1 unit along both $\displaystyle x-axis \text{ and } y-axis$). $\displaystyle ABCD$ is a quadrilateral whose vertices are $\displaystyle A(2,2), B(2, -2), C(0,-1) \text{ and } D(0,1)$.

(i) Reflect quadrilateral $\displaystyle ABCD$ on the y-axis and name it as $\displaystyle A'B'CD$

(ii) Write down the coordinate of $\displaystyle A' \text{ and } B'$

(iii) Name two points which are invariant under the above reflection

(iv) Name the polygon $\displaystyle A'B'CD$                                                                            [4]

(a) $\displaystyle 4^{th} \text{term} = 16, 7^{th} \text{term} = 128$

$\displaystyle T_n = ar^{n-1}$

$\displaystyle \text{Therefore } T_4: 16 = ar^3 \Rightarrow \frac{16}{a} =r^3$

$\displaystyle T_7: 128 = ar^6$

$\displaystyle \Rightarrow 128 = a(r^3)^2$

$\displaystyle \Rightarrow 128 = a. (\frac{16}{a})^2$

$\displaystyle \Rightarrow 128 = a . \frac{16^2}{a^2}$

$\displaystyle \Rightarrow a = \frac{256}{128} = 2$

$\displaystyle \text{Therefore } r^3 = \frac{16}{2} = 8 \Rightarrow r = 2$

(b) Face Value $\displaystyle (FV) = 50 \text{ Rs.}, \text{ Discount } = 10\%$

$\displaystyle \text{Market Value } (MV) = 50 - \frac{10}{100} \times 50 = 45 \text{ Rs.}$

Total Investment $\displaystyle = 22500 \text{ Rs.}$

$\displaystyle \text{(i) Number of shares bought } (n) = \frac{22500}{45} = 500$

$\displaystyle \text{(ii) Annual dividend received } = FV \times n \times \text{Dividend} \% = 50 \times 500 \times \frac{12}{100} = 3000 \text{ Rs.}$

$\displaystyle \text{(iii) Rate of return } = \frac{\text{Dividend Received}}{\text{Total Investment}} \times 100$

$\displaystyle = \frac{3000}{22500} \times 100 = 13.33\% \approx 13\%$

(c)

(i) Refer to the graph

(ii) $\displaystyle A'(-2, 2) \text{ and } B'(-2, -2)$

(iii) $\displaystyle C \text{ and } D$ are irrelevant to the reflection

(iv) $\displaystyle A'B'CD$ is a trapezium

$\displaystyle \\$

Question 6:

(a) Using the properties of proportion, solve for $\displaystyle x$. Given that $\displaystyle x$ is positive

$\displaystyle \frac{2x+\sqrt{4x^2-1}}{2x-\sqrt{4x^2-1}} = 4 \hspace{11.5cm} [3]$

$\displaystyle \text{(b) If } A = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} , B = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} , \text{ find } AC+B^2-10C \hspace{2.0cm} [3]$

$\displaystyle \text{(c) Prove that } (1+\cot \theta - \mathrm{cosec} \theta)(1 + \tan \theta + \sec \theta) = 2 \hspace{5.5cm} [4]$

$\displaystyle \text{(a) } \frac{2x+\sqrt{4x^2-1}}{2x-\sqrt{4x^2-1}} = 4$

Applying Componendo and Dividendo

$\displaystyle \frac{(2x+\sqrt{4x^2-1}) + (2x-\sqrt{4x^2-1})}{(2x+\sqrt{4x^2-1}) - (2x-\sqrt{4x^2-1})} = \frac{4+1}{4-1}$

$\displaystyle \frac{4x}{ 2 \sqrt{4x^2-1} } = \frac{5}{3}$

$\displaystyle 6x = 5\sqrt{4x^2-1}$

$\displaystyle 36x^2 = 25(4x^2 -1 )$

$\displaystyle 36x^2 = 100x^2 - 25$

$\displaystyle 25 = 64 x^2$

$\displaystyle x^2 = \frac{25}{64} \Rightarrow x = \frac{5}{8} \text{ or } \frac{-5}{8}$

(b) $\displaystyle AC+B^2-10C$

$\displaystyle = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} . \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} . \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} - 10 . \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2-3 & 12 \\ 5-7 & 28 \end{bmatrix} + \begin{bmatrix} -4 & 28 \\ -7 & -4+49 \end{bmatrix} - 10 \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix} + \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}$

(c) $\displaystyle (1+\cot \theta - \mathrm{cosec} \theta)(1 + \tan \theta + \sec \theta) = 2$

$\displaystyle \text{LHS } = (1+\cot \theta - \mathrm{cosec} \theta)(1 + \tan \theta + \sec \theta)$

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

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

$\displaystyle = \frac{(\sin \theta + \cos \theta)^2-1^2}{\sin \theta \cos \theta}$

$\displaystyle = \frac{\sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}$

$\displaystyle = \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$

$\displaystyle = 2 = \text{ RHS}$

Hence Proved

$\displaystyle \\$

Question 7:

(a) Find the value of $\displaystyle k$ for which the following equation has equal roots.

$\displaystyle x^2+4kx+(k^2-k+2)=0$                                                                                      [3]

(b) On a map drawn to a scale of $\displaystyle 1:50000$, a rectangular plot of land $\displaystyle ABCD$ has the following dimensions. $\displaystyle AB = 6 \text{ cm }$; $\displaystyle BC = 8 \text{ cm }$ and all angles are right angles. Find :

(i) the actual length of the diagonal $\displaystyle AC$ of the plot in km

(ii) the actual area of the plot in sq. km                                                              [3]

(c) $\displaystyle A(2,5), B(-1, 2) \text{ and } C(5,8)$ are the vertices of the triangle $\displaystyle ABC$, $\displaystyle M$ is a point on $\displaystyle AB$ such that $\displaystyle AM : MB = 1:2$. Find the coordinates of $\displaystyle M$. Hence find the equation of the line passing through the point $\displaystyle C \text{ and } M$.                                       [4]

(a) Given $\displaystyle x^2+4kx+(k^2-k+2)=0$

Therefore $\displaystyle a = 1, b = 4k \text{ and } c = (k^2-k+2)$

For equal roots, $\displaystyle b^2 - 4ac = 0$

$\displaystyle \Rightarrow 16k^2 - 4(k^2-k+2)= 0$

$\displaystyle \Rightarrow 16k^2- 4k^2 + 4k -8 = 0$

$\displaystyle \Rightarrow 12k^2 + 4k - 8 = 0$

$\displaystyle \Rightarrow 3k^2 + k - 2 = 0$

$\displaystyle \Rightarrow 3k^2 + 3k - 2k - 2 = 0$

$\displaystyle \Rightarrow 3k (k+1) - 2 (k+1) = 0$

$\displaystyle \Rightarrow (k+1)(3k-2) = 0$

$\displaystyle \Rightarrow k = -1 \text{ or } \frac{2}{3}$

(b) Scale is $\displaystyle 1:50000$

(i) $\displaystyle AC = \sqrt{AB^2 +BC^2} = \sqrt{64+36} = \sqrt{100} = 10 \text{ cm }$

Let the length of $\displaystyle AC$ be $\displaystyle x$ on the ground

$\displaystyle \text{Therefore } \frac{1}{50000} = \frac{10}{x} \Rightarrow x = 500000 \text{ cm }= 5 \text{ km}$

(ii) $\displaystyle AB$ on the ground $\displaystyle = 6 \times 50000 = 300000 \text{ cm }= 3 \text{ km}$

$\displaystyle BC$ on the ground $\displaystyle = 8 \times 50000 = 400000 \text{ cm }= 4 \text{ km}$

Therefore Area of $\displaystyle ABCD = 3 \times 4 \text{ km}^2 = 12 \text{ km}^2$

(c) $\displaystyle m_1 : m_2 = 1:2 \text{ and } A(2,5), B(-1, 2) \text{ and } C(5,8)$

Let $\displaystyle M(x, y)$

$\displaystyle x = \frac{m_1.x_2 + m_2.x_1}{m_1 + m_2} = \frac{1(-1) + 2(2)}{1+2} = \frac{3}{3} = 1$

$\displaystyle y = \frac{m_1.y_2 + m_2.y_1}{m_1 + m_2} = \frac{1(2) + 2(5)}{1+2} = \frac{12}{3} = 4$

Hence $\displaystyle M(1, 4)$

$\displaystyle \text{Slope of } CM = \frac{y_2-y_1}{x_2-x_1} = \frac{8-4}{5-1} = \frac{4}{4} = 1$

Therefore equation of $\displaystyle CM$

$\displaystyle y - y_1 = m(x- x_1)$

$\displaystyle \Rightarrow y - 4 = 1 (x- 1)$

$\displaystyle \Rightarrow y - x = 3$

$\displaystyle \\$

Question 8:

(a) Rs. 7500 was divided equally among a certain number of children. Had there been 20 less children each would have received Rs.100 more. Find the original number of students.                                                                                                          [3]

(b) If the mean of the following distribution is 24, find the value of $\displaystyle a$.                [3]

 Marks 0-10 10-20 20-30 30-40 40-50 No. of Students 7 $\displaystyle a$$\displaystyle a$ 8 10 5

(c) Using ruler and compass only, construct a $\displaystyle \triangle ABC$ such that $\displaystyle BC = 5 \text{ cm }\text{ and } AB = 6.5 \text{ cm }\text{ and } \angle ABC = 120^{\circ}$

(i) Construct a semi-circle of $\displaystyle \triangle ABC$

(ii) Construct a cyclic quadrilateral $\displaystyle ABCD$ , such that $\displaystyle D$ is equidistant from $\displaystyle AB \text{ and } BC$.                                                                                                    [4]

(a) Let the number of children be $\displaystyle = x$

Therefore if $\displaystyle x$ children received $\displaystyle 7500 \text{ Rs.}$

$\displaystyle \text{Then 1 child received } \frac{7500}{x} \text{ Rs.}$

Let the number of children be $\displaystyle = (x-20)$

Therefore if $\displaystyle (x-20)$ children received $\displaystyle 7500 \text{ Rs.}$

$\displaystyle \text{Then 1 child received } \frac{7500}{x-20} \text{ Rs.}$

Given $\displaystyle \text{ } \frac{7500}{x-20} - \frac{7500}{x} = 100$

$\displaystyle \Rightarrow \frac{75}{x-20} - \frac{75}{x} = 1$

$\displaystyle 75x-7x(x-20) = (x-20)x$

$\displaystyle x^2 - 20x - 1500= 0$

$\displaystyle (x+30)(x-50)=0 \Rightarrow x = -30 ( \text{not possible}) \text{ or } 50$

Therefore the number of children be $\displaystyle = 50$

(b)

 Marks Mid Term $\displaystyle (\overline{x})$$\displaystyle (\overline{x})$ No. of Students $\displaystyle (f)$$\displaystyle (f)$ $\displaystyle fx$$\displaystyle fx$ 0-10 5 7 35 10-20 15 $\displaystyle a$$\displaystyle a$ $\displaystyle 15a$$\displaystyle 15a$ 20-30 25 8 200 30-40 35 10 350 40-50 45 5 225 $\displaystyle \Sigma f = 30+a$$\displaystyle \Sigma f = 30+a$ $\displaystyle \Sigma fx= 810+15a$$\displaystyle \Sigma fx= 810+15a$

$\displaystyle \text{Mean } (\overline{x}) = \frac{\Sigma fx}{\Sigma f}$

$\displaystyle \Rightarrow 24 = \frac{810+15a}{30+a}$

$\displaystyle \Rightarrow 720 + 24a = 810 + 15a$

$\displaystyle \Rightarrow 9a = 90$

$\displaystyle \Rightarrow a = 10$

(c)

$\displaystyle \\$

Question 9:

(a) Priyanka has a recurring deposit account of Rs. 1000 per month at 10% per annum. If she gets Rs. 5550 as interest at the time of maturity, find the total time for which the account was held.                                                                         [3]

$\displaystyle \text{(b) In } \triangle PQR , MN \text{ is parallel to } QR \text{ and } \frac{PM}{MQ}=\frac{2}{3}$

$\displaystyle \text{(i) Find } \frac{MN}{QR}$

(ii) Prove that $\displaystyle \triangle OMN \text{ and } \triangle ORQ$ are similar

$\displaystyle \text{(iii) Find, Area of} \triangle OMN : \text{ Area of } \triangle ORQ$                                                       [3]

(c) The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and cone is each 4 cm. Find the volume of the solid.                                                                                                                                      [4]

(a) Monthly Income $\displaystyle (x) = 1000 \text{ Rs.}$

$\displaystyle r= 10\%, I = 5550 \text{ Rs. Number of months} = n$

$\displaystyle I = \frac{n(n+1)}{2} \times \frac{1}{12} \times x \times \frac{r}{100}$

$\displaystyle \Rightarrow 5550 = \frac{n(n+1)}{2} \times \frac{1}{12} \times 1000 \times \frac{10}{100}$

$\displaystyle \Rightarrow n(n+1) = \frac{5550 \times 2 \times 12}{100}$

$\displaystyle \Rightarrow n^2+n - 1332 = 0$

$\displaystyle \Rightarrow n^2 + 37n = 36n - 1332 = 0$

$\displaystyle \Rightarrow n(n+37)-36(n+37) = 0$

$\displaystyle \Rightarrow (n+37)(n-36) = 0$

$\displaystyle \Rightarrow n = -37 or 36$ (negative number is not possible)

Therefore $\displaystyle n = 36$

$\displaystyle \text{Therefore total time line} = 36 \text{ months or } 3 \text{ years }$

(b)

(i) In $\displaystyle \triangle PMN \text{ and } \triangle PQR$

$\displaystyle \angle P$ is common

SInce $\displaystyle MN \parallel QR$

$\displaystyle \Rightarrow \angle PMN = \angle PQR \text{ and } \angle PNM = \angle PRQ$ (alternate angles)

Therefore $\displaystyle \triangle PMN \sim \triangle PQR$ (by AAA Postulate)

$\displaystyle \Rightarrow \frac{PM}{PQ} = \frac{MN}{QR}$

$\displaystyle \Rightarrow \frac{PM}{PM+MQ} = \frac{MN}{QR}$

$\displaystyle \text{Given } \frac{PM}{MQ} = \frac{2}{3}$

$\displaystyle \Rightarrow \frac{2}{2+3} = \frac{MN}{QR}$

$\displaystyle \Rightarrow \frac{MN}{QR} = \frac{2}{3}$

(ii) Consider $\displaystyle \triangle OMN$ & $\displaystyle \triangle OQR$

Since $\displaystyle MN \parallel QR$

$\displaystyle \angle MNQ = \angle OQR \text{ (alternate angles)}$

$\displaystyle \angle NMO = \angle ORQ \text{ (alternate angles)}$

Therefore $\displaystyle \triangle OMN \sim \triangle OQR$ (AAA Postulate)

$\displaystyle \text{(iii) } \frac{\text{Area }\triangle OMN}{\text{Area } \triangle ORQ} = \frac{MN^2}{QR^2} = \Big( \frac{2}{5} \Big)^2 = \frac{4}{25}$

(c) Radius $\displaystyle (r) = 7 \text{ cm }$, Height of Cylinder $\displaystyle (h) = 4 \text{ cm }=$ Height of Cone $\displaystyle (h)$

Volume of total solid = Volume of Cone + Volume of Cylinder + Volume of Hemisphere

$\displaystyle = \frac{1}{3} \pi r^2 h + \pi r^2 h + \frac{2}{3} \pi r^3$

$\displaystyle = \pi r^2 ( \frac{h}{3} + h + \frac{2}{3} r)$

$\displaystyle = \pi r^2 ( \frac{4}{3} h + \frac{2}{3} r)$

$\displaystyle = \frac{22}{7} \times 7^2 ( \frac{4}{3} \times 4 + \frac{2}{3} \times 7)$

$\displaystyle = \frac{154}{3} \times (16+14) = 1540 \text{ cm}^3$

$\displaystyle \\$

Question 10:

(a) Use remainder theorem to factorize the following polynomial.

$\displaystyle 2x^3+3x^2-9x-10$                                                                                                               [3]

(b) In the figure given below $\displaystyle O$ is the center of the circle. If $\displaystyle OR=OP \text{ and } \angle ORP = 20^{\circ}$. Find the value of $\displaystyle x$ giving reasons.                             [3]

(c) The angle of elevation from a point $\displaystyle P$ of the top of the tower $\displaystyle QR$ , $\displaystyle 50 \text{ m}$ high is $\displaystyle 60^{\circ}$ and that the tower $\displaystyle PT$ from the point $\displaystyle Q$ is $\displaystyle 30^{\circ}$. Find the height of the tower $\displaystyle PT$, correct to the nearest meter.                                                                                   [4]

(a) $\displaystyle f(x) = 2x^3+3x^2-9x-10$

Let $\displaystyle x+1$ be a factor $\displaystyle \Rightarrow x = -1$

$\displaystyle f(-1) = 2(-1)^3+3(-1)^2-9(-1)-10 = -2 + 3 + 9 -10 = 0$

Therefore $\displaystyle x+1$ is a factor

$\displaystyle \begin{array}{r l l } x+1 ) & \overline{2x^3+3x^2-9x-10} & (2x^2+x-10 \\ (-) & \underline{2x^3+2x^2} & \\ & \hspace{1.0cm} x^2-9x-10 & \\ (-) &\hspace{1.0cm} \underline{x^2+x \ \ \ \ \ \ \ \ \ \ \ } & \\ & \hspace{1.5cm} -10x-10 & \\ (-) & \hspace{1.5cm} \underline {-10x-10} & \\ & \hspace{3.0cm} \times & \\ \end{array}$

$\displaystyle f(x) = (x+1) (2x^2+x-10)$

$\displaystyle = (x+1) (2x^2 + 5x-4x-10)$

$\displaystyle = (x+1) (2x+5)(x-2)$

(b) $\displaystyle OP = OQ =$ Radius

$\displaystyle QR = OP$ (given)

Therefore $\displaystyle QR = OQ$

In $\displaystyle \triangle OQR$ since $\displaystyle OQ = QR$ (Isosceles triangle)

$\displaystyle \angle QOR = \angle QRO = 20^{\circ}$

$\displaystyle \angle OQP = 180^{\circ}-140^{\circ} = 40^{\circ}$

Since $\displaystyle \triangle OPQ$ is Isosceles triangle, $\displaystyle \angle OPQ = \angle OQP = 40^{\circ}$

Therefore $\displaystyle \angle POQ = 180 - 40 -40 = 100^{\circ}$

Therefore $\displaystyle x + 100^{\circ} + 20^{\circ} = 180^{\circ}$ (straight line)

$\displaystyle \Rightarrow x = 60^{\circ}$

(c)  From $\displaystyle \triangle PRQ:$

$\displaystyle \tan 60^{\circ} = \frac{50}{x} \Rightarrow x = \frac{50}{\sqrt{3}}$

From $\displaystyle \triangle PTQ:$

$\displaystyle \tan 30^{\circ} = \frac{y}{\frac{50}{\sqrt{3}}} \Rightarrow y = \frac{50}{3} = 16.67 \text{ m } \approx 17 \text{ m}$

$\displaystyle \\$

Question 11:

(a) The $\displaystyle 4^{th}$ term of an A.P. is 22 and $\displaystyle 15^{th}$ term is 66. Find the first term and the common difference. Hence find the sum of the series to 8 terms.                 [4]

(b) Use graph paper for this question

A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded:

Taking 2 cm = height of 10 cm along one axis and 2 cm = 10 boys along the other axis draw an ogive of the above distribution. Use the graph to estimate the following:

 Height in cm 135-140 140-145 145-150 150-155 155-160 160-165 165-170 No of boys 4 8 20 14 7 6 1

(i) the median

(ii) the lower quartile

(iii) if above 158 cm is considered as a tall boy in a class, find the number of boys who are tall in the class.                                                                          [6]

(a) $\displaystyle T_n = a+ (n-1)d$

$\displaystyle T_4: 22 = a+(4-1) d \Rightarrow a+ 3d = 22$ … … … … … (i)

$\displaystyle T_15: 66 = a+(15-1) d \Rightarrow a+ 14d = 66$ … … … … … (ii)

Solving (i) and (ii)

$\displaystyle {\hspace{0.7cm} a+ 14d = 66} \\ (-) \ \underline{a+ 3d = 22} \\ {\hspace{1.25cm} 11d \ = 44 }$

$\displaystyle \Rightarrow d = 4$

Substituting in (i)

$\displaystyle a+ 3 \times 4 = 22 \Rightarrow a = 10$

$\displaystyle S_n = \frac{n}{2} \{ 2a+ (n-1)d \}$

$\displaystyle S_8 = \frac{8}{2} \{ 2\times 10+ (8-1) \times 4 \}$

$\displaystyle = 4 (20+28) = 192$

(b)

 Wages No. workers $\displaystyle (f)$$\displaystyle (f)$ Cumulative Frequency (c.f) $\displaystyle 135-140$$\displaystyle 135-140$ $\displaystyle 140-145$$\displaystyle 140-145$ $\displaystyle 145-150$$\displaystyle 145-150$ $\displaystyle 150-155$$\displaystyle 150-155$ $\displaystyle 155-160$$\displaystyle 155-160$ $\displaystyle 160-165$$\displaystyle 160-165$ $\displaystyle 165-170$$\displaystyle 165-170$ $\displaystyle 4$$\displaystyle 4$ $\displaystyle 8$$\displaystyle 8$ $\displaystyle 20$$\displaystyle 20$ $\displaystyle 14$$\displaystyle 14$ $\displaystyle 7$$\displaystyle 7$ $\displaystyle 6$$\displaystyle 6$ $\displaystyle 1$$\displaystyle 1$ $\displaystyle 4$$\displaystyle 4$ $\displaystyle 12$$\displaystyle 12$ $\displaystyle 32$$\displaystyle 32$ $\displaystyle 46$$\displaystyle 46$ $\displaystyle 53$$\displaystyle 53$ $\displaystyle 59$$\displaystyle 59$ $\displaystyle 60$$\displaystyle 60$

On the graph paper, we plot the following points:

$\displaystyle (4, 140), (12, 145), (32, 150), (46,155), (53, 160), (59,165), (60,170)$

$\displaystyle n = 60$

$\displaystyle \text{(i) Median } = (\frac{n}{2})^{th} \text{term} = \frac{60}{2} =30^{th} \text{term}$

From the graph $\displaystyle 30^{th} \text{term}=149$

$\displaystyle \text{(ii) Lower quartile } = (\frac{n}{4})^{th} \text{term} = \frac{60}{4} =15^{th} \text{term}$

From the graph $\displaystyle 15^{th} \text{term} =146$

(iii) The number of boys $\displaystyle \geq 158 \text{ cm }= 60-50 = 10$ students