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 $x$ and $y$ if

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

(b) Sonia had a recurring deposit in a bank and deposited Rs. 600 per month for $2$ $\frac{1}{2}$ 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]

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

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

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

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

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

Hence, $x =2 \ and \ y = 10$

(b)  Monthly investments $(x) = 600 \ Rs.$

$n = 30 \ months$, $r = 10\%$

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

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

$= 2325 \ Rs.$

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

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

$n(S) = 10$

(i) Prime numbers: $2$

$n(E) = 1$

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

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

$n(E) = 5$

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

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

$n(E) = 3$

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

(iv) Odd numbers: None

$n(E) = 0$

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

$\\$

Question 2:

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

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

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

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

(i) $\angle QRS$

(ii) $\angle RQS$

(iii) $\angle PRQ$     [4]

(a)   Circumference $= 132 \ cm$

Height $= 25 \ cm$

Circumference $= 2\pi r = 132$

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

Volume $= \pi r ^2 h$

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

$= 34650 \ cm^3$

(b)  $(k-3), (2k+1)$ and $(4k+3)$ are in AP

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

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

$\Rightarrow k = 2$

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

(c)   Join $PR$

(i) $\angle SPQ + \angle QRS = 180^o$

$\Rightarrow \angle QRS = 180^o - 73^o = 107^o$ (opposite angles of a cyclic quadrilateral)

(ii) $\angle PSR + \angle PQR = 180^o$

$\Rightarrow 82^o+55^o+\angle SQR = 180^o$

$\Rightarrow \angle SQR = 180^o -82^o-55^o = 43^o$

(iii) $\angle PSR = 55^o = \angle PQS$ (angles in the same segment)

Therefore $\angle SRQ = 180^o - 73^o = 107^o$

$\angle PRQ = 105^o - 55^o = 52^o$

$\\$

Question 3:

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

(b) Prove that $\sqrt{sec^2 \theta + csec^2 \theta} = tan \ \theta + cot \ \theta$     [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)  $f(x) = x^3+ax+b$

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

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

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

$\Rightarrow 2a-b = -8$ … … … … … (i)

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

$\Rightarrow -3a+b = 27$  … … … … … (ii)

Solving (i) and (ii)

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

$\Rightarrow a = -19$

Substituting in (ii)

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

$\Rightarrow b = -30$

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

$LHS =$ $\sqrt{sec^2 \theta + csec^2 \theta}$

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

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

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

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

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

$= RHS$. Hence proved.

(c)

Mode $= 4600$

$\\$

Question 4:

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

$-2+10x \leq 13x + 10 \leq 24 + 10x, x \in Z$     [3]

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

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

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

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

$\Rightarrow -12 \leq 3x$

$\Rightarrow -4 \leq x$

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

$\Rightarrow 3x < 14$

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

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

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

(b)  Two equations are:

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

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

Since the two lines are perpendicular to each other

$m_1 \times m_2 = -1$

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

(c)  $x^2+7x=7$

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

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

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

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

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

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

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

or $x = 0.885 \approx 0.89, -7.885 \approx 7.89$

$\\$

SECTION B [40 Marks]

(Attempt any four questions from this Section.)

Question 5:

(a) The $4^{th}$ term of a G.P is $16$ and the $7^{th}$ 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 $x-axis$ and $y-axis$). $ABCD$ is a quadrilateral whose vertices are $A(2,2), B(2, -2), C(0,-1)$ and $D(0,1)$.

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

(ii) Write down the coordinate of $A'$ and $B'$

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

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

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

$T_n = ar^{n-1}$

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

$T_7: 128 = ar^6$

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

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

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

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

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

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

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

Total Investment $= 22500 \ Rs.$

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

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

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

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

(c)

(i) Refer to the graph

(ii) $A'(-2, 2)$ and $B'(-2, -2)$

(iii) $C$ and $D$ are irrelevant to the reflection

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

$\\$

Question 6:

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

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

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

(c) Prove that $(1+cot \ \theta - cosec \ \theta)(1 + tan \ \theta + sec \ \theta) = 2$     [4]

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

Applying Componendo and Dividendo

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

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

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

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

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

$25 = 64 x^2$

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

(b)  $AC+B^2-10C$

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

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

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

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

(c)   $(1+cot \ \theta - cosec \ \theta)(1 + tan \ \theta + sec \ \theta) = 2$

$LHS = (1+cot \ \theta - cosec \ \theta)(1 + tan \ \theta + sec \ \theta)$

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

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

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

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

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

$= 2 = RHS$

Hence Proved

$\\$

Question 7:

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

$x^2+4kx+(k^2-k+2)=0$     [3]

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

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

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

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

(a)  Given $x^2+4kx+(k^2-k+2)=0$

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

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

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

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

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

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

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

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

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

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

(b)  Scale is $1:50000$

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

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

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

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

$BC$ on the ground $= 8 \times 50000 = 400000 \ cm = 4 \ km$

Therefore Area of $ABCD = 3 \times 4 \ km^2 = 12 \ km^2$

(c)  $m_1 : m_2 = 1:2$ and $A(2,5), B(-1, 2)$ and $C(5,8)$

Let $M(x, y)$

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

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

Hence $M(1, 4)$

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

Therefore equation of $CM$

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

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

$\Rightarrow y - x = 3$

$\\$

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 $a$     [3]

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

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

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

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

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

Therefore if $x$ children received $7500 \ Rs.$

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

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

Therefore if $(x-20)$ children received $7500 \ Rs.$

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

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

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

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

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

$(x+30)(x-50)=0 \Rightarrow x = -30 (not possible) \ or \ 50$

Therefore the number of children be $= 50$

(b)

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

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

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

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

$\Rightarrow 9a = 90$

$\Rightarrow a = 10$

(c)

$\\$

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]

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

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

(ii) Prove that $\triangle OMN$ and $\triangle ORQ$ are similar

(iii) Find, $Area \ of \ \triangle OMN : Area \ of \ \triangle ORQ$     [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 $(x) = 1000 \ Rs.$

$r= 10\%, \ \ \ I = 5550 Rs. \ \ \ Number \ of \ months = n$

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

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

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

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

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

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

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

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

Therefore $n = 36$

Therefore total time line$= 36$ months or $3$ years

(b)  (i) In $\triangle PMN$ and $\triangle PQR$

$\angle P$ is common

SInce $MN \parallel QR$

$\Rightarrow \angle PMN = \angle PQR$ and $\angle PNM = \angle PRQ$ (alternate angles)

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

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

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

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

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

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

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

Since $MN \parallel QR$

$\angle MNQ = \angle OQR$ (alternate angles)

$\angle NMO = \angle ORQ$ (alternate angles)

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

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

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

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

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

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

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

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

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

$\\$

Question 10:

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

$2x^3+3x^2-9x-10$     [3]

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

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

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

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

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

Therefore $x+1$ is a factor

Now dividing:

$x+1 ) \overline{2x^3+3x^2-9x-10} (2x^2+x-10$

$(-) \underline{2x^3+2x^2}$

$x^2-9x-10$

$(-) \underline{x^2+x}$

$-10x-10$

$(-) \underline {-10x-10}$

$\times$

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

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

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

(b)  $OP = OQ =$ Radius

$QR = OP$ (given)

Therefore $QR = OQ$

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

$\angle QOR = \angle QRO = 20^o$

$\angle OQP = 180^o-140^o = 40^o$

Since $\triangle OPQ$ is Isosceles triangle, $\angle OPQ = \angle OQP = 40^o$

Therefore $\angle POQ = 180 - 40 -40 = 100^o$

Therefore $x + 100^o + 20^o = 180^o$ (straight line)

$\Rightarrow x = 60^o$

(c) From $\triangle PRQ:$

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

From $\triangle PTQ:$

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

$\\$

Question 11:

(a) The $4^{th}$ term of an A.P. is 22 and $15^{th}$ 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)  $T_n = a+ (n-1)d$

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

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

Solving (i) and (ii)

$a+ 14d = 66 \\ (-) \underline{a+ 3d = 22} \\ 11d = 44$

$\Rightarrow d = 4$

Substituting in (i)

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

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

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

$= 4 (20+28) = 192$

(b)

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

On the graph paper, we plot the following points:

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

$n = 60$

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

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

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

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

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

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