Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2017)

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) If $b$ is the mean proportion between $a$ and $c$, show that;     [3]

$\frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}$ $=$ $\frac{a^2}{c^2}$

(b) Solve the equation $4x^2-5x-3=0$ and give your answer correct to two decimal.       [4]

(c) $AB$ and $CD$ are two parallel chords of a circle such that $AB=24 \ cm$ and $CD=10 \ cm$. If the radius of the circle is $13 \ cm$, find the distance between the two chords.     [3]

(a)  $\frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}$ $=$ $\frac{a^2}{c^2}$

Given $\frac{a}{b}$ $=$ $\frac{b}{c}$ $\Rightarrow b^2 = ac$

Now, LHS $=$ $\frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}$

$=$ $\frac{a^4+a^2(ac)+(ac)^2}{(ac)^2+(ac)c^2+c^4}$

$=$ $\frac{a^2(a^2+ac+c^2)}{c^2(a^2+ac+c^2)}$

$=$ $\frac{a^2}{c^2}$ $=$ RHS.

Hence Proved.

(b)  The equation is $4x^2-5x-3=0$

We know the roots of a quadratic equation are: $x =$ $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Here $a = 4, b = -5 \ and \ c = -3$

Substituting, we get

$x =$ $\frac{5 \pm \sqrt{25+48}}{8}$

$\Rightarrow x =$ $\frac{5 \pm \sqrt{73}}{8}$

$\Rightarrow x =$ $\frac{5 \pm 8.54}{8}$

$\Rightarrow x =$ $\frac{5 + 8.54}{8}, \ \frac{5 - 8.54}{8}$

$\Rightarrow x =$ $\frac{13.54}{8}, \ \frac{-3.54}{8}$

$\Rightarrow x = 1.69 \ or \ -0.44$

(c)   $ON = \sqrt{13^2-12^2} = \sqrt{25} = 5 \ cm$

$OM = \sqrt{13^2-5^2} = \sqrt{144} = 12 \ cm$

$MN = NO + OM = 5 + 12 = 17 \ cm$

$\\$

Question: 2

(a) Evaluate without using trigonometric tables,

$sin^2\ 28^o + sin^2\ 62^o+tan^2\ 38^o-cot^2\ 52^o + \frac{1}{4} sec^2\ 30^o$     [3]

(b) If  $A = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}$ and $A^2 - 5B^2= 5C$ Find matrix $C$ where $C$ is a $2 \ by \ 2$ matrix     [4]

(c) Jaya borrowed $Rs.50000$ for $2$ years . The rates of interest for two successive years are $12\%$ and $15\%$ respectively. She repays $Rs.33000$ at the end of the first year. Find the amount she must pay at the end of the second year to clear her debt.     [3]

(a) $sin^2\ 28^o + sin^2\ 62^o+tan^2\ 38^o-cot^2\ 52^o + \frac{1}{4} sec^2\ 30^o$

$=$ $sin^2\ 28^o + sin^2\ (90^o - 28^o)+tan^2\ 38^o-cot^2\ (90^o - 38^o) + \frac{1}{4} sec^2\ 30^o$

$=$ $sin^2\ 28^o + cos^2\ 28^o+tan^2\ 38^o-tan^2\ 38^o +$ $\frac{1}{4}$ $\times$ $\frac{4}{3}$

$=$ $1+$ $\frac{1}{3}$

$=$ $\frac{4}{3}$

(b) $A^2 - 5B^2= 5C$

$\Rightarrow \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} - 5 \times \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} = 5C$

$\Rightarrow \begin{bmatrix} 1+9 & 3+12 \\ 3+12 & 9+16 \end{bmatrix} - 5 \times \begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4 \end{bmatrix} = 5C$

$\Rightarrow \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} - 5 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 5C$

$\Rightarrow \begin{bmatrix} 5 & 15 \\ 15 & 20 \end{bmatrix} = 5C$

$\Rightarrow \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} = C$

(c)  We know $A = P (1+$ $\frac{r}{100})^n$

$A$ is the amount and $P$ is the principal.

For first year:

$P = 55000 \ Rs., r = 12\%, n = 1$

$A = 55000 (1+$ $\frac{12}{100})^1$ $= 550 \times 112 = 56000 \ Rs$

For second year:

$P = (55000 - 33000) = 23000 \ Rs, r = 15\%, n = 1$

$A = 23000 (1+$ $\frac{15}{100})^1$ $= 230 \times 115 = 26450 \ Rs$

$\\$

Question: 3

(a) The catalog price of a computer set is $Rs.42000$. The Shopkeeper gives a discount of $10\%$ on the listed price. He further gives an off-season discount of $5\%$ on the discounted price. However, sales tax at $8\%$ is charged on the remaining price after the two successive discounts. Find:

(i) The amount of sales tax a customer has to pay

(ii) The total price to be paid by the customer for the computer set.     [3]

(b) $P(1,-2)$ is a point on the line segment $A(3,-6)$ and $B(x,y)$ such that $AP:PB$ is equal to $2:3$. Find the coordinates of $B$    [4]

(c) The marks of 10 students of a class in an examination arranged in ascending order is as follows: $13, 35, 43, 46, x, x+4, 55, 61, 71, 80$. If the median marks is 48, find the value of $x$. Find the mode of the given data.    [3]

(a)  Cost Price $= 42000 \ Rs.$

1st Discount $= 10\%$

$10\% \ of \ 42000 Rs. =$ $\frac{10}{100}$ $\times 42000 = 4200 \ Rs.$

Price after discount $= 42000 - 4200 = 37800 \ Rs.$

2nd Discount $= 5\%$

$5\% \ of \ 37800 \ Rs. =$ $\frac{5}{100}$ $\times 37800 = 1890 \ Rs.$

Price after discount $= 37800 - 1890 = 35910 \ Rs.$

(i) Sales tax $(@ 8\%) =$ $\frac{8}{100}$ $\times 35910 = 2872.80 \ Rs$

(ii) Price to be paid by the customer $= 35910 + 2872.80 = 38782.80 \ Rs.$

(b)  $A(3, -6), P(1, -2)$ and let $B(x, y)$

$m:n = 2:3$

We know the ratio formula: $x =$ $\frac{mx_2+nx_1}{m+n}$ $, y =$ $\frac{my_2+ny_1}{m+n}$

Therefore

$1 =$ $\frac{2x+3 \times 3}{2+3}$ $\Rightarrow 5 = 2x+9 \Rightarrow x = -2$

Similarly

$-2 =$ $\frac{2y+3 \times (-6)}{2+3}$ $\Rightarrow -10 = 2y-18 \Rightarrow y = 4$

Hence the coordinate of $B(-2, 4)$

(c)   $13, 35, 43, 46, x, x+4, 55, 61, 71, 80$

Given: $Median = 48$ and $n(number \ of \ terms) = 10$

Median for even $n:$ $\frac{\frac{n^{th}}{2} term + \frac{(n+1)^{th}}{2} term}{2}$

Hence $48 =$ $\frac{x+x+4}{2}$ $\Rightarrow 96 = 2x+4 \Rightarrow 2x=92 \Rightarrow x = 46$

Hence the $5^{th} \ term = 46$ and the $6^{th} \ term = 50$

Looking at all the terms, we see that $46$ is repeated twice and hence the mode is $46$

$\\$

Question 4:

(a) What must be subtracted from $16x^3-8x^2+4x+7$  the resulting expressing has $2x+1$ as a factor?     [3]

(b) In the given figure $ABCD$ is a rectangle. It consists of a circle and two semi circles each of which are of radius $5 \ cm$. Find the area of the shaded region. Give your answer correct to three significant figures.     [4]

(c) Solve the following in equation and represent the solution set on a number line.

$-8\frac{1}{2} < -\frac{1}{2}-4x \leq 7\frac{1}{2}, \ \ x \in I$     [3]

(a)  $f(x) = 16x^3-8x^2+4x+7$

Given $2x+1$ is a factor $\Rightarrow 2x+1 = 0 \Rightarrow x = -$ $\frac{1}{2}$

$f($ $\frac{-1}{2})$ $= 16($ $\frac{-1}{2})^3$ $-8($ $\frac{-1}{2})^2$ $+4($ $\frac{-1}{2}$ $)+7= -2-2-2+7 = 1$

Therefore $1$ has to be subtracted from the given polynomial.

(b)  Radius of the circle $= 5 \ cm$

Therefore, as shown in the diagram: Breadth of the rectangle $= 10 \ cm$ and Length of the rectangle $= 20 \ cm$

Area of $ABCD = 20 \times 10 = 200 \ cm^2$

Taking $\pi = 3.14$

Area of the circles and semi circles $= 2 \times \pi r^2 = 2 \times 3.14 \times 5^2 = 157 \ cm^2$

Therefore the shaded area $= 200 = 157 = 43 \ cm^2$

(c)  Given $-8\frac{1}{2} < -\frac{1}{2}-4x \leq 7\frac{1}{2}$

$\Rightarrow$ $-\frac{17}{2}$ $<$ $\frac{-1-8x}{2}$ $\leq$ $\frac{15}{2}$

$\Rightarrow -17 < -1-8 \leq 15$

Therefore we have two equations:

 $-17 < -1-8x$ $\Rightarrow 17 > 1+8x$ $\Rightarrow 8x < 16$ $\Rightarrow x < 2$ $-1-8x \leq 15$ $\Rightarrow 1+8x \geq -15$ $\Rightarrow 8x \geq -16$ $\Rightarrow x \geq -2$

Hence $-2 \leq x < 2$

Since $x \in I$, the values of $x \ are \ -2, -1, 0, 1$

$\\$

Section B (40 Marks)

Attempt any four questions from this Section:

Question: 5

(a) Given matrix $B = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}$. Find the matrix $X$  if $X = B^2-4B$.  Hence solve for $a \ and \ b$ given that $X \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}$     [4]

(b) How much should a man invest in $Rs.50$ shares selling at $Rs.60$ to obtain an income of $Rs.450$, if the rate of dividend declared is $10\%$. Also find his yield percent, to the nearest whole number.     [3]

(c) Sixteen cards are labeled as $a,b,c, ..., m,n,o,p$. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card draw is:

(i) A Vowel

(ii) A Consonant

(iii) None of the letters of the word median     [3]

(a)  $X = B^2-4B$

$\Rightarrow X= \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} - 4 \times \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}$

$\Rightarrow X= \begin{bmatrix} 1+8 & 1+3 \\ 8+24 & 8+9 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix}$

$\Rightarrow X= \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}$

$\Rightarrow X= \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

Now $X \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 5a \\ 5b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}$

$\Rightarrow a = 1 \ and \ b = 10$

(b)  Face Value $(FV) = 50 \ Rs.$

Market Value $(MV) = 60 \ Rs.$

Dividend $= 10\%$

(i) Let the number of shares bought $= n$

Dividend Amount $= FV \times (no. \ of \ shares) \times Dividend \%$

$\Rightarrow 450 = 50 \times n \times$ $\frac{10}{100}$

$\Rightarrow n =90$

(ii) $Profit = 450 \ Rs., \ n = 90 \ MV = 60 \ Rs.$

Investment $= 90 \times 60 = 5400 \ Rs.$

Yield $\% =$ $\frac{450}{5400}$ $\times 100 = 8.33\%$

Therefore yield nearest to a whole number is $8\%$

(c) No of cards $= 16$

Cards: $a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p$

(i) No of vowels: $a, e, i, o$

Probability (vowels) $=$ $\frac{4}{16}$ $= 0.25$

(ii) No of consonants: $b, c, d, f, g, h, j, k, l , m, n, p$

Probability (consonants) $=$ $\frac{12}{16}$ $= 0.75$

(iii) Letters other than ‘median’ $= b, c, f, g, h, j, k, l, o, p$

Probability (other than median) $=$ $\frac{10}{16}$ $= \frac{5}{8}$ $= 0.625$

$\\$

Question: 6

(a) Using a ruler and a compass construct a $\triangle ABC$ in which $AB=7 \ cm$, $\angle CAB = 60^o$ and $AC = 5 \ cm$. Construct the locus  of:

(i) Points equidistant from $AB$ and $AC$

(ii) Point equidistant from $BA$ and $BC$

Hence construct a circle touching the three sides of the triangles internally.      [4]

(b) A conical tent is to accommodate $77$ persons. Each person must have $16m^3$ of air to breathe. Given the radius of the tent as $7 \ m$ find the height of the tent and also its curved surface area.     [3]

(c) If $\frac{7m+2n}{7m-2n}$ $=$ $\frac{5}{3}$, use the properties of proportion to find:

(i) $m:n$

(ii)  $\frac{m^2+n^2}{m^2-n^2}$     [3]

(a)

(b) Radius of the cone $= 7 \ m$

No of people $= 77$

Volume per person $= 16 \ m^3$

Therefore the volume of the tent $= 77 \times 16 = 1232 \ m^3$

(i) Volume of the tent $=$ $\frac{1}{3}$ $\pi r^2 h$

$\Rightarrow 1232 =$ $\frac{1}{3}$ $\times \frac{22}{7}$ $\times 7^2 \times h$

$\Rightarrow h =$ $\frac{1232 \times 3}{22 \times 7}$ $= 24 \ m$

(ii) Curved surface area of a cone $= \pi r l$

$l = \sqrt{24^2+7^2} = \sqrt{576 + 49} = 25$

Therefore curved surface area $=$ $\frac{22}{7}$ $\times 7 \times 25 = 550 \ m^2$

(c)  $\frac{7m+2n}{7m-2n}$ $=$ $\frac{5}{3}$

(i) Applying componendo and dividendo

$\frac{(7m+2n)+(7m-2n)}{(7m+2n) - (7m-2n)}$ $=$ $\frac{5+3}{5-3}$

$\frac{14m}{4n}$ $=$ $\frac{8}{2}$

$\frac{7m}{2n}$ $=$ $\frac{4}{1}$

$\frac{m}{n}$ $=$ $\frac{8}{7}$

(ii) $\frac{m^2+n^2}{m^2-n^2}$

$=$ $\frac{(\frac{m}{n})^2+1}{(\frac{m}{n})^2-1}$

$=$ $\frac{(\frac{8}{7})^2+1}{(\frac{8}{7})^2-1}$

$=$ $\frac{64+49}{64-49}$ $=$ $\frac{113}{49}$

$\\$

Question: 7

(a) A page from a Saving Bank Account Passbook is given below:

 Date Particular Amount Withdrawal Amount Deposited Balance Jan 07, 2016 B/F – – 3000.00 Jan  10, 2016 By Cheque – 2600.00 5600.00 Feb 08, 2016 To Self 1500.00 – 4100.00 Apr 06, 2016 By Cheque 2100.00 – 2000.00 May 04, 2016 By Cash – 6500.00 8500.00 May 27, 2016 By Cheque – 1500.00 10,000.00

(i) Calculate the interest for the 6 months from January to June 2016, at $6\%$ per annum.

(ii) If the account is closed on 1st July 2016, find the amount received by the account holder.     [5]

(b) Use a graph for this question (Take $2 cm =1$ unit on both $x \ and \ y$ axis)

(i) Plot the following points : $A(0,4), B(2,3), C(1,1)$ and $D(2,0)$

(ii) Reflect points $B, C, D$ on the $y-axis$ and write down their coordinates. Name the images as $B' C' D'$ respectively.

(iii) Join the points $A, B, C, D', D', C', B'$ and $A$ in order, so as to form a closed figure, Write down the equation of the line of symmetry of the figure formed.     [5]

(a) Qualifying principal for various months:

 Month Principal (Rs.) January 5600 February 4100 March 4100 April 2000 May 8500 June 10000 Total 34300

$P = Rs. \ 34300 R = 6\% \ and \ T=$ $\frac{1}{12}$

(i) $I = P \times R \times T = 34300 \times$ $\frac{6}{100}$ $\times$ $\frac{1}{12}$ $= Rs. \ 171.5$

(ii) Amount received by the holder on 1st July $= 10000 + 171.5 = 10171.5 \ Rs.$

(b)

(i) Please refer to the diagram above.

(ii) $B'(-2, 3), \ C'(-1, 1) \ and \ D'(-2, 0)$

(iii) The enclosed figure is a kite. The line of symmetry is: $y = 0$ or $y-axis$

$\\$

Question: 8

(a) Calculate the mean of the following distribution using step deviation method:     [4]

 Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of Students 10 9 25 30 16 10

(b) In the given figure $PQ$ is a tangent to the circle at $A, AB$ and $AD$ are bisectors of $\angle CAQ$ and $\angle PAC$, if $\angle BAQ=30^o$ prove that:

1. $BD$ is a diameter of the circle
2. $ABC$ is an isosceles triangle.     [3]

(c) The printed price of an air conditioner is $Rs.45000$. The wholesaler allows a discount of $10\%$ to the shopkeeper. The shopkeeper sells the article to customer at a discount of $5\%$ of the marked price. Sales tax (under VAT) is changed at the rate of $12\%$ at every stage. Find:

1. VAT paid by the shopkeeper to the government;
2. The total amount paid by the customer inclusive of tax.     [3]

(a) Table as follows:

 Marks Mid Term $(x)$ No. of Students $(f)$ $D = x-A$ $A = 25$ $t = \frac{x-A}{i}$ $f.t$ 0-10 5 10 -20 -2 -20 10-20 15 9 -10 -1 -9 20-30 25 25 0 0 0 30-40 35 30 10 1 30 40-50 45 16 20 2 32 50-60 55 10 30 3 30 $\Sigma f = 100$ $\Sigma ft = 63$

Mean $(\overline{x}) = A +$ $\frac{\Sigma ft}{\Sigma f}$ $\times i = 25 +$ $\frac{63}{100}$ $\times 10 = 25+6.3 = 31.3$

(b)  Consider the diagram as shown:

$\angle BAQ = 30^o$ (Given)

$\angle BAQ = \angle CAB = 30^o$

$\therefore \angle CAP = 180^o-60^o = 120^o$

$\therefore \angle CAD = \angle DAP = 60^o$

(i) $\angle DAB = 90^o$ (since DB is the diameter)

(ii) $\angle CAB = \angle CDB = 30^o$ (angles subtended by the  cord CB on the circumference of the circle)

(iii) $\angle DAC = \angle DBC = 60^o$ (angles subtended by the cord DC on the circumference of the circle)

When the chord bisects the tangent…

$\angle DAP = \angle ADC$

$\therefore \angle ADB = 30^o$

$\therefore \angle ABD = 180^o-90^o-30^o=60^o$

Also $\angle DCB = 90^o$

$\therefore \angle ACB = 180^o - 30^o - 120^o = 30^o$

$\therefore \triangle ACB$ is an isosceles triangle.

(c)   Price $= 45000 \ Rs.$

Wholesaler to Shopkeeper

Price $= 45000 \ Rs.$

Discount $= 10\%$

Therefore discount amount $=$ $\frac{10}{100}$ $\times 45000 = 4500 \ Rs.$

Therefore discounted price $= 45000 - 4500 = 40500 \ Rs.$

VAT $= 12\%$

Therefore VAT paid by the wholesaler $=$ $\frac{12}{100}$ $\times 40500 = 4860 \ Rs.$

Shopkeeper to Customer

Price $= 45000 \ Rs.$

Discount $= 5\%$

Therefore discount amount $=$ $\frac{5}{100}$ $\times 45000 = 2250 \ Rs.$

Discounted price $= 45000 - 2250 = 42750 \ Rs.$

VAT $= 12\%$

Therefore VAT paid by the retailer $=$ $\frac{12}{100}$ $\times 42750 = 5130 \ Rs.$

(i) VAT paid by the shopkeeper $= 5130 - 4860 = 270 \ Rs.$

(ii) Total price for the customer $= 42750 + 5130 = 47880 \ Rs.$

$\\$

Question: 9

(a) In the figure given, $O$ is the center of the circle. $\angle DAE=70^o$. Find giving suitable reasons, the measure of:      [4]

(i) $\angle BCD$

(ii) $\angle BOD$

(iii) $\angle OBD$

(b) $A(1, -3), B(4, 2)$ and $C(3, 2)$ are the vertices of a triangle.

(i) Find the coordinates of the centroid $G$ of the triangle.

(ii) Find the equation of the line through $G$ and parallel to $AC$.

(c) Prove that

$\frac{sin\ \theta - 2 sin^3 \theta}{2cos^3\theta - cos\ \theta}$ $= tan\ \theta$     [3]

(a)  $\angle DAE = 70^o$

$\angle DAB = 180^o-70^o = 110^o$

$\angle BAD + \angle BCD = 180^o$

$\angle \Rightarrow BCD = 180^o - 110^o = 70^o$

$\angle BOD = 2 BCD$

$\angle BOD = 140^o$

$\angle OB = OD$  (radius of the same circles)

$\angle \therefore OBD = ODB= x$

$\angle \therefore 2x = 180^o - 140^o \Rightarrow x = 20^o$

(i) $\angle BCD = 70^o$ ($ABCD$ is a cyclic quadrilateral)

(ii) $\angle BOD = 140^o$ (angle subtended by a chord at the center is twice that subtended on the circumference)

(iii) $\angle OBD = 20^o$ ( $\triangle OBD$ is isosceles)

(b)          $A(-1, 3), B(4, 2) \ and \ C(3, -2)$ are the given coordinates of the three vertices of the triangle

(i)    Let the centroid be $G(x, y)$

Therefore $x =$ $\frac{-1+3+4}{3}$ $= 2$

and $y =$ $\frac{3-2+2}{3}$ $= 1$

Hence centroid is $G(2, 1)$

(ii)  Slope of $AC =$ $\frac{-2-3}{3+1}$ $=$ $\frac{-5}{4}$

Therefore the slope of the line passing through $G$ and parallel to $AC =$ $\frac{-5}{4}$

Hence the equation of the line passing through $G$ and parallel to $AC$ is:

$y - 1 =$ $\frac{-5}{4}$ $(x-2)$

$4y - 4 = -5x+10$

or $4y+5x=14$

(c) $\frac{sin\ \theta - 2 sin^3 \theta}{2cos^3\theta - cos\ \theta}$ $= tan\ \theta$

LHS $=$ $\frac{sin\ \theta - 2 sin^3 \theta}{2cos^3\theta - cos\ \theta}$

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

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

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

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

$=$ $tan \theta$

$=$ RHS. Hence proved.

$\\$

Question: 10

(a) The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages is a year is 550. Find their ages.     [4]

(b) The Daily wages of 80 workers in a project are given below.     [6]

 Wages in Rs 400-450 450-500 500-550 550-600 600-650 650-700 700-750 No. of workers 2 6 12 18 24 13 5

Use a graph paper to draw an ogive for the above distribution. (Use a scale of $2 \ cm = \ Rs.50$ on $x-axis$ and $2 \ cm=10$ workers on $y-axis$). Use your ogive to estimate:

(i) The medium wage of the workers.

(ii) The lower quartile wage of workers.

(iii) The number of workers who can more than Rs.625 daily;

(a)  Let the age of Vivek $= x$

and Let the age of Amit $= y$

Given: $+y = 47$ and $xy = 550$

Hence $x(47-x) = 550$

$\Rightarrow x^2-47x+550=0$

$\Rightarrow x^2 -25x-22x+550 = 0$

$\Rightarrow x(x-25) - 22(x-25) = 0$

$\Rightarrow (x-22)(x-25) = 0$

$\Rightarrow x = 22 \ or \ 25$

When $x = 22, y = 47-22 = 25$

When $x = 25, y = 47-25 = 22$

Hence the age of the two is $22, \ 25$ years.

(b)

 Wages No. workers $(f)$ Cumulative Frequency (c.f) 400-450 450-500 500-550 550-600 600-650 650-700 700-750 2 6 12 18 24 13 5 2 8 20 38 62 75 80

On the graph paper, we plot the following points:

$(2, 450), (8, 500), (20, 550), (38,600), (62, 650), (75,700), (80,750)$

$n = 80$

(i) Median $=$ $(\frac{n}{2})^{th}$ $term =$ $\frac{80}{2}$ $=40^{th} term$

From the graph $40^{th} term=605$

(ii) Lower quartile $=$ $(\frac{n}{4})^{th}$ $term =$ $\frac{80}{4}$ $=20^{th} term$

From the graph $20^{th} term =550$

(iii) The number of workers earning more that 625 per day $= 80-49 = 31$ students

$\\$

Question: 11

(a) The angles of depression of two ships $A$ and $B$ as observed from the top of a light house $60 \ m$ high are $60^o$ and $45^o$ respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships, Give your answer correct to the nearest whole number.     [4]

(b) $PQR$ is a triangle, $S$ is a point on the side $QR$ of triangle $PQR$ such that $\angle QPR = \angle PSR$  Given $QP=8\ cm$, $PR=6 \ cm$ and $SR=3 \ cm$

(i) Prove $\triangle PQR \sim \triangle SPR$

(ii) Find the length of $QR$ and $PS$

(iii) $\frac{Area \ of \ \triangle PQR}{Area \ of \ \triangle SPR}$     [3]

(c) Richard has a recurring deposit account in a bank for 3 years at $7.5\%$ p.a. simple interest. If he gets $Rs.8325$ as interest at the time of maturity, find;

(i) The monthly deposit

(ii) The maturity Value     [3]

(a)  In $\triangle MNB: tan \ 45^o =$ $\frac{60}{NB}$ $\Rightarrow NB = 60 \ m$

In $\triangle MNA: tan \ 60^o =$ $\frac{60}{AN}$ $\Rightarrow AN =$ $\frac{60}{\sqrt{3}}$ $= 34.64 \ m$

Hence $AB = NB + NA = 60 + 34.64 = 94.64 \ m$

Rounding off the the nearest whole number, we get $AB = 95 \ m$

(b)

(i) Consider $\triangle PQR$ and $\triangle PSR$

$\angle QPR = \angle PSR$ (given)

$PR$ is common

$\angle R$ is common.

Therefore by AAA postulate, $\triangle PQR \sim \triangle PSR$

(ii) Since $\triangle PQR \sim \triangle PSR$

$\therefore \frac{PR}{SR}$ $=$ $\frac{QR}{PR}$

$\Rightarrow \frac{6}{3}$ $=$ $\frac{QR}{6}$

$\Rightarrow QR = 12 \ cm$

(iii) $\frac{Area \ of \ \triangle PQR}{Area \ of \ \triangle SPR}$ $=$ $\frac{6^2}{3^2}$ $=$ $\frac{35}{9}$ $=$ $4:1$

(c)   $I = 8325 \ Rs.$

$n = 3 \times 12 = 36$ months

$R = 7.5\%$

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

$8325 =$ $\frac{P \times 36 \times 37}{2}$ $\times$ $\frac{7.5}{100}$ $\times$ $\frac{1}{12}$

$\Rightarrow P =$ $\frac{8325 \times 2 \times 100 \times 12}{36 \times 37 \times 7.5}$ $= 2000 \ Rs.$

(ii) Maturity Value $= P \times n + I = 2000 \times 36 + 8325 = 80325 \ Rs.$

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