December 2nd 2017

MATHEMATICS (ICSE 2014)

Two and Half Hour. Answers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.

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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 B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the Answer. Omission of essential working will result in 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) Ranbir borrows $Rs. 20,000$ at $12\%$ per annum compound interest. If he repays $Rs. 8400$ at the end of the first year and $Rs.9680$ at the end of the second year, find the amount of loan outstanding at the beginning of the third year. [3]

(b) Find the value of $x$ , which satisfy the in equation $-2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\le 2, x \in W$ . Graph the solution set on the number line. [3]

(c) A die has 6 faces marked by the given numbers as shown below: $1, 2, 3, -1, -2, -3$ . The die is thrown once. What is the probability of getting?

(i) A positive integer

(ii) An integer greater than $-3$

(iii) The smallest integer [4]

Answer.

(a) Given: Principal for the first year $(P)= Rs. \ 20000, \ Rate (r)=12\%$

We known that $A=P( 1+\frac{1}{100})^n$ [Reference Link]

Amount after the 1st year $= 20000 (1+\frac{12}{100})^1 = 20000 (\frac{112}{100}) = Rs. \ 22400$

Money repaid at the end of 1st year $Rs. \ 8400$

Principle for the 2nd year $22400 - 8400 = Rs. \ 14000$

Amount after 2nd year $14000 ( 1+\frac{12}{100} )^1= Rs. \ 15680$

Money repaid at the end of the second year $= Rs. \ 9680$

The loan amount outstanding at the beginning of the third year $15680-9680 = Rs. \ 6000$

(b) Given $-2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\le 2$

$\Rightarrow - \frac{17}{6} < \frac{3-4x}{6} \le 2$

Multiplying throughout by 6

$- 17 <3-4x \le 12$

$-17<3-4x \ and \ 3-4x \le 12$

$\Rightarrow 4x<3+17 \ \ \ \ \ \ \ \ \ \ \Rightarrow 3-12 \le 4x$

$\Rightarrow 4x < 20 \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow - 9 \le 4x$

$\Rightarrow x<5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow - \frac{9}{4}<x$

Therefore $(5>x\ge \frac{-9}{4})$

Hence the solution set is $\{x \in W, - \frac{9}{4} \le x<5 \}$

Therefore the values of $x$ are $\{0,1,2,3,4 \}$

The graph of the solution set is shown by dots on the number line.

(c) No. of sample space $n(S) =6$

A positive integer $= \{1,2,3\}$

No. of favorable cases $n(E) = 3$

Probability $= \frac{n(E)}{n(S)}=\frac{3}{6}= \frac{1}{2}$

An integer greater than $-3 = \lbrace 1,2,3,-1,-2\rbrace$

No. of favorable cases $n(E)=5$

Probability $= \frac{n(E)}{n(S)}=\frac{5}{6}$

Smallest integer $= -3$

Probability of smallest integer $= \frac{n(E)}{n(S)}=\frac{1}{6}$

$\\$

Question 2:

(a) Find $x,y$ if $\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}$ [3]

(b) Sharukh opened a Recurring Deposit Account in a bank and deposited Rs. 800 per month for $1\frac{1}{2}$ years. If he received Rs. 15,084 at the time of maturity. Find the rate of interest per annum. [3]

(c) Calculate the ratio in which the line joining $A(-4,2) \ and \ B(3,6)$ is divided by point $P(x,3)$ Also find (i) $x$ (ii) Length of $AP$ . [4]

Answer:

(a) Given $\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} -2 \times -1+0 \times 2x \\ 3 \times - 1 + 1 \times 2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2 \\ -3+2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2-6 \\ -3+2x+3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}$

$2y = 4 \ or \ y = -2 \ and \ 2x = 6 \ or \ x = 3$

(b) Here, $Principal (P)$ = money deposited per month $=Rs. \ 800$

$N=$ Time for which the money is deposited = $1\frac{1}{2} \ years =18 \ months$

Let the rate of interest be $r\%$ per annum, then

$Interest=P\times$ $\frac{n(n+1)}{2\times 12}\times \frac{r}{100}$ [Reference Link]

$= 800 \times$ $\frac{18\times 19}{2\times 12}\times \frac{r}{100}$ $= 114r \ Rs.$

Total money deposited $= 18 \times 800= Rs. \ 14400$

Since money deposited +Interest = Maturity value

$14400+114r=15084$

$114r=15084-14400$

$114r=684$

$r=\frac{684}{114}=6$

Hence rate of interest $= 6\% p.a.$

(c) Let $P(x, 3)$ divide the line segment joining the points $A (-4, 2)$ and $B (3, 6)$ in the ratio $k :1$

Coordinate of $P$ is: [Reference Link]

$( \frac{m_1. x_2+m_2.x_1}{m_1+m_2}, \frac{m_1. y_2+m_2.y_1}{m_1+m_2} = \frac{3k-4}{k+1}, \frac{6k+2}{k+1})$

But Coordinate of $P \ is \ (x,3)$

$\frac{6k+2}{k+1}=3$

$6k+2=3k+3$

$3k=1 \Rightarrow k=\frac{1}{3}$

The required ratio is $\frac{1}{3}:1 \ i.e. \ 1:3$ , (Divides Internally)

(i) Therefore $x=$ $\frac{3k-4}{k+1}$

Substituting $k$ $=\frac{1}{3}$

$x=$ $\frac{3\times \frac{1}{3}-4}{\frac{1}{3}+1}= \frac{1-4}{\frac{1+3}{3}} = \frac{-3}{\frac{4}{3}}= \frac{-9}{4}$

(ii) Coordinate of $P$ is $(\frac{-9}{4}, 3)$

Length of $AP = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$= \sqrt{(-\frac{9}{4}+4)^2+(3-2)^2}$

$= \sqrt{(\frac{-9+16}{4})^2+(3-2)^2} = \sqrt{\frac{49}{16}+1}$

$= \sqrt{\frac{49+16}{16}}= \sqrt{\frac{65}{16}}=\sqrt{\frac{65}{4}}$

$\\$

Question 3:

(a) Without using trigonometric tables, evaluate

$sin^2 \ 34^o + sin^2 \ 56^o + 2 \ tan \ 18^o tan \ 72^o - cot ^2 \ 30^o$ [3]

(b) Using the remainder and factor theorem, factor the following polynomial: $x^{3}+10x^{2}-37x+26$ [3]

(c) In the figure given below $ABCD$ is a rectangle $AB=14 \ cm, \ BC=7\ cm$ . From the rectangle a quarter circle $BFEC$ and a semicircle $DGE$ are removed Calculate the area of the remaining piece of the rectangle (Take $\pi = \frac{22}{7}$ ) [4]

Answers:

(a) Given $sin^2 \ 34^o + sin^2 \ 56^o + 2 \ tan \ 18^o tan \ 72^o - cot ^2 \ 30^o$

$= sin^2 \ 34^o + sin^2 \ (90^o - 34^o) + 2 \ tan \ 18^o \ tan \ (90^o-18^o) - cot ^2 \ 30^o$

$= sin^2 \ 34^o + cos^2 \ 34^o + 2 \ tan \ 18^o \ cot \ 18^o - (\sqrt{3})^2$

$= 1+ 2 \ tan \ 18^o \times \frac{1}{tan \ 18^o}-3$

$= 1+2-3 = 0$

(b) Let $f(x)=x^3+10x^2-37x+26$

Putting $x =1$ , we get

$f(1)=1+10-37+26=0$

By factor theorem $x-1$ is actor of $f(x)$

$x-1 ) \overline {x^3+10x^2-37x+26} (x^2+11x-26$
$(-) \ \ \underline {x^3-x^2}$
$11x^2-37x+26$
$(-) \ \ \underline{11x^2-37x}$
${ -26x+26}$
$(-) \ \ \underline{ -26x+26}$
$\times$

On dividing $x^{3}+10x^{2}-37x+26$ by $(x-1)$ , we get $x^{2}+11x-26$ as the quotient and remainder $= 0$

Therefore the other factor of $f(x)$ are the factor of $x^{2}+11x-26$

Now, $x^{2}+11x-26$

$= x^{2}+13x-2x-26$

$= x(x+13)-2(x+13)$

$= (x+13)(x-2)$

Hence $x^{3}+10x^{2}-37x+26=(x-1)(x-2)(x+13)$

(c) Area of rectangle $ABCD= 14 \times 7 = 98cm$

Area of quarter circle $BFEC=\frac{1}{4}\pi (7^{2})=\frac{49}{4}\pi$

Area of semicircle $DGE= \frac{1}{2}\pi (\frac{7}{2})^2=\frac{1}{2}\times \frac{49}{4}\pi$

Area of remaining piece of rectangle $= 98-[\frac{49}{4}\pi +\frac{1}{2}\times \frac{49}{4}\pi ]$

$= 98-\frac{49}{4}\pi \lbrack 1+\frac{1}{2}\rbrack$

$= 98-\frac{49}{4}\times \frac{22}{7}\times \frac{3}{2}=98-\frac{231}{4}$

$= 98-57.75 = 40.25 cm^2$

$\\$

Question 4:

(a) The number $6,8,10,12,13$ and $x$ are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of $x$ [3]

(b) In the figure, $\angle DBC=58^o. BD$ is a diameter of the circle. Calculate: [3]

(i) $\angle BDC$

(ii) $\angle BEC$

(iii) $\angle BAC$

(c) Using graph Paper to answer the following questions. (Take $2 \ cm = 1$ unit on both axis)

(i) Plot the points $A (-4,2)$ and $B (2,4)$

(ii) $A'$ is the image of $A$ when reflected in the $y-axis$ . Plot it on the graph paper and write the coordinates of $A'$

(iii) $B'$ is the image of $B$ when reflected in the line $AA'$ write the coordinates of $B'$

(iv) Write the geometric name of the figure $ABA'B'$

(v) Name a line of symmetry of the figure formed. [4]

Answers:

(a) Arrange numbers in ascending order are $6,8,10,12,13, x.$

Mean $= \frac{6+8+10+12+13+x}{6}=\frac{49+x}{6}$

No. of terms $(n) = 6 \ (even)$

Median $= \frac{ (\frac{n}{2})^{th} \ term + (\frac{n}{2}+1)^{th} \ term }{2}$

Median $= \frac{ (\frac{6}{2})^{th} \ term + (\frac{6}{2}+1)^{th} \ term }{2} = \frac{3^{rd}+4^{th}}{2} = \frac{10+12}{2} = \frac{22}{2}$ $= 11$

According to given condition

$\frac{49+x}{6}=11 \Rightarrow 49+x=66$

or $x=17$

(b) In $\triangle BCD , \angle DBC = 58^o \ (given)$

(i) $\angle BCD = 90^o$ (angle in the semi circle)

$\therefore \angle DBC + \angle BCD + \angle BDC = 180^o$

$58^o+90^o+\angle BDC = 180^o$

$\Rightarrow \angle BDC = 180^o-148^o = 32^o$

(ii) $\angle BEC + \angle BDC = 180^o$ (cyclic quadrilateral)

$\angle BEC = 180^o-32^o = 148^o$

(iii) $\angle BAC = \angle BDC$ (angles in the same segment)

$\angle BAC = 32^o$

(i) Coordinate of $A' = (4,2)$

(ii) Coordinate of $B' = (2, 0)$

(iii) Geometric name $= kite$

(iv) $AA'$ is the symmetric line.

SECTION B [40 Marks]

(Answer any four questions in this Section.)

Question 5:

(a) A shopkeeper bought a washing machine at a discount of $20 \%$ from a wholesaler, the printed price of the washing machine being $Rs. \ 18,000$ . The shopkeeper sells it to a consumer at a discount of $10 \%$ on the printed price. If the rate of sales tax is $8 \%$ find:

(i) the VAT paid by the shopkeeper

(ii) the total amount that the consumer pays for the washing machine. [3]

(b) If $\frac{x^2+y^2}{x^2-y^2} = \frac{17}{8}$ , then find the value of

(i) $x:y$

(ii) $\frac{x^{3}+y^{3}}{x^{3}-y^{3}}$ [3]

(c) In $\triangle ABC, \angle ABC = \angle DAC. AB = 8 \ cm, AC = 4 \ cm, AD = 5 \ cm$

(i) Prove that $\triangle ACD$ is similar to $\triangle BCA$

(ii) Find $BC \ and \ CD$

(iii) Find area of $\triangle ACD : \ area \ of \ \triangle ABC$ [4]

Answers:

(a) Given: Printed price of washing machine $= Rs.\ 18000$

Rate of discount $=20\%$

(i) Amount of discount to shopkeeper $= \frac{20}{100}\times 18000=Rs.\ 3600$

Shopkeeper’s Price $= 18000-3600 = Rs.\ 14400$

Sales Tax paid by shopkeeper $= \frac{8}{100}\times 14400=Rs.\ 1152$

Discount for consumer $= \frac{10}{100}%$

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