Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2014)

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) 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 $\displaystyle x$ , which satisfy the in equation $\displaystyle -2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\le 2, x \in W.$ 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]

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

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

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

Money repaid at the end of 1st year $\displaystyle = \text{ Rs. } 8400$

Principle for the 2nd year $\displaystyle = 22400 - 8400 = \text{ Rs. } 14000$

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

Money repaid at the end of the second year $\displaystyle = \text{ Rs. } 9680$

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

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

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

Multiplying throughout by 6

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

$\displaystyle -17<3-4x \text{ and } 3-4x \le 12$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Smallest integer $\displaystyle = -3$

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

$\displaystyle \\$

Question 2:

$\displaystyle \text{(a) Find } x, \ y \text{ if } \begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix}. \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix} \hspace{6.0cm} [3]$

(b) Sharukh opened a Recurring Deposit Account in a bank and deposited Rs. 800 per month for 1.5 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 $\displaystyle A(-4,2) \text{ and } B(3,6)$ is divided by point $\displaystyle P(x,3)$ Also find (i) $\displaystyle x$ (ii) Length of $\displaystyle AP.$                                                              [4]

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

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

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

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

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

$\displaystyle 2y = 4 \text{ or } y = -2 \text{ and } 2x = 6 \text{ or } x = 3$

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

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

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

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

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

Total money deposited $\displaystyle = 18 \times 800= \text{ Rs. } 14400$

Since money deposited +Interest = Maturity value

$\displaystyle 14400+114r=15084$

$\displaystyle 114r=15084-14400$

$\displaystyle 114r=684$

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

Hence rate of interest $\displaystyle = 6\% \text{ p.a. }$

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

Coordinate of $\displaystyle P$ is: [Reference Link]

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

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

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

$\displaystyle 6k+2=3k+3$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 3:

(a) Without using trigonometric tables, evaluate

$\displaystyle \sin^2 34^{\circ} + \sin^2 56^{\circ} + 2 \tan 18^{\circ} \tan 72^{\circ} - \cot ^2 30^{\circ}$                                                                [3]

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

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

(a) Given $\displaystyle \sin^2 34^{\circ} + \sin^2 56^{\circ} + 2 \tan 18^{\circ} \tan 72^{\circ} - \cot ^2 30^{\circ}$

$\displaystyle = \sin^2 34^{\circ} + \sin^2 (90^{\circ} - 34^{\circ}) + 2 \tan 18^{\circ} \tan (90^{\circ}-18^{\circ}) - \cot ^2 30^{\circ}$

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

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

$\displaystyle = 1+2-3 = 0$

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

Putting $\displaystyle x =1$ , we get

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

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

$\displaystyle \begin{array}{r l l } x-1 ) & \overline {x^3+10x^2-37x+26} & (x^2+11x-26 \\ (-) & \underline {x^3-x^2} & \\ & \hspace{1.0cm} 11x^2-37x+26 & \\ (-) &\hspace{1.0cm} \underline{11x^2-37x} & \\ & \hspace{2.0cm} {-26x+26} & \\ (-) & \hspace{2.0cm} \underline{ -26x+26} & \\ & \hspace{3.0cm} \times & \\ \end{array}$

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

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

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

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

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

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

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

(c) Area of rectangle $\displaystyle ABCD= 14 \times 7 = 98 \text{ cm }$

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

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

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

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

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

$\displaystyle = 98-57.75 = 40.25 \text{ cm}^2$

$\displaystyle \\$

Question 4:

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

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

(i) $\displaystyle \angle BDC$

(ii) $\displaystyle \angle BEC$

(iii) $\displaystyle \angle BAC$

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

(i) Plot the points $\displaystyle A (-4,2) \text{ and } B (2,4)$

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

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

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

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

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

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

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

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

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

According to given condition

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

or $\displaystyle x=17$

(b) In $\displaystyle \triangle BCD , \angle DBC = 58^{\circ} (given)$

(i) $\displaystyle \angle BCD = 90^{\circ}$ (angle in the semi circle)

$\displaystyle \therefore \angle DBC + \angle BCD + \angle BDC = 180^{\circ}$

$\displaystyle 58^{\circ}+90^{\circ}+\angle BDC = 180^{\circ}$

$\displaystyle \Rightarrow \angle BDC = 180^{\circ}-148^{\circ} = 32^{\circ}$

(ii) $\displaystyle \angle BEC + \angle BDC = 180^{\circ}$ (cyclic quadrilateral)

$\displaystyle \angle BEC = 180^{\circ}-32^{\circ} = 148^{\circ}$

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

$\displaystyle \angle BAC = 32^{\circ}$

(c) As shown in the graph below:

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

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

(iii) Geometric name $\displaystyle = kite$

(iv) $\displaystyle AA'$ 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]

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

(i) $\displaystyle x:y$

$\displaystyle \text{(ii) } \frac{x^{3}+y^{3}}{x^{3}-y^{3}} \hspace{11.0cm} [3]$

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

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

(ii) Find $\displaystyle BC \text{ and } CD$

(iii) Find area of $\displaystyle \triangle ACD : \text{area of } \triangle ABC$ [4]

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

Rate of discount $\displaystyle =20\%$

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

Shopkeeper’s Price $\displaystyle = 18000-3600 = \text{ Rs. } 14400$

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

$\displaystyle \text{Discount for consumer }= \frac{10}{100}\times 18000=\text{ Rs. } 1800$

Price of consumer $\displaystyle =18000-1800= \text{ Rs. } 16200$

$\displaystyle \text{Tax charged by the shopkeeper }= \frac{8}{100}\times 16200=\text{ Rs. } 1296$

Since, Tax paid by the shopkeeper $\displaystyle = \text{ Rs. } 1152$

VAT paid by the shopkeeper=Tax charged – Tax Paid $\displaystyle =1296-1152=\text{ Rs. } 144$

(ii) Total amount paid by the consumer for washing machine $\displaystyle = 1620+1296 = \text{ Rs. } 17496$

$\displaystyle \text{(b) Given: } \frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{17}{8}$

(i) Applying componendo and dividend

$\displaystyle \frac{(x^{2}+y^{2}) +(x^{2}-y^{2})}{(x^{2}+y^{2})-(x^{2}-y^{2})}=\frac{17+8}{17-8}$

$\displaystyle \Rightarrow \frac{2x^{2}}{2y^{2}}=\frac{25}{9}= \frac{x^{2}}{y^{2}}=\frac{25}{90}$

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

$\displaystyle \Rightarrow x:y=5:3$

(ii) Cubing both sides we get

$\displaystyle \frac{x^{3}}{y^{3}}=\frac{5^{3}}{3^{3}}=\frac{125}{27}$

Applying componendo and dividend

$\displaystyle \frac{x^{3}+y^{3}}{x^{3}-y^{3}}=\frac{125+27}{125-27}$

$\displaystyle \Rightarrow \frac{x^{3}+y^{3}}{x^{3}-y^{3}}=\frac{152}{98}$

$\displaystyle \Rightarrow \frac{x^{3}+y^{3}}{x^{3}-y^{3}}=\frac{76}{49}$

(c)

(i) In $\displaystyle \triangle ACD \text{ and } \triangle BCA$

$\displaystyle \angle C$ (common angle)

$\displaystyle \angle ABC = \angle CAD$ (given)

Therefore $\displaystyle \triangle ACD \sim \triangle BCA$ (AAA Postulate)

(ii) Since $\displaystyle \triangle ACD \sim \triangle BCA$

$\displaystyle \Rightarrow \frac{AC}{BC}=\frac{CD}{CA}=\frac{AD}{BA}$ (corresponding sides are proportional)

$\displaystyle \frac{4}{BC}=\frac{CD}{4}=\frac{5}{8}$

$\displaystyle \Rightarrow \frac{4}{BC}=\frac{5}{8} \text{ and } \frac{CD}{4}=\frac{5}{8}$

$\displaystyle \Rightarrow BC=\frac{4\times 8}{5}=\frac{32}{5}=6.4 \text{ cm} \text{ and } CD=\frac{5}{8}\times 4=\frac{5}{2}=2.5 \text{ cm}$

(iii) Since $\displaystyle \Delta ACD \sim \Delta ABC$

$\displaystyle \frac{\text{area } (\Delta ACD)}{\text{area } (\Delta ABC)}=\frac{AC^{2}}{AB^{2}} = \frac{4^{2}}{8^{2}}=\frac{16}{64}=\frac{1}{4}$

$\displaystyle \text{area } (\Delta ACD ) : \text{area } (\Delta ABC)=1:4$

Question 6:

(a) Find the value of $\displaystyle 'a'$ for which the following points $\displaystyle A (a,3), B(2,1) \text{ and } C (5,a)$ are collinear. Hence find the equation of the line.                                                    [3]

(b) Salman invests a sum of money in Rs. 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is Rs. 600 , calculate;

(i) the number of shares he bought

(ii) his total investment

(iii) the rate of return on his investment                                                          [3]

(c) The surface area of a solid metallic sphere is $\displaystyle 2464 \text{ cm}^2.$ it is melted and recast into solid right circular cones of radius  3.5 cm  and height 7 cm.  calculate:

(i) the radius of the sphere.

$\displaystyle \text{(ii) the number of cones recast. (Take } \pi = \frac{22}{7} )$                                               [4]

$\displaystyle \text{(a) Given: } A (a,3), B(2,1) \text{ and } C(5,a) \text{ are collinear.}$

$\displaystyle \text{Slope of } AB= \text{Slope of } BC$

$\displaystyle \frac{1-3}{2-a}=\frac{a-1}{5-1}$

$\displaystyle \frac{-2}{2-a}=\frac{a-1}{3}$

$\displaystyle \Rightarrow -6=(2-a)(a-1)$

$\displaystyle \Rightarrow -6=2a-2-a^{2}+a$

$\displaystyle \Rightarrow a^{2}-3a-4=0$

$\displaystyle \Rightarrow a^{2}-4a+a-4=0$

$\displaystyle \Rightarrow (a-4)(a+1)=0$

$\displaystyle a=4, -1$

Rejecting $\displaystyle a=-1$

$\displaystyle \text{Slope of } BC = \frac{a-1}{5-2}=\frac{4-1}{3}=\frac{3}{3} =1$

$\displaystyle \text{Equation of} BC; (y-1)=1 (x-2)$

$\displaystyle y-1=x-2$

$\displaystyle x-y=1$

(b) Nominal value of share $\displaystyle = \text{ Rs.} 50$

$\displaystyle \text{Dividend on 1 share } = \frac{15}{100}\times 50=\text{ Rs.} 7.50$

Total dividend of Salman $\displaystyle =\text{ Rs.} 600$

$\displaystyle \text{(i) No. of shares Salman bought } = \frac{600}{7.50}=80$

$\displaystyle \text{(ii) Premium on 1 share } =\frac{20}{100}\times 50=Rs. 10$

Market value of 1 share $\displaystyle = 50+10=\text{ Rs.} 60$

Total investment for 80 shares $\displaystyle = 80 \times 60 = \text{ Rs.} 4800$

$\displaystyle \text{(iii) Rate of return } = \frac{600}{4800}\times 100=12.5\%$

(c)

(i) Let the radius sphere $\displaystyle = r \text{ cm}$

Surface area of sphere $\displaystyle = 4\pi r^{2}=2464 \text{ cm}^{2}$

$\displaystyle r^{2}=\frac{2464}{4\pi }$

$\displaystyle r^{2}=\frac{2464\times 7}{4\times 22}=196$

$\displaystyle r=14\text{ cm}$

$\displaystyle \text{(ii) Volume of sphere } = \frac{4}{3}\pi r^{2}=\frac{4}{3}\pi (14)^{3}$

$\displaystyle \text{Volume of cone } = \frac{1}{3}\pi ^{2}h=\frac{1}{3}\pi (3.5)^2 \times 7$

$\displaystyle \text{No. of cones recast } = \frac{\text{Volume of sphere}}{\text{Volume of cone}}$

$\displaystyle = \frac{\frac{4\pi }{3}(14)^{3}}{\frac{1}{3}\pi (3.5)2\times 7} = \frac{4\times 14\times 14\times 14}{3.5\times 3.5\times 7}=\frac{3200}{25} = 128$

Question 7:

(a) Calculate the mean of the distribution given below using the short cut method;                                                                                                                            [3]

 Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80 No. of students 2 6 10 12 9 7 4

(b) In the figure given below, diameter $\displaystyle AB \text{ and } CD$ of a circle meet at $\displaystyle P. PT$ is a tangent to the circle at $\displaystyle T. CD=7.8 \text{ cm}, PD=5 \text{ cm } PB=4 \text{ cm}$ find:

(i) $\displaystyle AB$

(ii) the length of tangent $\displaystyle PT.$                                                                            [3]

$\displaystyle \text{(c) Let } A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} , B = \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \text{ and } C = \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}. \\ \\ \text{ Find } A^2+AC-5B. \hspace{11.0cm} [3]$

(a) Table as follows

 $\displaystyle Marks \\ (C.I)$$\displaystyle Marks \\ (C.I)$ $\displaystyle f$$\displaystyle f$ Mean Value $\displaystyle x$$\displaystyle x$ $\displaystyle A=45.5 d=x-A$$\displaystyle A=45.5 d=x-A$ $\displaystyle f \times d$$\displaystyle f \times d$ 11-20 21-30 31-40 41-50 51-60 61-70 71-80 2 6 10 12 9 7 4 15-5 25-5 35-5 45-5 55-5 65-5 75-5 -30 -20 -10 0 10 20 30 -60 -120 -100 0 90 140 120 $\displaystyle \Sigma f=50$$\displaystyle \Sigma f=50$ $\displaystyle \Sigma fd=70$$\displaystyle \Sigma fd=70$

$\displaystyle Mean = A + \frac{ \Sigma fd}{\Sigma f} = 45.5+\frac{70}{50} = 45.5 + 1.4 = 46.9$

(b)

(i) Since chord $\displaystyle CD$ and tangent at point $\displaystyle T$ intersect each other at $\displaystyle P .$

$\displaystyle PC \times PD=PT^{2} .$……….. (i)

Since chord $\displaystyle AB$ and tangent at point $\displaystyle T$ interested each other at $\displaystyle P$ ,

$\displaystyle PA\times PB= PT^{2} .$……….. (ii)

From (i) $\displaystyle \&$ (ii) $\displaystyle PC\times PD=PA\times PB$

Given; $\displaystyle CD=7.8 \text{ cm }, PD=5 \text{ cm }, PB=4 \text{ cm }$

$\displaystyle PA=PB+AB=4+AB \\ \\ PC=PD+CD=5+7.8=12.8 \text{ cm }$

Putting these values in eq. (3)

$\displaystyle 12.8\times 5=(4+AB)\times 4$

$\displaystyle \Rightarrow 4+AB= \frac{12.8\times 5}{4}$

$\displaystyle \Rightarrow 4+AB=16$

$\displaystyle \Rightarrow AB=12 \text{ cm }$

Hence, $\displaystyle AB = 12 \text{ cm }$

(ii) From (i), $\displaystyle PT^{2}=PA\times PB=12.8\times 5$

$\displaystyle PT^{2}=64$

Length of tangent $\displaystyle PT=8 \text{ cm}$

(c) $\displaystyle A^2+AC-5B$

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

$\displaystyle = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} - \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -23 & 3 \\ 17 & 6 \end{bmatrix}$

$\displaystyle \\$

Question 8:

(a) The compound interest, calculate yearly, on a certain sum of money for the second year is Rs. 1320 and for the third year is Rs. 1452. Calculate the rate of interest and the original sum of money.                                                                [3]

(b) Construct a $\displaystyle \triangle ABC$ with $\displaystyle BC=6.5 \text{ cm}, AB=5.5 \text{ cm}, AC=5 \text{ cm}$ , Construct the in circle of the triangle measure and record the radius of the in circle. [3]

(c) Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below;

 Pocket expenses (in Rs.) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 Number of students (frequency) 10 14 28 42 50 30 14 12

Draw a histogram representing the above distribution and estimate the mode from the graph.                                                                                                               [4]

(a) Compound Interest for the third year $\displaystyle = \text{Rs. }1452$

Compound Interest for the second year $\displaystyle = \text{Rs. } 1320$

Simple Interest on $\displaystyle \text{Rs. } 1320$ for one year $\displaystyle = 1452-1320=\text{Rs. } 132$

$\displaystyle \text{Rate of interest } = \frac{132\times 100}{1320}=10\%$

Let the original money be $\displaystyle \text{Rs. } 'P'$

Amount after 2 year – amount after one year = Compound Interest for second year.

$\displaystyle \Rightarrow P (1+\frac{10}{100})^2-P(1+\frac{10}{100})=1320$

$\displaystyle \Rightarrow P[(\frac{110}{100})^2-(\frac{110}{100})]=1320$

$\displaystyle \Rightarrow P\lbrack (\frac{11}{10})^2-\frac{11}{10}\rbrack =1320$

$\displaystyle \Rightarrow (\frac{121}{100}-\frac{11}{10})=\text{Rs. } 1320$

$\displaystyle \Rightarrow P\times \frac{11}{100}=\text{Rs. }1320$

$\displaystyle \Rightarrow P= \frac{1320\times 100}{11}=\text{Rs. } 12,000$

Rate of interest $\displaystyle 10\%$

Original sum of money $\displaystyle = \text{Rs. } 12,000$

(b) Steps of construction:

1. Construct a $\displaystyle \triangle ABC$ with the given data:
1. Draw a line BC of 6.5 cm length using a ruler
2. The make an arc of 5.5 cm from B and similarly, make an arc of 5 cm from C
3. The place where the two arcs intersect, join that point to B and C and complete the triangle.
2. Draw the internal bisectors of $\displaystyle \angle B \text{ and } \angle C.$ Let these bisectors cut at $\displaystyle O$
1. Draw an arc from point B so that it cuts the two sides of the angle ABC
2. From the point of intersections, draw two arcs
3. Join the point B and the point of intersection and draw a line. This line bisects the angle ABC.
4. Repeat the above 3 steps for point C.
5. The two bisectors will intersect at the point O which is the center of the in-circle.
3. Taking $\displaystyle O$ as center and touching the side of the circle as the radius, Draw a in-circle which touches all the sides of the $\displaystyle \triangle ABC$
4. From $\displaystyle O$ draw a perpendicular to side $\displaystyle BC$ which cut at $\displaystyle N .$
5. Measure $\displaystyle ON$ which is required radius of the in circle. $\displaystyle ON=1.5 \text{ cm} .$

(c)

Question: 9

(a) If $\displaystyle (x-9): (3x+6)$ is the duplicate ratio of $\displaystyle 4:9$ , find the value of $\displaystyle x.$             [3]

(b) Solve for $\displaystyle x$ using the quadratic formula. Write your answer correct to two significant figures. $\displaystyle (x-1)^2-3x+4=0$                                                                      [3]

(c) A page from the saving bank account of Priyanka is given below:                [4]

 Date Particular Amount Withdrawal Amount Deposited Balance 03/04/2006 05/04/2006 18/04/2006 25/05/2006 30/05/2006 20/07/2006 10/09/2006 19/09/2006 B/F By cash By Cheque To Cheque By Cash By Self By Cash To Cheque – – – 5,000.00 – 4,000.00 – 1,000.00 – 2,000.00 6,000.00 – 3,000.00 – 2,000.00 – 4,000.00 6,000.00 12,000.00 7,000.00 10,000.00 6,000.00 8,000.00 7,000.00

If the interest earned by Priyanka for the period of ending September, 2006 is Rs.175, find the rate of interest.

(a) Given: $\displaystyle (x-9): (3x+6)$ is the duplicate ratio of $\displaystyle 4:9$

$\displaystyle \Rightarrow \frac{x-9}{3x+6}=(\frac{4}{9})^2$

$\displaystyle \Rightarrow \frac{x-9}{3x+6}=(\frac{16}{81})$

$\displaystyle \Rightarrow 81x-729=48x+96$

$\displaystyle \Rightarrow 81x-48x=96+729$

$\displaystyle \Rightarrow 33x=825$

$\displaystyle \Rightarrow x=\frac{825}{33}=25$

(b) Given: $\displaystyle (x-1)^2-3x+4=0$

$\displaystyle \Rightarrow x^{2}+1-2x-3x+4=0$

$\displaystyle \Rightarrow x^{2}-5x+5=0$

Comparing $\displaystyle x^{2}-5x+5=0$ with $\displaystyle ax^{2}+bx+c=0$ , we get $\displaystyle a=1, b=-5, C=5$

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

$\displaystyle x= \frac{-(-5)\pm \sqrt{(-5)^{2}-4(1)(5)}}{2 \times 1}$

$\displaystyle = \frac{5\pm \sqrt{25-20}}{2} =\frac{5\pm \sqrt{5}}{2}$

$\displaystyle = \frac{5\pm 2.236}{2}= \frac{5+2.236}{2} \text{ and } \frac{5-2.236}{2}$

$\displaystyle = \frac{7.236}{2} \text{ and } \frac{2.764}{2}$

$\displaystyle =3.618 \text{ and } 1.382$

(c) Qualifying principal for various months:

 Month Principal (Rs.) April 6000 May 7000 June 10000 July 6000 August 6000 September 7000 Total for 1 month 42000

$\displaystyle \text{Now, } P =\text{Rs. } 42,000, I=\text{Rs. }175, T=\frac{1}{12}\text{years, Rate}= R$

$\displaystyle \text{Interest } = \frac{P\times R\times T}{100}$

$\displaystyle 175 = \frac{42,000\times R\times 1}{100\times 12}$

$\displaystyle R = \frac{175\times 100\times 12}{42,000\times 1} = \frac{2100}{420}=5\%$

$\displaystyle \\$

Question: 10

(a) A two digit number is such that the product of its digits is 6, If 9 is added to the number, the digits interchange their places. Find the number.                    [4]

(b) Marks obtained by 100 students in a Mathematics test are given below;   [6]

 Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of Students 3 7 12 17 23 14 9 6 5 4

Draw an ogive for the distribution on a graph sheet. (Use a scale of 2 cm = 10 units on both axis)

Use the ogive to estimate the:

(i) median

(ii) lower quartile

(iii) number of students who obtained more than 85% marks in the test.

(iv) number of students who did not pass in the if the pass percentage was 35

(a) Let the required two digit number be $\displaystyle 10x+y$

Given: $\displaystyle xy=6 \text{ and } 10x+y+9=10y+x$

$\displaystyle 10x-x+y-10y+9=0$

$\displaystyle \Rightarrow 9x-9y+9=0$

$\displaystyle \Rightarrow x-y+1=0$

$\displaystyle \Rightarrow y=x+1$

Since $\displaystyle xy = 6$ (given)

$\displaystyle \Rightarrow x(x+1)=6$

$\displaystyle \Rightarrow x^{2}+x-6=0$

$\displaystyle \Rightarrow x^2 +3x-2x-6=0$

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

$\displaystyle x=-3 or 2$

$\displaystyle x=-3$ (not possible)

When $\displaystyle x=2, y=x+1=2+1=3$

The required two digit number $\displaystyle = 10x+y = 10 \times 2+3 = 23$

(b)

 Marks No. of Students Cumulative Frequency (c.f) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 8 7 12 17 23 14 9 6 5 4 3 10 22 39 62 76 85 91 96 100

On the graph paper, we plot the following points:

$\displaystyle (10, 3), (20, 10), (30, 22), (40,39), (50, 62), (60,76), (70,85), \\ (80, 91), (90, 96),(100, 100)$

$\displaystyle \text{(i) Medium } = (\frac{n}{2})^{th} \text{term} = \frac{100}{2}=50^{th} \text{term}$

From the graph $\displaystyle 50^{th}\text{term}=44$

$\displaystyle \text{(ii) Lower quartile } = (\frac{n}{4})^{th} \text{term} = \frac{100}{4}=25^{th} \text{term}$

From the graph $\displaystyle 25^{th}\text{term} =32$

(iii) The number of students who obtained more than 85%  marks in test $\displaystyle = 100-94 = 6$ students

(iv) The number of students who did not pass in the test if test if the pass percentage was 35 $\displaystyle = 30$ students.

$\displaystyle \\$

Question 11:

(a) In the figure given below, $\displaystyle O$ is the center of the circle, $\displaystyle AB \text{ and } CD$ are two chords of the circle. $\displaystyle OM \perp AB \text{ and } ON \perp CD .$

$\displaystyle AB=24 \text{ cm}, OM=5 \text{ cm}, ON=12 \text{ cm}$ Find the:

(ii) length of chord $\displaystyle CD$                                                                                          [3]

(b) Prove the identity:

$\displaystyle (\sin \theta + \cos \theta) (\tan \theta + \cot \theta) = \sec \theta + \mathrm{cosec} \ \theta$                                                       [3]

(c) An airplane at an altitude of $\displaystyle 250 \text{m}$ observed of depression of two boats on the opposite banks of a river to be $\displaystyle 45^{\circ} \text{ and } 60^{\circ}$ respectively. Find the answer correct to the nearest whole number.                                                                         [4]

(a) Given $\displaystyle AB=24\text{ cm}; OM=5 \text{ cm}, ON=12 \text{ cm}$

$\displaystyle OM \perp AB$

$\displaystyle M$ is mid point of $\displaystyle AB$

$\displaystyle AM=12 \text{ cm}$

(i) Let radius of circle $\displaystyle = r$

From $\displaystyle \triangle AMO$

$\displaystyle AO^{2}=AM^{2}+OM^{2}$

$\displaystyle r^{2}=12^{2}+5^{2}=144+25=169$

$\displaystyle r=13 \text{ cm}$

(ii) Now from $\displaystyle \triangle CNO; CO^{2}=ON^{2}+CN^{2}$

$\displaystyle r^{2}=(12^{2})+CN^{2}$

$\displaystyle 13^{2}-12^{2}=CN^{2}$

$\displaystyle 169-144=CN^{2}$

$\displaystyle \Rightarrow CN^{2}=25$

$\displaystyle \Rightarrow CN=5$

As $\displaystyle ON \perp CD, N$ is mid point of $\displaystyle CD$

$\displaystyle \therefore CD=2 CN=2 \times 5=10 \text{ cm}$

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

LHS $\displaystyle = (\sin \theta + \cos \theta)(\tan \theta+ \cot \theta)$

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

$\displaystyle = \Big(\sin \theta + \cos \theta \Big) \Big(\frac{\sin^2 \theta+ \cos^2 \theta}{\sin \theta. \cos \theta} \Big)$

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

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

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

$\displaystyle = \sec \theta+\mathrm{cosec} \theta = RHS$

Hence Proved.

(c) Let $\displaystyle AD =250 \text{ m}$ height of airplane

Two boats are at $\displaystyle B \text{ and } C .$

Let $\displaystyle BD =x \text{ and } DC=y$ as shown in the diagram.

$\displaystyle \text{From } \triangle ADB; \frac{x}{250}=\cot 45^{\circ}$

$\displaystyle \frac{x}{250}=1 \Rightarrow x=250 \text{ m}$

$\displaystyle \text{From } \triangle ADC; \frac{y}{250}=\cot 60^{\circ}$

$\displaystyle \frac{y}{250}=\frac{1}{\sqrt{3}}$

Width of river $\displaystyle BC = BD+DC = x + y$

$\displaystyle = 250 +\frac{250}{\sqrt{3}}$

$\displaystyle = 250 \Big(1+\frac{1}{\sqrt{3}} \Big)=250 \Big(\frac{\sqrt{3}+1}{\sqrt{3}} \Big)$

$\displaystyle = 250 \Big(\frac{1.732+1}{1.732} \Big)=250\Big(\frac{2.732}{1.732} \Big)$

$\displaystyle = 250 \times 1.577$

$\displaystyle = 394.25 \text{ m}= 394 \text{ m}$