Maximum Marks: 40
Time allowed: One and a half hours
Answers to this Paper must be written on the paper provided separately.
You will not be allowed to write during the first 10 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 from Section A and any three questions from Section B.
The intended marks for questions or parts of questions are given in brackets [ ].

Section A [16 Marks] [16×1]

Question 1. If matrix A is of order 3 x 2 and matrix B is of order 2 x 2 then the matrix AB is of order

(a) 3 x 2          (b) 3 x 1         (c) 2 x 3         (d) 1 x 3

$\fbox{A}$

If matrix A is of order p × q  and matrix B is of order q × r , then matrix AB is of order p × r.

Hence in the given question, matrix AB is of order 3 × 2.

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Question 2. The percentage share of SGST of total GST for an Intra-State sale of an article is

(a) 25%         (b) 50%         (c) 75%         (d) 100%

$\fbox{B}$

GST is Goods and service Tax Applicable on all the goods and services which comes under GST regime.

There are different rate of GST for different category of product and services 5% , 12% , 18%  and 28%

GST for Intra-State Transactions  ( Transactions with in same state) are Equally divided in  2 categories CGST and SGST.

CGST = SGST = GST/2

CGST = SGST = 50% of GST

CGST = Central GST

SGST = State GST

The percentage share of SGST of total GST for an Intra-State sale of an article is 50% .

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Question 3. ABCD is a trapezium with AB parallel to DC. Then the triangle similar to $\triangle AOB$ is

(a) $\triangle ADB$         (b) $\triangle ACB$         (c) $\triangle COD$         (d) $\triangle COB$

$\fbox{C}$

Given that, ABCD is trapezium with AB parallel to DC .

Therefore , In $\triangle AOB$ and $\triangle COD$ we have,

$\angle AOB = \angle COD$       ( Vertically opposite angles. )

$\angle OAB = \angle OCD$   ( since $AB \parallel DC$ , alternate angles. )

Therefore $\triangle AOB \sim \triangle COD$ ( By AA similarity )

Hence, $\triangle AOB$ is similar to $\triangle COD$  .

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Question 4. The mean proportion between 9 and 16 is

(a) 25         (b) 144         (c)7         (d) 12

$\fbox{ D }$

When four numbers say $A,B,C,D$ are in similar quantities such that $A:B= C:D$ then they are said to be in proportion.

When $A,B,C$ are in continued proportion then $B$ is said to be the mean proportional.

$\text{Let us take } A=9 \text{ and } C=16 \text{ and } B \text{ as } x. \text{ Then } A:x : : x:C$

$9:x : : x:16 \Rightarrow x^2 = 9 \times 16 \Rightarrow x^2 = 144 \Rightarrow x = 12$

So, the mean proportional of 9 and 16 is 12.

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Question 5. A man deposited Rs. 500 per month for 6 months and received Rs. 3300 as the maturity value. The interest received by him is:

(a) 1950         (b) 300         (c) 2800         (d) none of these

$\fbox{ B }$

Total money deposited $= 500 \times 6 = 3000 \text{ Rs.}$

Money received at maturity $= 3300 \text{ Rs.}$

Hence the interest received by him $= 3300 - 3000 = 300 \text{ Rs.}$

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Question 6. The solution set representing the following number line is

(a) $\{ x: x \in R, -3 \leq x < 2 \}$       (b) $\{ x: x \in R, -3 < x < 2 \}$

(c) $\{ x: x \in R, -3 < x \leq 2 \}$      (d) $\{ x: x \in R, -3 \leq x \leq2 \}$

$\fbox{ A }$

The convention is that a hollow circle marks the end of the range with a strict inequality $( \text{ i.e. } < \text{ or } > )$ and a darkened circle marks the end of a range involving an equality as well $( \text{ i.e. } \leq \text{ or } \geq ).$

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Question 7. The first three terms of an arithmetic progression (A. P.) are 1, 9, 17, then the next two terms are

(a) 25 and 35      (b) 27 and 37     (c) 25 and 33    (d) none of these

$\fbox{ C }$

$a = 1$

$d = 9-1 = 8$

$4^{th} \text{ term } = a + (n-1)d = 1 + ( 4-1) (8) = 25$

$5^{th} \text{ term } = a + (n-1)d = 1 + ( 5-1) (8) = 33$

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Question 8. If $\triangle ABC \sim \triangle QRP$ then the corresponding proportional sides are

$\displaystyle \text{(a) } \frac{AB}{QR} = \frac{BC}{RP} \hspace{0.5cm} \text{(b) } \frac{AC}{QR} = \frac{BC}{RP} \hspace{1.0cm} \text{(c) } \frac{AB}{QR} = \frac{BC}{QP} \hspace{1.0cm} \text{(d) } \frac{AB}{PQ} = \frac{BC}{RP}$

$\fbox{ A }$

When two triangles are similar, their corresponding angle are equal and corresponding sides are proportional.

For example: If $\triangle ABC$ is similar to $\triangle QRF$

$\text{.i.e } \triangle \sim \triangle QRP$

$\angle A = \angle Q, \hspace{0.5cm} \angle B = \angle R, \hspace{0.5cm} \angle C = \angle P$

$\displaystyle \text{and } \frac{AB}{QR} = \frac{BC}{RP} = \frac{AC}{QP}$

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Question 9. If $x \in W$ , then the solution set of the inequation $-x > -7$, is

(a) $\{ 8,9,10, \cdots \}$    (b) $\{0,1,2,3,4,5,6\}$     (c) $\{ 0,1,2,3, \cdots \}$    (d) $\{ -8, -9, -10, \cdots \}$

$\fbox{ A }$

Given: $x \in W$                 To find: the solution set of $-x > - 7$

Simplify the given inequality:        $-x > - 7 \Rightarrow x > 7$

Know that the symbol W represents a set of whole number that range from $0 \text{ to } \infty.$

Therefore, The solution set of $-x > -7$ will be $8, 9, 10, \cdots . \infty$

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Question 10. The roots of the quadratic equation $4x^2 - 7x + 2 = 0$ are 1.390, 0. 359.The roots correct to 2 significant figures are

(a) 1.39 and 0.36          (b) 1.3 and 0.35        (c) 1.4 and 0.36         (d) 1.390 and 0.360

$\fbox{ C }$

Root of the equation are :

$\displaystyle x = \frac{7\pm\sqrt{49-32}}{2 \times 4} = \frac{7 \pm \sqrt{17}}{8} = \frac{7 \pm 4.12}{8}$

$\displaystyle \Rightarrow x = 1.4 \text{ or } 0.36$

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Question 11. $1.5, 3, x$ and $8$ are in proportion, then $x$ is equal to

(a) 6           (b) 4          (c) 4.5           (d) 16

$\fbox{ B }$

$\displaystyle \frac{1.5}{3} = \frac{x}{8} \Rightarrow 3x = 12 \Rightarrow x = 4$

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Question 12. If a polynomial $2x^2 - 7x - 1$ is divided by $(x + 3)$, then the remainder is

(a) – 4           (b) 38            (c) -3           (d) 2

$\fbox{ B }$

$\begin{array} {r c c } x+3 ) & \overline{\ \ \ 2x^2 - 7x - 1 } & ( 2x-13 \\ (-) & \underline{ 2x^2 + 6x } & \\ & {\hspace{1.6cm} -13x - 1 } & \\ & {\hspace{1.7cm} \underline{ -13x - 39 } } & \\ & {\hspace{3.1cm} 38 } & \end{array}$

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Question 13. If 73 is the nth term of the arithmetic progression 3, 8, 13, 18…, then ‘n’ is

(a) 13         (b) 14         (c) 15         (d) 16

$\fbox{ C }$

So, AP is $3, 8, 13, 18, \cdot$

We know that, common difference is the difference of the consecutive terms.

$d = 8 - 3 = 13 - 8 = 5 \Rightarrow d = 5$

Now, $a = 3, T_n = 73, d = 5$

Applying the given formula,

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

$\Rightarrow 73 = 3 + ( n-1) ( 5)$

$\Rightarrow 70 = ( n-1) (5)$

$\Rightarrow n-1 = 14$

$\Rightarrow n = 15$

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Question 14. The roots of the quadratic equation $x^2+2x+1 = 0$ are

(a) Real and distinct  (b) Real and equal  (c) Distinct   (d) Not real/imaginary

$\fbox{ B }$

Here $a=1, b=-2, c=1$

$b^2-4ac=(-2)^2-4 \times 1 \times 1=4-4=0$

Therefore the roots are real and equal.

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Question 15. Which of the following statement is not true?

(a) All identity matrices are square matrix

(b) All null matrices are square matrix

(c) For a square matrix number of rows is equal to the number of columns

(d) A square matrix all of whose elements except those in the leading diagonal are zero is the diagonal matrix

$\fbox{ B }$

A matrix which has equal number of rows and columns is called a square matrix.

If each of the elements of a matrix is zero, it is call a null matrix. Examples could be:

$\displaystyle \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \text{ or } \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

In the first case it is a square matrix while in the other case it is not a square matrix..

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Question 16. If $(x - 2)$ is a factor of the polynomial $x^3 + 2x^2 - 13 x + k$, then  $'k'$ is equal to

(a) -10           (b) 26           (c) -26           (d) 10

$\fbox{ D }$

Since $( x - 2 )$ is a factor of the polynomial $f(x),$ then $f(2) = 0$

$\Rightarrow 2^3 + 2(2)^2 - 13 (2)x + k = 0$

$\Rightarrow 8 + 8 - 26 + k = 0$

$\Rightarrow k - 10 = 0$

$\Rightarrow k = 10$

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Section B [12 Marks] [6×2]

Question 17. A man deposited Rs.1200 in a recurring deposit account for 1 year at 5% per annum simple interest. The interest earned by him on maturity is

(a) 14790           (b) 390          (c) 4680           (d) 780

$\fbox{ B }$

$\displaystyle \text{Interest } = P \times \frac{n(n+1)}{2 \times 12} \times \frac{R}{100} = 1200 \times \frac{12 \times 13}{12 \times 2} \times \frac{5}{100} = 6 \times 13 \times 5 = 390$

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Question 18. If $x^2- 4$ is a factor of polynomial $x^3+ x^2- 4x- 4$, then its factors are

(a) $(x-2) (x+2) (x+1)$         (b) $(x-2) (x+2) (x-1)$

(c) $(x-2) (x-2) (x+1)$           (d) $(x-2) (x-2) (x-1)$

$\fbox{ A }$

$\begin{array} {r c c } x^2- 4 ) & \overline{x^3+ x^2- 4x- 4 } & ( x+1 \\ (-) & \underline{ x^3 -4x } & \\ & {\hspace{1.6cm} x^2-4 } & \\ (-) & {\hspace{1.7cm} \underline{ x^2-4 } } & \\ & {\hspace{3.1cm} 0 } & \end{array}$

$\therefore x^3+ x^2- 4x- 4 = (x^2- 4)(x+1) = (x-2)(x+2)(x+1)$

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Question 19. The following bill shows the GST rates and the marked price of articles A and B:

$\displaystyle \begin{array} {|l | c| c| } \hline & \text{ BILL: GENERAL STORE } & \\\hline \text{Articles } & \text{Marked Price } & \text{Rate of GST } \\ \hline \text{A } & \text{Rs. } 300 & 12\% \\ \hline \text{B } & \text{Rs. } 1200 & 5\% \\ \hline\end{array}$

The total amount to be paid for the above bill is: –

(a) 1548          (b) 1596              (c) 1560           (d) 1536

$\fbox{ B }$

$\displaystyle \text{Total bill paid } \\ \\ = \Big( 300 + \frac{12}{100} \times 300 \Big) + \Big( 1200 + \frac{5}{100} \times 1200 \Big) = (300+ 36) + ( 1200 + 60) = 1596$

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Question 20. The solution set for the linear inequation $-8 \leq x - 7 < - 4, \ \ x \in I$ is

(a) $\{ x: x \in R, -1 \leq x < 3 \}$           (b) $\{ 0, 1, 2, 3 \}$

(c) $\{-1, 0, 1, 2, 3 \}$           (d) $\{ -1, 0, 1, 2 \}$

$\fbox{ D }$

$-8 \leq x - 7 < - 4, \ \ x \in I$

$\Rightarrow -1 \leq x < 3, \ \ x \in I$

$\Rightarrow \{ -1, 0, 1, 2 \}$

$\\$

$\displaystyle \text{Question 21. If } \frac{5a}{7b} = \frac{4c}{3d} \text{ then by Componendo and dividendo }$

$\displaystyle \text{(a) } \frac{5a+7b}{5a-7b} = \frac{4c-3d}{4c+3d} \hspace{2.0cm} \text{(b) } \frac{5a-7b}{5a+7b} = \frac{4c+3d}{4c-3d} \\ \\ \\ \text{(c) } \frac{5a+7b}{5a-7b} = \frac{4c+3d}{4c-3d} \hspace{2.0cm} \text{(d) } \frac{5a+7b}{5a+7b} = \frac{4c-3d}{4c-3d}$

$\fbox{ C }$

$\displaystyle \text{Given } \frac{5a}{7b} = \frac{4c}{3d}$

Applying Componendo and dividendo

$\displaystyle \frac{5a+7b}{5a-7b} = \frac{4c+3d}{4c-3d}$

$\\$

$\displaystyle \text{Question 22. If } A = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \text{ then } A^2 \text{ is }$

$\displaystyle \text{(a) } \begin{bmatrix} 4 & 0 \\ 1 & 49 \end{bmatrix}$        $\displaystyle \text{(b) } \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix}$        $\displaystyle \text{(c) } \begin{bmatrix} 4 & 0 \\ 9 & 49 \end{bmatrix}$        $\displaystyle \text{(d) } \begin{bmatrix} 1 & 9 \\ -9 & 48 \end{bmatrix}$

$\fbox{ B }$

$\displaystyle \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 2 \times 2 + 0 \times (-1) & 2 \times 0 + 0 \times 7 \\ -1 \times 2 + 7 \times (-1) & -1 \times 0 + 7 \times 7 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix}$

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Section C [12 Marks] [3×4]

Question 23. The distance between station A and B by road is 240 km and by train it is 300 km. A car starts from station A with a speed x km/hr whereas a train starts from station B with a speed 20 km/hr more than the speed of the car.

(i) The time taken by car to reach station B is

$\displaystyle \text{(a) } \frac{240}{x} \hspace{1.0cm} \text{(b) } \frac{300}{x} \hspace{1.0cm} \text{(c) } \frac{20}{x} \hspace{1.0cm} \text{(d) } \frac{300}{x+20}$

$\fbox{ A }$

$\displaystyle \text{Time taken by car }= \frac{\text{Distance between A and B}}{\text{Speed}} = \frac{240}{x}$

$\\$

(ii) The time taken by train to reach station A

$\displaystyle \text{(a) } \frac{240}{x} \hspace{1.0cm} \text{(b) } \frac{300}{x} \hspace{1.0cm} \text{(c) } \frac{20}{x} \hspace{1.0cm} \text{(d) } \frac{300}{x+20}$

$\fbox{ D }$

$\text{Speed of car } = x \text{ km/hr }$

$\text{Therefore speed of train } = (x+20) \text{ km/hr }$

$\displaystyle \text{Time taken by car }= \frac{\text{Distance between A and B}}{\text{Speed}} = \frac{300}{x+20}$

$\\$

(iii) If the time taken by train is 1 hour less than that taken by the car, then the quadratic equation formed is

(a) $x^2 + 80x - 6000=0$          (b) $x^2 + 80x -4800=0$
(c) $x^2 + 240x - 1600 =0$        (d) $x^2-80x +4800 = 0$

$\fbox{ A }$

$\displaystyle \frac{240}{x} - \frac{300}{x+20} = 1$

$\displaystyle \Rightarrow 240 ( x+20) - 300 x = x( x+ 20)$

$\displaystyle \Rightarrow 240 x + 4800 - 300 x = x^2 + 20 x$

$\displaystyle \Rightarrow x^2 + 80x - 4800 = 0$

$\\$

(iv) The speed of the car is

(a) 60 km/hr           (b) 120 km/hr            (c) 40 km/hr            (d) 80 km/hr

$\fbox{ C }$

$x^2 + 80x - 4800 = 0$

$x^2 + 120x - 40x - 4800 = 0$

$x( x + 120) - 40 ( x+120) = 0$

$(x+120) ( x- 40) = 0$

$\Rightarrow x = - 120 \text{ or } x = 40$

$x \text{ cannot be negative}$

$\text{Hence } x = 40 \text{ km/hr}$

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Question 24. In the given triangle $PQR, AB \parallel QR, QP \parallel CB$ and AR intersects CB at O.

Using the given diagram answer the following question:

(i) The triangle similar to $\triangle ARQ$ is

(a) $\triangle ORC$        (b) $\triangle ARP$        (c) $\triangle OBR$        (d) $\triangle QRP$

$\fbox{ A }$

$\text{Consider } \triangle ARQ \text{ and } \triangle ORC$

$\angle OCR = \angle AQR$  ( Alternate angles)

$\angle RAQ = \angle ROC$  ( Alternate angles)

$\therefore \triangle ARQ \sim \triangle ORC$ ( by AAA axiom)

$\\$

(ii) $\triangle PQR \sim\triangle BCR$ by axiom

(a) SAS       (b) AAA       (c) SSS       (d) AAS

$\fbox{ B }$

$\text{Consider } \triangle PQR \text{ and } \triangle BCR$

$\angle PQR = \angle BCR$  ( Alternate angles)

$\angle QPR = \angle CBR$  ( Alternate angles)

$\therefore \triangle PQR \sim \triangle BCR$  ( by AAA axiom)

$\\$

(iii) If QC =6 cm, CR = 4 cm, BR = 3 cm. The length of RP is

(a) 4.5 cm       (b) 8 cm       (c) 7.5 cm       (d) 5 cm

$\fbox{ C }$

$\displaystyle \frac{RP}{BR} = \frac{RQ}{RC} \Rightarrow RP = 3 \times \frac{10}{4} = 7.5 \text{ cm }$

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(iv) The ratio PQ: BC is

(a) 2 : 3         (b) 3 : 2         (c) 5 : 2         (d) 2 : 5

$\fbox{ C }$

$\displaystyle \frac{PQ}{BC} = \frac{RQ}{RC} \Rightarrow \frac{PQ}{BC} = \frac{10}{4} = \frac{5}{2}$

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Question 25. The nth term of an arithmetic progression (A.P.) is (3n + 1)

(i) The first three terms of this A. P. are

(a) 5, 6, 7         (b) 3, 6, 9         (c) 1, 4, 7         (d) 4, 7, 10

$\fbox{ D }$

Given $T_n = 3n + 1$

$\therefore T_1 = 3(1) + 1 = 4 \hspace{0.5cm} T_2 = 3(2) + 1 = 7 \hspace{0.5cm} T_3 = 3(3) + 1 = 10$

$\\$

(ii) The common difference of the A.P. is

(a) 3         (b) 1         (c) -3         (d) 2

$\fbox{ A }$

$T_{n+1}-T_n = [ 3(n+1)+1] - [ 3n + 1] = (3n+1) + 3 - ( 3n + 1) = 3$

$\\$

(iii) Which of the following is not a term of this A.P.

(a) 25         (b) 27         (c) 28         (d) 31

$\fbox{ B }$

$\text{The A.P. is } \{ 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 33, \cdots \}$

Hence 27 is not the term of this A.P.

$\\$

(iv) Sum of the first 10 terms of this A.P. is

(a) 350          (b) 175         (c) -95         (d) 70

$\fbox{ B }$
$\displaystyle S_{10} = \frac{10}{2} \Big[ 2 \times 4 + (10-1) (3) \Big] = 5 [ 35] = 175$
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