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• Please check that this question paper consists of 30 questions.
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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose center is $(2, -3)$ and $B$ is the point $( 1, 4)$.

Let the coordinates of $A$  be $(x, y)$

Since $(2, -3)$ is the center, it is also the midpoint of $AB$

$\therefore \frac{x+1}{2}$ $= 2 \Rightarrow x+1 = 4 \Rightarrow x = 3$

Similarly, $\frac{y+4}{2}$ $= -3 \Rightarrow y+4 = -6 \Rightarrow y = -10$

$\therefore A is (3, -10)$

$\\$

Question 2: For what value of $k$, the roots of the equation $x^2 + 4x + k = 0$ are real?

Or

Find the value of $k$, for which the roots of the equation $3x^2 - 10x + k = 0$ are reciprocal of each other.

Given $x^2 + 4x + k = 0$

For roots to be real, $b^2 - 4ac = 0 \Rightarrow 4^2 - 4(1)(k) = 0 \Rightarrow k = 1$

Or

Given $3x^2 - 10x +k = 0 \Rightarrow a = 3, b = -10, c = k$

Roots are : $\frac{10 \pm \sqrt{10^2-12k}}{6}$

Therefore Roots are $\frac{10 + \sqrt{100-12k}}{6}$ and $\frac{10 - \sqrt{100-12k}}{6}$

$\Rightarrow \frac{10 + \sqrt{100-12k}}{6}$ $=$ $\frac{6}{10 - \sqrt{100-12k}}$

$\Rightarrow 100 - (100-12k) = 36$

$\Rightarrow 100 - 100 + 12 k = 36$

$\Rightarrow k = 3$

$\\$

Question 3: Find $A$ if $\tan 2A = \cot (A-24^o)$

Or

Find the value of $( \sin^2 33^o + \sin^2 57^o)$

$\tan 2A = \cot (A- 24^o)$

$\Rightarrow \tan 2A = \tan (90^o - ( A - 24^o))$

$\Rightarrow 2A = 90^o -A + 24^o$

$\Rightarrow 3A = 114 \Rightarrow A = 38^o$

Or

$\sin^2 33^o + \sin^2 57^o$

$= \sin^2 33^o + \sin^2(90^o-33^o)$

$= \sin^2 33^o + \cos^2 33^o$

$= 1$

$\\$

Question 4: How many two digits number are divisible by $3$?

Numbers divisible by $3$ are $3, 6, 9, 12, \cdots$

Lowest $2$ digit number divisible by $3$ is $12$

The highest $2$ digit number divisible by $3$ is $99$

So the series starts with $12$ and ends with $99$

Difference between the numbers is $3$

So the AP will be $12, 15, 18, \cdots , 99$

We need to find $n$

$\therefore a = 12, a_n = 99, d = 3$

Now $a_n = a + (n-1) d$

$\Rightarrow 99 = 12 + (n-1) (3)$ $\Rightarrow 99 = 9 + 3n$ $\Rightarrow 90 = 3n$ $\Rightarrow n = 30$

Therefore there are 30 two digit numbers divisible by 3

$\\$

Question 5: In Fig. 1, $DE \parallel BC, AD = 1$ cm and $BD = 2$ cm. What is the ratio of the $ar( \triangle ABC)$ to the $ar ( \triangle ADE)$ ?

Given $DC \parallel BC$

Consider $\triangle ABC$ and $\triangle ADE$

$\angle ADE = \angle ABC$ (alternate angles)

$\angle A$ is common

$\angle AED = \angle ACD$ ( alternate angles)

$\therefore \triangle ADE \sim \triangle ABC$ ( AAA criterion)

Since $DE \parallel BC$, by Thales Theorem

$\frac{AD}{DB}$ $=$ $\frac{AE}{EC}$ $=$ $\frac{1}{2}$

Since triangles are similar,

$\frac{ar(\triangle ABC)}{ar(\triangle ADE)}$ $=$ $\frac{3^2}{1}$ $=$ $\frac{9}{1}$

$\\$

Question 6: Find a rational number between $\sqrt{2}$ and $\sqrt{3}$.

We know $\sqrt{2} = 1.412$ and $\sqrt{3} = 1.732$

Therfore $1.5$ is between $\sqrt{2}$ and $\sqrt{3}$

$\Rightarrow 1.5$ $=$ $\frac{15}{10}$ $=$ $\frac{3}{2}$

Hence $\frac{3}{2}$ is between $\sqrt{2}$ and $\sqrt{3}$

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Find the HCF of $1260$ and $7344$  using Euclid’s algorithm.

Or

Show that every positive odd integer is of the form $(4q+1)$ or $(4q+3)$ , where $q$ is some integer.

$1260 ) \overline{7344} ( 5 \\ \hspace*{1cm} \underline{6300} \\ \hspace*{1cm} 1044 ) \overline{1260} ( 1 \\ \hspace*{2cm} \underline{1044} \\ \hspace*{2cm} 216 ) \overline{1044} ( 4 \\ \hspace*{3cm} \underline{864} \\ \hspace*{3cm} 180 ) \overline{216} ( 1 \\ \hspace*{4cm} \underline{36} \\ \hspace*{4cm} 36 ) \overline{180} ( 5 \\ \hspace*{4.5cm} \underline{180} \\ \hspace*{4.75cm} \times$

According to Eculid’s division theorem, any positive number can be expressed as $a = bq+r$ where $q$ is the quotient, $b$ is the divisor and $r$ is the remainder and $0 \leq r < b$

$\therefore 180 = 36 \times 5 + 0$

So HCF of $1260$ and $7344$ is $36$

Or

Let $a$ be any positive integer.

Eculid’s division theorem, any positive number can be expressed as $a = bq+r$ where $q$ is the quotient, $b$ is the divisor and $r$ is the remainder and $0 \leq r < b$

Take $b = 4 \Rightarrow a = 4q + r$

Since $0 \leq r < 4$, the possible remainders are $0, 1, 2, 3$

That is $a$ can be $4q, 4q+1, 4q+2$ or $4q+3$

Since $a$ is odd, $a$ cannot be $4q$ or $4q+2$

Therefore any odd integer is of the form $4q+1$ or $4q+3$.

$\\$

Question 8: Which term of $AP$  $3, 15, 27, 39, \cdots$ will be $120$ more than its $21^{st}$ term?

Or

If $S_n$, the sum of the first $n$ terms of an $AP$ is given by $S_n = 3n^2 - 4n$, find the $n^{th}$ term.

Given $AP$ is $3, 15, 27, 39, \cdots$

Here $a = 3, d = 15-3 = 12$

Therefore $21^{st}$ term is given by

$T_{21} = a + (21-1)d = 3 + 30 \times 12 = 243$

Required term $= 243 + 120 = 363$

Let it be the $n^{th}$ term, then

$T_n = 363$

$\Rightarrow a + (n-1)d = 363$

$\Rightarrow 3 + (n-1) \times 12 = 363$

$\Rightarrow 12n = 372$

$\Rightarrow n = 31$

Hence $31^{st}$ term is the required term.

Or

Given $S_n = 3n^2 - 4n$

Put $n = 1 , S_1 = 3(1)^2 - 4(1) = -1$

$n = 2 , S_1 = 3(2)^2 - 4(2) = 12-8 = 4$

$n = 3 , S_1 = 3(3)^2 - 4(3) = 27 - 12 = 15$

$S_1 = -1 \Rightarrow a = -1$

$S_2 - S_1 = 4 - ( -1) = 5$

$S_3 - S_2 = 15 - 4 = 11$

Hence the common difference is $d = 11 - 5 = 6$

Therefore $n^{th}$ term $t_n = a + (n-1) d = -1 + ( n-1) 6$

$\Rightarrow t_n = 6n - 7$

$\\$

Question 9: Find the ratio in which the segment joining the points $(1, -3)$ and $( 4, 5)$  is divided by $x-axis$. Also find the coordinates of this point on $x-axis$.

Given Points $A(1, -3)$ and $B(4, 5)$

Let the point $P(x, 0)$ be the point which divides the segment joining $A$ and $B$ in the ratio $m:1$

The using the section formula, we get

$(x, 0) = \Big($ $\frac{m \times 4 + 1 \times 1}{m+1}$ $,$ $\frac{m \times 5 + 1 \times (-3)}{m+1}$ $\Big)$

$\Rightarrow 0 =$ $\frac{m \times 5 + 1 \times (-3)}{m+1}$

$\Rightarrow 5m = 5$

$\Rightarrow m =$ $\frac{3}{5}$

Therefore $P$ divides $(1, -3)$ and $(4, 5)$ in the ratio $3:5$

$\therefore x =$ $\frac{m \times 4 + 1 \times 1}{m+1}$

$=$ $\frac{\frac{3}{5} \times 4 + 1 \times 1}{\frac{3}{5}+1}$

$=$ $\frac{12+5}{3+5}$ $=$ $\frac{17}{8}$

Therefore the point is $\Big($ $\frac{17}{8}$ $, 0 \Big)$

$\\$

Question 10: A game consists of tossing a coin three times and noting the outcome each time. If getting the same result in all the tosses is success, find the probability of losing the game.

Possible outcomes of tossing a coin three times is $2 \times 2 \times 2 = 8$

i.e. Total number of events $= 8$

The favorable events are $HHH$ and $TTT$

Therefore Probability of success is $\frac{2}{8}$ $=$ $\frac{1}{4}$

Hence the probability of losing is $1 -$ $\frac{1}{4}$ $=$ $\frac{3}{4}$

$\\$

Question 11: A die is thrown once. Find the probability of getting a number which i) is a prime number ii) lies between $2$ and $6$.

The possible outcomes when a dice is thrown one are $1, 2, 3, 4, 5, 6$

Therefore Total No. of events $= 6$

No of prime events $= 3 ( 2, 3, 5$ are prime numbers  $)$

No of events when the number lies between $2$ and $6 = 3$

Therefore

i) Probability $($ is a prime number $) =$ $\frac{3}{6}$ $=$ $\frac{1}{2}$

ii) Probability $($ number lies between $2$ and $6) =$ $\frac{3}{6}$ $=$ $\frac{1}{2}$

Note: 1 only has one positive divisor $(1$ itself  $)$. So it is not a prime number. A prime number is a positive integer whose positive divisor are exactly $1$ and the number itself.

$\\$

Question 12: Find $c$ if the system of equations $cx+3y + ( 3-c) = 0 ;$ $12x + cy - c = 0$ has infinitely many solutions.

Given: $cx + 3y + (3-c) = 0$ and $12x + cy - c = 0$

If the system of equations are $ax_1 + b_1y + c_1 = 0$ and $ax_2 + b_2y + c_2 = 0$ and they have infinitely many solutions then it satisfy the following:

$\frac{a_1}{a_2}$ $=$ $\frac{b_1}{b_2}$ $=$ $\frac{c_1}{c_2}$

From the given system

$a_1 = c, b_1 = 3, c_1 = 3-c$

$a_2 = 12, b_2 = c, c_2 = -c$

Therefore

$\frac{c}{12}$ $=$ $\frac{3}{c}$ $=$ $\frac{3-c}{-c}$

Taking the first two

$c^2 = 36 \Rightarrow c = \pm 6$   … … … … … (i)

Taking the last two

$-3c = 3c - c^2 \Rightarrow c^2 = 6c \Rightarrow c = 0, 6$   … … … … … (ii)

From (i) and (ii) we get $c = 6$

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that $\sqrt{2}$ is an irrational number.

Assume $\sqrt{2}$ is a rational number

i.e. it can be expressed as a rational fraction of the form $\frac{b}{a}$ where $a, b$ are relatively prime numbers.

Since $\sqrt{2} =$ $\frac{b}{a}$

we have $2 =$ $\frac{b^2}{a^2}$ or $b^2 = 2 a^2$

$b^2$ must be even and since $b^2$ is even, $b$ should be even

Let $b = 2c$

We have $4c^2 = 2a^2 \Rightarrow a^2 = 2c^2$

Since $2c^2$ is even, $a^2$ is even and since $a^2$ is even, $a$ must be even

However two even numbers cannot be relatively prime so $\sqrt{2}$ cannot be expresses as a rational fraction.

Hence $\sqrt{2}$ is irrational.

$\\$

Question 14: Find the value of $k$ such that the polynomial $x^2 - ( k+6)x + 2 ( 2k-1)$ has sum of its zeros  equal to half of their products.

Polynomial is $x^2 - (k+6)x + 2(2k-1) = 0$

Let $\alpha$ and $\beta$ be the zeros of the equation

On comparing with $ax^2 + bx+ c = 0$ we get

$a = 1, b = -(k+6) c = 2(2k-1)$

Therefore $\alpha + \beta =$ $- \frac{b}{a}$ $=$ $\frac{k+6}{1}$

$\Rightarrow \alpha + \beta = k+6$   … … … … … i)

$\alpha . \beta =$ $\frac{c}{a}$ $=$ $\frac{2(2k-1)}{1}$

$\Rightarrow \alpha . \beta = 4k-2$   … … … … … ii)

Given $\alpha + \beta =$ $\frac{1}{2}$ $\alpha . \beta$

$\Rightarrow k+6 =$ $\frac{1}{2}$ $(4k-2)$

$\Rightarrow k+6 = 2k - 1 \Rightarrow k = 7$

$\\$

Question 15: A father’s age is three times the sum of the ages of his two children. After $5$ years his age will be two times the sum of their ages. Find the present age of the father.

Or

A fraction becomes $\frac{1}{3}$ when $2$ is subtracted from the numerator  and it becomes $\frac{1}{2}$ when $1$ is subtracted from the denominator. Find the fraction.

Let the ages of sons be $x$ and $y$

Given that father’s age is $3$ times the ages of his two sons

Therefore present age of father $= 3 (x+y)$   … … … … … i)

Given $5$ years hence, father’s are will be twice the sum of the ages of his sons

Father’s age after $5$ years $= 3(x+5) + 5$

Ages of his sons $5$ year hence $= (x+5)$ and $(y+5)$

$\therefore 3(x+y) + 5 = 2 [ (x+5) + (y+5) ]$

$\Rightarrow 3x + 3y + 5 = 2x + 10 + 2y + 10$

$\Rightarrow x + y = 15$    … … … … … ii)

Using i) and ii) we get that the age of father is $3 ( 15) = 45$ years.

Or

Let the fraction be $\frac{x}{y}$

$\therefore \frac{x-2}{y}$ $=$ $\frac{1}{3}$

$\Rightarrow 3x - 6 = y$

$\Rightarrow 3x - y = 6$  … … … … … i)

Also $\frac{x}{y-1}$ $=$ $\frac{1}{2}$

$\Rightarrow 2x = y - 1$

$\Rightarrow 2x - y = -1$  … … … … … ii)

Subtracting ii) from i) we get $x = 5$

From ii) $y = 2x+1 = 2(5) + 1 = 11$

Hence the fraction is $\frac{5}{11}$

$\\$

Question 16: Find the point on $y-axis$ which is equidistant from the point $(5, -2)$ and $(-3, 2)$.

Or

The line segment joining the points $A(2, 1)$ and $B(5, -8)$ is trisected at points $P$ and $Q$ such that $P$ is nearer to $A$. If $P$ also lies on the line given by $2x-y+k = 0$, find the value of $k$.

Let $P(0, y)$ be equidistant from $A(5, -2)$ and $B(-3, 2)$

$\therefore AP = \sqrt{(5-0)^2 + (-2-y)^2} = \sqrt{25 + (-2-y)^2}$

$BP = = \sqrt{(-3-0)^2 + (2-y)^2} = \sqrt{9 + (2-y)^2}$

Since $AP = BP \Rightarrow AP^2 = BP^2$

$\therefore 25 + (-2-y)^2 = 9 + (2-y)^2$

$\Rightarrow 25 + 4 + y^2 + 4y = 9 + 4 + y^2 - 4y$

$\Rightarrow 29 + 8y = 13$

$\Rightarrow 8y = - 16$

$\Rightarrow y = -2$

Hence the point is $(0, -2)$

Or

$AP : AB = 1:3$ (trisects)

$\frac{AP}{AB}$ $=$ $\frac{1}{3}$

$\Rightarrow \frac{AP}{AP+PB}$ $=$ $\frac{1}{3}$

$\Rightarrow 3AP = AP + PB$

$\Rightarrow 2AP = PB$

$\Rightarrow \frac{AP}{PB}$ $=$ $\frac{1}{2}$

$\Rightarrow AP : PB = 1:2$

Applying section formula

Coordinates of $P = \Big($ $\frac{1 \times 5 + 2 \times 2}{1 + 2}$ $,$ $\frac{1 \times (-8) + 2 \times (1)}{1 + 2}$ $\Big)$

$\Rightarrow P = \Big($ $\frac{5+4}{3}$ $,$ $\frac{-8+2}{3}$ $\Big)$

$\Rightarrow P = (3, -2)$

Since $P(3, -2)$ lies on $2x - y + k = 0$

$\therefore 2(3) - (-2) + k = 0$

$\Rightarrow 6 + 2 + k = 0$

$\Rightarrow k = -8$

$\\$

Question 17: Prove that $(\sin \theta + \mathrm{cosec} \theta)^2 + ( \cos \theta + \sin \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta$.

Or

$(1 + \cot A - \mathrm{cosec} A)( 1 + \tan A + \sec A) = 2$

To Prove: $( \sin \theta + \mathrm{cosec} \theta )^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta$

LHS $= ( \sin \theta + \mathrm{cosec} \theta )^2 + (\cos \theta + \sec \theta)^2$

$= \sin^2 \theta + 2 \sin \theta \ \mathrm{cosec} \theta + \mathrm{cosec}^2 \theta + \cos^2 \theta + \sec^2 \theta + 2 \cos \theta \ \sec \theta$

$= 1 + 2 + 2 + \mathrm{cosec}^2 \theta + \sec^2 \theta$

$= 5 + ( 1+ \cot^2 \theta) + ( 1 + \tan^2 \theta)$

$= 7 + \tan^2 \theta + \cot^2 \theta$

$=$ RHS. Hence proved.

Or

To Prove: $(1 + \cot A - \mathrm{cosec} A)(1 + \tan A + \sec A) = 2$

LHS $= (1 + \cot A - \mathrm{cosec} A)(1 + \tan A + \sec A)$

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

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

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

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

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

$= 2 =$ RHS. Hence proved.

$\\$

Question 18: In Fig. 2, $PQ$ is a chord of length $8$ cm of a circle of radius $5$ cm and center $O$. The tangent at $P$ and $Q$ intersect at point $T$. Find the length of $TP$.

$TP = TQ$ (Length of the tangents from an external point to a circle are equal)

In $\triangle TPQ TP = TQ$

$OT$ is the bisector of $\angle PTQ$

$\therefore OT \perp PQ$

Since $OT \perp PQ$

$PR = RQ$ ( perpendicular from center to a chord bisects the chord)

$\therefore PR = QR =$ $\frac{1}{2}$ $PQ = 4$ cm

In Right triangle $ORP$

$OP^2 = OR^2+ PR^2$

$\Rightarrow 5^2 = OR^2 + 4^2$

$\Rightarrow OR^2 = 25 - 16 = 9$

$\Rightarrow OR = 3$ cm

Let $TP = x$

In Right triangle $PRT$

$TP^2 = PR^2 + RT^2$

$\Rightarrow x^2 = 4^2 + RT^2$   … … … … … i)

Since $TP$ is a tangent, $OP \perp TP$

$\angle OPT = 90^o$

In Right triangle $OPT$

$OT^2 = OP^2 + TP^2$

$\Rightarrow OT^2 = 5^2 + x^2$

$\Rightarrow (OR+RT)^2 = 5^2 + x^2$

$\Rightarrow (3+RT)^2 = 5^2 + x^2$

$\Rightarrow 9 + RT^2 + 6 RT = 25 + x^2$

From i) $x^2 = 16+RT^2$

$9 + RT^2 + 6 RT = 25 + 16+RT^2$

$\Rightarrow 6RT = 32$

$\Rightarrow RT =$ $\frac{32}{6}$ $=$ $\frac{16}{3}$

$\therefore x^2 = 16 + \Big($ $\frac{16}{3}$ $\Big)^2 =$ $\frac{400}{9}$

$\Rightarrow x =$ $\frac{20}{3}$

Hence $TP$ is $\frac{20}{3}$ cm

$\\$

Question 19: In Fig. 3, $\angle ACB = 90^o$ and $CD \perp AB$, prove that $CD^2 = BD \times AD$

Or

If $P$ and $Q$ are point of sides $CA$ and $CB$ respectively, of $\triangle ABC$, right angled at $C$, prove that $(AQ^2+BP^2) = ( AB^2 + PQ^2)$

Given: $\angle ACB = 90^o, CD \perp AB$

In $\triangle CAD, CA^2 = CD^2 + AD^2$   … … … … … i)

In $\triangle CDB, CB^2 = CD^2 + DB^2$   … … … … … ii)

Adding i) and ii)

$CA^2 + CB^2 = 2 CD^2 + AD^2 + DB^2$

$\Rightarrow AB^2 = 2 CD^2 + AD^2 + DB^2$

$\Rightarrow AB^2 - AD^2 = DB^2 + 2CD^2$

$\Rightarrow ( AB - AD)(AB+AD) - DB ^2 = 2CD^2$

$\Rightarrow DB(AB + AD) - DB^2 = 2 CD^2$

$\Rightarrow DB [ AB + AD - DB ] = 2 CD^2$

$\Rightarrow CD^2 = DB. AD$. Hence proved.

Or

$AQ^2 = AC^2 + CQ^2$   … … … … … i)

$PB^2 = PC^2 + CB^2$   … … … … … ii)

Adding i) and ii)

$AQ^2 + PB^2 = AC^2 + CQ^2 + PC^2 + CB^2$

$\Rightarrow AQ^2 + PB^2 = AB^2 + PQ^2$. Hence proved.

$\\$

Question 20: Find the area of the shaded region in Fig. 4, if $ABCD$ is a rectangle with sides $8$ cm and $6$ cm and $O$ is the center of the circle. (Take $\pi = 3.14$ )

$\angle ADC =90^o$

(angle subtended by the diameter on the circumference is $90^o$ )

$\therefore ABCD$ is a rectangle

$AC = \sqrt{8^2+6^2} = 10$ cm

Therefore the radius of the circle $= 5$ cm

Area of the circle $= \pi (5)^2 = 25\pi$

Area of $ABCD = 6 \times 8 = 48 \ cm^2$

Hence the shaded area $= 25 \pi - 48 = 30.5 \ cm^2$

$\\$

Question 21: Water in a canal, $6$ m wide and $1.5$ m deep, is flowing at the speed of $10$ km/hour. How much area will it irrigate in $30$ minutes; if \$latex 8 cm standing water is needed?

Speed of water $= \frac{10 \ km}{hr}$ $=$ $\frac{10 \times 1000 \ m}{60 \ min}$ $=$ $\frac{1000 \ m}{6 \ min}$

Rate of flow $= 6 \times 1.5 \ m^2 \times$ $\frac{1000 \ m}{6 \ min}$ $=$ $\frac{1500 \ m^3}{min}$

Therefore volume of water in $30$ mins $= 1500$ $\frac{m^3}{min}$ $\times 30 \ min = 45000 \ m^3$

Let Area of irrigation $= x$

$\therefore x \times$ $\frac{8}{100}$ $\ m = 45000 m^3$

$x =$ $\frac{45000 \times 100}{8}$ $\ m^2 = 562500 \ m^2$

$\\$

Question 22: Find the mode of the following frequency distribution

 Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 8 10 10 16 12 6 7

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: Two water taps can together can fill in a tank in $1$ $\frac{7}{8}$ hours. The tap with longer diameter  takes $2$ hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Or

A boat goes $30$ km upstream and $44$ km downstream in $10$ hours. In $13$ hours it can go $40$ km upstream and $55$ km downstream. Determine the speed of the stream and that of the boat in still water.

Let the time taken by the smaller tap to fill up the tank completely $= x$ hours

So the volume of tank filled by the smaller tap in $1$ hours $=$ $\frac{1}{x}$

Also it is given that the time taken by the larger tap is $2$ hours less

Time taken by larger tap to fill the tank completely $= x - 2$

So the volume of tank filled by the larger tap in $1$ hours $=$ $\frac{1}{x-2}$

Given that both the taps fill the tank in $1$ $\frac{7}{8}$ hours $=$ $\frac{15}{8}$ hours

Tank filled by smaller tap in $\frac{15}{8}$ hours $=$ $\frac{15}{8x}$

Similarly, tank filled by larger tap in $\frac{15}{8}$ hours $=$ $\frac{15}{8(x-2)}$

Therefore $\frac{15}{8x}$ $+$ $\frac{15}{8(x-2)}$ $= 1$

$15[8(x-2)] + 15[8x] = 8x[8(x-2)]$

$120x - 240 + 120 x = 64 x^2 -16 x$

$240 x - 240 = 64x^2-16x$

$64x^2 - 256x + 240 = 0$

$4x^2 - 16x + 15 = 0$

$x =$ $\frac{16 \pm \sqrt{16^2 - 4 \times 4 \times 15}}{8}$ $=$ $\frac{16 \pm 4}{8}$

$\therefore x = 2.5$ hours or $1.5$ hours

$x$ cannot be $1.5$ hours as the larger tap takes $2$ hours less.

Hence $x = 2.5$ hours

Therefore small tap will fill the tank in $2.5$ hours and the larger tap will fill the tank in $0.5$ hours.

Or

Let the speed of the boat in still water $= x$ km/hr

Let the speed of the stream $= y$ km/hr

Speed of the boat downstream $= (x+y)$ km/hr

Sped of the boat upstream $= ( x- y)$ km/hr

Boat goes $30$ km upstream and $44$ km downstream in $10$ hours

$\Rightarrow$ $\frac{30}{x-y}$ $+$ $\frac{44}{x+y}$ $= 10$    … … … … … … i)

Boat goes $40$ km upstream and $55$ km downstream in $13$ hours

$\Rightarrow$ $\frac{40}{x-y}$ $+$ $\frac{55}{x+y}$ $= 13$    … … … … … … ii)

Let $\frac{1}{x-y}$ $= u$ and $\frac{1}{x+y}$ $= v$

Substituting in i) and ii)

$30u + 44v = 10$    … … … … … … iii)

$40u + 55v = 13$   … … … … … … iv)

From iii) we get $u =$ $\frac{10-44v}{30}$

Substituting in iv) we get

$40($ $\frac{10-44v}{30}$ $) + 55v = 13$

$\Rightarrow 40-176v+ 165v = 39$ $\Rightarrow 11v = 1$ $\Rightarrow v =$ $\frac{1}{11}$

Substituting in iii) we get

$30 u + 44 ($ $\frac{1}{11}$ $) = 10$ $\Rightarrow 30u + 4 = 10$ $\Rightarrow u =$ $\frac{6}{30}$ $=$ $\frac{1}{5}$

Now solving for $x$ and $y$

$\frac{1}{x-y}$ $=$ $\frac{1}{5}$ $\Rightarrow x - y = 5$    … … … … … … v)

$\frac{1}{x+y}$ $=$ $\frac{1}{11}$ $\Rightarrow x + y = 11$    … … … … … … vi)

Adding v) and vi) we get $2x = 16 \Rightarrow x = 8$ km/hr

From vi) $y = 11-8 = 3$ km/hr

Therefore speed o boat in still water is $8$ km/hr and speed of stream is $3$ km/hr

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Question 24: If the sum of the first four terms of an $AP$ is $40$ and that of the first $14$ terms is $280$. Find the sum of the first $n$ terms.

$S_n =$ sum of $n$ terms of an A.P. $=$ $\frac{n}{2}$ $[ 2a + (n-1)d]$

$S_4 \Rightarrow 40 =$ $\frac{4}{2}$ $[2a + 3d] = 2a + 3d = 20$   … … … … … i)

$S_{14} \Rightarrow 280 =$ $\frac{14}{2}$ $[2a + 13d] = 2a + 13d = 40$   … … … … … ii)

solving i) and ii)

gives $a = 7$    and $d= 2$

so $S_n =$ $\frac{n}{2}$ $[ 2(7) + (n-1)2] = 7n + n^2 - n = n^2 + 6n$

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Question 25: Prove that $\frac{\sin A - \cos A + 1}{\sin A + \cos A + 1}$ $=$ $\frac{1}{\sec A - \tan A}$

LHS $=$ $\frac{\sin A - \cos A + 1}{\sin A + \cos A - 1}$

$=$ $\frac{\sin A - \cos A + 1}{\sin A + \cos A - 1}$ $\times$ $\frac{\sin A + \cos A + 1}{\sin A + \cos A + 1}$

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

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

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

$=$ $\frac{2 \sin A (\sin A + 1)}{2 \sin A \cos A}$

$=$ $\frac{\sin A + 1}{\cos A}$

$=$ $\frac{\sin A}{\cos A}$ $+$ $\frac{1}{\cos A}$

$= \tan A + \sec A$

$= (\sec A + \tan A)$ $\Big($ $\frac{\sec A - \tan A}{\sec A - \tan A}$ $\Big)$

$=$ $\frac{\sec^2 - \tan^2 A}{\sec A - \tan A}$

$=$ $\frac{1}{\sec A - \tan A}$

$=$ RHS. Hence proved.

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Question 26: A man in a boat is rowing away from the light house at $100$ m high takes $2$ minutes to change the angle of elevation of the top of the light house from $60^o$ to $30^o$. Find the speed of the boat in meters per minute. [ Use $\sqrt{3} = 1.732$ ]

Or

Two poles of equal height are standing opposite each other on either side of the road, which is $80$ m wide. From a point between them on the road, the angle of elevation of the top of the poles are $60^o$ to $30^o$ respectively. Find the height of the poles and the distance of the point from the poles.

$\tan 60^o =$ $\frac{AB}{BD}$ $= \sqrt{3}$

$\therefore BD =$ $\frac{AB}{\sqrt{3}}$ $=$ $\frac{100}{\sqrt{3}}$

$\tan 30^o =$ $\frac{AB}{CB}$ $=$ $\frac{1}{\sqrt{3}}$

$\therefore CB = \sqrt{3} AB = \sqrt{3} (100)$

$\therefore CD = CB - DB = 100\sqrt{3} -$ $\frac{100}{\sqrt{3}}$ $=$ $\frac{200}{\sqrt{3}}$

Therefore Speed $=$ $\frac{distance}{time}$ $=$ $\frac{\frac{200}{\sqrt{3}}}{2}$ $=$ $\frac{100}{\sqrt{3}}$ m/min

Or

$\tan 60^o =$ $\frac{h}{EB}$ $= \sqrt{3} \Rightarrow EB =$ $\frac{h}{\sqrt{3}}$

$\tan 30^o =$ $\frac{h}{CE}$ $=$ $\frac{1}{\sqrt{3}}$ $\Rightarrow CE = \sqrt{3} h$

Given $CE + EB = 80$

$h\sqrt{3} +$ $\frac{h}{\sqrt{3}}$ $= 80$

$\Rightarrow h =$ $\frac{80 \sqrt{3}}{3 + 1}$ $= 20 \sqrt{3}$ m

$\therefore CE = \sqrt{3} (20 \sqrt{3}) = 60$ m

Hence $EB =$ $\frac{20 \sqrt{3}}{\sqrt{3}}$ $= 20$ m

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Question 27: Construct a $\triangle ABC$ in which $CA$ is $6$ cm, $AB = 5$ cm and $\angle BAC = 45^o$. The construct a triangle whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle ABC$.

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Question 28: A bucket open at the top is in the form of a frustum of a cone with a capacity $12308.08 \ cm^3$. The radii of the top and bottom of the circular ends of the bucket and $20$ cm and $12$ cm respectively. Find the height of the bucket, and also the area of the metal sheet  used in making it. ( Use $\pi = 3.14$ )

Volume $= 12308.80 \ cm^3$

$r_1 = 20$ cm  $r_2 = 12$ cm  $h = ?$

Volume of the bucket $=$ $\frac{\pi}{3}$ $h ( {r_1}^2 + {r_2}^2 +r_1. r_2 )$

$\Rightarrow 12308.80 = 3.14 \times$ $\frac{h}{3}$ $( 20^2 + 12^2 + 20 \times 12)$

$\Rightarrow h = 12308.80 \times$ $\frac{3}{3.14 \times 784}$ $= 15$ cm

Therefore height of bucket $= 15$ cm

$l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{15^2 + 8^2} = 17$ cm

Therefore surface are of the metal sheet used

$=$ curved surface area $+$ base area

$= \pi ( r_1 + r_2) l + \pi r^2$

$= 3.14 \times 32 \times 17 + 3.14 \times 12^2 = 2160.32 \ cm^2$

Hence the height of the bucket is $15$ cm and surface are of the metal sheet is $2160.32 \ cm^2$

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Question 29: Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of square of the other two sides.

Given: A right angled $\triangle ABC$, right angled at $B$

To prove: $AC^2 = AB^2 + BC^2$

Draw: $BD \perp AC$

Proof: In $\triangle ADB \& \angle ABC$

$\angle ADB = \angle ABC = 90^o$

$\angle BAD = \angle CAB$ (common angle)

$\therefore \triangle ADB \sim \triangle ABC$ ( by AA criterion)

Therefore $\frac{AD}{AB}$ $=$ $\frac{AB}{AC}$ (corresponding sides are proportional)

$AB^2 = AD \times AC$   … … … … … i)

Similarly, $\triangle BDC \sim \triangle ABC$

and $BC^2 = CD \times AC$   … … … … … ii)

Adding i) and ii)

$AB^2 + BC^2 = AD \times AC + CD \times AC$

$\Rightarrow AB^2 + BC^2 = AC ( AD + CD)$

$\Rightarrow AB^2 + BC^2 = AC^2$. Hence proved.

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Question 30: If the median of the following frequency distribution is $32.5$, find the values of $f_1$ and $f_2$.

 Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total Frequency $f_1$ 5 9 12 $f_2$ 3 2 40

Or

The marks obtained by $100$ students in a class in an examination are given below.

 Marks No. of Students 0-5 2 5-10 5 10-15 6 15-20 8 20-25 10 25-30 25 30-35 20 35-40 18 40-45 4 45-50 2

 Class Interval Frequency Cumulative Frequency 0-10 $f_1$ $f_1$ 10-20 5 $f_1 +$ 5 20-30 9 $f_1+$ 14 30-40 12 $f_1 +$ 26 40-50 $f_2$ $f_1+f_2+$ 26 50-60 3 $f_1+f_2+$ 29 60-70 2 $f_1+f_2+$ 31 $N=40$

Given, Median $= 32.5$

The median class $= 30-40$

$L = 30, h = 40-30 = 10, f= 12, F = 14 + f_1$

Median $= l +$ $\frac{\frac{N}{2} - F}{f}$ $\times h$

$\Rightarrow 32.5 = 30 +$ $\frac{20-(14+f_1)}{12}$ $\times 10$

$\Rightarrow 32.5 - 30 =$ $\frac{20-(14+f_1)}{12}$ $\times 10$

$\Rightarrow 2.5 =$ $\frac{6 - f_1}{12}$ $\times 10$

$\Rightarrow 2.5 =$ $\frac{6 - f_1}{6}$ $\times 5$

$\Rightarrow 2.5 \times 6 = (6-f_1) \times 5$

$\Rightarrow 3 = 6 - f_1$

$\Rightarrow f_1 = 3$

Given sum of frequencies $= 40$

$f_1 + f_2 + 31 = 40$

$f_2 + 34 = 40$

$f_2 = 6$

Hence $f_1 = 3$ and $f_2 = 6$

Or

i)

 Marks Number of Students Less than 5 2 Less than 10 7 Less than 15 13 Less than 20 21 Less than 25 31 Less than 30 56 Less than 35 76 Less than 40 94 Less than 45 98 Less than 50 100

ii) $N = 100$, $\frac{N}{2}$ $= 50$

Hence Median $= 29$

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