(a) Capital employed - Goodwill + current liabilities

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47 b₁ + 54.10 = 2.47 b₂ + 54.10
2.47b₁ = 2.47b₂ ⇒ b₁ = b₂
Thus, h(b₁) = h(b₂) ⇒ b₁ = b₂, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) = 147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.

A = { 1, 2, 3, 7}, B = { 3, 0, -1, 7}
A x B = {(1, 3), (1, 0), (1, - 1), (1, 7),(2,3), (2, 0), (2, -1), (2, 7), (3,3), (3, 0), (3, - 1), (3, 7), (7, 3)., (7, 0), (7, -1), (7 ,7)}
R₄ = {(7, - 1), (0, 3), (3, 3), (0, 7)}
In this (0, 3) and (0, 7) ∈ R₄
But (0, 3) and (0, 7) are not the elements of A x B.
Hence R₄ is not a relation from A to B.

A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

A table

x	0	1	2	3
f(x)	1	3	5	7

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

\({ x }^{ 2 }-\left( -\frac { 7 }{ 2 } \right) x+\frac { 5 }{ 2 } =0\) gives 2x² + 7x + 5 = 0

Here t₂ - t₁ = t₃ - t₂
13 - 6 = 20 - 13 = 7
It is an Arithmetic series
Sum \(\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l)\)
a = 6 : l = 97
a + (n - 1) d = 97
6 + (n - 1) (7) =  97
(n - 1) (7) = 97 - 6
(n - 1) (7) = 91
\(n-1=\frac{91}{7}=13\)
n = 13 + 1 = 14
Now \(S_{n}=\frac{14}{2}(6+97)=7 \times 103\)
6 + 13 + 20 +...+ 97 = 721

\(7x-\frac { 3 }{ 17 } \) = 0
Comparing with ax + by + c = 0
\(\text { Slope } =-\frac{a}{b} \)
\(=-\frac{7}{0}=\text { undefined } \)

Given points (2, 3) and (- 7, - 1)
Equation of the line passing through (x₁ , y₁) and (x₂, y₂) is
\( \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-3}{-1-3}=\frac{x-2}{-7-2} \)
9 (y - 3) - 4 (x - 2)
9y - 27 = 4 x - 8
4 x - 9y +19 = 0

(–10, –4), (–8, –1) and (–3, –5)

Area of triangle \(=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+\right. \left.x_{3}\left(y_{1}-y_{2}\right)\right] \text { sq. units } \)
\(=\frac{1}{2}[-10(-1+5)-8(-5+4)-3(-4+1)] \)
\(=\frac{1}{2}[-10(4)-8(-1)-3(-3)] \)
\(=\frac{1}{2}[-40+8+9] \)
\(=\frac{1}{2}[-40+17]=-\frac{23}{2}=-11.5 \)
[ Area cannot be negative ]
Area of triangle = 11.5 sq. units.

(i) Given P(A) = 0.42
P(not A) = 1 - P(A)
\(\mathrm{P}(\bar{A})=1-0.42=0.58\)
(ii) Given P(B) = 0.48
P(not B) = 1 - P(B)
\(\mathrm{P}(\bar{B})=1-0.48=0.52\)
(iii) P(A or B) = \(P(A \cup B)\)
\(=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B)\)
= 0.42 + 0.48 - 015
= 0.90 - 0.16
P(A or B) = 0.74

leap year has 366 days. So it has 52 full weeks and 2 days. 52 Saturdays must be in 52 full weeks.
The possible chances for the remaining two days will be the sample space.
S = {(Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun)}
n(S) = 7
Let A be the event of getting 53^rd Saturday.
Then A = {Fri-Sat, Sat-Sun}; n(A) = 2
Probability of getting 53 Saturdays in a leap year is P(A0 = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 7 } \).

Height of the barrel = h = 7 cm
Diameter = 5 mm
radius r = \(\frac{5}{2}=2.5 \mathrm{~mm}=0.25 \mathrm{~cm}\)
Volume of cylindrical barrel = \(\pi r^{2} h\)
\(=\frac{22}{7} \times 0.25 \times 0.25 \times 7\)
= 1.375 cm²
Given 1.375 cm³ of ink is used for writing 330 words
Number of words that can be written with one - fifth of a litre
\({\left[\because 1000 \mathrm{~cm}^{3}=1 \mathrm{ltr} ; \frac{1}{5} \times 1000 \mathrm{~cm}^{3} ; 200 \mathrm{~cm}^{3}\right]} \)
\(=\frac{330}{1.375} \times 200=48000 \text { words } \)

2m² + 19m + 30 = 2m² + 4m + 15m + 30 = 2m(m + 2) + 15(m + 2)
= (m + 2)(2m + 15)
Now, equating the factors to zero we get,
(m + 2)(2m + 15) = 0
m + 2 gives, m = -2 or 2m + 15 = 0 we get, m = \(\frac {-15}{2}\)
Therefore the roots are -2, \(\frac {-15}{2}\)
Some equations which are not quadratic can be solved by reducing them to quadratic equations by suitable substitutions. Such examples are illustrated below.

Given PQ is the tangent from the point P outside the circle and OQ is the radius.
We know that tangent meet the radius perpendicularly
\( \therefore \angle P Q O =90^{\circ} \)
\(\angle P O Q =180-120=60^{\circ} \)
\( [\because\ \angle\ P O Q \ and\ \angle P O R\ are \ linear\ pair \ of\ angles]\)
\( In \ \triangle P O O \angle P O O+\angle P O O+\angle O P Q=180^{\circ}\)
[sum of angles of a triangle]
60^o + 90^o + \(\angle\)OPQ = 180o
\(\angle\)OPQ = 180^o - 150^o
\(\angle\)OPQ = 30^o

Radius of conical flask = 'r' units
Height of conical flask = 'h' units
Volume of conical flask = Volume of water
\(=\frac{1}{3} \pi r^{2} h \text { cu. units }\)
Since, water is poured into the cylindrical flask
Volume of cylinder = Volume of water
\(\pi(\mathrm{xr})^{2} H=\frac{1}{3} \pi r^{2} h\)
[xr - radius of cylinder, H - height]
\(\mathrm{X}^{2} \mathrm{r}^{2} \mathrm{H}=\frac{r^{2}}{3} h\)
Height of the water in cylinder flask
\(\mathrm{H}=\frac{h}{3 x^{2}}\)

\( \frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1} =\tan ^{2} \theta \)
\(\text { LHS } =\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1} \)
\(=\frac{1-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1} \)
\(=\frac{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\cos ^{2} \theta}}{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}\)
\(=\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta}{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)} \)
\(=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\tan ^{2} \theta=\text { RHS }\)

Given n = 28
t_n = 4n - 3
t₁ = 4(1) - 3 = 4 - 3 = 1 = a
t₂ = 4(2) - 3 = 8 - 3 = 5
d = t₂ - t₁ = 5 - 1 = 4
\(\mathrm{S}_{\mathrm{n}} =\frac{n}{2}[2 a+(n-1) d] \)
\(=\frac{28}{2}[2(1)+(28-1)(4)] \)
= 14 [2 + 27 (4)] = 14 [2 + 108]
= 14 x 110 = 1540
Sum of first 28 terms of the given A.P. = 1540

f = {(-1, 3), (0, -1), 2, -9)
Since 'f' is a linear function
f(x) = (ax) + b
in (0 , -1) , when x = 0 , f(0) = -1
a(0) + b = -1 , b = -1
in (-1 , 3) , when x = -1 , f(-1) = 3
a( -1) + b = 3 ,a - 1 = 3
-a = 4, a = -4
f(x) = - 4x - 1

Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

1 + \(\frac { co{ t }^{ 2 }\theta }{ 1+cosec\theta +1 } \) = 1+ \(\frac { cose{ c }^{ 2 }\theta -1 }{ cosec\theta +1 } \) [since cosec²-1 = cot²\(\theta \)
= 1+\(\frac { (cosec\theta +1)(cosec\theta -1) }{ cosec\theta +1 } \) 
1 +( cosec\(\theta \)-1) = cosec\(\theta \)

Construction
Step 1 : Draw a line segment BC = 8 cm.
Step 2 : At B, draw BE such that \(\angle\)CBE = 60^o.
Step 3 : At B, draw BF such that \(\angle\)EBF = 90^o .
Step 4 : Draw the perpendicular bisector to BC, which intersects BF at O and BC at G.
Step 5 : With O as centre and OB as radius draw a circle.
Step 6 : From B, mark an arc of 6 cm on BC at D.
Step 7 : The perpendicular bisector intersects the circle at I. Joint ID.
Step 8 : ID produced meets the circle at A. Now join AB and AC. Then \(\triangle\)ABC is the required triangle.

Given a triangle PQR we are required to construct another triangle whose sides are \(\frac{3}{5}\) of the corresponding sides of the triangle PQR.



Steps of construction
1. Construct a \(\triangle\) PQR with any measurement
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 5 (the greater of 3 and 5 in \(\frac { 3 }{ 5 } \)) points.
Q₁Q_2, Q₃, Q₄ and Q₅ on QX so that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅
4. Join Q₅R  and draw a line through Q₃ (the third point, 3 being smaller of 3 and 5 in \(\frac { 3 }{ 5 } \)) parallel to Q₅R to intersect QR at R'.
5. Draw line through R' parallel to the line RP to intersect QP at P'.
Then, \(\triangle\)P'QR' is the required triangle each of whose sides is three-fifths of the corresponding sides of \(\triangle\) PQR.

      

A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.

Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30^o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45^o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

Team A
x1	d1 = x - 28	d12
50	22	484
20	-8	64
10	-18	324
30	2	4
30	2	4
140	Σd = 0	880

\(\bar { { x }_{ 1 } } \) =\(\frac { 140 }{ 5 } \) =28
σ₁ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 880 }{ 5 } } \)
= \(\sqrt { 176 } \)
= 13.27
CV₁ = \(\frac { \sigma }{ \bar { x_{ 1 } } } \) x 100
CV₁ = \(\frac { 13.27 }{ 28 } \) x 100
= 47.39%
CV₁ < CV₂

Team B
x2	d2 = x - 28	d2
40	10	100
60	30	900
20	-10	100
20	-10	100
10	-20	400
150	Σd = 0	1600

\(\bar { { x }_{ 2 } } =\frac { 150 }{ 5 } \) =30
σ₂ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 1600 }{ 5 } } \)
= \(\sqrt { 320 } \)
 =17.89
CV₁ = \(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100
CV₂  = \(\frac { 17.89 }{ 30 } \) x 100
= 59.63%
∴ Team A is more consistent.

Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
X²+x-20 = 0
(x + 5)(x - 4) = 0
x=-5,4
∴ x=4
(∵ x≠-5, x is a natural number
The next number = 4 + 1= 5
Two consecutive numbers are 4,5.

Let the numbers be ∝, β
Sum of the roots = ∝ + β = 15          ...(1)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { 3 }{ 10 } \quad \quad \quad ...(2)\)
\(\\ \frac { +\alpha }{ \alpha \beta } =\frac { 3 }{ 10 } \)
10(∝+ β)= 3∝β    ....(3)
30∝β=10x15=150
Products of the roots =∝β=50 ....(4)
∴ From (1) & (4), we have
x²-15x+50=0
(x-10)(x-5)=0⇒x=10,5

Here a = 24, d = 21-24 = -3, S_n = 78. We need to find n.
We know that,
S_n= \(\frac{n}{2}\) (2a + (n - 1)d)
78 = \(\frac{n}{2}\) 48 + (n -1)(-3))
78 = \(\frac{n}{2}\) 2(51-3n)
or 3n² - 51n + 156 = 0
n² -17n + 52 = 0
(n - 4)(n - 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

1,1,1,2,2,2,3,3,3
t₂ - t₁ = 1-1 = 0
t₃ - t₂ = 1-1 = 0
t₄ - t₃ = 2-1 = 1
Here t₂ - t₁ ≠ t₃ - t₂
ஃ It is not an A.P.

Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.,
= \(\frac { 1 }{ 2 } \) [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
= \(\frac { 1 }{ 2 } \) [-4k] = 0
k = 0 
Area of ΔABC
= \(\frac { 1 }{ 2 } \)[2(0 + 3) + 4(-3 - 3) + 6(3 - 0)] = 0

\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \)
For f(1) & f(3)
Since 1 & 3 lies between -5 ≤ x ≤2, We take, f(x) = x +5
f(1) = 1 + 5 = 6
f(-3) = -3 +5 = 2
For f(4): Since 4 lies between 2 < x < 6 We take, f(x) = x -1
f(4) = 4 -1 = 3
For f(-6): Since -6 lies between -7≤ x < -5 ,We take, f(x) = x² + 2x +1
f(-6) = (-6)² + 2(-6) + 1 = 36 -12 + 1 = 24 +1 = 25
Now ,\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) =\(\frac{ 4(2) +2(3) }{ 25 - 3(6)}\)
= \(\frac{ 8 + 6 }{ 25 -18}\)
= \(\frac{14 }{ 7}\) = 2
Hence, the value of \(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) = 2

2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 x 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

\(\frac { 16{ x }^{ 2 }-2x-3 }{ 3{ x }^{ 2 }-2x-1 } \div \frac { 8{ x }^{ 2 }+11x+3 }{ 3{ x }^{ 2 }-11x-4 } =\frac { 16{ x }^{ 2 }-2x-3 }{ 3{ x }^{ 2 }-2x-1 } \times \frac { 3{ x }^{ 2 }-11x-4 }{ 8{ x }^{ 2 }+11x+3 } \)
= \(\frac { \left( 8x+3 \right) \left( 2x-1 \right) }{ \left( 3x+1 \right) \left( x-1 \right) } \times \frac { \left( 3x+1 \right) \left( x-4 \right) }{ \left( 8x+3 \right) \left( x+1 \right) } =\frac { \left( 2x-1 \right) \left( x-4 \right) }{ \left( x-1 \right) \left( x+1 \right) } =\frac { 2{ x }^{ 2 }-9x+4 }{ { x }^{ 2 }-1 } \)

King spade, Queen spade, Jack spade are removed
∴ total number of cards = 52 - 3 = 49.
(i) Probability (diamond)= \(\frac { 13 }{ 49 } \)
(ii) Probability (queen) \(\frac { 4-1 }{ 49 } =\frac { 3 }{ 49 } \)
(iii) Probability (spade) = \(\frac { 13-3 }{ 49 } =\frac { 10 }{ 49 } \)
(iv) Probability (heart bearing number 5) = \(\frac { 13-5 }{ 49 } =\frac { 8 }{ 49 } \)

Total number of balls = 5 red + 6 white + 7 green + 8 black
n(S) = 25
(i) Let A be the event of getting white ball
n(A) = 6
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{6}{26}=\frac{3}{13}\)
(ii) Let B be the event of getting black or red
n(B) = 5 + 8 = 13
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{13}{26}=\frac{1}{2}\)
(iii) Since A is the event of getting white ball, A is the event of not getting white ball
P(A ) = 1- P(A)
\(=1-\left(\frac{3}{13}\right)=\frac{13-3}{13} \)
\(P(\bar{A}) =\frac{10}{13} \)
(iv) Let 'C' be the event of getting neither white nor black.
C = 5 red + 7 green
n(C) = 12
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{12}{26}=\frac{6}{13}\)

Sample space S = {1, 2, 3, 4,.... 1000}
n(S) = 1000
Let A be the event of selecting a card that is perfect square greater than 500
A = {23², 24², 25², 26², 27², 28², 29², 30², 31²}
n(A) = 9
(i) Probability of the first player wins a prize By picking one of the cards from A he may win a prize
\(P(A)=\frac{n(A)}{n(S)}=\frac{9}{1000}\)
(ii) Probability of the second player winning a prize, if the first has won
Let 'B' be the event of second player wins a prize.
Since the card picked in (i) is not replaced
n(B) = 3
n(S) = 1000 -1 = 999
\(P(B)=\frac{n(B)}{n(S)}=\frac{8}{999}\)

Let AB be the lamp post and CD be the apartment given CD = 66 m = EB.
\( \angle A C E=60^{\circ} \)
\(\angle E C B=\angle C B D=30^{\circ} \)
In the right triangle \(\triangle\)BDC
\(\tan 30^{\circ}=\frac{C D}{B D}\)
\( \frac{1}{\sqrt{3}}=\frac{66}{B D} \)
\(B D=66 \sqrt{3}\)
= 66 x 1.732 = 114.312 m
The distance between the lamp post and the apartment = 114.31 m
Now BD = EC = 114.31 m
In the right triangle \(\triangle\)ACE
\( \tan 60^{\circ} =\frac{A E}{C E} \)
\(\sqrt{3} =\frac{A E}{66 \sqrt{3}} \)
\(A E =66 \sqrt{3} \times \sqrt{3}[\text { From (1)] }\)
= 66 x 3 =198m
Height of the lamp post = AB
= AE + EB = 198 + 66 = 264m

Let a man is standing on the deck of a ship at a point E
Such that AE = 40 m.
AE = BD = 40m
Also AB = ED
Let BC be the height of the hill.
(i) In right triangle \(\triangle\)ABE
\( \tan 30^{\circ} =\frac{A E}{A B} \)
\(\frac{1}{\sqrt{3}} =\frac{40}{A B} \)
\(A B =40 \sqrt{3} m \)
AB = 40 x 1.732 - 69.28 m
Distance of the hill from the ship = 69.28 m
(ii) In the right triangle \(\triangle\)CDE
\( \tan 60^{\circ} =\frac{C D}{E D} \)
\(\sqrt{3} =\frac{C D}{40 \sqrt{3}} \quad[\because A B=E D=40 \sqrt{3} \mathrm{~m}]\)
\(C D=40 \sqrt{3} \times \sqrt{3}=40 \times 3=120 \mathrm{~m}\)
Now height of the hill = BC = BD + DC
= 40 + 120 = 160m
Height of the hill = 160 m
Distance of the hill from ship = 69.28 m.

Let D and C be the positions of two ships AB be the light house of height 'h'm.
o
o
In right triangle BAC
\(\tan 60^{\circ} =\frac{A B}{A C} \)
\(\sqrt{3} =\frac{h}{A C} \)
\(\mathrm{AC} =\frac{h}{\sqrt{3}} \)
In right triangle BAD
\(\tan 30^{\circ} =\frac{A B}{A D} \)
\(\frac{1}{\sqrt{3}} =\frac{h}{A D} \)
\(\mathrm{AD} =h \sqrt{3} \)
\((1)+(2) \Rightarrow \mathrm{AC}+\mathrm{AD}=\frac{h}{\sqrt{3}}+h \sqrt{3} \)
\(\mathrm{DC}=\frac{h+h \sqrt{3} \sqrt{3}}{\sqrt{3}} \)
\(\mathrm{DC}=\frac{h+3 h}{\sqrt{3}}=\frac{4 h}{\sqrt{3}} m \)
Distance between the ships is \(\frac{4 h}{\sqrt{3}} m\)

Given AB || CD || EF
AB = 6 cm, BD = 5 cm, EF = 4 cm, CD = x cm, DE = y cm
\(\text { In } \triangle E C D \text { and } \triangle E A B\)
\( \angle C E D=\angle A E B \)            [common]
\(\angle E C D=\angle E A B \)             [corresponding angles]
\(\triangle E C D \sim \triangle E A B\)
[By AA similarity criteria]      ...(1)
\(\therefore \ \frac{E C}{E A}=\frac{C D}{A B}\)
[ Corresponding parts of similar triangles are proportional]
\(\frac{E C}{E A}=\frac{x}{6}\)            ....(2)
In \(\triangle A C D\ and\ \triangle A E F \)
\(\angle C A D=\angle E A F\)          [Common]
\( \angle A C D =\angle A E F \)        [Corresponding angles]
\(\triangle A C D \sim \triangle A E F \)      [By AA similarity]
\(\frac{A C}{A E} =\frac{C D}{E F}\)
[Corresponding parts of similar triangle are proportional]
\(\therefore \quad \frac{A C}{A E}=\frac{x}{4}\)          ....(3)
Adding (2) and (3)
\( \frac{E C}{E A}+\frac{A C}{A E} =\frac{x}{6}+\frac{x}{4} \)
\(\frac{E C+A C}{A E} =\frac{4 x+6 x}{24} \)
\(\frac{A E}{A E} =\frac{10 x}{24} \)
\(1 =\frac{10 x}{24} \)
\(\mathrm{x} =\frac{24}{10}=\frac{12}{5} \mathrm{~cm}\)
From (1) \(\triangle E C D \sim \triangle E A B\)
\( \frac{D C}{A B} =\frac{E D}{E B} \)
\(\frac{x}{6} =\frac{y}{5+y} \)
\(\because \mathrm{x}=\frac{12}{5} \Rightarrow \frac{12 / 5}{6} =\frac{y}{5+y} \)
\(\frac{12}{5 \times 6} =\frac{y}{5+y} \)
12(5 + y) = 30 y
60 + 12y = 30 y
60 = 30y - 12y
18y = 60
\( y=\frac{60}{18} \)
\(y=\frac{10}{3} \mathrm{~cm}\)

Volume of a cone = 1005 \(\frac{5}{7}\) cu.cm
\(\text { i.e., } \frac{1}{3} \pi r^{2} h=1005 \frac{5}{7}\)
area of base = area of circle
\( =201 \frac{1}{7} \text { sq. units } \)
\(i.e \ \pi r^{2}=201 \frac{1}{7} \Rightarrow r^{2}=64 \)
Substituting in (1), r = 8 cm
\(\frac{1}{3}\left(201 \frac{1}{7}\right) \mathrm{h}=1005 \frac{5}{7} \)
\(\frac{1}{3}\left(\frac{1408}{7}\right) \mathrm{h}=\frac{7040}{7} \)
\(h=\frac{7040}{7} \times \frac{7}{1408} \times 3=15 \mathrm{~cm}\)
Slant height of cone \(l =\sqrt{h^{2}+r^{2}} \)
\(=\sqrt{15^{2}+8^{2}}=\sqrt{225+64} \)
\(=\sqrt{289}=17 \mathrm{~cm} \)

Given that intercepts are in the ratio 2 : 5
\(\frac{a}{b} =\frac{2}{5} \)
\(a =\frac{2 b}{5} \)
Equation of the line in Intercepts form is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\frac{x}{\left(\frac{2 b}{5}\right)}+\frac{y}{b}=1 \)
\(\frac{5 x}{2 b}+\frac{y}{b}=1 \)
5x + 2y = 2b
This passes through ( 1, - 4)
5(1) + 2(-4) = 2b
\(5-8=2 b \Rightarrow b=-\frac{3}{2} \)
\(a =\frac{2 b}{5}=\frac{2\left(-\frac{3}{2}\right)}{5}=-\frac{3}{5} \)
Equation of a straight line is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{\left(-\frac{3}{5}\right)}+\frac{y}{\left(-\frac{3}{2}\right)}=1\)
\(\frac{5 x}{-3}+\frac{2 y}{-3}=1\)
5x + 2y + 3 = 0.

b=-42
a=49

\(f(x)=\frac { x-1 }{ x+1 } ,x\neq 0\)
\(f(f(x))=f\left( \frac { x-1 }{ x+1 } \right) =\frac { \left( \frac { x-1 }{ x+1 } \right) -1 }{ \left( \frac { x-1 }{ x+1 } \right) +1 } \)
\(=\frac{\frac{\not x-1-x-1}{(\not x+1)}}{\frac{\not x-1+x+1}{(\not x+1)}}=\frac{-2}{2 x}=\frac{-1}{x}\)
Hence it is proved.

Let r₁ and r₂ be the radii of sphere and hemisphere respectively.
Given TS.A of sphere = T.S.A of hemisphere
\(4 \pi r_{1}^{2} =3 \pi r_{2}^{2} \)
\(\frac{r_{1}^{2}}{r_{2}^{2}} =\frac{3}{4} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{\sqrt{3}}{2} \)
Ratio of their volumes : \(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{2}{3} \pi r_{2}^{3}}=2\left(\frac{r_{1}}{r_{2}}\right)^{3}\)
\(=2\left(\frac{\sqrt{3}}{2}\right)^{3} \)
\(=\frac{3 \sqrt{3}}{4}=3 \sqrt{3}: 4 \)
Ratio of their volumes = \(3 \sqrt{3}: 4\)

Clearly the ladder AC make a right triangle with the wall AB force at a distance BC. \(\angle\)B - 90^o

By Pythagoras theorem
AC² = AB² + BC²
5² = 4² + BC²
BC² = 25 - 16
BC² = 9
BC = 3m
If C moves 1.6 m towards the wall BC becomes
3m - 1.6m = 1.4m
Now in \(\triangle\)ABC
AC² = AB² + BC²
5² = AB2 +(1.4)²
25 - 1.96 = AB²
AB² = 23.04
AB = 4.8m
The new height of the wall = 4.8 m
Difference =4.8 - 4 = -0.8m
The ladder would be placed 0.8 m upward the wall.

Let r and h be the radius and height of the cylinder respectively.
Given that, volume of the tank = 1.078 x 10⁶ = 1078000 litre
1078 m³ (since 1l = \(\frac{1}{1000}m^3\))
diameter = 7m gives radius = \(\frac{7}{2}\)m
volume of the tank = \(\pi\)r h 2 cu. units
1078 = \(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times h\)
Therefore, height of the tank is 28 m

From the figure,
slope = \(\frac { change\quad in\quad y\quad coordinate }{ change\quad is\quad x\quad coordinate } \)
=\(\frac { 68-32 }{ 20-0 } =\frac { 36 }{ 20 } =\frac { 9 }{ 5 } \)= 1.8
The line crosses the Y axis at (0, 32)
So the slope is \(\frac { 9 }{ 5 } \) and y intercept is 32.

\(\frac { 2{ a }^{ 2 }+5a+3 }{ 3{ a }^{ 2 }+7a+6 } \div \frac { { a }^{ 2 }+6a+5 }{ -5{ a }^{ 2 }-35a-50 } \)
\(=\frac { { 2a }^{ 2 }+5a+3 }{ { 2a }^{ 2 }+7a+6 } \times \frac { -5{ a }^{ 2 }-35a-50 }{ { a }^{ 2 }+6a+5 } \)