Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have 
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 75 }{ 5 } \)=15
d'=\(\frac { x-\bar { x } }{ c } =\frac { x-A }{ c } \)
A is assumed mean c is common factor.
Here A= 15, C = 5
σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x c
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
If 3 is added to each value, we get 8, 13, 18,23, 28 as new values.

∴ σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
S.D. doesn't change when a number is added or subtracted to the values.

x + y + z = 6            ....(1)
2x + 3y + 4z = 20     ...(2)
3x + 2y + 5z = 22   ....(3)

Sub. z = 3 in (5) ⇒ y - 2(3) =-4
y=2
Sub. y = 2, z = 3 in (1), we get
x+2+3=6
x=1
x= 1,y = 2, z = 3

We have 6 = 2¹ x 3¹ and
20 = 2 x 2 x 5 = 2² x 5¹
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 x 2 x 3 x 5 = 60.
As done in your earlier classes. Note that HCF (6, 20) = 21 = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 2² x 3¹ x 5¹ = 60.
= Product of the greatest power of each prime factor, involved in the numbers.

It is not a function as the vertical line PQ cuts the graph at two points

Let P(x, y) be equidistant from the points A (7, 1) and B (3, 5).
We are given that AP = BP. So, AP² = BP²
(x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
x²- 14x + 49 + y - 2y + 1 = x²- 6x + 9 + y -10y + 25
x - y = 2
Which is the required relation.

Through O, draw PQIIBC so that P lies on AB and Q lies on DC
Now, PQ II BC
\(PQ\bot AB\quad PQ\bot OC\)
\(\left( \because \angle B={ 90 }^{ 0 }and\angle C={ 90 }^{ 0 } \right) \)
So, \(\angle BPQ={ 90 }^{ 0 }\quad \angle CQP={ 90 }^{ 0 }\)
Therefore BPQC and APQD are both rectangles. Now from \(\Delta OPB\)
OB² = BP² + OP²
Similarly from \(\Delta OQD\)
OD² = OQ² + DQ²
From \(\Delta OQC\)
OC²  = OQ² + CQ²
\(\Delta OAP\) we have
OA² = AP² +OP²
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ²
(As BP = CQ and DQ = AP)
= CQ² + OP² + OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC²+ OA
[From (3) and (4)]

cot 85° + cos 75°
= cot(90° - 5°) + cos(90° - 15°)
= tan 5° + sin 15°

From \(\Delta ADC\) we have
AC² = AD² + CD²       ....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB² = AD² + BD²     ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² - AC² = BD² - CD²
AB² + CD² = BD² + AC²

We are given that sin 3A = cos (A - 26°) ...(1)
Since sin 3A = cos(90° - 3A) we can write (1) as
cos (90° - 3A) = cos (A - 26°)
Since 90° - 3A and A - 26° are both acute angles
90° - 3A = A - 26°
which gives A = 29°

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

σ  = 21.2, \(\bar { x } \)  = 36.6
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 21.2 }{ 36.6 } \) x 100 = 57.92%

No. of coins required \(=\frac{Volume\ of\ the\ cylinder}{Volume\ of\ 1\ coin}\)
\(=\frac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { \pi \times \frac { 42 }{ 20 } \times \frac { 45 }{ 20 } \times 10 }{ \pi \times \frac { 15 }{ 20 } \times \frac { 15 }{ 20 } \times \frac { 2 }{ 10 } } \)

= 450

Mean \(\bar { x } \)  = \(\frac { 16+13+17+21+18 }{ 5 } =\frac { 85 }{ 5 } \)  = 17

σ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 34 }{ 5 } } =\sqrt { 638 } \)
σ = 2.61
Co-efficient of variation
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 2.61 }{ 17 } \) x 100
= 15.35%

Let the tens digits be x. Then the uints digits=\(\frac{18}{x}\)
∴ Number =10x+\(\frac{18}{x}\)
and number obtained by interchanging the digits =10x+\(\frac{18}{x}\)
\(\\ \therefore \left( 10x+\frac { 18 }{ x } \right) -\left( 10\times \frac { 18 }{ x } +x \right) =63\)
\(\Rightarrow 10x+\frac { 18 }{ x } -\frac { 180 }{ x } -x=0\)
\(\Rightarrow 9x-\frac { 162 }{ x } -63=0\)
⇒ 9x²-63x-162=0
⇒ x²-7x-18=0
⇒ (x-9)(x+2)=0⇒ x=9,-2
But a digit can never be (-ve), so x = 9.
So, the required number \(=10\times 9+\frac { 18 }{ 9 } =92\)

Let the length of the side of the chess board be x cm. Then
Area of 64 squares = (x - 4)²
(x - 4)² = 64 x 6.25
⇒ x²-8x+ 16=400
⇒X²- 8x - 384 =0
⇒ X²- 24x + 16x - 384 = 0
⇒(x - 24)(x + 16) = 0
⇒ x=24 cm.

a_n = 3 +2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 x 2 = 7
a₃ = 3 + 2 x 3 = 9
List of numbers becomes 5, 7, 9, 11,.........
Here, 7 - 5 = 9 - 7 = 11 - 9 = 2 and so on. So, it forms an A.P. with common difference d = 2.
To find S₂₄ we have n = 24, a = 5, d = 2.
S₂₄ = \(\frac{24}{2}\) [2 x 5 + (24 - 1) x ]
= 12 [10 + 46] = 672.
So, sum of first 24 terms of the list of numbers is 672.

-2, 2, -2, 2, -2
t₂ - t₁ = 2-(-2) = 4
t₃ - t₂ = -2 -2 = -4
t₄ - t₃ = 2 - (-2) = 4
It is not an A.P.

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

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

The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula above, is given by
= \(\frac { 1 }{ 2 } \)[1(6 + 5) +(-4) (-5 + 1) + (-3)(-1 - 6)]
= \(\frac { 1 }{ 2 } \)[11 + 16 + 21] = 24 square units.