we have \([{5^6\over 5^3}]\times 5^2\)  =[5^6-3]x5²      \((\because {a^m\over a^n}=a^{m-n})\)
= 5³ X 5² 
= 5^{3 + 2}                                         \((\because a^m \times a^n =a^{m+n})\)
= 5⁵
Thus,\([{5^6\over 5^3}]\times 5^2=5^5\)

(a) Sequence \(\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}\)
L.C.M.of 5, 10,15,30 = 30
Sequence be
\(\frac{2\times6}{5\times6},\frac{7\times3}{10\times3},\frac{8\times2}{15\times2},\frac{13}{30}\)
or \(\frac{12}{30},\frac{21}{30},\frac{16}{30},\frac{13}{30}\)
Its ascending order is
\(\frac{12}{30}<\frac{13}{30}<\frac{16}{30}<\frac{21}{30}\)
or \(\frac{2}{5}<\frac{13}{30}<\frac{18}{15}<\frac{7}{10}\)
(b) L.C.M: and to find ascending order.
(c) In a class, the students should stand in ascending order of height.

Since,
AB=AC
BD= CE
∴ AB + BD = AC + CE
⇒ AD=AE
Now in ∆ADC and ΔAEB
AD=AE
AC=AB
\(\angle\)A = \(\angle\) A
So, by S.A.S. congruency we have
ΔADC≅ΔAEB
\(\Rightarrow\) by C.P.C.T., CD = BE

WZ =XY = 6cm
ZY = XW = 4cm
We observed that opposite sides of rectangle are equal and each angles of rectangle is 90°.

Faces = 5, edges = 9, vertices = 6

The given figure is a hexagon, So, the order of rotation is 6 and angle of rotation of hexagon is \(\frac { 360 }{ 6 } =60°\)

Given, side of the square PQRS = 20 cm
\(\therefore\) Area of the square = (Side)2
                               = (20)² = 20 x 20 = 400 cm²
Now, area of  \(\Delta\)QPT = \(\frac{1}{2}\times\) Base x Height
                               = \(\frac{1}{2}\times\) PQ x PT = \(\frac{1}{2}\times\) 20 x 10
                               = \(\frac{200}{2}\) = 100 cm²
Area of  \(\Delta\)TSU = \(\frac{1}{2}\times\) Base x Height
                       = \(\frac{1}{2}\times\) SU x TS = \(\frac{1}{2}\times\) 10 x 10 = \(\frac{100}{2}\) = 50 cm²
and area of  \(\Delta\)QRU = \(\frac{1}{2}\times\) Base x Height
                             = \(\frac{1}{2}\times\) UR x QR = \(\frac{1}{2}\times\) 10 x 20 = \(\frac{200}{2}\) = 100 cm²
Area of shaded portion = Area of rectangle PQRS - (Area of \(\Delta\)QPT + Area of \(\Delta\)TSU + Area of \(\Delta\)QRU)
                                 = 400 - (100 + 50 + 100) cm² = 400 - 250 = 150 cm²
Hence, the area of the shaded portion is 150 cm².

Apples used on Monday = \(\frac{4}{5}\) of 5 kg
=\(\frac{4}{5}\times 5=\frac{4\times 5}{5}=\)4kg
Apples left = (5 - 4) kg = 1kg
Next day, apples used = \(\frac{1}{3}\)of 1 kg=\(\frac{1}{3}\)kg
Left apples =\((1-\frac{1}{3})kg=\frac{3\times 1-1\times 1}{3}=\frac{3-1}{3}=\frac{2}{3}\)kg

Let the sum borrowed by Meena be Rs x.
Rate of interest, R = 9%; Time, T = 1yr; Interest, I = Rs45
\(\therefore I={P\times R\times T\over 100}\Rightarrow 45={P\times9\times1\over100}\)
\(\Rightarrow 9P=45\times100\Rightarrow P={4500\over 9}\Rightarrow P=Rs500\)
Hence, the sum borrowed by Meena is Rs 500.

We have  x² -y² -1, y² -1 - x²,1 - x² -y^{2
\(\therefore\)}  required sum
= =x² -y² -l+y² -l-x² + l-x² -y²
=x² -x² -x² -y² +y² -y² -1-1+1 [rearranging terms]
=x²(1- 1- 1)+ y²(- 1+ 1- 1)-1
= x²(-l) + y²(-1)-1 = x² - y² -1

88°

In \(\triangle OMP\),
\(\angle OMP=\angle OPM\)
\(\therefore\) OM=OP.................1
[Since, sides opposite to equal angles are equal]
Also, in \(\triangle OMN\),
\(\angle OMN=\angle ONM\Rightarrow OM=ON\Rightarrow OP=ON\)  [from Eq.1]
Hence, OM is the median of \(\triangle MPN\).

We have, 16 = 4 +3{t + 2)
Given equation can be written as
         4 + 3 (t + 2) = 16
On transposing (+4) from LHS to RHS, we get
        3(t + 2) = 16 - 4   \(\Rightarrow\)  3(t + 2) = 12
On dividing both sides by 3, we get
        \(\frac{3(t+2)}{3}=\frac{12}{3}\quad \Rightarrow\) t + 2 = 4
Again, transposing (+2) from LHS to RHS, we get
         t = 4 - 2   \(\Rightarrow\)  t = 2
Hence, t = 2 is a solution of the given equation.

On rearranging the heights in ascending order, we get
160,160,160,160,161,162,162,162,162,163,163, 163,163,163,163, 163,163,163,164,164,164,165, 165,165,168.

Here, we observe that 163 is occurring nine times, i.e. more frequently.
Hence, mode of their heights is 163.
By the mode calculated above, we understand that most of the children have height 163 ern.

We have,

Sum of the digits along
Ist row = 5 + (-1) + (-4) = 5 - 5 = 0
Ilnd row = (-5) + (-2) + 7= -7 + 7=0
IIIrd row = 0 + 3 + (- 3) = 3 + (-3) = 0
Similarly, sum of the digits along
Ist column = 5 + (-5) +0=5 + (-5)=0
IInd column = (-1) + (-2) + 3= (-3) +3 =0
IIIrd column = (-4) + 7 + (- 3) = 7 + (-7) = 0
Sum of the digits along
Ist diagonal =5 + (-2) + (-3)= 5 + (-5)=0
IInd diagonal = (-4) + (-2) + 0 = - 6 \(\neq\) 0
Since, the sum of digits along the Ilnd diagonal \(\neq\) 0, so it is not a magic square.
(ii)Magic Square:
Sum of the digits along:
1st row = 1+ (-10) + 0 = 1 - 10
= -9
2nd row = (-4) + (-3) + (-2)
= (-7) + (-2) = -9
3rd row = (-6) + 4 + (-7) = (-13) + 4
= -9
1st column = 1 + (-4) + (-6) = 1 - 10
= -9
2nd column = (-10) + (-3) + 4
= (-13) + 4 = -9
3rd column = 0 + (-2) + (-7) = 0 + (-9)
=-9
One diagonal = 1 + (-3) + (-7) = 1 + (-10)
= -9
Second diagonal = (-6) + (-3) + 0
= (-9) + 0 = -9
∵ Each row, column and diagonal have the same sum.
∴ The square (ii) is the magic square.

Let us join AC.
In \(\triangle ABCand\triangle ADC\)
\(\overline { AB } =5cm,\overline { AD } =5cm\quad \Rightarrow \overline { AB } =\overline { AD } \)
\(\overline { BC } =7.5cm,\overline { DC } =7.5cm\quad \Rightarrow \overline { BC } =\overline { DC } \)
\(\overline { AC } =\overline { AC } \)  [common]
\(\therefore \triangle ABC\cong \triangle ADC\)  [SSS congruency]

\(2y+\frac { 5 }{ 2 } =\frac { 37 }{ 2 } \)
\(\Rightarrow \quad 2y=\frac { 37 }{ 2 } -\frac { 5 }{ 2 } \)
\(\Rightarrow \quad 2y=\frac { 37-5 }{ 2 } \)
\(\Rightarrow \quad 2y=\frac { 32 }{ 2 } \)
\(\Rightarrow\) 2y = 16
\(\frac { 2y }{ 2 } =\frac { 16 }{ 2 } \)  [Dividing by 2]
Thus, y = 8

(i) ➝ Side, (ii) ➝ Top, (iii) ➝ Front

5^-6

0 (zero) is neither a positive nor a negative rational number.

The multiple lines of symmetry of each figure are shown below:
This figure has 2 lines of symmetry which are shown by dotted lines.

12xz, 12x² z² are unlike terms, since algebraic factors are not same. [in 12xz, algebraic factors are x, z and in 12x²z², algebraic factors are x, x, z, z]

We have, 0.24 
\(\frac{0.24}{1}\) on removing decimal, \(\frac{024}{100}\), In simplest form =\(\frac{6}{25}\)

No, the given triangle does not exist, since it doesn't satisfy the triangle inequality.
In the given \(\triangle ABC\)
\(\overline{AB}=24cm,\overline{AC}=12cm,\overline{BC}=11cm\)
Now \(\overline{AC}+\overline{BC}=12+11=23\) and \(\overline{AB}=24cm\)
\(\therefore \overline {AC}+\overline{BC}<\overline{AB}.\) Hence, the required triangle doesn't exist.

44%

Given weights (in kg) are 39, 36, 35, 36, 41, 49 and 36.
Mode is the most occurring observations.
Here, 36 kg occurs 3 times.
\(\therefore\)  Mode = 36 kg

33cm

Reeta went wrong in solving + (- 6) and took it as (+ 6).

We know that, the two angles are supplementary, if their sum is 180°.
\(\therefore\)  \(\angle \)1 + \(\angle \)2 = 180° [given]
Now, if \(\angle \)1 is decreased, then \(\angle \)2 should be increased, so that both the angles still remain supplementary.
i.e. \(\angle \)1 + \(\angle \)2=180°

9.5853 x 10³

\(\therefore\) Area of rectangle = I x b and area of circle = \(\pi\)r²
\(\therefore\) Area of total rectangular land = 15 m x 10 m
                                              =150 m²
Area of land covered by plants = 9 m x 1 m = 9 m²
Area of land covered by camel = 5 m x 3 m = 15 m²
Region of land covered by ox is circular area.
So, the diameter is given, d = 2.8 m
\(\therefore\)  Radius = \(\frac{d}{2}=\frac{2.8}{2}\) = 1.4 m.
\(\therefore\) Area of land covered by ox = \(\pi\)r²
                                           = \(\frac{22}{7}\) x 14 x 14 = 616 m²
Total area covered by plants, camel and ox
                   = 9 + 15 + 6.16 = 30.16 m²
Remaining land for living = Total area - Area covered by plants and animals
                                    = (150 - 30.16 ) m² = 119.84 m²
\(\therefore\) Ratio of areas kept for animals and plants to the living area
                      = 30.16: 119.84 = 3016: 11984 = 377: 1498
The value depicted here is that we should save our environment and balance environment.

Steps of construction
Step I Firstly, we draw a rough sketch of triangle with given measures marked on it.

Step II Draw a line segment AB of length 3 cm.
Step III Draw an angle of 45° on point B and produce it to ray Y.
Step IV With B as centre, draw an arc of 3 cm which intersects ray BYat C.
Step V Join AC.

Thus, \(\triangle ABC\) is the required isosceles triangle.

48

The net of the given solid figure will be

Required expression is
89a³ - 64a² + 6a + 16 - (21a³ -17a²)
= 89a³ - 64a² + 6a + 16 - 21a³ + 17a²
= 89a³ - 21a³ - 64a² + 17a² + 6a + 16
= 68a³ - 47a² + 6a + 16
So, 21a³ -17a² is 68a³ - 47a² + 6a + 161ess than
89a³ - 64a2 + 6a + 16.

30 men

For a pair of integers whose product is -36 and whose difference = 15.
So, first integer = -3 and second integer = 12
Their product = (-3) \(\times\) 12 = -(3 \(\times\) 12)= -36
and the difference between these two integer is 15.

Given, a II d, b II e, c II f, a, b, c, d, e and f are line segments.
Give them points A, B, C, 0, E, F, G, H, I, J and K.

\(\because\) \(\angle\)AKE =120°
\(\therefore\)\(\angle\)JKL = \(\angle\)AKE = 120°
[vertically opposite angles)
Now, a ll d
\(\therefore\) \(\angle\)JKL + \(\angle\)KCH=180°          [cointerior angles]
\(\Rightarrow\) 120° + \(\angle\)z = 180°
\(\Rightarrow\) \(\angle\)z = 180° - 120° = 60°
Also, \(\angle\)FGH + \(\angle\)KCH = 180°   [cointerior angles]
\(\Rightarrow\) LFGH = 180° - 60° = 120°
\(\angle\)FGH= \(\angle\)BGD      [vertically opposite angles]
\(\Rightarrow\) \(\angle\)BGD =120° \(\Rightarrow\) \(\angle\)y=120°
and \(\angle\)x + \(\angle\)FGH = 180°  [cointerior angles]
\(\therefore\) \(\angle\)x = 180°-120° = 60°
Hence, \(\angle\)x = 60°, \(\angle\)y = 120° and \(\angle\)z = 60°.

Given, \(\\ \frac { 1 }{ 2 } +\frac { 1 }{ 6 } +\frac { 1 }{ 8 } \)
For common/same denominator
LCM of 2,6 and 8 is 24.
\(\frac { 1 }{ 2 } =\frac { 1\times 12 }{ 2\times 12 } =\frac { 12 }{ 24 } \frac { 1 }{ 6 } =\frac { 1\times 4 }{ 6\times 4 } =\frac { 4 }{ 24 } \\ \frac { 1 }{ 8 } =\frac { 1\times 3 }{ 8\times 3 } =\frac { 3 }{ 24 } \\ \therefore \frac { 1 }{ 2 } +\frac { 1 }{ 6 } +\frac { 1 }{ 8 } =\frac { 12 }{ 24 } +\frac { 4 }{ 24 } +\frac { 3 }{ 24 } \\ =\frac { 12+4+3 }{ 24 } =\frac { 19 }{ 24 } \)

and given. = 3
From (a), y + y + y + y + y = x + x + 3 + 3
\(\Rightarrow\)  5y = 2x + 6   \(\Rightarrow\)   5y - 2x = 6
\(\Rightarrow\)  2x - 5y = - 6 ...(i)
From (b), x + x = y + y + 3 + 3
\(\Rightarrow\)  2x = 2y + 6   \(\Rightarrow\)   2x - 2y = 6
\(\Rightarrow\)  x - y = 3 [dividing both sides by 2] ...(ii)
From (c), y + y + y + 3 + 3 + 3 = x + x + x
\(\Rightarrow\)  3 y + 9 = 3x  \(\Rightarrow\)  3x - 3y = 9
\(\Rightarrow\)  x - y = 3 [dividing both sides by 3] ...(iii)
From Eq. (iii), x - y = 3  \(\Rightarrow\) x = y + 3
On putting x = y + 3 in Eq. (i). we get
2(Y + 3) - 5Y = -6   \(\Rightarrow\)   2y + 6 - 5Y = -6
-3y + 6 = - 6   \(\Rightarrow\)   -3 y = - 6 - 6 = -12
             \(y=\frac{12}{3}=4\)
On putting y = 4 in Eq. (ii), we get
x - y = 3  \(\Rightarrow\)  x - 4 = 3
\(\Rightarrow\)  x = 3 + 4 = 7   \(\Rightarrow\)  x = 7
:. The value of  \(\Delta\)  = x = 7 and the value of • = y = 4.

(i) Yes, OP+OQ>PQ because on joining OP and OQ,we get a \(\triangle OPQ\) and in a triangle, sum of the lengths of any two sides is always greater than the third side.

(ii) Yes, OQ+OR>QR because on joining OQ and OR,we get a \(\triangle OQR\) and in a triangle, sum of the lengths of any two sides is always greater than the third side.
(iii) Yes, OR+OP>RP because on joining OR and OP, we get a \(\triangle OPR\) and in a triangle, sum of the lengths of any two sides is always greater than the third side.

(a) The above double bar graph compares the sale of brands A and 8during the months of January to June.
(b) We can clearly see from the double bar graph that sales for brand A reduced in the month of March compared to that of February.
(c) Sales of brand A in June = Rs. 57 lakh
and sales of brand 8 in June = Rs. 54 lakh
Difference in sales = 57 - 54 = Rs. 3 lakh
(d) Average sales of brand 8
=\(\frac { total\ sales\ of\ brand\ b\ in\ sin\ months\ from\ january\ to\ june }{ 6 } \)
= \(\frac { 36+38+43+35+45+54 }{ 6 } \)
= \(\frac { 251 }{ 6 } \) Rs.41.83 lakh
(e) We can clearly see from the double bar graph that sales of brand 8is less than sales of brand A in the month of April and June
(f) Sales of brand A in January = 31
and sales of brand 8 in January = 36
Required ratio = \(\frac { 31 }{ 36 } \)or 31:36

2x+3y=220°

\(\frac{-8}{-9}=\frac{-8\times2}{-9\times2}=\frac{-16}{-18}\)

Required percentage =\({50\over 100}\times 100\%\)
=50%.

90° - 90° = 0°.

\(1kg=1000gm;150g=\frac{150 gm}{1000}kg=0.15\ kg\)

3m + 7 = 16 \(\Rightarrow\) 3m = 16 -7 = 9\(\Rightarrow m={9\over3}=3\)

400 - 200 = 200; 200 - 100 = 100;
200 = 2\(\times\)100

(i) x + y = 1200                                  ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 180⁰
\(\Rightarrow\)x + 2y = 1800                                  ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°

ABCD is the rhombus such that its diagonals AC and BD intersect at O. Here BD = 12 cm.

Suppose AC = x cm.
Since, the area of a rhombus =\(\frac{1}{2}\)\(\times\)[Product of 2 its diagonals]
\(\therefore\)Area of the rhombus ABCD =\(\frac{1}{2}\)\(\times\) 12 \(\times\) x cm²
2 But the area of the rhombus ABCD = 96 cm²
\(\Rightarrow\)\(\frac{1}{2}\)\(\times\)12 \(\times\) x = 96
\(\Rightarrow\)x=\(\frac{96\times2}{12}\)cm \(\Rightarrow\)x=8\(\times\)2cm
\(\Rightarrow\) x=16 cm
Since, the diagonals of a rhombus, bisect each other at right angles.
\(\therefore\)\(\angle \)COD = 90°, OC =\(\frac{1}{2}\)\(\times\) 16 cm=8 cm
and Od=\(\frac{1}{2}\)\(\times\) 12 cm =6 cm
Now, in right \(\Delta\)COD, we have
\(\Rightarrow\)OD² + OC² = CD² \(\Rightarrow\) 6² + 8² = CD²
\(\Rightarrow\)36 + 64 = CD² \(\Rightarrow\)100 = CD²
\(\Rightarrow\) 10² = CD² \(\Rightarrow\) CD = 10 cm.
Since, all the sides of rhombus are equal and perimeter of a rhombus = 4 x Side.
\(\therefore\) Perimeter ofthe rhombus = 4 x 10 em = 40 cm

A = 2 + 4x + 8x²
B = -3-5x + x²
C = 1 + 3x-7x²
A+B+C=?
A + B + C = (2 + 4x + 8x²) + (- 3 - 5x + x²)+ (1 + 3x-7x²)
= 2 + 4x + 8x² - 3 - 5x + x² + 1 + 3x -7x²
= x²(8 + 1 - 7) + x( 4 - 5 + 3) + 2 - 3 + 1
=2x²+2x+0
=2x²+2x
= 2x(x + 1).

(25)^n-1 + 100 = 5^(2n-1)
⇒ (5²)^n-1 + 100 = 5^(2n-1)
⇒ 5^2n-2 + 100 = 5^2n-1
⇒ 5^2n-2 - 5^2n-1 = - 100
⇒ 5^{2n - 1} - 5^{2n- 2} = 100
⇒ 5^2n-2 x (5 -1) = 100
⇒ 5^2n-2 x 4 = 100
\(⇒\ 5^{2n-2}={100\over 4}=25\)
Thus, 5^{2n- 2} = 5²
As base is same on both the sides
∴ 2n-2 = 2
⇒ 2n=2+2
⇒ 2n = 4
\(⇒\ n={4\over 2}=2\)

Let the total marks = x
Then, marks for accent and presentation
= 20% of x
=\(\frac { 20 }{ 100 } \times x=\frac { x }{ 5 } \)
Remaining marks = x - \(\frac { x }{ 5 } ={5x-x\over5}={4x\over 5}\)
Now, marks for subject matter =60% of \(\frac { 4x }{ 5 } \)
\(\frac { 60 }{ 100 } \times \frac { 4x }{ 5 } =\frac { 12x }{ 25 } \)
Remaining marks \(={4x\over 5}-{12x\over 25}={5\times4x-12x\times1\over25}\)
\(={20x-12x\over25}={8x\over 25}\)
According to the question, there are 8 marks for rebuttal.
so,  \(\frac { 8x }{ 25 } =8\Rightarrow 8x=25\times 8\)
\(\Rightarrow \quad x=\frac { 25\times 8 }{ 8 } 25\)

Age of Ram's father = 49 years
Let the age of Ram be x years
\(\therefore\) 3x + 4 = 49
\(\Rightarrow\) 3x = 49 - 4
\(\Rightarrow\) 3x = 45
\(\Rightarrow\) \(x=\frac { 45 }{ 3 } \)
x = 15
\(\Rightarrow\) Ram's age = 15 years

Arranging the observations (marks) in ascending order:
2, 2, 5, 5, 5, 8, 8, 8, 8, 13, 13, 13, 13, 16, 16, 16, 17, 17, 18,18,18,18,18,19,20
Sum of observations
\(2\times2=4\\5\times3=15\\8\times4=32\\13\times4=52\\16\times3=48\\17\times2=34\\18\times5=90\\19\times1=19\\20\times1=20\\\quad\quad\quad\quad\_\_\_\_\\\ Total=314\\ \quad\quad\quad\quad\_\_\_\_\)Number of students = 25
\(\therefore\) Mean marks \(=\frac{Sum\ of\ the\ observations}{Total\ number\ of\ students}\)
\(=\frac{314}{25}=12.5\)
So, the number of students who scored marks more than mean marks is
4 + 3 + 2 + 5 + 1 + 1 = 16.

Sequence is \(\frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,\frac { -5 }{ 6 } \\ \)
L.C.M.of 5, 10 and 6=30
\(\Rightarrow \frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,\frac { -5 }{ 6 } \)
\(\Rightarrow -\frac { 3\times 6 }{ 5\times 6 } ,\frac { 7\times 3 }{ -10\times 3 } ,-\frac { 5\times 5 }{ 6\times 5 } \)
\(\Rightarrow -\frac { 18 }{ 30 } ,\frac { 21 }{ 30 } ,\frac { 25 }{ 30 } \)
Since \(-\frac { 25 }{ 30 } <\frac { 7 }{ -10 } <\frac { -3 }{ 5 } \)
Hence sequence in ascending order is
\(\frac { -5 }{ 6 } <\frac { 7 }{ -10 } <\frac { -3 }{ 5 } \).

In the above figure, AB and CD are two poles whose heights are 9 m and 14m respectively.
\(\Rightarrow\) AB = EC = 9m
and BD = 13m
DE = 14 - 9
= 5m
Now in right ΔBDE, by Pythagoras
BD² = BE² + DE²
13² = BE² + 5²
\(\Rightarrow\) BE² = (13)² - (5)²
= 169-25
BE² = 144
\(\Rightarrow\) BE = \(\sqrt { 144 } \)
\(\Rightarrow\) BE = 12m.
Hence, distance between their feet = 12 m.