Let ABCDHEFG be plasticine cube. Let a part xyz be sliced off from it. Let x, y, z lie on edge FG, FE and FB respectively.Case I. In cube ABCDHEFG,
F = 6
V = 8
E = 12
Clearly, F + V = E + 2 (= 14)
Thus, Euler's Formula is true.

Case II. When a part XYZ is sliced
Then, F = 7
V = 10
E = 15
Clearly, F + V = E + 2 (- 17)
Thus, Euler's Formula is again true

35°

In order to know who worked longer, we will compare fractions \(\frac { 7 }{ 12 }\)and​\(\frac { 3 }{ 4 } \)
We have,
(LCM of 12 and 4) = 12
Converting each fraction into an equivalent fraction with 12 as denominator, we have
\(\frac { 7 }{ 12 } =\frac { 7\times 1 }{ 12\times 1 } =\frac { 7 }{ 12 } \) and \(\frac { 3 }{ 4 } =\frac { 3\times 3 }{ 4\times 3 } =\frac { 9 }{ 12 } \)
∵        7 < 9
∵      \(\frac { 7 }{ 12 } <\frac { 9 }{ 12 }\) or \( \frac { 7 }{ 12 } <\frac { 3 }{ 4 } \)
Thus, Vaibhav finished colouring in longer time
Now, \(\frac { 3 }{ 4 } -\frac { 7 }{ 12 } =\frac { 9 }{ 12 } -\frac { 7 }{ 12 } =\frac { 9-7 }{ 12 } =\frac { 2 }{ 12 } =\frac { 1 }{ 6 } \)
Hence, Pramod finished colouring in\(\frac { 1 }{ 6 } \)hour more  time than Suval.

Given, 1029
∵ 1029 = 3 x 7 x 7 x 7 = 3 x 7³

The exponent form of 1029 is 3 x 7³.

∠ABC = 90°
So, ΔABC is a right angled triangle.

We have, \(\frac { -3 }{ 7 } =\frac { \boxed { } }{ 14 } =\frac { 9 }{ \boxed { } }= \frac { -6 }{ \boxed { } } \)
Here, the rational numbers equivalent to rational number \(\frac { -3 }{ 7 } \) are as follow:
\(-\frac { 3 }{ 7 } \) \(=\frac { -3\times 2 }{ 7\times 2 } =\frac { -6 }{ 14 } \Rightarrow \frac { -3 }{ 7 } =\frac { -3\times (-3) }{ 7\times (-3) } =\frac { 9 }{ -21 } \)
Also, \(\frac { -3 }{ 7 } =\frac { -3\times 2 }{ 7\times 2 } =\frac { -6 }{ 14 } \)
So, the missing integers in the boxes are filled as
\(\\ \frac { -3 }{ 7 } =\frac { \boxed { -6 } }{ 14 } =\frac { 9 }{ \boxed { -21 } } =\frac { 6 }{ \boxed { 14 } } \)

Given, radius = 7 cm, diameter = 14 cm
According to the question,
Area of the triangle   = \(\frac{1}{2}\) x Base x Height
                               = \(\frac{1}{2}\) 14 x 7 = 7 x 7 = 49 cm²
Area of a semi-circle = \(\frac{\pi r^2}{2}=\frac{\frac{22}{7}\times (7)^2}{2}\)
                              = 11 x 7 = 77 cm²
\(\therefore\) Required area = Area of the semi-circle - Area of the largest triangle
                        = 77 - 49 = 28 cm².

No, it doesn't have a rotational symmetry.

Given, 2x²y + 3xy²
Total terms = 2, i.e. 2x²y and 3xy²
Tree diagram

Given, SP of an article = Rs 250 ; Profit % = 5%
Let CP of an article be Rs x.
Then, profit = 5% of CP of an article = \({5\over100}\times x={5x\over100}\)
Now, SP of an article = CP + Profit
\(\therefore\)   \(250=x+{5x\over 100}\Rightarrow {250 \over 1}\)
\(\Rightarrow\)   \(1065x=250\times 100\)
\(\Rightarrow\)   \(x={250\times100\over105}\Rightarrow x={5000\over 21}\)= Rs 238.10
Hence, CP of an article is Rs. 238.10.

Since, 40° + 30° + x =180°            [linear pair]
\(\Rightarrow \) 70° + x =180° \(\Rightarrow\) x =180° -70°  \(\Rightarrow\) x =110°
\(\Rightarrow \) \(\angle\)z = 40°          [vertically opposite angles]
\(\therefore\) \(\angle\)y + \(\angle\)z =180°       [linear pair]
\(\Rightarrow\) \(\angle\)y + 40° =180°
\(\Rightarrow\) \(\angle\)y =180 - 40°\(\Rightarrow\)  \(\angle\)y =140°
So, \(\angle\)x =110°, \(\angle\)y =140° and \(\angle\)z = 40°

a = 5

Given, 41 = 44
Taking LHS, \(4l=\frac{4l}{4}=l\)
Taking RHS, \(44=\frac{44}{4}=11\)
\(\therefore\) l = 11,which is the required solution.

ClassVIII

We have,

Now, completing the given number line is as follows:

Hence, the position of required letters is shown in the following table:

C³ X C⁴ = (C X C X C) X (C X C X C X C)
=​​​​​​​\({ C }^{ \boxed { 7 } }\)

Base = 8 cm
Height = 2.5 cm 
∴ Area of the parallelogram
=base \(\times\)height
=8\(\times\)2.5 = 20 cm'²

In \(\triangle \)PLO and \(\triangle \)PMO, we have
\(\angle PLO=\angle PMO=\)90°[Given]
\(\overline { OP } =\overline { OP } \)  [Hypotenuse]
PL=PM      [Given]
\(\therefore \)Using RHS congruency, we get 
\(\triangle \)PLO \(\cong \)\(\triangle \)PMO

S.P. =Rs. 540
P% = Rs.20%
Let C.P be Rs.100.
Profit is Rs.20.
S.P = 100 + 20 = 120
Now, when S.P' is Rs.120, the C.P. is Rs.100
. . When S.P. is Rs.540
C.P =\(\frac { 100 }{ 120 } \times 540\)
= Rs.450.

In ΔLMN, we have, mㄥL = 60°, mㄥN = 120° and LM= 5 cm
Here, mㄥL + mㄥN = 60° + 120° = 180°
But the sum of two angles of a triangle should be less than 180°. So, the triangle with the given measures can not be constructed.

The isometric sketch of the given cuboid will be

It has an order of 1.

In the expression 7x +xy², the terms which contains x are 7x and xy² , so the coefficient of x are 7 and y² respectively

We have, \(\frac{5}{6}\times 2\frac{3}{7}=\frac{5}{6}\times (\frac{2\times7+3}{7})=\frac{5}{6}\times (\frac{14+3}{7})\)
=\(\frac{5}{6}\times \frac{17}{7}=\frac{5\times 17}{6\times 7}=\frac{85}{42}=2\frac{1}{42}\)

No, the sum of any two angles of a triangle is not always greater than the third angle.

Given rational number are \(\frac { 4 }{ 5 } \) and \(\frac { 6 }{ 7 } \)
LCM of 5 and 7 is 35.
\(\therefore \frac { 4\times 7 }{ 5\times 7 } =\frac { 28 }{ 35 } \) and \(\frac { 6\times 5 }{ 7\times 5 } =\frac { 30 }{ 35 } \)
\(\because\)30 > 28 so \(\frac { 6 }{ 7 } >\frac { 4 }{ 5 } \)

13

Given, angle 30°
Complement of 30° = 90° - 30° = 60°

We have, (-25) \(\times\) (-72)
First integer (x) = - 25                                    [given]
Second integer (y) = - 72                               [given]
\(\therefore\) Product (xy) = (-25) \(\times\) (-72) = 1800           [\(\because\) (-a) \(\times\)(-b) = ab]

25.3

(i) F = 7
V = 10
E = 15
F + V = 7 + 10 = 17
E + 2 = 15 + 2 = 17
So, F + V = E + 2
Hence, Euler's Formula is verified.
(ii) F = 9
V = 9
E = 16
F + V = 9 + 9 = 18
E + 2 = 16 + 2 = 18
So, F +V = E + 2
Hence, Euler's Formula is verified.

580000

Let us draw a rectangle ABCD, such that BD is a diagonal
\(\therefore\)Each angle of ABCD is a right angle.
\(\therefore\) \(\angle C\)= 90°
\(\Rightarrow\) \(\triangle\)BCD is a right triangle
Since the side opposite to 90° is hypotenuse and (Hypotenuse)² = [Sum of the squares of the legs of the triangle]
\(\therefore\) BD² = BC² + CD²
\(\Rightarrow\) (x)² = (8)² + (6)²
\(\Rightarrow\) x² = 64 + 36
\(\Rightarrow\) x² = 100
x² ​​​​​​​ = 10²\(\Rightarrow\)X = 10
Thus, the required length of the diagonal is 10 cm.

xy² Z, 2xy² Z, y² xz, 3xzy²

(a) (i) Given, breadth of the family room = 5.48 m and length of the family room = 4.57 m
\(\therefore\) Perimeter of the family room
                  = 2 (Length + Breadth)
                  = 2(5.48 + 4.57) = 2 x 10.05 = 20.10 m
(ii) Given, length of the living room = 7.53 m and breadth of the living room = 3.81 m
\(\therefore\) Perimeter of the living room
                  = 2 (Length + Breadth)
                  = 2(7.53 + 3.81) = 2 x 11.34 = 22.68 m
(iii) Given, breadth of the dining room = 5.41 m and length of the dining room = 5.48 m
\(\therefore\) Perimeter of the dining room
                  = 2 (Length + Breadth)
                  = 2(5.41 + 5.48) = 2 x 10.89 = 21.78 m
(iv) Given, length of bedroom 1= 3.04 m and breadth of bedroom 1= 3.04m
\(\therefore\) Perimeter of the bedroom 1
                  = 2 (Length + Breadth)
                  = 2(3.04+ 3.04) = 2 x 6.08 = 12.16 m
(v) Given, breadth of bedroom 2= 3.04 m and length of bedroom 2= 2.43 m
\(\therefore\) Perimeter of the bedroom 2
                  = 2 (Length + Breadth)
                  = 2(3.04 + 2.43) = 2 x 5.47 = 10.94 m.
(b) For bedroom 1,
Given, length of bedroom 1= 3.04 m and breadth of bedroom 1 = 3.04 m
We know that, area of bedroom 1 = Length x Breadth
\(\therefore\) Area of bedroom 1= 3.04 x 3.04 = 9.2416 sq m
\(\therefore\) Cost of carpeting 1 sq m = Rs 200
\(\therefore\) Cost of carpeting 9.2416 m²
                 = 9.2416x 200=~1848
For bedroom 2,
Given, length of bedroom 2= 3.04 m breadth of bed rom = 2.43 m
\(\therefore\) Area of bedroom 2 = Length x Breadth
                               = 3.04 x 2.43 = 7.3872 m²
\(\therefore\) Cost of carpeting 1 m² = Rs 200
\(\therefore\) Cost of carpeting 7.3872 m²
                             = 7.3872 x 200 = Rs 1477
For living room,
Given, length of living room = 7.53 m and breadth of living room = 3.81 m
\(\therefore\) Area of living room = 3.87 x 7.53 = 28.6893 m²
Cost of carpeting of living room 1 m² = Rs 200
\(\therefore\) Cost of carpeting 28.6893 m²
                          = Rs 200 x 28.6893 = Rs 5737.86
For dining room,
Given, length of dining room = 5.48 m and breadth of dining room = 5.41 m
\(\therefore\) Area of dining room = 5.41 x 5.48
                                 = 29.6468 m²
\(\therefore\) Cost of carpeting 1 m² = Rs 200
\(\therefore\) Cost of carpeting 29.6468 m²
                               = 29.6468 x 200 = Rs 5929.36
For family room,
Given, length of family room = 5.48 m and breadth of family room = 4.57 m
\(\therefore\) Area of family room = 5.48 x 4.57 = 25.0436 m²
so, Cost of carpeting family room
                               = 25.0436 x 200 = Rs 5008.72
(c) Total perimeter of all the five rooms
                              = 20.10 rn + 22.68 m + 21.78 m + 12.16 m+ 10.94 mm
                              = 87.66
\(\therefore\) Given, cost of moulding each room = Rs 500 per m
\(\therefore\) Total cost of moulding all five rooms
                             = 87.66 x 500 = Rs 43830.

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.

Let mark a point A on each figure and also, indicate the angle through which is to be rotated as shown below:

Now, to find the rotational symmetry,we proceed as follows:
For figure (a), it requires two rotations, each through an angle of 180°, about the marked point (X) to come back to its original position. So, it has a rotational symmetry of order 2.
For figure (b), it requires two rotations, each through an angle of 180°, about the marked point (X) to come back to its original position. So, it has a rotational symmetry of order 2.
For figure (c), it requires three rotations, each through an angle of 120° about the marked point (X) to come back to its original position. So, it has a rotational symmetry of order 3.
For figure (d), it requires four rotations, each through an angle of 90°, about the marked point (X) to come back to its original position. So, it has a rotational symmetry of order 4.
For figure (e), it requires four rotations, each through an angle of 90° about the marked point (X) to come back to its original position. So, it has a rotational symmetry of order 4.
For figure (f), it requires five rotations, each through an angle of 72°, about the marked point (X) to come back its original position. So, it has rotational symmetry of order 5.
For figure (g), it requires six rotations, each through an angle of 60° about the marked point (X) to come back its original position. So, it has rotational symmetry of order 6.
For figure (h), it requires three rotations, each through an angle of 120°, about the marked point to come back to its original position. So, it has rotational symmetry of order 3.

0.21

Reciprocal of x and y is \(\\ \frac { 1 }{ x } \)and  \(\frac { 1 }{ y } \)
\(\therefore \) sum of reciprocal = \(\\ \frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 1 }{ \frac { -4 }{ 9 } } +\frac { 1 }{ \frac { 5 }{ 12 } } \)
\(\\ =\frac { -9 }{ 4 } +\frac { 12 }{ 5 } =\frac { -45+48 }{ 20 } =\frac { 3 }{ 20 } \\ \)

Given, AB II ML, \(\angle \)LOB =142°
\(\angle \)A = 32°
\(\angle \)MOA and \(\angle \)A are alternate angles.
So, \(\angle \)MOA = 32°
Also, (\(\angle \)MOA + x) and \(\angle \)142° form a linear pair. So,
\(\angle \)MOA + x + 142° =180°
32° + x + 142° = 180°
x =180° -142° - 32°
x = 180° -174°
x = 6°
Anil's answer in the test was not correct. The value of x was 6°.
Hence, Anil does not understand the concept of alternate angles and linear pair. So, he was still confused about the lines and angles chapter.

(i) Given, total number of students = 32
Number of absent students = 8
\(\therefore\) Percentage of absent students
\(({Number \ of \ absent \ students \over Total \ number \ of \ students}\times 100)\% \\ ({8\over32}\times 100)\%=25\%\)
Hence, 25% of the students are absent.
(ii) Total number of radios = 25
Number of radios which are out of order = 16
\(\therefore\) Percentage of radios which are out of order
\(=({Number \ of \ radios \ which \ are \ out \ of \ order\over Total \ number \ of \ ratios}\times 100)\% \\=({16\over25}\times 100)\%=(16\times4)\%=64\%\)
Hence, 64% of radios are out of order.
(iii)Total number of parts = 500
Number of defective parts = 5
\(\therefore\) Percentage of defective parts
\(=({Number \ of \ defective\ parts\over Total \ number \ of \ parts}\times 100)\% \\=({5\over500}\times 100)\%=1\%\)
Hence, 1% parts are defective.
(iv) Total number of voters = 120
Number of voters voted yes = 90
\(\therefore\) Percentage of voters voted yes
\(=({Number \ of \ voters \ voted \ yes \over Total \ number \ of \ votes }\times 100)\% \\=({90\over120}\times 100)\%=(3\times 25)\%=1\%\)
Hence, 75% of voters voted yes.

Given, temperature at 12 noon = + 10°C
Rate of change in temperature = - 2°C/h
Total time from 12 noon to mid-night = 12 h
Change in temperature in 12 h = 12°C \(\times\) (- 2) = - 24°C
Temperature at mid-night = 10 + (- 24) = 10 - 24= -14° C
Now, temperature difference between 10°C and - 8°C
= 10 - (- 8) = 10 + 8 = 18° C = 18/2 = 9
So, temperature change of 18° C will take place in 9 h from 12 noon
Thus, the temperature 8° C below 0° (- 8° C) would be at 9 pm.

Congruent

Let x be the age of Amit.
So, age of Sohan Lal, the father of Amit = 4x yr
If the difference of their ages is 27 yr.
Then, 4x- x = 27 \(\Rightarrow\) 3x = 27 \(\Rightarrow x=\frac{27}{3}=9\)
Hence, age of Amit is 9 yr.

(a) Possible situations to happen are as follows:
(i) On tossing a coin, getting either a head or a tail
(ii) On drawing one card from a pack of 52 cards one side will appear.
(iii) Getting a number from 1 to 6 by throwing a die.
(b) Impossible to happen are as follows:
(i) A girl in the boy's school.
(ii) Getting a number 8 by throwing a die.
(iii) A person of height 3 m.
(c) Mayor may not happen situations are as follows:
(i) To toss a coin and get tail.
(ii) Probably it may rain.
(iii) An ant rising to 4 m height

4: 9 = 40: x
\(\Rightarrow {4\over9}={40\over x}\Rightarrow x=90\).

7>2

(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°

∵ The sum becomes Rs.32,500 in 3 years.
∴ Amount after 3 years = Rs.32,500
Amount after 2 years = Rs.30,000
∴ Interest for 1 year = Rs.32,500 - Rs.30,000
= Rs.2,500
⇒ Interest for 2 years= Rs.2,500 x 2 = Rs.5,000
∴ Principal = [Amount after 2 years] - [Interest for 2 years]
= Rs.30,000 - Rs.5,000 = Rs.25,000
Now, let rate of interest be x% p.a.
∴ Principal (P) = Rs.25,000
Time (T) = 2 years
Interest (S.l.) = Rs.5,000
Rate (R%) = x% p.a.
S.I = \({P\times R\times T\over100}⇒5000={25,000\times x \times2\over100}\)
\(⇒\ x={5,000\times100\over25,000\times2}={1\times10\over1\times1}=10\)
Thus, the required rate = 10% p.a.

Here, we observe that the parallelogram having same perimeters ean have different areas.

Total number of students = 8
Sum of the marks obtained =85 + 62 + 36 + 48 + x + 75 + 39 + 72=417+x
\(\therefore \)Mean marks =\(\frac { (417\quad +x) }{ 8 } \)
\(\Rightarrow \)\(\frac { (417\quad +x) }{ 8 } \)=60
\(\Rightarrow \)417 + x = 60 x 8 = 480
\(\Rightarrow \)x = 480 - 417 = 63
Thus, the required value of x = 63

Substituting a = 3 and b = 2 in
(a) a + b, we get
a + b = 3 + 2 = 5
(b) 7a - 4b,  we get
7a - 4b = 7 x 3 - 4 x 2
= 21 - 8 = 13
(c) a² + 2ab + b², we get
a² + 2ab + b² = 3² + 2 x 3 x 2 + 2²
= 9 + 2 x 6 + 4
= 9 + 12 + 4 = 25.
(d) a3 - b3, we get
a³ - b³ = 3³ - 2³ = 3 x 3 x 3 - 2 x 2 x 2
= 27 -8 = 19.

Let the number (- 15)^-1 should be divided by x to get the quotient (-5)^-1.
\(\therefore (-15)^{-1}\div x=(-5)^{-1}\)
\(\Rightarrow \frac{1}{-15}\div x=\frac{1}{-5}\)
\(\Rightarrow \frac{1}{-15}\times \frac{1}{x}=\frac{1}{-5}\)
\(\Rightarrow \frac{1}{-15x}=\frac{1}{-5}\)
\(\Rightarrow -15x\times 1=-5\times 1\)
\(\Rightarrow -15x = -5\)
\(\Rightarrow x=-\frac{5}{-15}\)
\(\Rightarrow x=-\frac{1}{-3}\)
\(\Rightarrow x=\frac{1}{3}\)

Let one part of 184 be x.
\(\therefore\) Other part be (184 - x)
Now, according to question,
\(\Rightarrow \quad \frac { 1 }{ 3 } x-\frac { 1 }{ 7 } (184-x)=8\)
\(\Rightarrow \quad \frac { x }{ 3 } +\frac { x }{ 7 } -\frac { 184 }{ 7 } =8\)
\(\Rightarrow \quad \frac { 7x+3x }{ 21 } =8+\frac { 184 }{ 7 } \)
\(\Rightarrow \quad \frac { 10x }{ 21 } =\frac { 56+184 }{ 7 } \)
\(\Rightarrow \quad \frac { 10x }{ 21 } =\frac { 240 }{ 7 } \)
\(\Rightarrow \quad x=\frac { 21\times 240 }{ 7\times 10 } \)
= 72
Hence, parts are 72 and 184 - 72 = 112.

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 ΔAOB,
OA + OB > AB                                     ....(i)
Similarly, in ΔBOC,
OB + OC > BC                                   ....(ii)
and in ΔAOC,
OA + OC > AC                                 ....(iii)
By adding (i), (ii) and (iii),we get
OA + OB + OB + OC + OA + OC > AB + BC + CA
\(\Rightarrow\) 2O A + 2OB + 2OC > AB + BC + CA
\(\Rightarrow\) 2(OA + OB + OC) > AB + BC + CA