5 X 10³ + 9 X 10²+ 0 x 10¹ + 1 x 10⁰

(a) = (b) > (b) <

Given, side of square plot = m
Triangular plot has base and height = m
Now, Area of square plot = (Side)² = m²
Area of triangular plot = \(\frac { 1 }{ 2 } \)x base x height
=  \(\frac { 1 }{ 2 } \)  x m xm = \(\frac { { m }^{ 2 } }{ 2 } \)
\(\therefore\)  Total area of both plots
= Area of square plot + Area of triangular plot
= \({ m }^{ 2 }+\frac { { m }^{ 2 } }{ 2 } =\frac { 2{ m }^{ 2 }+{ m }^{ 2 } }{ 2 } =\frac { { 3m }^{ 2 } }{ 2 } \)

Given, perimeter of \(\Delta\)ABC = perimeter of \(\Delta\)POR
So, perimeter of \(\Delta\)POR = 14 + 6 + 10 = 30 cm
Now, perimeter of \(\Delta\)ABC,
        30 = AB + BC + AC  \(\Rightarrow\) 30 = AB + 5 + 13
\(\Rightarrow\)    30 = AB + 18           \(\Rightarrow\) AB = 30 - 18 = 12
\(\therefore\) Area of the \(\Delta\)ABC = \(\frac{1}{2}\) x Base x Height
                               = \(\frac{1}{2}\) x 5 x 12 = 5 x 6 = 30 cm².

As per the given information in the question,
the strength of school = 2000
Percentage of girls in school = 40%
Percentage of boys in school = 100% - 40% = 60%
Number of boys in school =\({60\over100}\times 2000\)
= 60 x 20 = 1200
Hence, number of boys in school is 1200.

Given, PQ = 5 cm, mㄥPQR = 105° and m ㄥQRP = 40°
In ΔPQR, by angle sum property, we have
ㄥPQR + ㄥQRP + ㄥRPQ = 180°
⇒ 105° + 40° + ㄥRPQ = 180°
⇒ ㄥRPQ = 180° -145° = 35°
Thus, we have PQ = 5 cm, ⇒ ⇒P = 35° and ⇒Q = 105°
Now, to draw ΔPQR, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch of traingle with measures marked on it.

Step II Draw a line segment PQ = 5 cm.

Step III At point P, draw a ray PX making an angle of 350 with PQ i.e., ㄥQPX = 35⁰.

Step IV At point Q, draw a ray QY making an angle of 105° with PQ i.e. ㄥPQY = 105°.

Step V Extend the ray PX and QY. Let rays PX and QY intersect at R.

Thus, ΔPQR is the required triangle.

129⁰

Rajan's weight 8 months ago was 82 kg. After regular schedule and exercise his weight reduced to \(\frac { 3 }{ 20 } \) of his original weight.
Reduced weight =  \(\frac { 3 }{ 20 } \) \(\times \) 82 
= 3\(\times \) 4.1=12.3Kg
=12 kg and 300g
New weight =82-12.3kg=69.7kg
=69 kg and 700g 
The value depicted from Rajan is that he properly follows the physician's diet chart and do exercise regularly. So Rajan's is very conscious about his physical fitness.

Since, lines EF and GH are parallel to each other.
Where, \(\angle\)EAB = 70° and \(\angle\)ACH =120°
\(\angle\)EAB and \(\angle\)CBA are alternate angles.
So, \(\angle\)CBA = 70°
\(\because\) (\(\angle\)EAB + \(\angle\)BAC) and \(\angle\)ACH are alternate angles.
so, \(\angle\)EAB + \(\angle\)BAC = 120°
\(\Rightarrow\) 70° + \(\angle\)BAC =120°
 \(\angle\)BAC =120° \(\Rightarrow\)-70° = 50°
\(\because\)\(\angle\)CAE = \(\angle\)EAB + \(\angle\)BAC
\(\therefore\) \(\angle\)CAE = 70° + 50° = 120°
Also, \(\angle\)FAC + \(\angle\)CAE =180°
\(\Rightarrow\)\(\angle\)FAC = 180° -120°
\(\therefore\) \(\angle\)FAC = 60°

(i) We know that, 3 \(\div\) 1= 3, so 3 \(\div\) (- 1)= - 3
On comparing with a \(\div\) b = - 3, we get a = 3 and b = -1
Hence, the required pair of integers is (3, -1).
(ii) We know that, 12 \(\div\) 4 = 3, so 12 \(\div\)(- 4) = - 3
On comparing with a \(\div\) b = - 3, we get a = 12 and b = -4
Hence, the required pair of integers is (12, - 4).
(iii) We know that, 9 \(\div\) 3 = 3, so 9 \(\div\) (- 3) = - 3
On comparing with a \(\div\) b = - 3, we get a = 9 and b = -3
Hence, the required pair of integers is (9, - 3).
(iv) We know that,12 \(\div\) 4 = 3, so (-12) \(\div\) 4 = -3
On comparing with a \(\div\) b = - 3, we get a = -12 and b=4
Hence, the required pair of integers is (-12, 4).
(v) We know that, 15 \(\div\) 5 = 3, so 15 + (- 5) = -3
On comparing with a \(\div\) b = - 3, we get a = 15 and b = -5
Hence, the required pair of integers is (15, - 5).

(i) Let the number be x.
Now, 1 is subtracted from a number, then
(x -1), the difference is multiplied by \(\frac{1}{2}\)
Here, \(\frac{1}{2}\) (x - 1) gives result 7
\(\therefore \frac{1}{2}(x-1)=7\)
(ii) Let age of Soh an be x yr. Then, the age of
Mohan is (x + 3) yr.
\(\therefore\) Sum of their ages = 43
\(\Rightarrow\)   x+(x+ 3)=43
(iii) Let the number be x. Then, five times of number be 5x.
Since it is increased by 7.
So, 5x + 7 gives result 27.
Hence, the equation is 5x + 7 = 27.

(i) Mode = Heighest frequency observation = 300
For median, we have to arrange the data in ascending order, 150,200,300,300,500,600, 650, 800, 1500
Median = 500
(ii) Value He is very socialised student and also
very kind towards his socities.

Not congruent

Time taken by Michael to finished colouring the picture \(\frac{7}{12}\)hr
Time taken by Vaibhav to finished colouring the picture = \(\frac{3}{4}\)hr

LCM of 12 and 4 = 3 X 4 = 12
On converting these fractions into like fractions, we get
\(\frac{3}{4}=\frac{3\times 3}{4\times 3}=\frac{9}{12} and \frac{7}{12}=\frac{7}{12}\)
It is clear, 9>7 [numerators of fractions]
So, \(\frac{9}{12}>\frac{7}{12} i.e., \frac{3}{4}>\frac{7}{12}\)
So, Vaibhav takes longer time to finish colouring the picture. Difference between time taken by Vaibhav and Michael
\(=\frac{3}{4}-\frac{7}{12}\)

LCM of 4 and 12 = 2 x 2 x3 = 12
\(=\frac{3}{4}-\frac{7}{12}=\frac{3\times 3-7\times 1}{12}\)
=\(\frac{9-7}{12}=\frac{2}{12}=\frac{1}{6}\) [dividing numerator and denominator by 2]
Hence, Vaibhav takes \(\frac{1}{6}\) h more than Michael.

\(S.I=\frac { P\times R\times T }{ 100 } \)
\(=\frac { 50,000\times 12\times 2 }{ 100 } \)
S.I =Rs.12,000

Circle

2⁵ x 2³ = 2⁵⁺³ = 2⁸

528 cm²

With respect to the given line(s) of symmetry, the other hole(s) are marked in the figures given below:

We have, 7 rupees 7 paise = 7 rupees + 7 paise
=\(7+\frac{7}{100}\)=7+0.07=7.07       [∵ 1 paise= Rs \(\frac{1}{100}\)]

In the expression 1.2ab - 2.4b + 3.6 a, terms are 1.2ab, -2.4b and 3.6a and factors of l.2ab are 1.2, a, b, factors of -2.4b are -2.4, b and factors of 3.6a are 3.6, a.

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.

We have, (-100) \(\div\) 5 = - (100 \(\div\) 5) = -100/5 = -20

x=40°

Given, equation 3(2x + 1) = 6
Dividing both sides by 3, we get
       \((2x+1)=\frac{6}{3}\Rightarrow\) 2x + 1 = 2
      2x = 2 - 1  \(\Rightarrow\)  2x = 1  \(\Rightarrow\)  x = \(\frac{1}{2}\)
\(\therefore x=\frac{1}{2}\) is the required solution.

In ΔCAB and ΔDAB, ㄥC = ㄥD = 90°
Hypotenuse AB = Hypotenuse AB     [common]
CA= DB= 2 cm
So, both triangles are congruent by RHS congruence rule.
The correspondence is A ↔️ B, C ↔️ D, B ↔️ A.
In symbolic form, ΔACB ≅ ΔBDA

x = 21

\(\frac { 75 }{ 182 } \)

(a)Total amount deposited in account = (5x² + 3x-2) + (x²-x + 5)
=5x² + 3x-2+ x²-x + 5
=5x² + x²+ 3x-x-2+ 5
=6x²+2x+3
Balance amount
= (6x²+2x+3)-(2x²-x-3)
= 6x²+2x+3-2x²-x-3
= 6x²-2x²+ 2x + x + 3 + 3
=4x²+3x+6.
(b) Addition of algebraic expressions.
(c) Value: Economy is everywhere.

Steps of Construction:
(a) Draw a line I and mark a point A on it.
(b) Construct an angle of 90° at A to draw AB perpendicular at l.
(c) Mark a point X on AB such that AX = 3 cm.
(d) At X, construct an angle of 90° to draw a perpendicular to AB. Thus, 'm' is the required line through X such that m II l.

Total of parts = 3 + 2 = 5
\(\Rightarrow \) Son got = \(\frac { 3 }{ 5 } \)part
\(\Rightarrow \) So his percentage share =\(\frac { 3 }{ 5 } \)X 100=60%
His daughter got=\(\frac { 2 }{ 5 } \) part
So, her percentage share =\(\frac { 2 }{ 5 } \)X100=40%

Given,\(\frac { -7 }{ 8 } \boxed { } \frac { 14 }{ -16 } \) or \(\frac { -7 }{ 8 } \boxed { } \frac { 14\times (-1) }{ -16\times (-1) } \) or \(\frac { -7 }{ 8 } \boxed { } \frac { -14 }{ 16 } \)
Here, both rational numbers are negative. 
On ignoring the negative sign, we have \(\frac { 7 }{ 8 } \) and \(\frac { 14 }{ 16 } \).
Now, Converting these rational numbers into the
 rational numbers having same denominator, we get 
\(\frac { 7 }{ 8 } =\frac { 7\times 2 }{ 8\times 2 } =\frac { 14 }{ 16 } \) and \(\frac { 14 }{ 16 } =\frac { 14 }{ 16 } \) [ multiplying the numerator and denominator by 2]
Since,       14=14 
\(\\ \therefore \frac { 14 }{ 16 } =\frac { 14 }{ 16 } \Rightarrow \frac { 7 }{ 8 } =\frac { 14 }{ 16 } \)
\(\Rightarrow \frac { -7 }{ 8 } =\frac { -14 }{ 16 } \Rightarrow \frac { -7 }{ 8 } =\frac { 14 }{ -16 } \)
Hence, \(\frac { -7 }{ 8 } =\frac { 14 }{ -16 } \)

Given, 1024
∵ 1024 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2¹⁰

The exponent form of 1024 is 2¹⁰.

Net for the given cone figure will be

Given, atomic weight of Helium = 4.0030, Hydrogen = 1.0080 and Oxygen = 16.0000
Difference between atomic weights of Oxygen and Helium
 16.0000
-04.0030
________
 11.9970
________

Let ABCD represents the rectangular field of length AB = 90 m and breadth AD = 60 m. Area of shaded portion i.e. area of the rectangle PQRS and the area of rectangle EFGH represents the area of crossroads, but here the area of square KIMN is taken twice, so it will be subtracted.

Now, PQ = 3 m, PS = 60 m, EH = 3 m, EF = 90 m, KL = 3 m and KN = 3 m
(i) Area covered by the road = Area of rectangle PQRS + Area of rectangle EFGH - Area of square KIMN
                                         = PQ x PS + EH x EF - (KL)²
                                         = 3 x 60 + 3 x 90 - 3 x 3 = 180 + 270 - 9 = 441 m²
(ii) Cost of constructing 1 m² road = Rs 110
\(\therefore\) Cost of constructing 441 m² road
               = Rs (441 x 110) = Rs 48510
Hence, the cost of constructing the road is Rs 48510.

In ΔABD and ΔADC, we have
AB= AC =35 cm, BD= DC = 25 cm and AD = AD [common]
Therefore, the three sides of MBD are equal to three sides of ΔADC.
So, two triangles are congruent by SSS congruence rule.
From the above equality relations, we have
A ↔️ A, B ↔️ C, and D ↔️ D
In symbolic form, ΔABD ≅ ΔACD

As per the given information, we can draw the following figure, which is a right angled triangle at B.

Distance from starting point is the hypotenuse of \(\triangle ABC\),
(6)² +(8)² = (Distance)²
\(\Rightarrow\) 36 + 64 = (Distance)²
\(\therefore\ Distance=\sqrt 100=10 km\)

As per the given information in the question,
Room temperature = 30° C
Change in temperature per hour = - 5° C
Change in temperature in 12 h = 12 x (-5) = - 60° C
Room temperature after 12 h = 30° C + ( - 60° C)
= -30°C

Radha's monthly salary and overtime = Rs 17480
Let x be the monthly salary.
Then, over-time = Rs (x -10000)
\(\therefore\) 17480 - x = x -10000
           2x = 27480 \(\Rightarrow\) x=13740
Hence, her monthly salary is Rs 13740.

From the bar graph, we can see students which scored marks above 90 are in the range of 90-92. Hence, 4 students scored above 90 marks

Given, \(\angle \)1= 30°
Since, \(\angle \)1 and  \(\angle \)2 form a linear pair.
\(\therefore\) \(\angle \)1+ \(\angle \)2 = 180° \(\Rightarrow\) 30° + \(\angle \)2 = 180°
[transposing 30° to RHS]
\(\Rightarrow\)  \(\angle \)2 = 180° - 30°\(\Rightarrow\) \(\angle \) 2 = 150°
Now, \(\angle \)3 = \(\angle \)1= 30° [vertically opposite angles]
Hence, \(\angle \)2 = 150° and \(\angle \)3 = 30°

1000 = 10 x 10 x 10 = 10³

Side =\(\sqrt { 1600 } \)=40 m

ABC ↔️ PQR, ABC ↔️ PRQ,
ABC ↔️ QRP, ABC ↔️ QPR,
ABC ↔️ RPQ, ABC ↔️ RQP.

Definition of an isosceles triangle.

600 -100 = 500

(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°

Let the radii of the given circles be r₁ and r₂. Let their circumference be 'C₁' and 'C₂' respectively.
\(\therefore\)C₁ = 2\(\pi\)r₁ and C₂ = 2\(\pi\)r₂
Since, C₁ : C₂ = 5 : 6
\(\therefore\)(2\(\pi\)r₁):(2\(\pi\)r₂)=5:6
\(\Rightarrow\)\(\frac { 2\pi r_{ 1 } }{ 2\pi r_{ 2 } } =\frac { 5 }{ 6 } \Rightarrow \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 5 }{ 6 } \)
Now, let the area of the given circles be A₁ and A₂
\(\therefore\)A₁ = \(\pi\)r₁² and A₂ = \(\pi\)r₂²
\(\Rightarrow\)\(\frac { { A }_{ 1 } }{ A_{ 2 } } =\frac { { { \pi r }_{ 1 } }^{ 2 } }{ { { \pi r }_{ 2 } }^{ 2 } } =\frac { { { r }_{ 1 } }^{ 2 } }{ { { r }_{ 2 } }^{ 2 } } \Rightarrow \frac { A_{ 1 } }{ A_{ 2 } } =\left( \frac { { { r }_{ 1 } } }{ { { r }_{ 2 } } } \right) ^{ 2 }\)
\(\Rightarrow\)\(\frac { { A }_{ 1 } }{ A_{ 2 } } =\left( \frac { 5 }{ 6 } \right) ^{ 2 }=\frac { 25 }{ 36 } \Rightarrow \)A₁:A₂
= 25:36
Thus, the areas of the two given circles are in the ratio 25 : 36.

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

A = 2a - 3b
B=-3a+4b
C = -a + b
A+B+C=?
A + B + C = (2a-3b) + (-3a + 4b) + (-a + b)
= 2a-3b-3a + 4b-a + b
= a(2 - 3 -1) + b(- 3 + 4 + 1)
= a(2 - 4) + b(1 + 1)
=-2a+2b
A+B-C=?
A + B - C = (2a- 3b) + (- 3a + 4b) - (- a + b)
= 2a-3b-3a + 4b + a-b
= a(2-3 + 1) + b(-3 + 4-1)
= a(0) + b(-4 + 4)
= a(0) + b(0)
= 0.

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

Quantity of pure alcohol in 400 ml. of 15% = 400 \(\times \frac { 15 }{ 100 } \)
= 60 ml
Now, we add x ml of pure alcohol to the sample.
So, total pure alcohol = (60 + x) ml.
But volume of new sample = (400 + x) ml.
\(\therefore\) Percentage of pure alcohol in new sample \(=\frac { (60+x) }{ (400+x) } \times 100\)
which is equal to 32%
\(\Rightarrow \quad \frac { 60+x }{ 400+x } \times 100=32\)
\(\Rightarrow \quad \frac { 60+x }{ 400+x } =\frac { 32 }{ 100 } \)
\(\Rightarrow\) 100(60 + x) = 32 (400 + x)
\(\Rightarrow\) 100x + 6000 = 32x + 12800
\(\Rightarrow\) 100x-32x = 12800 - 6000
\(\Rightarrow\)  68x = 6800
\(\Rightarrow \ x=\frac { 6800 }{ 68 } =100\ ml\)

Let Akhil's income be Rs.100
\(\Rightarrow \) Nikhil's income = 100 - 20
=Rs.80
Now Nikhil's income be Rs.80
\(\Rightarrow \)Akhil's income is Rs.20 more than that of Nikhil
\(\because \)When Nikhil's income is Rs.80. than Akhil's income is Rs.20 more
When Nikhil's income is Rs.100, then Akhil's income
=\(\frac { 20 }{ 80 } \times \)100
= 25% more
Hence Akhil's income is 25% more than Nikhil's income.

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.