Here, we can take two factors 4 and 5 such that,
4 X 5 = 20 and 4 + 5 = 9.
Now, put these value in given expression
x² +(4+5)x + 4 X5=x² + 4x+5x + 4 X 5=x(x + 4) +5 (x+ 4)=(x + 4)(x +5).

13 h

By prime factorisation,
we have
175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7x 7 x 7 = 2 x 2 x 2 x 7 = 56

Thus, the cube root of 175616 is 56.

\(\frac{27}{8}\)

∵ A = P\((1+\frac{R}{100})^{n}\)
∴ A = 500\((1+\frac{12}{100})^{2}=500 (\frac{112}{100})^{2}\)
= 500\(\times \frac{12544}{10000}\) = Rs 627.2
∴ CI = A - P = Rs (627.2-500) = Rs 127.2

\(\pm \sqrt { 21 } \)

Given, area of trapezium = 34 cm²
and height of trapezium = 4 cm
Let length of one parallel side be x cm, then another parallel side is 10 cm.
Now, area of trapezium =\(\frac { 1 }{ 2 } \times \) sum of parellel sides\(\times\)Perpendicular distance between the parallel sides
⇒  34 =\(\frac { 1 }{ 2 } \times \) (x+10) \(\times\)4 ⇒ 34 =2x(x+10)
⇒ 17 = x + 10
⇒ x = 17-10 = 7
Hence, the length of other parallel side is 7 cm

In the trapezium ABCD, we have
AB II CD
Also,sum of interior angles B and C is 180°
∴ (x + 20)° + (x - 30)° =180°
⇒ 2x⁰ -10⁰ =180°
⇒ 2x⁰ = 190°
⇒ x⁰ = 95°

In case of a quadrilateral, it is necessary to have atleast the knowledge of five parts to be able to construct it uniquely. We shall need data about specified parts of the quadrilateral as follows:
(a) When four sides and one diagonal are given.
(b) When two diagonals and three sides are given.
(c) When two adjacent sides and three angles are given.
(d) When three sides and two included angles are given.
(e) When some special properties are given.

\(x=\frac { -8 }{ 5 } \) 

Here, side of the square = 6 cm
\(\therefore\) Area of the square = Side x Side
= 6 cm x 6 cm = 36 cm²
Since, [Area of the rhombus]
= [Area of the square]
\(\therefore\) Area of the rhombus = 36 cm²
One of the diagonal of the rhombus = 4 cm
Let the other diagonal of the rhombus be
\(\therefore\) Area of rhombus
\(\therefore\frac{1}{2}\times Product \ of\ the\ diagonals\)
\(\therefore\) Area of the rhombus \(=\frac{1}{2}\times 4\times d\)
Now, \(\frac{1}{2}\times 4\times d=36\Rightarrow d=\frac{36\times 2}{4}=18\)
Thus, the required length of the rhombus = 18 cm.

Here, it is clear that, more the number of winners, less is the prize money for each winner. So, it is a case of inverse proportion.
Let x₂ = 2,y₂ = 50000 and x₃ = 4, y₃ =?
We know that, x₂y₂ = x₃ y₃
∴ 2 x 50000 = 4 x y₃
\({ y }_{ 3 }=\frac { 2\times 50000 }{ 4 } =\frac { 50000 }{ 2 } =25000\)
Now, x₃ = 4, y₃ = 25000 and x 4 = 5, y₄ =?
∵ x₃y₃ = x₄y₄
4 X 25000 = 5 x y₄
\({ \Rightarrow y }_{ 4 }=\frac { 4\times 25000 }{ 5 } =4\times 5000=20000\)
Now x₄ =5,y₄ =20000 and x₅= 8,y₅= ?
∵ x₄y4 = x₅y₅
∴5 x 20000 = 8 x y₅
\({ y }_{ 5 }=\frac { 5\times 20000 }{ 8 } =5\times 2500=12500\)
Now, x₅ = 8, y₅ = 12500 and x₆ = 10, y₆ =?
∵ x₅y₅ = x₆y₆
∴ 8 x 12500 = 10 x y₆
\({ y }_{ 6 }=\frac { 8\times 12500 }{ 10 } =10000\)
Now, x₆ = 10, y₆ = 10000 and x₇ = 20, y₇ =?
∵ x₆y₆ = x₇ y₇
∴ 10 x 10000 = 20 x y₇
\({ y }_{ 7 }=\frac { 10\times 10000 }{ 20 } =5000\)
Thus, the complete table is given below:

∴ 1 x 100000 = 2 x 50000 = 4 x 25000 = 5 x 20000 = 8 x 12500 = 10 x10000 = 20 x 5000
100000 = 100000 = 100000 = 100000 = 100000 = 100000 = 100000
x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄=x₅y₅ =x₆y₆ =x₇y₇
Hence, the prize money given to an individual winner is inversely proportional to the number of winners.

Let the number of gold coins initially be x 
Then, the number of non-gold coins be 3x. When,
10 more gold coins added
Then, the total number of gold coins = (10 + x)
Then, according to the question , \(\frac { (10+x) }{ 3x } =\frac { 1 }{ 2 } \)
\(\Rightarrow\) 2 (10 + x) = 3x  \(\Rightarrow\) 20 + 2x = 3x \(\Rightarrow\) x = 20
Then, total number of coins at last = 3x + 10 + x  = 4x + 10 = 4 x 20 + 10 = 90

A = 1, B = 6 and C = 1

62

(a) Number of teachers teaching Chemistry = \(\frac{23}{100}\times\) 1800 = 414
     Number of teachers teaching English     = \(\frac{27}{100}\times\) 1800 = 496
     Number of teachers teaching Biology     =  \(\frac{12}{100}\times\) 1800 = 216
Total numbers of teachers teaching Chemistry, English and Biology
= 414 + 486 + 216
= 1116
(b) Number of teachers teaching English = 486
Number of teachers teaching Physics = \(\frac{17}{100}\times\) 1800 = 306
Number of teachers teaching Mathematics = \(\frac{13}{100}\times\) 1800 = 234
Number of teachers teaching Biology = 216
\(\therefore\) Difference between the total number of teachers teaching English and Physics together and the total number of teachers who teach Mathematics and Biology together
= ( 486 + 306) - (234 + 216) = 792 - 450 = 342.

Let ABCD is a parallelogram and the obtuse angles of the parallelogram are ㄥA and ㄥC.

Let AM is the altitude drawn from the vertex A on DG and CN is the altitude drawn from the vertex C on AB.
∴ ㄥDAM = 45° and ㄥNCB = 45°    [given]
In ΔAMD, we know that, sum of the interior angles of a triangle is 180°.
ㄥD + 45° + 90° = 180°
 ㄥD =180° - 45° - 90°
=180° -135° = 45°
Similarly, in ΔCNB
ㄥB = 45°
Now, ㄥA + ㄥD =180°
[sum of adjacent angles In a paralleloprarn.is 180°]
∴ ㄥA + 45° = 180°          [∵ㄥD = 45°]
⇒ ㄥA =180° - 45° =135°
Also, ㄥA = ㄥC
 [∵ opposite angles of a parallelogram are equal]
ㄥC =135°

Here, principal (P) = Rs 62500,
time(n) = 1\(\frac{1}{2}\)yr = \(\frac{3}{2}\)yr and rate (R) = 8% 
For half-yearly,
\(n=\frac{3}{2}\times 2\)=  3yr and R = \(\frac{8}{2}\)% = 4%
∴ Amount (A) =P \((1+\frac{R}{100})^{n}=62500(1+\frac{4}{100})^{3}\)
= 62500\((\frac{100+4}{100})^{3}=62500(\frac{104}{100})^{3}\)
= 62500\(\times \frac{104}{100}\times \frac{104}{100}\times \frac{104}{100}\)
= 104ｘ26ｘ26 = Rs 70304
∴ Compound interest =Amount (A) - Principal (P)
= Rs (70304 - 62500) = Rs 7804  

(i) 15 (ii) 8 (iii) 9 (iv) 8 (v) 8 (vi) 17

Firstly, we draw a rough sketch of rectangle OKAY, which helps us in deciding steps of construction.

We know that, in a rectangle, opposite sides are equal and parallel and angles between two adjacent sides is 90°.
In rectangle OKAY,
OK = AY = 7 cm, KA = OY = 5 cm
and \(\angle \)OKA=\(\angle \)KOY = 90°
Steps of construction
Steps IDraw OK = 7 cm.
Step IIAt K, draw a ray KX making an \(\angle \)OKX = 90°
Step III Cut KA = 5 cm from ray KX.
Step IV With A as centre and radius 7 cm, draw an arc.
Step V With O as centre and radius 5 cm, draw another arc which intersects the arc drawn in Step IV at Y.
Step VI Join AY and OY.

\(2c{ a }^{ 2 }b,\frac { 2 }{ 3 } bc{ a }^{ 2 },3{ a }^{ 2 }bc\)

We have 31944 = 2 x 2 x 2 x 3
x11x11x11
Since, the prime factors of 31944 do not appear in triples as 3 is left over.
\(\therefore\)31944 is not a perfect cube.
Obviously, 31944 \(\div\) 3 will be a perfect cube
i.e., [31944] \(\div\) 3
= [2 x 2 x 2 x 3 x 11 x 11 x11] \(\div\) 3
or 10648 = 2 x 2 x 2 x 11 x 11 x 11
\(\therefore\) 10648 is a perfect cube.
Thus, the required least number = 3.

5 x 10^-4

\(9\over7\)

6\(\frac{1}{4}\)yr

Let the present age of granddaughter be x yr.
Then, the present age of grandfather =10x yr
According to the question, 10x = x + 54
\(\Rightarrow\) 10x - x = 54 [transposing x to LHS]
\(\Rightarrow\) 9x = 54 \(\Rightarrow\) x = \(\frac { 54 }{ 9 } \) or x = 6
Present age of granddaughter = x = 6 yr and present age of grandfather = 10x = 10 X 6 = 60 yr Hence, the present age of granddaughter and grandfather are 6 yr and 60 yr.

Perimeter = 4 + 3 + 5 = 12 cm.

3 x 3 = 9.

The length of the rectangle on 20-25 is maximum.

(7x) (-12x) = - 84x²

\({10\over 1000}={4\over ?}⇒?=400\)

10x² - 18x³ + 14x² = 2x² (5 - 9x + 7x²).