56.4

\(\frac { 1 }{ 2 } \times \frac { 5 }{ 8 } =\frac { 1\times 5 }{ 2\times 8 } =\frac { 5 }{ 16 } \)

44875 \(\times\) 99- (- 44875) = 44875 \(\times\) 99- 44875 \(\times\) - 1
= 44875 \(\times\) [99- (-1)]
[using distributively over addition]
= 44875 \(\times\) (99 + 1)
= 44875 \(\times\) 100
= 4487500.

\(\angle \)x = 55°            [vertically opposite angles]
\(\angle \)y + 55° = 180°          [by linear pair]
\(\therefore \)      \(\angle \)y=180°-55°=125°
\(\angle \)z = \(\angle \)y = 125° [verticallyopposite angles]
Hence, \(\angle \)x = 55° , \(\angle \)y = 125° and \(\angle \)z = 125°

\(1.44\div1.2=\frac{144}{100}\div\frac{12}{10}=\frac{144}{100}\times\frac{10}{12}=\frac{12}{10}=1.2\)
or
\(\frac{1.44}{1.2}=\frac{144}{120}=\frac{12}{10}=1.2\)

Given,\(\angle 1=(2a+b)^0\) and  \(\angle 6=(3a-b)^0\)
so, \(\angle 1=\angle 7\) [alternate exterior angles]
So, we have \(\angle 7=(2a+b)^0\)
\(\therefore \angle 6+\angle 7=180^0\)              [linear pair]
\(\Rightarrow\) (3a-b)⁰ + (2a + b)⁰ = 180°
\(\Rightarrow\) 3a- b + 2a + b = 180°
\(\Rightarrow\) 5a = 180°
on dividing both sides by 5, we get
\(\Rightarrow a=\frac{180^0}{5}=36^0\)
\(\because \angle 1+\angle 2=180^0\)                    [linear pair]
\(\Rightarrow 2a+b+\angle 2=180^0\)
\(\Rightarrow 2\times36^0+b+\angle 2=180^0\)
\( \Rightarrow b+\angle 2=180^0-72^0\)
\(\Rightarrow b+\angle 2=180^0\)
\(\therefore \angle 2=(108-b)\)

No, PQ²> (PR)² + (QR)²

(i) Take a = 3, b = 4, c = 5
a \(\times\) (b - c) = 3 \(\times\) (4 - 5)
= 3 \(\times\) (- 1) = - 3 
a\(\times\)b-a\(\times\)c=3\(\times\)4-3\(\times\)5
= 12 -15 =-3
\(\therefore\) a \(\times\) (b - c) = a \(\times\) b - a \(\times\) c
(ii) Take a = - 3, b = - 4, c = - 5
(iii) Take a = 3, b = 4, c = - 5
(iv) Take a = 3, b = - 4, c = 5
(v) Take a = - 3, b = 4, c = 5
(vi) Take a = 3, b = - 4, c = - 5
Verify the remaining parts yourself exactly as in (i).

x + 2x = 180°     (Co-interior angle)
3x = 180°
x = 60°

Sum of the observations (heights)
= [142 + 150 + 146 + 154 + 148] cm =740 cm
Number of observations = 5
\(\therefore \)Mean height = \(\frac { Sum\ of\ observations }{ Number\ of\ observations } \)
\(\frac { 140 }{ 5 } \) cm = 148 cm
Thus, the required mean height = 148 cm

Age of Prashant = 40 years.
Let the age of son be x years.
 \(\therefore\)According to the condition,
5 x (Age of son) + 5 = Prashant's age
\(\Rightarrow \)5[x] + 5 = 40
\(\Rightarrow \)5x + 5 = 40
Transposing 5 to R.H.S., we have
5x = 40 - 5
\(\Rightarrow \)5x = 35
\(\Rightarrow \) x=\(\frac{35}{5}\)=7
Hence, the required age of son is 7 years.

Total weight of the fruits purchased by Sameera is \(\left( 3\frac { 1 }{ 2 } +4\frac { 3 }{ 4 } \right) \)kg
Now, \(3\frac { 1 }{ 2 } +4\frac { 3 }{ 4 } =\frac { 7 }{ 2 } +\frac { 19 }{ 4 } \)
= \(\frac { 7\times 2 }{ 2\times 2 } =\frac { 19\times 1 }{ 4\times 1 } \)
= \(\frac { 14 }{ 4 } +\frac { 19 }{ 4 } =\frac { 14+19 }{ 4 } =\frac { 33 }{ 4 } =8\frac { 1 }{ 2 } 2\)
Hence, total weight 8\(\frac { 1 }{ 4 } \) is kg

Distance towards East denoted (+ve) sign and Distance towards West denoted (-ve) sign.
Since first Ridhaan goes toward East 25 km
So distance covered by him = + 25 km
Again, he goes towards West.
So, distance covered by him on same road
= -35 km.
So,position of Ridhaan from A= + 25 - 35
= -10 km.
(b) Value: Addition of Integers.
(c) Value: A rolling stone can gather no moss.

Given, \((1\div \frac{2}{9})+(1\div 3\frac{1}{5})+(1\div 2 \frac{2}{3})\)
Now, \(3 \frac{1}{5}= \frac{(3\times 5)+1}{5}= \frac{15+1}{5}= \frac{16}{5}\)
2\( \frac{2}{3}= \frac{(2\times 3)+2}{3}= \frac{6+2}{3}= \frac{8}{3}\)
ஃ \((1\div \frac{2}{9})+(1 \div \frac{16}{5})+(1\div \frac{8}{3})\) 
[∵ reciprocal of \( \frac{2}{9}\)is\( \frac{9}{2}\), reciprocal of \( \frac{16}{5}\) is \( \frac{5}{16}\) and reciprocal of \( \frac{8}{3}\)is\( \frac{3}{8}\)]
=\((1\times \frac{9}{2})+(1\times \frac{5}{16})+(1\times \frac{3}{8})\)
=\(( \frac{1\times 9}{1 \times 2}))+( \frac{1\times 5}{1 \times 16})+( \frac{1\times 3}{1 \times 8})\)
=\(( \frac{9}{2})+( \frac{5}{16})+( \frac{3}{8})\)      [∵ LCM of 2, 16 and 8 is 16]
=\( \frac{(9\times 8)+(5\times 1)+(3\times 2)}{16}\)
=\( \frac{72+5+6}{16}= \frac{72+11}{6}= \frac{83}{16}\)=5.1875

In a right angled triangle, atleast one angle should be 90°. Let the third angle be x.
We know that, the sum of all three angles in a triangle is equal to 180°.
\(\therefore\) 30°+90°+x=180° 
\(\Rightarrow\) 120°+x=180°
\(\Rightarrow\) x = 180° - 120° = 60°

(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 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.

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.