Let consecutive numbers be x and x + 1.
Hence, according to question,
x + x + 1 = 53
\(\Rightarrow\) 2x + 1 = 53
\(\Rightarrow\) 2x = 53-1
[On transposing 1 to RHS]
\(\Rightarrow\) 2x = 52
\(\Rightarrow \quad \frac { 2x }{ 2 } =\frac { 52 }{ 2 } \)
\(\Rightarrow\) x = 26
\(\therefore\) Smaller number is 26

We have,  (-49) \(\div\) (49) = - (49 \(\div\) 49) = - (49/49) = -1

We have, 7, - 5, 4, 0 and - 4
Since every positive integer is greater than 0 and every negative integer is less than O. So, rearrange the given numbers in ascending order i.e. -5, - 4, 0, 4, 7.
In order to represent the given integers on the number line, we draw a line and mark a point on it almost in the middle (say0) of the line.
Now, starting from 0 and count 4 units on the right of the middle point, we get 4 and further count 3 units from 4 to the right, we obtain 7.
Again, starting from 0 and count 4 units to the left of it, we obtain - 4 and further count 1 unit from its left, we get -5.

The top view of the glass is a pair of concentric circles because its top is a circle as well as its base is a concentric circle of smaller radius.
Due to symmetry, the side view will not appear different if taken from some other direction.

Here, radius (r) = 14 cm
\(\because\)Circumference of circle =2\(\pi\)r
\(\therefore\) Circumference of the semi-circular disc =\(\frac{1}{2}\)(2\(\pi\)r)=\(\pi\)r
=\(\frac{22}{7}\)\(\times\)14== 44 cm
Since,   diameter = 2 x r
= 2 x 14 em = 28 em
\(\therefore\)Perimeter of the semi-circular disc = 44 cm + 28 cm = 72 cm

We have: 5x +\(\frac{1}{3}\)=2-3x
\(\Rightarrow \) 5x+3x=2-\(\frac{1}{3}\)
[Transposing (-3x) to L.H.S. and 1. to R.H.S.]
\(\Rightarrow \) 8x=\(\frac{6-1}{3}=\frac{5}{3}\)
\(\Rightarrow \) x=\(\frac{5}{3}\times \frac{1}{8}=\frac{5}{24}\)
Thus, x=\(\frac{5}{24}\)
is the required solution of the given equation.

(a) Score of Mohan= 6 \(\times\) (5) + 4 \(\times\) (- 2)
= 30 + (-8)
=30 - 8 = 22
(b) Rekha's score = 5 \(\times\) (5) + 3 \(\times\) (-2) + 2 \(\times\) 0
= 25 + (-6) + 0
= 25 - 6 = 19
(c) Mayank's score = 5 \(\times\) (5) + 5 \(\times\) (- 2)
= 25 + (-10) = 25 -10 = 15
(d) Jeet's score = 3 \(\times\) (5) + 5 \(\times\) (- 2)
= 15 + (-10) = 15-10 = 5

In ratio, we would write this as
Rice : Urad dal = 2 : 1
= 2 + 1 = 3 total parts
Now, thi s means \(\frac { 2 }{ 3 } \)X100%
=\(\frac { 200 }{ 3 } =66\frac { 2 }{ 3 } \)%
and,  urad dal in % =\(\frac { 1 }{ 3 } \times 100\)% = 33 \(\frac { 1 }{ 3 } \)%

The given figure have order of rotation is 3 and angIe of rotation will be \(\frac { 360° }{ 3 } =120°\)

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

\(⇒\ \left(1\over 5\right)^{5+19}=\left(1\over 5\right)^{8x}\)
[∵ a^m x aⁿ = a^m+n]
\(⇒\ \left(1\over 5\right)^{24}=\left(1\over 5\right)^{8x}\)
When bases are equal, then by equating their exponents, we get
8x = 24
\(∴\ x={24\over 8}=3\)

(a) 9% of Rs.700= \(\frac { 9 }{ 100 } \times \)700
= 9 X 7
= Rs.63
(b) 6 \(\frac { 2 }{ 3 } \)% of 75m =\(\frac { 20 }{ 3 } \)% of 75 m
=\(\frac { 20 }{ 300 } \times \)75 m
=\(\frac { 1 }{ 15 } \times \)75 m
=5 m
(c) 7.5 % of 80 kg =\(\frac { 7.5 }{ 100 } \times \)80 kg
=\(\frac { 3 }{ 40 } \times \)80 = 6 kg
(d) 8% of 5 liters =\(\frac { 8 }{ 100 } \times \)5 litres
=\(\frac { 40 }{ 100 } \times \)0.4 litre.