72 + (-5) - x = 72
- 5 - x = 72 - 72 = 0
- 5 - x = 0
= x -5
x = -5

From this figure
<CDE = <EAB
(Alternate interior angles)
x^o = 48^o
<ABE = <DCE
(Alternate interior angles)
y^o = 60^o
We know that z^o = y^o + 48^o
= 60^o + 48^o
= 108^o

Let the angles be x and y. Given that one angle exceeds the twice of the other angle by 48^o

y = 2x + 48^o
x + y = 180^o (x and y are interior angles)
x + 2x + 48 = 180^o
3x + 48 = 180^o
3x = 132
x = 44^o
y = 2x + 48^o
= 2 x 44 + 48
= 88 + 48
y = 136

From the figure
<1 and <2 are interior angles
<1 + <2 = 180^o
53^o + 12 = 180^o
<2 = 127^o

From the figure
y = 42^o [vertically opposite angles]
z = 42^o [Alternate interior angles]
x + 63^o + z = 180^o
x + 63 + 42 = 180^o
x + 105 = 180^o
x = 75^o

Let the angles be 11x and 25x. Since the angles are linear pair.
11x + 25x = 180^o
36x = 180^o
x = 5
The angles are 11x = 11 x 5 = 55^o and
25x = 25 x 5 = 125^o

(i)  positive integer.
(ii)  negative integer.

Since E is the mid point of DC
DE = 12 m ; EC = 12 m
ADEB is a trapezium of parallel side
a = 12 m, b = 24 m and h = 18 m
Area of trapezium = \(\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b}) \text { sq.units }\)
\(=\frac{1}{2} \times 18(12+24)\)
= 9 x 36
Cultivated Area = 324 m²

Given : One of the diagonals of a rhombus is thrice the other
\(\Rightarrow \ \mathrm{d}_{2}=3 \mathrm{~d}_{1}\)       ............(1)
Sum of the length of the diagonals = 24 cm
d₁+ d₂ = 24cm
3 d₁ + 3d₂ = 24
4d₁ = 24
\(d_{1}=\frac{24}{4}=6 \mathrm{~cm}\)
d₁ = 6 cm
d₂  = 3d₁
= 3 x  6
d₂ = 18 cm
Area of the rhombus \(=\frac{1}{2} \mathrm{~d}_{1} \times \mathrm{d}_{2} \text { sq.units } \)
\(=\frac{1}{2} \times 6 \times 18 \)
= 54 sq.cm

Given : Area of a parallelogram = Area of a square
b x h = a x a
b x 18 = 48 x 48
\(b=\frac{48 \times 48}{18}\)
Base (b) = 128 m

Given : A = 1080 sq.cm, a = 55.6 cm, b = 34.4 cm
Distance between them, h = ?
Area of trapezium = 1080 sq.cm
\(\frac{1}{2} \times h(a+b) =1080 \)
\(\frac{1}{2} \times h(55.6+34.4) =1080 \)
\(\frac{1}{2} \times h \times 90 =1080 \)
\(h =\frac{1080 \times 2}{90} \)
Distance, h = 24 cm.

Parallel sides, a = 24 cm
b = 20 cm
Distance between them (Height) h = 15 cm
Area of trapezium \(=\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b}) \text { sq.units } \)
\(=\frac{1}{2} \times 15(24+20) \)
\(=\frac{1}{2} \times 15 \times 44 \)
Area of the trapezium = 330 cm²

Let x be the number.
-17 + x = -19
x = -19 + 17
x =  -2
-2 should be added to -17 to get -19.

From the figure
∴ ∠JlL = ∠JIK + ∠KIL
= 38° + 27°
= 65°
∴ ∠JlL = 65°

(11 + 7) + 10 = 18 + 10 = 28
11 + (7 + 10) = 11 + 17 = 28
Yes, (11 + 7) + 10 and 11 + (7 + 10) are equal.
This is Associative property. a+ (b + c) = (a + b) + c.

Scoring marks for each incorrect answer = -1
Marks reduced for 25 incorrect answers in paper I = -25
Marks reduced for 13 incorrect answers in paper II = -13
Total reduction of marks = -25 + (-13)
=-25 -13 = -38
= -38 marks
Total reduction of marks in both papers = 38

Price increases every year for item x = Rs.10.
Price decreases every year for item y = Rs. -15
Price of x in 2018 = Rs. 50
Price of x in 2020 = Rs. 50 +2 x Rs. 10 = Rs. 70 
= Rs. 50 + Rs. 20 = Rs. 70
Price of y in 2018 = Rs. 90
Price of y in 2020 = Rs. 90 + 2 x Rs. 15 = Rs. 60
Item x is costlier in the year 2020.

(i) 13x + 12y - 5 from 27x + 5y - 43
27x + 5y-43 - (13x + 12y- 5)
= 27x + 5y - 43 - 13x - 12y + 5
= (27 - 13) x + (5 - 12) y + (- 43) + 5
= 14x + (-7)y + (-38)
= 14x - 7y - 38
(ii) 3p + 5 from p - 2q + 7
= P - 2q + 7 + (- 3p -5)
= (1-3)p-  2q + 2
= -2p - 2q  + 2
(iii) m + n from 3m - 7n
Additive inverse of (m + n) is -m -n,
= 3m -7n - (m + n) = 3m -7n + (-m -n)
= (3m - m) + (-7n - n)
= (3 - 1)m + (-7 -1) n
= 2m - 8n
(iv) 2y + Z from 6z - 5y
Additive inverse of 2y + z = -2y - z.
6z - 5y-(2y + z)
= ( 6z -5y) +(-2y - z)
= (62 - z) + (-5y - 2y)
= (6 - 1)z + (-5 -2)y
= 5z + (-7)y = 5z - 7y

Lift in the ground floor denotes 0.
It goes 5 floors down = -5
It moves up to 10 floors = +10
Position of the lift = 10 - 5 = 5
Lift will be in the 5m floor