For subtraction, write the like terms one below the other, we get
12a -9ab +5b -3
+4a - 7ab + 3b + 12
(-)     (+)     (-)    (-)
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8a - 2ab + 2b -15
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Thus, the required answer is 8a - 2ab + 2b - 15.

First expression = 2x(z - x - y)
= (2x) x z - (2x) x (x) - (2x) xy
= 2xz - 2x² - 2xy
Second expression = 2y(z - y - x)
= (2y)x(z)-(2y) x (y)-(2y)x(x)
= 2yz - 2y² - 2yx
On adding above expressions, we get
2xz - 2x² - 2xy
+              -2yx + 2yz - 2y²
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2xz - 2x² - 4xy + 2yz - 2y²
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(7m - 8n)² + (7m + 8n)²
= [(7m)² - 2(7m)(8n) + (8n)²] + [(7m)² + 2(7m)(8n) + (8n)²]
and (a-b)²= a² -2ab+b²
= (49m² -112mn + 64n²)+(49m² +112mn + 64n²)
= 49m² -112mn + 64n² + 49m² +112mn + 64n²
= 98m² + 128n²

[(4m)² + 2(4m)(5n) + (5n)²] + [(5m)² + 2(5m)(4n) +(4n)²]
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²
=  (16 + 25)m²+ (40 + 40)mn + (25 + 16​​​​​)²
= 41m² + 80mn + 41n²

Let ABCDEFGH be the regular octagonal surface as shown below in the figure:

Given, each side of octagon = 5m,AD = 11 m
perpendicular from B to AD i.e. BP =4m .Join HE
Since, it is a regular octagon, so trapeziums ABCD and HEFG will be congruent and ADEH is a rectangle.
Now, area of trapezium ABCD = Area of trapezium HEFG
 = (AD+BC)BP
 = \(\frac { 1 }{ 2 } \)(11+5)\(\times\) 4 =\(\frac { 1 }{ 2 } \) \(\times\)16\(\times\)4
 = 8 \(\times\)4 =32 m²
and area of rectangle ADEH = AD\(\times\) DE =11 \(\times\)5 =55m²
Now, area of octagon ABCDEFGH
= Area of trapezium ABCD +Area of rectangle ADEH +Area of trapezium HFGH
= 32 + 55 + 32= 119 m²
Hence, required area of octagonal surface is 119 m²

Steps of construction
StepI Draw MT = 6 cm and draw \(\angle\)TMX = 45°.
StepII Cut MA from MX, so that MA = 10 cm.
StepIII Draw an arc of 5 cm with centre T and draw an arc of 3 cm with centre A.
StepIV Mark the intersection point of both the arcs as E. Join TE and AE.

Thus, quadrilateral TEAM is constructed.

Letus draw a rough sketch of the quadrilateral.

Steps of construction
Step I Draw AM = 5 cm.
Step II Construct \(\angle\)MAX = 60° and \(\angle\)AMY = 90°.
StepIII Cut the length of AG = 4 cm from AX and the length of EM = 4.5 cm from MY.
StepIV Join GE.

Thus, we get the quadrilateral GAME by construction.

Rs.450

Let the number of winners be x.
Then, the number of participants who did not win = 36-x
Amount spent on x prizes = Rs. 500 x x = Rs. 500x
Amount spent on (36 - x) prizes = Rs. 100 x (36 - x) =
= Rs. (3600 -100x)
But 500x + (3600 -100x) = 4800
\(\Rightarrow\) 500x + 3600 -100x = 4800
\(\Rightarrow\) 400x = 4800 - 3600
\(\Rightarrow\) x = 1200 x \(\frac { 1 }{ 400 } \Rightarrow x=3\)
So, the number of winners is 3

We have, perimeter = 77 cm 
\(\therefore\) (x + 1)+ (x + 2) + (2x + 2) + (2x + 1)+ (x + 1) = 77
7x + 7= 77  7x = (77 - 7)
7x = 70  x = 70 x\(\frac { 1 }{ 7 } \)
Hence, the required value of x is 10 cm 

Number of persons added = 5
\(\therefore\) Present number of persons = 20 + 5 = 25
Since, for more persons, the food will last less number of days
\(\therefore\) It is a case of inverse variation.
\(\therefore \ 25\times x=20\times 10\Rightarrow x=\frac { 20\times 10 }{ 25 } =8\)
\(\therefore\) The food will now last for 8 days.

More is the number of persons, less is the time to complete the task.
\(\therefore\) It is a case of inverse variation,
Now, we have:

\(\therefore \ 5\times 4=4\times x\Rightarrow x=\frac { 5\times 4 }{ 4 } =5\)
Thus, the required number of days = 5.

Present duration each period = (50 - 5) minutes = 45 minutes.
Let the present number of period be 'x'.
Since, more the number of periods, less is the duration of a period.
\(\therefore\) It is a case of inverse variation.
We have:

\(\therefore \ 9\times 50=x\times 45\Rightarrow x=\frac { 9\times 50 }{ 45 } =10\)

Thus the required number of periods = 10.

Cost price of 200 bulbs = RS. 200 x 10 = RS. 2000
5 bulbs were fused. Hence, number of bulbs left = 200 – 5 = 195
These were sold at RS. 12 each.
The SP of 195 bulbs = RS. 195 x 12 = RS. 2340
He obviously made a profit (as SP > CP).
Profit = RS. 2340 – RS. 2000 = RS. 340
On RS. 2000, the profit is RS. 340. 
Profit \(=\frac{340}{2000} \times 100 \%=17 \%\)