(a) The area of two bedrooms and kitchen are respectively equal to
(i) 5x, 5y (ii) 10x, 5y (iii) 5x, 10y (iii) x, y
(b) Find the length of the outer boundary of the layout.
(i) 27 m (ii) 15 m (iii) 50 m (iv) 54 m
(c) The pair of linear equation in two variables formed from the statements are
(i) x + y = 13, x + y = 9
(ii) 2x + y = 13, x+y=9
(iii) x + y = 13, 2x + y = 9
(iv) None of the above
(d) Which is the solution satisfying both the equations formed in (iii)?
(i) x = 7, y = 6
(ii) x = 8, y = 5
(iii) x = 6, y = 7 (iv) x = 5, y = 8
Answers
Answer:
a) 10x, 5y
b) 54m
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Answer:
below
Step-by-step explanation:
(a)
length of bedroom =x m
Breadth of bedroom =5 m
And, length of kitchen =15−(x+2)=13−x m
Breadth of kitchen =5 m
So, Area of both kitchen and bedrooms =2× times of area of bedroom + area of kitchen
=2(5×x)+(13−x)×5
=10x+(65−5x)
=(65+5x)m
2
(b)
From figure,
Perimeter of the living room
=15+2+x+5+(15−x)+5+2=44m
(c)
From figure,
Total area of both living the room and the bedrooms =(5×x)+(7×15)=(5x+105)m
2
The cost of carpeting is Rs. 50/m
2
Therefore, total cost of carpeting
=(5x+105)×50=Rs.250(x+21)
(d)
Total area, bathroom and kitchen
=(15−x)×5m
2
The cost of tiling is Rs. 30/m
2
Therefore, total cost of tilling
=(15−x)×5×30=Rs.150(15−x)
(e)
Given, Area of floor of each bedroom 35 m
2
So, Area of one bedroom
=5xSq.m
∴5x=35
⇒x=7m