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22 May 2020, 12:03
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E
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The first floor of a building is partitioned into four rectangular rooms I, II, III, and IV, as shown in the figure above, with perimeters 60, 140, 100, and 80 meters, respectively. Assuming that the widths of the walls are negligible, what is the perimeter in meters of the first floor QRST ?
A. 760
B. 400
C. 380
D. 200
E. 190
PS20439
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1.png [ 3.46 KiB | Viewed 12080 times ]
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yashikaaggarwal
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22 May 2020, 21:47
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Perimeter= 2(l+b)
Let the perimeter of 1st room be => 2(L1+B1) = 60
=> (L1+B1) = 30 -----(1)
Let the perimeter of 2nd room be => 2(L2+B2) = 140
=> (L2+B2) = 70 -----(2)
Let the perimeter of 3rd room be => 2(L3+B3) = 100
=> (L3+B3) = 50 -----(3)
Let the perimeter of 4th room be => 2(L4+B4) = 80
=> (L4+B4) = 40 -----(4)
Now (L1+B1)+(L2+B2)+(L3+B3)+(L4+B4) Are the corners or the sum of edge of first floor.
30+70+50+40=190
Answer is E
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Let the perimeter of 1st room be => 2(L1+B1) = 60
=> (L1+B1) = 30 -----(1)
Let the perimeter of 2nd room be => 2(L2+B2) = 140
=> (L2+B2) = 70 -----(2)
Let the perimeter of 3rd room be => 2(L3+B3) = 100
=> (L3+B3) = 50 -----(3)
Let the perimeter of 4th room be => 2(L4+B4) = 80
=> (L4+B4) = 40 -----(4)
Now (L1+B1)+(L2+B2)+(L3+B3)+(L4+B4) Are the corners or the sum of edge of first floor.
30+70+50+40=190
Answer is E
Posted from my mobile device
General Discussion
22 May 2020, 21:23
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Bunuel
When you add the perimeter of all the rooms, we add the internal wall twice, so we will have to subtract the length of internal wall from the total, and that is nothing but length and breadth of complete floor.
Length=x+y=x+p+q
Breadth=a+b
So the total perimeter =\( I+II+III+IV=2(x+a)+2(b+(x+p)+2(a+y)+2(q+b)\)
\(=2(x+a+b+(x+p)+a+y+q+b)\)
\(=2(x+y+b+q+p+a+b+a+b)=2(2L+2B)=2(2(L+B))\)
=> \(60+140+100+80=380 =2(2(L+B))....2(L+B)=190\)
E
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