All the three sides of a room are integer values in meters. What is th

Question

All the three sides of a room are integer values in meters. What is the maximum possible length of a rod, which can be kept inside the room?

(1) The area of the room’s floor is 64 sq. m
(2) The sum of areas of adjacent walls of the room is 320 sq. m

RajatVerma1392 · Answer

Hi,

As we are given three sides are integers, let them be l,b,h. We need to find longest rod that means we need to find the diagonal of the block.
We cannot decide whether it`s cube or cuboid till now.
So,
1. The area of room = 64 
l*b = 64
There can be possibilities about l and b => 4*16, 8*8 , 16*4
Insufficient.

2. Sum of adjacent walls = 320
2(l*b + b * h + h* l) = 320 i.e. Surface area of cuboid
or 6  {l^2}  = 320

but we do not know whether it is cube or cuboid.
Insufficient.

3. if we combine them, then we can decide that block is cuboid, as all values are not equal.
taking any possible combination, such as: l= 8 , b= 8 :
2(64 + 8*h + 8*h) = 320
=> 16*h = 96
=> h = 6
we have unique values for l,b,h.
Thus, diagonal =  \sqrt{(l^2 + b^2 + h^2)}

Thus combining both is sufficient.

Please give kudos if you like the solution.

MahmoudFawzy · Answer

statement (1):  it means that L*W = 64 
so (l,W) pair can have 4 values: (1,64), (2,32), (4,16), (8,8) .. (plus lacking info about the height) --->  Insufficient

statement (2):  it means that the sum of the areas of the  4  walls = 320 
2(L*H) + 2(W*H) = 320
L*H + W*H = 160
H(L+W) = 160 --> which means that (L+W) must be a factor of 160 (so that H remains an integer),
so (l,W) pair can have multiple values: (1,159), (2,158), (4,4), (4,8), ... --->  Insufficient 
 
 combining (1) and (2) , 
from statement (1), there is two valid (L,W) pair whose sum is a factor of 160, the pairs are (4,16)=20 and (8,8)=16
in case (L,W) pair is (4,16), H will be 8 ---> the diagonal =  \sqrt {L^2+W^2+H^2} = \sqrt {4^2+16^2+8^2} = \sqrt{336}  
in case (L,W) pair is (8,8), H will be 10 ---> the diagonal =  \sqrt {L^2+W^2+H^2} = \sqrt {8^2+8^2+10^2} = \sqrt{228}  
so  Insufficient  ( E )

saukrit · Answer

We need to know 3 sides of the room to find its diagonal(highest length of the rod possible)

Statement (1)  The area of the room’s floor is 64 sq. m
 Height can be anything, Not sufficient at all

Statement (2)  The sum of areas of adjacent walls of the room is 320 sq. m 
Multiple combinations of floor area possible. No unique answer.  Again Insufficient

St (1) + St(2)

Now we have the possibility of (l,w) combinations for the floor area to be 64 as below:

(1,64) 
(2,32)
(4,16)
(8, 8)

Also, we know that all sides have to be integers. Let the height be h
Therefore we must have for each case 2(l*h + w*h) =320 such that h is an integer

Case 1: 2h +128h=320 -->h won't be an integer so (1,64) rejected
Case 2: 4h + 64h=320 -->h won't be an integer so (2,32) rejected
Case 3: 8h+ 32h= 320 -->h is an integer =8 so (4,16) is a valid combo for (l,w) with h=8
Case 4: 16h+16h=320 -->h is an integer =10 so (8,8) is a valid combo for (l,w) with h=10

For case 3 and Case 4 we get 2 different values of diagonal = \sqrt{16+256+64} or \sqrt{64+64+100} 
 \sqrt{336} or \sqrt{228}

No unique answer , Hence E

All the three sides of a room are integer values in meters. What is th

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All the three sides of a room are integer values in meters. What is th

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