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01 Jul 2019, 23:42
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E
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Difficulty:
65%
(hard)
Question Stats:
57% (02:39) correct
43%
(01:30)
wrong
based on 37
sessions
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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
(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
02 Jul 2019, 00:20
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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.
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.
02 Jul 2019, 06:18
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Harshita6789
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)
