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Math Expert V
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 69% (01:37) correct 31% (01:37) wrong based on 48 sessions

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Competition Mode Question

A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters

B. 20.22 meters

C. 21.65 meters

D. 22.34 meters

E. 23.32 meters

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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters

B. 20.22 meters

C. 21.65 meters

D. 22.34 meters

E. 23.32 meters

The longest possible rod which can be placed in the room is = sq. root of (15^2+12^2+10^2)
= 21.65 meters C

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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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The longest rod that can placed in a room which is 15m long, 12m wide, and 10m tall can be gotten by first computing the diagonal of the length and width of the room and and computing the diagonal the computed diagonal forms with the height of the building.

So (12*12+15*15)^0.5 = 369^0.5
Longest possible rod = (369+10*10)^0.5 = (469)^0.5 = 21.65.

The answer is therefore C in my view.
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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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the longest possible rod which can be placed in the room in a cuboid shape... should be placed at the diagonal
Diagonal length in cuboid = sqrt(15^2+12^2+10^2)= sqrt(469) = 21.65
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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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Given: length l = 15m, breadth b=12m and height h=10m

Longest rod that can be placed in a room is by placing it diagonally.

Length of diagonal of a cuboid(Length of longest rod)= √(l^2 + b^2 + h^2)

= √(225 + 144 + 100) m

= √(469)m

20^2=400 so √(469) would be a bit greater than 20.

Thus the length of the longest rod would be 21.65 meters (OPTION C)

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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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we know that hypotenuse is the longest side in a triangle therefore the hypotenuse of the triangle with the larger sides will be largest

therefore the answer is L= under-root(15^2+12^2+10^2) ---space diagonal formula
=21.6

therefore C is correct
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Originally posted by sampriya on 12 Sep 2019, 22:13.
Last edited by sampriya on 13 Sep 2019, 20:55, edited 2 times in total.
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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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longest possible rod ; √15^2+12^2+10^2 ; √469 ; 21.65 mtrs
IMO C

A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters

B. 20.22 meters

C. 21.65 meters

D. 22.34 meters

E. 23.32 meters
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Posts: 1228
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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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Quote:
A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters
B. 20.22 meters
C. 21.65 meters
D. 22.34 meters
E. 23.32 meters

longest possible rod placed in a cuboid is a diagonal through the middle:
$$d^2=l^2+w^2+h^2…d^2=15^2+12^2+10^2…d^2=225+144+100…d^2=469$$
$$d^2=469…(400=20^2,…441=21^2,…484=22^2)…441<d^2=469<484…21<d<22$$

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Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo  [#permalink]

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It would be sqrt(15^2 + 12^2 + 10^2)= sqrt(225+144+100)= sqrt(469)
Then POE will come, as the calculation is cumbersome.
21^2=441
22^2=484
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" I CAN AND I WILL" Re: A room is 15 meters long, 12 meters wide & 10 meters in height. The lo   [#permalink] 13 Sep 2019, 03:48
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