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05 Aug 2021, 06:15
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B
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Difficulty:
95%
(hard)
Question Stats:
40% (02:37) correct
60%
(02:34)
wrong
based on 53
sessions
History
Date
Time
Result
Not Attempted Yet
The least number of square tiles required to cover the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad, is:
(A) 902
(B) 814
(C) 738
(D) 656
(E) 556
(A) 902
(B) 814
(C) 738
(D) 656
(E) 556
05 Aug 2021, 06:40
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area=(1517x902)cm^=41^2 x2x11x37
let p be the no of square tiles required. and a^2 be the area of each required square tile.
as per question,
for p to be min , a^2 needs to be maximum.
so from above a^=41^2
so ,
p=2x11x37=814
B is the answer.
let p be the no of square tiles required. and a^2 be the area of each required square tile.
as per question,
for p to be min , a^2 needs to be maximum.
so from above a^=41^2
so ,
p=2x11x37=814
B is the answer.
07 Aug 2021, 07:21
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Bunuel
15 m 17 cm = 1517 cm.
9 m 2 cm = 902 cm.
We can start by finding the GCF of 1517 and 902.
Since the numbers are really big we should factorize them, \(1517 = 37*41\) and \(902 = 2*451 = 2*11*41\)
Next, we can see 41 is the GCF. Using tiles with sides of 41, on one side we would have 37 tiles and on the other would have 2*11 tiles.
Therefore we would have 37 * 22 tiles, this number has a unit digit of 4 so we choose B.
Ans: B
