what will be the least possible number of planks if 3 pieces of timber 42m, 49m and 63 m long have to be divided into planks of same length.
1>7
2>8
3>22
4>none
answer- c
planks
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- bblast
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- manpsingh87
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HCF of 42,49,63 =7;bblast wrote:what will be the least possible number of planks if 3 pieces of timber 42m, 49m and 63 m long have to be divided into planks of same length.
1>7
2>8
3>22
4>none
answer- c
also total length of timber which is to be divided into planks of same lenght=42+49+63=154;
hence total no. of planks would be 154/7=22 hence C
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Factorize 42,49 and 63..
42 = 2 x 3 x 7
49 = 7 x 7
63 = 7 x 3 x 3
Common Factor ( Here there is only one common factor viz. 7, however if there were more than one factors then we consider the HCF as we need the least no. of planks) = 7
42 is 6 x 7
49 is 7 x 7
63 is 9 x 7
Therefore least no. of planks = 6 + 7 + 9 = 22.
Hope this helps.
42 = 2 x 3 x 7
49 = 7 x 7
63 = 7 x 3 x 3
Common Factor ( Here there is only one common factor viz. 7, however if there were more than one factors then we consider the HCF as we need the least no. of planks) = 7
42 is 6 x 7
49 is 7 x 7
63 is 9 x 7
Therefore least no. of planks = 6 + 7 + 9 = 22.
Hope this helps.
-
- Junior | Next Rank: 30 Posts
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- Joined: Sun Jan 16, 2011 10:45 pm
Factorize 42,49 and 63..
42 = 2 x 3 x 7
49 = 7 x 7
63 = 7 x 3 x 3
Common Factor ( Here there is only one common factor viz. 7, however if there were more than one factors then we consider the HCF as we need the least no. of planks) = 7
42 is 6 x 7
49 is 7 x 7
63 is 9 x 7
Therefore least no. of planks = 6 + 7 + 9 = 22.
Hope this helps.
42 = 2 x 3 x 7
49 = 7 x 7
63 = 7 x 3 x 3
Common Factor ( Here there is only one common factor viz. 7, however if there were more than one factors then we consider the HCF as we need the least no. of planks) = 7
42 is 6 x 7
49 is 7 x 7
63 is 9 x 7
Therefore least no. of planks = 6 + 7 + 9 = 22.
Hope this helps.