A is twice as fast as B and B is thrice as fast as C the journey covered by C is 54 mins will be covered by B in how many minutes?
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S = D/T
where S means Speed, D means Distance, & T means Time
Distance is common for all;
Let S(c) mean Speed of C
Let T(c) mean Time of C
Let S(c) mean Speed of C
Let T(b) mean Time of B
T(c)= 54mins
S(b)= 3 S(c) (Eq 1)
T(b)= D/S(c)
I.e,
54= D/S(c)
I.e,
D= 54 S(c) (Eq 2)
T(b)= D/S(b)
From Eq 1 & Eq 2
T(b)= 54 S(c) / 3 S(c)
So,
T(b)= 18mins
where S means Speed, D means Distance, & T means Time
Distance is common for all;
Let S(c) mean Speed of C
Let T(c) mean Time of C
Let S(c) mean Speed of C
Let T(b) mean Time of B
T(c)= 54mins
S(b)= 3 S(c) (Eq 1)
T(b)= D/S(c)
I.e,
54= D/S(c)
I.e,
D= 54 S(c) (Eq 2)
T(b)= D/S(b)
From Eq 1 & Eq 2
T(b)= 54 S(c) / 3 S(c)
So,
T(b)= 18mins
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