If a person walks 25% more than of his usual speed, reaches his distance 90 minutes before. If the destination is 495 km away, then the usual speed of a person is (in km/hr)?
Let the usual speed be x km/hr New speed = 125% of x = 5x/4 km/hr Distance = 495 km According to the question 495/x – 495/(5x/4) = 90/60 495 x [(5 – 4)/5x] = 3/2 Therefore, x = 495 x (1/5) x (2/3) = 66 km/hr
How many male members are there in the family?
How is H related to L?
Who is husband of mother of Z?
How is R related to N?
Who among the following person is the son of A?
In a family of six members, W is the daughter of F’s daughter only brother. V is the daughter-in-law of L, who is husband of F. Z is brother of Y, the...
In a family, I is sister of G. L is father-in-law of N. N is mother of G. H is daughter of G’s mother. K is the only son of L. How is I related to L?
_____ is the father of _____.
I. V, U
II. P, Q
III. U, T
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Who is the husband of T?