Question { Infosys, 71625 }

a dog takes 4 leaps for every 5 leaps of a hare but 3
leaps of a dog are equal to 4 leaps of the hare .compare
their speeds(.......... please give explanation....)

Answer

let us assume



speed of dog D speed of hare H
leap of dog X leap of hare Y

now
a dog takes 4 leaps for every 5 leaps of a hare
let time T
in time T

dog =4X/D=T hare=5Y/H=T


3 leaps of a dog are equal to 4 leaps of the hare

3Y=4X

so

3Y/D=5Y/H=T

so D/H=3/5

Question { 7218 }

What is the factor of x-3.

Answer

factors are

1) 1
2) x-3
3) sqrt x-sqrt 3
4) sqrt x+sqrt 3

Question { CTS, 40282 }

A software engineer has the capability of thinking 100
lines of code in five
minutes and can type 100 lines of code in 10 minutes. He
takes a break for
five minutes after every ten minutes. How many lines of
codes will he
complete typing after an hour?

Answer

60min=10 work 5 break 10 work 5 break 10 work 5 break 10
work 5 break



total work =40 min

in 40 min= 5 min think 10 min type 5 min think 10 min type 5
min think 5 min type

so 100+100+50=250 lines