part 2 immediatel please
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let T is original time in which he reachs office without late or early .
we know,
time = distance/speed
let distance between his house to office is D
a/c to question,
D/25 = T + 10/60 ---------(1)
agin,
D/30 =T -5/60 -------------(2)
solve both equation ,
D/25 -D/30 =(10+5)/ 60
D{5/25 × 30} = 1/4
D/150 = 1/4
D =150/4 =75/2
put D =75/2 in equation (1)
75/2 ×25 = T + 1/6
3/2 =T + 1/6
T = 3/2 -1/6 =16/12 =4/3 hours
hence,
T = 1 hour 20minutes is perfect time
hence ,
9 : 35 am does his office starts
we know,
time = distance/speed
let distance between his house to office is D
a/c to question,
D/25 = T + 10/60 ---------(1)
agin,
D/30 =T -5/60 -------------(2)
solve both equation ,
D/25 -D/30 =(10+5)/ 60
D{5/25 × 30} = 1/4
D/150 = 1/4
D =150/4 =75/2
put D =75/2 in equation (1)
75/2 ×25 = T + 1/6
3/2 =T + 1/6
T = 3/2 -1/6 =16/12 =4/3 hours
hence,
T = 1 hour 20minutes is perfect time
hence ,
9 : 35 am does his office starts
abhi178:
is it right ??
now see answer .
have you any problems to understand this
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