Answer:
His time going 4 hours , his time returning 4.5 hours
Step-by-step explanation:
At first let the distance between his home and New York is D
Second let the time going is T1 and the time returning is T2
∵ The time returning exceeds the time going bu 30 min
∴ T2 - T1 = 30/60⇒we must change the minutes to hours because his rate is mile per hour
∴ T2 - T1 = 1/2⇒(1)
∵ His rate going is 45 mph
∴ T1 = D/45⇒(2)
∵His rate returning is 40 mph
∴ T2 = D/40⇒(3)
Substituting (2) and (3) in (1)
[tex]\frac{D}{40}-\frac{D}{45}=\frac{1}{2}[/tex]
[tex]\frac{45D-40D}{(40)(45)}=\frac{1}{2}[/tex]
[tex]45D-40D=900[/tex]
[tex]5D=900[/tex]
D = 180 m
Use this value of D in (2) and (3)
∴ T1 = 180/45 = 4 hours⇒time going
∴ T2 = 180/40 = 4.5 hours⇒time returning
His time going was 4 hours and his time returning was 4 and a half hours.
Given that Mr. Sawyer drove his car from his home to New York at the rate of 45 mph and returned over the same road at the rate of 40 mph, to determine, if his time returning exceeded his time going by 30 min, his time going and his time returning, the following calculation must be performed:
Therefore, his time going was 4 hours and his time returning was 4 and a half hours.
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