One day, Mr. Richards started 30 minutes late from home and reached hi

Question

One day, Mr. Richards started 30 minutes late from home and reached his office 50 minutes late, while driving 25% slower than his usual speed. How much time in minutes does Mr. Richards usually take to reach his office from home?

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

Ukaul · Accepted Answer

Let time taken to reach office everday = t
Let usual speed = s
Extra Time = 50 - 30 = 20 mins
Speed s is 25% slower = s - .25s = .75s
Now, D= S*T

st = (t+20) * (.75s) 
Solving Further ..
.75st + .75s*20 = st
Solving for t brings,
t= 60 mins. Option C.

Archit3110 · Answer

let total time taken daily be 1 hr to reach office
so if he left 30 mins late and reached 50 mins late ; 80 mins or say 1 hr 20 mins or say 20 mins over usual time
d = distance
s = speed
new speed = 0.75s
so time taken 
d/0.75s-d/s = 20
solve for d/s = 60 mins
IMO C

exc4libur · Answer

rule: distance = rate • time
usual time: d=tr
late time: d=(t+20)(0.75r)… {50-30=20 driving late}
equate distance: tr=0.75tr+15r… 0.25tr=15r… tr/4=15r… t=60

Answer (C).

ScottTargetTestPrep · Answer

We can let r = his usual rate and t = his usual time (in minutes). Since he started 30 minutes late from home and reached his office 50 minutes late, he actually spent 20 minutes more than his usual time traveling from his home to office.

rt = 0.75r(t + 20)

Dividing both sides by r, we have:

t = 0.75t + 15

0.25t = 15

t = 60

Answer: C

effatara · Answer

Let his usual speed and the time he takes to reach his office at that speed be denoted by 'v' and 't' respectively.

Since the ratio of his speeds {v and (3/4)v} is the inverse of the ratio of the times he takes at each respective speed: 
v/(3/4)v = (t+20)/t....> 4/3 = (t+20)/t.....> t = 60. ANS: C

vaibhav1221 · Answer

Assume the usual speed to be X
Distance to be Y
and the usual time to be Z

Usual speed and time
 X=\frac{Y}{Z} ... (1)

25% slower and 20 minutes more time (He reached office 50 minutes late because he left 30 minutes late and drove at a speed 25% slower than his usual speed implying that he was essentially 20 minutes late.)
 \frac{3}{4}X=\frac{Y}{(Z+20)} ... (2)

Equate  (1)  and  (2)  to get the value of Z as 60.

Hence, answer is C, 60 minutes.

AkmKawser · Answer

S = 4:3
T = 3:4 Thus (4-3) = 1 unit time difference

1 unit ≡ 20 min  
3 unit ≡ 20*3 = 60 min.

Abhishek009 · Answer

Let usual speed be 4

Usual Speed : New Speed = 4 : 3
Usual Time : New Time= 3 : 4  (Ratio of time is inversely related to Speed)

Difference in time in ratio is 1 unit and Actual difference in time is 20 Min...

So, Usual time will be 20*3 = 60 Min, Answer must be (C)

Alexdb10 · Answer

Ok so I was really confused with this question I think the syntax is wrong - I also could very well be wrong.

if Mr Richard started 30 minutes late from his home and reached his office 50 minutes late - does that not insinuate that it takes 20 minutes to drive from his home to his office at a rate 25% slower?

Bunuel · Answer

No. The 20-minute difference is not the travel time. It is the  extra delay caused by driving slower .

He left 30 minutes late but arrived 50 minutes late, so the slower speed added  20 extra minutes to the usual travel time . That 20 minutes is the  increase  due to driving 25% slower, not the total driving time.

Please check solutions above for more.

One day, Mr. Richards started 30 minutes late from home and reached hi

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