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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
For a difference of 30 mins the corresponding change in speed is 5, next we find that the change is proportional throughout ie t/6. so the speed of the train after time t is 40+ t/6 (D).
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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
animesh mohanty wrote:
For a difference of 30 mins the corresponding change in speed is 5, next we find that the change is proportional throughout ie t/6. so the speed of the train after time t is 40+ t/6 (D).

Pls. Note the time is in minutes not hours.
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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
HG90 wrote:
animesh mohanty wrote:
For a difference of 30 mins the corresponding change in speed is 5, next we find that the change is proportional throughout ie t/6. so the speed of the train after time t is 40+ t/6 (D).

Pls. Note the time is in minutes not hours.

change of 5 mph =0.083 miles per min which is achieved in 30 mins. This is the same as change of 0.041 miles per min that is achived in 15 mins. so the logic is correct.
P.S. i didn't mark the difference initially. Thank!
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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
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calculate the speed by substituting time at t= 30 minutes.
t= 30 minutes = 1/2 hrs
E- 40+10t = 40+10*1/2 = 40+5 = 45 mph

hence E is the answer.
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At time t measured in minutes after the beginning of the timed period, [#permalink]
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Bunuel wrote:

At time t measured in minutes after the beginning of the timed period, which of the following gives the speed of the train in accordance with the table?

(A) t/6
(B) 10t
(C) 40 + t
(D) 40 + t/6
(E) 40 + 10t

I think elimination might be quicker. Find one pair of correlated numbers (time, speed), to disprove each answer. I varied it just to be careful.

(A) t/6? S = 45, t = 30
45\(\neq\)(30/6) = 5: NO

(B) 10t? S = 50, t = 60
50 \(\neq\)(10 * 60) = 600: NO

(C) 40 + t? S = 55,t = 90
55 \(\neq\) (40 + 90) = 130: NO

(D) 40 + t/6? S = 60, t = 120
60 = (40 + 120/6) = (40 + 20) = 60
First test works. Come back

(E) 40 + 10t? S = 70, t = 180
70 \(\neq\)(40 + 180) = 220: NO

By POE, D is the answer. Test it:
50 S, 60t: (40 + 60/6) = (40 + 10) = 50
45 S, 30t: (40 + 30/6) = (40 + 5) = 45

That's enough.

Answer D
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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
time is given in minutes and speed in m.p.h
60 min = 1 hr
in the table 60 min gives speed as 55
Answer E satisfies the condition
40 + 10(1) = 50

validation: for 90 min: t = 3/2
40 + 10(3/2)= 55
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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
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Bunuel wrote:

At time t measured in minutes after the beginning of the timed period, which of the following gives the speed of the train in accordance with the table?

(A) t/6
(B) 10t
(C) 40 + t
(D) 40 + t/6
(E) 40 + 10t


Let’s plug in values from the table. In the first column, we have t = 0 for time and speed = 40.

We see that C, D, and E yield a speed of 40 when using 0 for t.

Testing the second column, we have t = 30 and speed = 45. Thus, we see that answers C and E do not yield 45 when t = 30. Answer choice D yields 40 + 30/6 = 45.

Answer: D
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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
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Re: At time t measured in minutes after the beginning of the timed period, [#permalink]
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