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If Mary always takes the same route to work, how long did it

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If Mary always takes the same route to work, how long did it

by BTGmoderatorDC » Fri Aug 24, 2018 12:56 pm

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If Mary always takes the same route to work, how long did it take Mary to get to work on Friday?

(1) It took Mary 20 minutes to get to work on Thursday.

(2) Mary's average speed on her trip to work was 25 percent greater on Thursday than it was on Friday.

OA C

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by ceilidh.erickson » Mon Aug 27, 2018 12:04 pm

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Given that the distance to work is always the same, we want to know the TIME that it took to get to work on Friday.

(1) It took Mary 20 minutes to get to work on Thursday.
We are given no relationship between Thursday and Friday - this might be faster, slower, or the same. Insufficient.

(2) Mary's average speed on her trip to work was 25 percent greater on Thursday than it was on Friday.
This gives a proportional relationship between Thursday and Friday's rates, but no actual values. We cannot solve for time.

(1) and (2) Together:
Since (rate)(time) = distance, and the distance to work is constant, then knowing a proportional difference in rate is enough to infer proportional difference in time:
[(5/4)(rate)][(4/5)(time)] = distance

If Thursday's rate was 25% greater than (125% of or 5/4) Friday's, then Thursday's time must have been 20% less than (80% of or 4/5 of) Friday's time.

Since we have a value for Thursday's time, this will be sufficient to calculate Friday's time: 20 min is 4/5 of Friday's time --> Friday's time = 20 min.
(We didn't actually have to calculate that last part once we recognized that we would get a value).

The answer is C.
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by fskilnik@GMATH » Mon Aug 27, 2018 12:55 pm

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BTGmoderatorDC wrote:If Mary always takes the same route to work, how long did it take Mary to get to work on Friday?

(1) It took Mary 20 minutes to get to work on Thursday.

(2) Mary's average speed on her trip to work was 25 percent greater on Thursday than it was on Friday.
? = time took on Friday (route is fixed)

(1) Insufficient

Mary could have taken 20min on Friday (same as the day before) or 30min (if she got traffic, for instance)

(2) Insufficient

Imagine the distance (route) was 100 km.

> Mary´s speed could have been 100 km/h on Friday (if 125 km/h on Thursday) and in this case ? = 1h
> Mary´s speed could have been 80 km/h on Friday (if 100 km/h on Thursday) and in this case ? = 100/80 h is different from 1h.

(1+2) When distance is fixed, we have (average speed may be considered constant) speed and time inversely proportional, hence

(*) Vthursday = (5/4) Vfriday ==> Tthursday = (4/5) Tfriday.

We have Tthursday (20min) , hence Tfriday is a unique numerical value, hence (1+2) is Sufficient.

(*) We usually prefer "V" for speed (although velocity is a vector, in which speed is only its magnitude), because in many countries "S" is reserved for space ("distance").

The above follows the notations and rationale taught in the GMATH method.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Portuguese-speakers :: https://www.gmath.com.br
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by deloitte247 » Sat Sep 01, 2018 12:09 pm

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Average speed = distance / time taken
Mary takes the same route to work. Hence, distance is constant.
So, how long did it take Mary to get to work on Friday?
Thus, time taken = distance / average speed
20 mins = distance/(time taken)
If mary's speed for Friday changes, time taken will also change.
$$average\ speed=\frac{dis\tan ce}{20\ \min s}$$
Statement 1 does not provide sufficient information to calculate the average speed. Hence, statement 1 is INSUFFICIENT.
Statement 2; Mary's average speed on her trip to work was 25% greater on Thursday than it was on Friday.
$$\left(average\ speed\ on\ Thursday-Friday\right)=25\%=\frac{dis\tan ce}{20\ \min s}$$
$$25\%=\frac{dis\tan ce}{time},\ and\ time=\frac{dis\tan ce}{25\%}$$
No specific information about the distance covered. Hence, statement 2 is INSUFFICIENT.

Combining statement 1 and 2 together;
Mary's speed on Thursday = distance / 20 minutes

Mary's speed on Friday = distance / (time * 25%)

$$\frac{d}{20}=\frac{d}{t\cdot25\%}$$
$$\frac{d}{20}=\frac{d}{t\cdot0.25}$$
$$\frac{\left(t\cdot0.25\cdot d\right)}{d}=\frac{\left(20\cdot d\right)}{d}$$
$$t=\frac{20}{0.25}=80\min s$$
Neither statement 1 nor statement 2 is SUFFICIENT alone but both statement combined together is SUFFICIENT

Therefore, ANSWER = OPTION C
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