12 Days of Christmas GMAT Competition - Day 2: Jingle, an elf

Let distance travelled on day 1, 2 and 3 is d1, d2, d3 respectively.
d1 d2 d3
Given a) d1+d2+d3 = 1080
Average = 360
Is d2< 360 ?

Statement 1: The range of the distances Jingle covered over these days was equal to his daily average (arithmetic mean) distance.
d3-d1 = 360
so we have,
d1 d2 d1+360
we have to find range of possible value of d2 for which we have to find possible max and min value of d2
for minimizing d2, d2=d1
3d2+360=1080 from a)
d2= 240 which is less than 360 hence supports our question
for maximizing d2, d2=d3
3d2-360 = 1080 from a)
d2= 480 which is more than 360 hence doesn’t support our question

d2 has range of [240,480], hence we can’t ascertain weather it will be less than 360

So A,D is out

Statement 2: On one of the days, Jingle covered 360 miles
So either of d1,d2,d3 is 360 miles and remaining 2 must have sum of 720 miles.
3 possible scenarios,
d2+d3 = 720, d1=360
d1+d3 = 720, d2=360
d1+d2 = 720, d3 = 360
Again for finding range of d2, we can find max and min for d2
For min d2=d1, max d2=d3
For minimum value i.e. d1=d2 in all 3 possible scenarios we get values as follows:
d2+d3 = 720, d1=360 --------> d2 =360
d1+d3 = 720, d2=360 --------> d2 =360
d1+d2 = 720, d3 = 360--------> d2 =360
For max value i.e. d2=d3 in all 3 possible scenarios we get values as follows:
d2+d3 = 720, d1=360 --------> d2 =360
d1+d3 = 720, d2=360 --------> d2 =360
d1+d2 = 720, d3 = 360--------> d2 =360

so possible value of d2 is 360 only, hence we can clearly ascertain that d2 can’t be less than 360 and statement 2 is sufficient

Hence answer is (B)

Daily average = 1080 / 3 = 360

1. Case 1: 180 360 540 ... Here daily average is equal to median. So, No
Case 2: 185 350 545 ... Here daily average is more than median. So, Yes

2. If 360 travelled on day 1, then average will be less than or equal to median, but not greater. If 360 travelled on Day 2, then average is equal to median. It is possible to travel 360 on day three if all the three days are travelled at 360. Hence, in any case average will not be greater than median. So, Yes

Ans: Option B

First we find Avg Daily = 1080/3 = 360

Statement 1 : Avg daily = range, so we found day 1 360, 2 360, 3 360. Avg Speed = Median
so statement 1 is enough to answer the question

Statement 2 : It is not enough to answer

So the answer is A

We need to find if arithmetic mean is greater than median of distance or not. As per the question, arithmetic means should be 1080/3=> 360
Now as per Option A, Daily range is equal to arithmetic mean, => This implies the distance travelled is equal daily. In that case median and arithmetic mean is equal.

As per option B, one of the days he travelled, 360 miles. But we are not sure of remaining days,
1. it might be number greater than 360 (700) then the last day travel will be the number smaller than 360 (20) miles subsequently=> in that case median is equal to arithmetic mean
2. it might be number smaller than 360 (340) then the last day travel will be the number greater than 360 (380) miles subsequently=> in that case also median is equal to arithmetic mean

The correct option is D

With statement 1 :

Keep in mind 3rd no. =2 x 1st no.

Take middle no. To be
Case (1) above 360 :
Then it will 220 420 440

Case (2) below 360 :
Then it will 260 300 520 so statement 1 not sufficient.
Eliminate A D

Statement 2: 1 day it is 360 , for 360 (1080/3) to be the average. The other numbers can't be above 360 ,they should be on either side of 360 to bring the sum to 1080 or else the sum will exceed 1080. So the mean is equal to median and no way greater.
So statement 2 provides the answer.

Answer choice : B

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Jingle, an elf, traveled a total of 1,080 miles over the course of three days. Was his daily average (arithmetic mean) distance greater than the median of the distances he covered on these days?

(1) The range of the distances Jingle covered over these days was equal to his daily average (arithmetic mean) distance.
(2) On one of the days, Jingle covered 360 miles.

Let's say that:
- dx= miles covered on day x, x could be 1 (day 1), 2 (day 2) or 3 (day 3).
- dm = daily average distance covered(arithmetic mean)
- M= median distance covered
If dmin = miles covered the day he did the least miles and dmax= miles covered the day he did the most miles
Then the set can be represented as {dmin, M, dmax}
futhermore we know that:
d1+d2+d3= 1080 ; then dm = 1080/3 = 360.
So the question is: is 360> M or is M<360?


(1) Range = dm
Range is dmax - dmin
Let's take some examples where d1+d2+d3= 1080 (dm always 360) and range is equal to 360 :
- d1= 140; d2=440; d3=500 => dmin=140; M=440; dmax=500 => M>dm
- d1=200; d2=320; d3=560 => dmin= 200; M=320; dmax=560 => M<dm
=> (1) is not sufficient to conclude

(2) Amongst d1, d2 and d3 or amongst dmin, M and dmax one is equal to 360
- as 360 is the mean, it can only be dmin or dmax if 360 miles were covered every other day. The set will be then {360, 360,360} and in that case M=dm and we can have a conclusion...
- otherwise M=360 with dmin and dmax different (we can have {180; 360;540} as an example ) and we will still have a conclusion since M=dm.

So (2) is sufficient

So Answer B

Jingle, an elf, traveled a total of 1,080 miles over the course of three days. Was his daily average (arithmetic mean) distance greater than the median of the distances he covered on these days?

(1) The range of the distances Jingle covered over these days was equal to his daily average (arithmetic mean) distance.
(2) On one of the days, Jingle covered 360 miles.

Solution:
Total distance traveled in 3 days = 1080
Hence, daily average distance = 1080/3 = 360 miles
Let the Median distance be M. Then we need to find if M < 360

Statement 1: The range of the distances Jingle covered over these days was equal to his daily average (arithmetic mean) distance.
Let a , b and c miles be the distance covered each day for 3 days in ascending order.
Hence Median (M) = b
Then Range = c - a

Given that c - a = 360
or c = 360 + a

Also a + b + c = 1080
or 2a + b = 720
or b = 720 - 2a
or M = 720 - 2a

say a = 200, then c = 560
and b or M = 320 which is < 360

say a = 170, then c = 530
and b = 380 which is > 360

Hence Statement 1 is NOT SUFFICIENT

Statement 2: On one of the days, Jingle covered 360 miles.
Let a , b and c miles be the distance covered each day for 3 days.
Given one of the days distance covered is 360 miles say = a
Then b + c = 720
if a = b = c, then median = average daily distance
if a = 360, say b = 1, then c = 719
in this case median is 360 which is = average daily distance

say if b = 359, c = 361
in this case also median is 360 which is = average daily distance

Similarly for all the cases either the median will be = or < average daily distance.

Hence statement 2 is SUFFICIENT

Correct answer is option B

Answer D - each alone is sufficient
Explanation:
Inference from Question
Average Distance traveled per day -> 1080 / 3 = 360 miles
We need to identify whether median of the distance travelled is greater than mean

Statement Analysis
as per statement 1: the range of distances covered is equal to daily average => the median value is equal to mean
as per statement 2 one of the days distance covered is 360 , but for the other 2 days it's not evident whether the distance covered is greater that 360 mies or less than 360 miles
if for day 1 -> 360
day 2 -> less than 360 then day 3 will be greater than 360 thus median is 360, which is equal to the mean

(1) x, y, x+360 where y is median ; y can have other values including 360
(2) if one of the value is 360 which is average, then one should be less than 360 and other should be greater than 360 making 360 as median

so, B alone must be the answer