12 Days of Christmas GMAT Competition - Day 9: In the first step of a

In the first step of a scientific experiment, 400 milliliters of a certain solution were divided among four empty containers.

What was the median volume of solution among the containers?

Total volume of solution = 400 millilitres
Number of containers = 4
Let's assume that volume of solutions in 4 containers = {x,y,z,w}; where x<=y<=z<=w
x+y+z+w = 400
Average volume = (x+y+z+w)/4 = 400/4 = 100 milliliters

(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
w - x = 2 (w-100) = 2w - 200
x + w = 200
y + z = 400 - x+w = 200
Medium volume = (y + z)/2 = 100
SUFFICIENT

(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
Median volume = (y + z)/2
w = (y+z)/2 + 50
2w = y+z + 100
x+y+z+w = 400
Value of Median = (y+z)/2 can not be derived from the given information
NOT SUFFICIENT

IMO A

Let the distribution be a, b, c, 400 - a - b - c, sorted increasingly

Case 1:
400 - 2a - b - c = 2(400-a-b-c - 100)
b + c = 200
b + c/2 = 100 => median volume,

Case 2:
400-a-b-c = 50 + (b+c)/2
800 - 2a -2b - 2c = 100 + b + c
700 = 2a + 3b + 3c

We can't determine a from here

Hence A

Let's analyze each statement to determine if we can find the median volume of solution among the four containers.

Given information: - Total volume: 400 mL - Number of containers: 4

We need to find the median volume

Statement (1): The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
Let's analyze this statement - Average volume = 400 mL ÷ 4 = 100 mL - Let the greatest volume be x mL - Range = 2(x - 100) mL This statement alone doesn't give us enough information to determine the median volume. We know the relationship between the range and the greatest volume, but we can't pinpoint the exact volumes in each container.

Statement (2): The container with the greatest volume of solution had 50 milliliters more than the median volume.
This statement directly relates the greatest volume to the median: - Greatest volume = Median + 50 mL However, this statement alone is not sufficient to determine the exact median volume. We need more information about the other volumes or their relationships.

Now, let's combine both statements: From (2), we know that the greatest volume is 50 mL more than the median. From (1), we can set up an equation: 2(x-100) =x - median, where x is the greatest volume Substituting the information from (2) into this equation: 2(median + 50) - 100) = (median + 50) - median 2(median - 50) = 50 median -50 - 25 median = 75 mL

Therefore, the median volume is 75 mL.

Conclusion: The correct answer is (C) - Statements (1) and (2) TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient

Total initial solution: 400 mL
Divided into 4 empty containers

T.F. median volume of solution among the containers

I. Highest volume - lowest volume = 2 (highest volume - average volume)
=> 2 average volume - lowest volume = highest volume
=> average volume = (lowest volume + highest volume)/2
=> this means the volumes are in a sequence - this is of the form (a+l)/2

So, a-d, a , a+d, a+2d = 400
4a + 2d = 400
a + d = 200
Average = 400/4 = 100
2*100 = a+2d + a-d = 2a + d

Median = (2a + d)/2 = 200/2 = 100
=> Sufficient

II. Greatest = median + 50
=> median = greatest - 50
=> not sufficient

>> Answer is A). Statement I alone is sufficient

We know since the overall sum = 400ml the avg = 100ml.
let H be highest and L be lowest

Considering st 1)
(H-L)=2*(H-100)
i.e H+L=200. Do we know anything about median?
WELL YES! since there are 4 terms median would be (2nd + 3rd)/2

since 1st and 4th term sum to 200, and total is 400 then (2nd + 3rd)=200.
i.e median =100. SUFFICIENT

St 2) let (2nd + 3rd)=y

L + y + H = 400
L + y + (y/2 + 50) = 400 as given H =(y/2 + 50)
on simplification 2L+3y=700 and there can be more than one value for y=> more than one value for median.
eg. L,y = 50,200
L,y=47,202
Hence NOT SUFFICIENT

Let H be highest
L be lowest
Avg as average
A,B,C,D as liquid in containers with d>c>b>a

Statement 1

H-L = 2(H-Avg) .. eqn 1
Avg is 400/4 = 100
Median is (B+C)/2

eqn 1 gives 200 = H+L
Then B & C container has 200 ml liquid
Median is 100

Sufficient

Statement 2
(A+B)/2 + 50 = H

We don't know the values of these variables
Hence, not sufficient

IMO A

Let the solutions in 4 container be a,b,c,d where a<=b<=c<=d
a+b+c+d = 400
Mean = 400/4 = 100
Median is (b+c)/2, so we need to know b+c

Statement 1:
d-a = 2(d-100)
a+d = 200
This means b+c = 200
Statement is sufficient

Statement 2:
(b+c)/2 = d-50
b+c = 2d - 100

a+2d-100+d = 400
a+3d = 500
On solving it, we get a = 50, d = 150 (given a<=d, other values do not fit)
Then b+c = 400-50-150=200
Statement is sufficient

Answer is C

let order be a,b,c,d
a+b+c+d= 400
avg=100
Median= b+c/2

(1) D-a= 2(d-100)
= d-a= 2d-200
d+a=200
a+b+c+d= 400
So b+c=200
Median= b+c/2
median=100

A sufficient

(2) let median=X
A,X,X+50
A+X+X+50=400
A+2X=350

If A= 50 X=150
If A= 10, X= 170

Not sufficient

Ans A

Lowest -- x -- y -- Highest

Median = ( x + y ) / 2

1)
Range = H - L

H - L = 2(H - 100)

H = 200 - L

H + L = 200

Hence, we can conclude that x + y = 400 - 200 = 200

Median = 100

The statement alone is sufficient. We can eliminate B, C, and E.

2) The container with the greatest volume of solution had 50 milliliters more than the median volume.

We can have various combinations

50 100 100 150 -- median = 100

85 85 90 140 -- median some other value than 100

The statement alone is not sufficient.

Option A

Let's call the container's volumes a,b,c,,d and a<b<c<d.

So the median is (b+c)/2. and we know a+b+c+d =400.

Stmt (1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.


Range = d-a,
So according to this stmt.

d-a = 2(d - (a+b+c+d)/4) => b/2+c/2 = a/2+d/2. Which means a+d = b+c.

We already know a+b+c+d = 400 so a+d=200, b+c=200. So the median is 100. Stmt is sufficient.

Stmt (2) The container with the greatest volume of solution had 50 milliliters more than the median volume.

d = 50+b/2+c/2. => d=50+m/2. & a+2m+d = 400 We have 2 equations and 3 variables. Hence Stmt 2 is not sufficient.


IMO A

Total solution = 400ml
Divided into 4 containers, calculate median?

Statement 1 - R = 2*(G - Avg)

R = G - L (greatest - lowest)
Avg = 400/4 = 100

G - L = 2*(G - 100)
G + L = 200

This means that the sum of other 2 solutions is 200, and median is the avg of middle two solutions which would be (400 - 200)/2 = 100

Hence this statement is sufficient.

Statement 2 - G = Median + 50

Let's take few cases to validate this -

50 100 100 150
80 90 90 140

As we can have multiple median values, this is insufficient.

Answer: A

In the first step of a scientific experiment, 400 milliliters of a certain solution were divided among four empty containers. What was the median volume of solution among the containers?

Assume A, B, C, and D are the four containers in ascending order of Volume.

Median Volume = (B+C)/2;

(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.

Avg Volume = 400/4 = 100

Range of Volumes: D - A = 2 * (D - 100);

D - A = 2D - 200; A+D = 200;

It means B+C = 200; And Median (B+C)/2 = 100; Sufficient.

(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.

D = 50 + (B+C)/2; 2D = 100 + B + C;

A+B+C+D = 100;

However, with these equations also we won't be able to find the values of B+C, hence Not Sufficient.

Let say 4 containers: A B C D (In increasing order)
A + B + C + D = 400

(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
D - A = 2 * (D - (400/4)) = 2D - 200
=> 200 = D + A
=> B + C = 200
=> Median = (B + C)/2 = 100
Statement (1) alone is sufficient.

(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
D = M + 50
B + C = 2M

A + B + C + D = 400
=> A + 3M = 350
=> A = 350 - 3M

Now, A >= 0
=> 3M <= 350
=> M <= 116

Now, D - A >= 0
=> M + 50 >= A
=> M + 50 >= 350 - 3M
=> 4M >= 300
=> M >= 75

=> 75 <= M <= 116
So, we can not get exact value
Statement (2) alone is insufficient

Answer: A

400 ml of certain solution is divided among 4 containers. What was the median volume of solution among the containers?

Let the containers be L,M1,M2 and G, be Least,middle-1, middle-2 and greatest respectively.

Statement-1: (1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
Range=G-L=2(G-Average)

We know that average is 400/4-- 100. Therefore the above equation changes to
--G-L=2(G-100)
--G-L=2G-200
--G+L=200 ---> From this greatest and least are 200, so M1 and M2 will be 200, for median we just need to calculate M1+M2/2 as the no. of observations are 4(even). Median is 100. Sufficient.

Statement 2: The container with the greatest volume of solution had 50 milliliters more than the median volume.

G=(M1+M2/2)+50.
But we don't know what the largest container has. Not sufficient.

Answer is A

Let the volumes in the container be a, b, c, and d, where a to d is in ascending order.

We have to find the median = (b+c)/2

Statement 1

Range is d - a, and the avg. volume is 400/4 = 100

d-a = 2*(d-100)
a = d-200

if d = 250, a = 50 and b + c = 100 (to ensure total vol. of 400)
if d = 275, a = 75 and b + c = 50

Multiple values possible. INSUFFICIENT

Statement 2

d = [(b+c)/2] + 50
if d = 150, b+c = 200 and if d = 100, b+c = 100

Multiple values possible. INSUFFICIENT

Both Together

a=d-200 and d = [(b+c)/2] + 50 => b+c = 2*(d-50)

Subs. in a+b+c+d=400

2d+b+c=600
4d = 700
d=175

b+c=250

Median = 125 SUFFICIENT

Answer C.

Step 1: Restate the Problem
We need to determine the median volume of solution among four containers. Let the volumes of the containers be
a,b,c,d, where: a≤b≤c≤d.
The total volume is: a+b+c+d=400.
The median is: Median= (b+c)/2

Statement (1)
The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
The range is d−a.
The average volume is 400/4 =100
The statement says:
d−a=2(d−100).

Simplify:
d−a=2d−200⟹a=d−200

This relates a and d, but it provides no information about b or c, which are necessary to calculate the median (b+c)/2
Thus, Statement (1) alone is insufficient.

Statement (2)
The container with the greatest volume of solution had 50 milliliters more than the median volume.

The greatest volume is d.
The median is (b+c)/2
​
The statement says: d= (b+c)/2 +50
Rearrange: b+c = 2d−100

While this relates
b+c to d, there is no information about a or d individually.
Without knowing all four variables or their relationships, we cannot calculate the median (b+c)/2
Thus, Statement (2) alone is insufficient.

Combine Statements (1) and (2)
From Statement (1):
a=d−200.

From Statement (2):
b+c=2d−100.

Using the total volume equation:
a+b+c+d=400,
substitute
a=d−200 and b+c =2d−100:

(d−200)+(2d−100)+d=400.

Simplify:
⟹ 4d−300=400⟹4d=700⟹d=175.

Substitute
d=175 into a=d−200:
a=175−200=−25
This result is invalid because
a, being a volume, cannot be negative. The assumptions or constraints of the problem cannot produce a valid distribution of volumes.

Conclusion
The information provided in both statements is inconsistent or insufficient to determine the median volume reliably.
The answer is (E).

The answer is C, together is sufficient.

We have total vol = 400 so average is 400/4 = 100

Statement 1: Is 4 containers are a,b,c,d we have range = d-a

d-a = 2(d-100) ; a=d-200. Insufficient to find the median

Statement 2: greatest volume d=median+50
d= b+c/2+50. which is again insufficient.

Combining two we have,
a+b+c+d = 400
d-200+b+c+d = 400

2d+b+c-200 = 400
2d+b+c = 600
from statement 2
2(b+c/2 + 50 )+b+c =600
2b+2c = 600 - 100
b+c = 500/2 = 250 .

Median b+c/2 = 250/2 = 125. So we have the answer.