M31-29

Question

The range of the heights of the female students in a certain class is 30 centimeters, and the range of the heights of the male students in the class is 40 centimeters. What is the range of the heights of all the students in the class?

(1) The average (arithmetic mean) height of the male students in the class is 10 centimeters greater than the average height of the female students in the class.
 
(2) The shortest female student in the class is 15 centimeters shorter than the shortest male student in the class.

Bunuel · Accepted Answer

If you think about it carefully, you’ll understand that this is not possible.

Statement (2) says: The shortest female student in the class is 15 centimeters shorter than the shortest male student in the class.

Let the shortest female student be x cm. 
 The tallest female student is x + 30 cm (since the range for females is 30 cm). 
 From (2), the shortest male student is x + 15 cm. 
 The tallest male student is (x + 15) + 40 = x + 55 cm (since the range for males is 40 cm).

Clearly, the tallest male student (x + 55) is taller than the tallest female student (x + 30).

Bunuel · Answer

Official Solution:

The range of the heights of the female students in a certain class is 30 centimeters, and the range of the heights of the male students in the class is 40 centimeters. What is the range of the heights of all the students in the class? 
 
(1) The average (arithmetic mean) height of the male students in the class is 10 centimeters greater than the average height of the female students in the class.
 
The averages cannot give the range for the class. Not sufficient.
 
(2) The shortest female student in the class is 15 centimeters shorter than the shortest male student in the class. 
 
This statement implies that the range of heights for all students is equal to the range of heights for male students plus 15 centimeters. Therefore, the height range for all students in the class is 55 centimeters. To clarify this concept further, check the diagram below:
 
 -(Shortest \ female)---(Shortest \ male)---(Tallest \ female)-----(Tallest \ male)- 
 
Sufficient.

Answer: B

dsheth7 · Answer

Hi Buenel.

Please correct me if I am wrong.

Since the shortest girl is 15 cm shorter that the shortest boy and the range of height of boys on 10 greater than the range of girl, thus the tallest student in the class is a boy.

Many thanks.

broall · Answer

You are right. Also, the tallest boy in that class is 25 cm taller than the tallest girl in that class.

manishtank1988 · Answer

My 2 cents:
Hopefully this will help visualize the issue:
1) (w1+…+wf)/f + 10 = (w1+…+wm)/m - Not sufficient

2) Sufficient
|--------fs-----------Ms----------------------fl-----------------------------------Ml------------------
0--------5-----15---20----------------------35----------------------------------60
 |--------------------30---------------|
 |---------------------------------------40-------------------| 
Thanks

shubhgmat · Answer

Nice question that can actually make you take more time if you just keep setting up the equations. It is important to realize that averages don't help you get the minimum and maximum values in a set. Averages only talk about the total sum and the total number of elements.

hydr01 · Answer

Shouldn't Statement 1 be sufficient? If we know that the male's average height is 10 cm taller, and that the male's range is 10 cm taller, then the range of the whole class should be 45 cm.

Examples:
1) Male Range = 70 cm - 30 cm = 40 cm, Average = 50 cm
 Female Range = 55 cm - 25 cm, Average = 30 cm, Average = 40 cm.
 Range of the Whole Class = 70 cm - 25 cm = 45 cm.

2) Male Range = 80 cm - 40 cm = 40 cm, Average = 60 cm
 Female Range = 65 cm - 35 cm = 30 cm, Average = 50 cm
 Range of the Whole Class = 80 cm - 35 cm = 45 cm

3) Male Range = 90 cm - 50 cm, Average = 70 cm
 Female Range = 75 cm - 45 cm, Average = 60 cm
 Range of the Whole Class = 90 cm - 45 cm = 45 cm.

Bunuel · Answer

Males and females in your examples seem to be quite short

OK. Let's assume we have toys. Two males, 50 and 10 cm tall. The range in this case = 50 - 10 = 40 cm and the average = 30. Three females 10, 10, and 40 cm tall. The range in this case = 40 - 10 = 30 and the average = 20. The range of the whole group = 50 - 10 = 40. Two males, 50 and 10 cm tall. The range in this case = 50 - 10 = 40 and the average = 30. Three females 5, 20, and 35cm tall. The range in this case = 35 - 5 = 30 and the average = 20. The range of the whole group = 50 - 5 = 45. As you can see (1) gives the different total range for the class.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Keshav307 · Answer

What if the tallest student in the class is a female. Then Adding 15 cm to the range of male student will not be helpful as the tallest student is not a male.
Please help!

Dbrunik · Answer

I think this is a  high-quality  question and I agree with explanation.

idlch · Answer

I like the solution - it’s helpful.

aayushjha0128 · Answer

How I went about solving this:
Shortest Male (SM), Shortest Female (SF), Tallest Male (TM), Shortest Male (SM)

To find: Range of heights => TM - SF or TF - SM

TM - SM = 40 (i)
TF - SF = 30 (ii)

Option 1: Average = Sum of all (male/female) / total (Male/female). Not related so ignore.
Option 2:
SF = SM - 15 (iii)

(i) + (ii)

TM - SM + TF - SF = 70
TM - SF = 70 + SM - TF
TM - SF = 70 + SF + 15 - TF (substituting iii)
TM - SF = 85 - (TF - SF)
TM - SF = 85 - 30 (Substituting ii)
TM - SF = 55
Therefore range = 55 (Sufficient)

yrapeti · Answer

I don’t quite agree with the solution. How can we assure the tallest male is taller than tallest female

Bunuel · Answer

From (2), F(min) = M(min) - 15.

Female range 30 gives F(max) = M(min) + 15.
 
Male range 40 gives M(max) = M(min) + 40.
 
Then M(max) - F(max) = 25, so the tallest male is taller.
 
Overall range = M(max) - F(min) = 55.

DarkAster · Answer

I like the solution - it’s helpful.

aditibatraa · Answer

I like the solution - it’s helpful.

Fonso72838 · Answer

I like the solution - it’s helpful.

repellatquas · Answer

If you want to try algebraically without forming cases, you can try this -
Suppose the tallest boy and girl are B and G, the shortest boy and girl are b and g, respectively.

We are given that,
G-g= 30 ... (i)
B-b= 40 ... (ii)

Option1. Averages here don't give anything about range. Not Sufficient.

Option 2. g = b-15
Substituting values in (i) to find out whether B is greater or G to find range,

G- (b-15) = 30
G-b = 15 and from (ii), we have B-b =40
Solving both, we get 
B-G = 25 (which implies that B is greater than G). 
Now, we need to find out B-g = B-(b-15) 
= B-b+15 
=55

Sufficient. B

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