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Let's denote the tallest male, tallest female, shortest male and shortest female as Tm,Tf,Sm & Sf.

From the question stem it can be seen that :-
Tm - Sm = 40
Tf - Sf = 30

Statement 1:
The average height of the male students in the class is 10 centimeters greater than the average height of the female students in the class.

Case I:
Now think about a scenario where you have just two males having height 190 and 150cms so as to satisfy the condition of range being 40; Avg height = 170cms
And you have just two females in class with height 175 and 145 so as to satisfy the condition of range being 30; Avg height = 160cms.

Now we can see that the average height of the male students in the class is 10 centimeters greater than the average height of the female students in the class.

The range of the class in this case will be 190 - 145 = 45

Case II:
But the condition of average height of male being 10 more than average height of female can be achieved in many ways.
For example there are two males 190 and 150 cms.Range is 40 Avg height is 170.
there are 4 females of heights 180,155,155,150 cms. Range is 30. Avg height is 160.
But now the range of the entire class will be 190 -150 = 40

So the range is now different from previous example.
Since we do not know the number of male and female students we cannot determine range of the entire class! So statement 1 is not sufficient.

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


Sm - Sf = 15
This means the tallest female is 15cms taller than shortest male. And we know the longest male is 40 cms taller than shortest male.
This makes it clear that the tallest student in class is Tm and shortest student in class is Sf
So the range of the class will be Tm - Sf

Now from question stem we know that

Tm - Sm = 40
Tm - (15+ Sf) = 40
Tm - Sf = 55

Range of the class is 55. Statement 2 is sufficient.

Answer B

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Bunuel
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.

Ans: B
Solution: well here we only know the range not the actual values for age.
so lets say shortest Female is F cm and Shortest Male is M cm tall. now because we know the range.
Tallest Female is F+30 cm and Tallest Male is M+40.

1) from statement 1 we can say that only by knowing the arithmetic mean we can not comment on the actual range comparison of the height. plus we have no information about the number of students (total, male, female) so we can not comment on the actual range. [insufficient]

2) from statement 2 we know that F=M-15
now if we put this value in the F+30 we get tallest female as M-15+30= M+15
so now we know the range in terms of M
shortest girl=M-15 and tallest person is the boy who is M+40
range of the class is 55cm. [Sufficient]
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Most of the Value DS questions based on Mean, Median, Mode, and Range intrigues the students in an unusual way. However, following a basic approach and some visual inspection can make work a lot easier.

Let's start by writing down what we know from the question and what the question is asking to find out (Yes, it is important to put down the info on the scratch paper. You can not keep everything in your head and assume that you will be as cool in the last set of questions as you were in the initial set of questions).

Know: Range of Female = 30 Range of All students = ?
Range of Male = 40

Now, think of what we need to get the answer. We need the smallest and the largest height (from the data of all students).

1) Av Male = Av Female + 10
This will not provide us the largest and smallest value. So, insufficient.

2) Shortest Female = Shortest Male - 15

Don't jump on the solution. That's a bad strategy. Writing equations can be one way. But that makes it complicated to comprehend. Let's use number line. It will help improve visual inspection.
You must use superscripts and subscripts while doing these kind of questions. Since, it is not possible here, I'll use following terminology and will show you step by step addition of information on the diagram, so that it is easier for you to understand:
SF = Shortest Female
SM = Shortest Male
TF = Tallest Female
TM = Tallest Male

|-----15-----|
|_________|________________________
SF SM


|----------30--------|
|---15---|-----15---|
|_______|________|________________
SF SM TF


|------------40---------|
|----------30--------|
|---15---|-----15---|
|_______|________|__________|______
SF SM TF TM



|--------------40--------|
|----------30--------|
|---15---|-----15---||------10-----|
|_______|________|__________|______
SF SM TF TM
|------------------55-----------------|

So, the range is 55. Thus, Statement 2 is sufficient.

So, the answer is B. :)

Sorry for the distorted diagrams. This is the best I can do in typing on this forum. :/
Note: the bar for 40 is starting from SM and ends on TM.

Give kudos if you liked the solution. :)


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Averages wont tell you anything about range.

Eg
Male student ages are
10 , 30 , 50
Number = 3 , avg =30 and range is 40

Female students
10, 10, 40
i.e. Number = 3 , avg = 20 and range is 30

Here overall average would be 40

But If the numbers for female students are
5, 20, 35 ---> Overall avg becomes 45 hence not suff

Statement 2 ---> This is anyway sufficient since the values are not overlapping

Hence B
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