CAT Averages,

Question

1.       5. Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90 and 91-100. The number of students who have scored in each range is given below

51 -60 - 3 students

61 -70 - 8 students

71 -80 - 7 students

81 -90 - 4 students

91 -100 - 3 students

Furthermore, we know that at least as many students scored 76 or more as those who scored below 75. What is the minimum possible average overall of this class?

1. 72 
2. 71.2 
3. 70.6 
4. 69.2 

Answer: Choice (C)

Explanatory Answer:

Let's employ the idea of a total of 25 students (all of the same weight) sitting on a see-saw, which has numbers from 51 to 100 marked on it. At least as many students are sitting on 76 (or to its right), as there are sitting to the left of 75. Now this means that you can have only one person sitting to the left of 75 and all the rest sitting beyond 76. But you can't do that, as you have other constraints as well.

First of all, you have to seat 3 students from 51 to 60, and 8 students from 61 to 70. Secondly, you also have to make sure that the average is the least. This means that the see-saw should be tilting as much to the left as possible, which in turn means that the number of people sitting to the left of 75 should be the highest possible.

This makes it 12 students to the left of 75, and the remaining 13 students on 76 or to its right.

Next, how do you ensure that the average is least, i.e. how do you ensure that the balance tilts as much as possible to the left? Make each student score as little as possible given the constraints.

So, the first 3 students only score 51 each. The next 8 students score only 61 each. 11 Students are now fixed. The 12^th student has to be below 75, so seat him on 71. The remaining 6 students (who are in the 71 to 80 range) have to score 76. The next 4 score 81 and the next 3 score 91.

This would give you the least average.

The lowest possible average would be:

= [(3 * 51) + (8 * 61) + 71 + (6 * 76) + (4* 81) +  (3 * 91)]/25 

= 70.6 

Answer Choice (C)

Mean Median - Final few questions

Few more questions on Mean, median. Have just tried to construct a few Yes/No questions to try to build some intuition. These questions are better-answered by constructing examples/counter-examples.

1. The median of n distinct numbers is greater than the average, does this mean that there are more terms above the average than below it?

2. In a sequence of 25 terms, can 20 terms be below the average? Can 20 terms be between median and average?

3. From 10 numbers, a,b,c,...j, all sets of 4 numbers are chosen and their averages computed. Will the average of these averages be equal to the average of the average of the 10 numbers?

There is one further question that does not come under this classification, but an interesting one nevertheless

Q. The 64 squares of a chessboard are filled with natural numbers from 1 to 100. Multiples cells can have the same number (Any number of cells can have the same number). If the value in any square is equal to the average of the numbers in all the squares around it (either 3, 5 or 8 squares, depending on the position of the square), in how many ways can the chessboard be filled?

This is a fairly difficult question, and not exactly a question for CAT prep, but a fun question nevertheless. (This question is an adaptation from one of the Olympiad questions. Have not been able to locate the exact one yet.Will send in a post once I locate the correct Olympiad)

Happy cracking.

Mean Median (Averages) - Solutions

Have given below the solutions to the questions on mean and median .

Q. Consider 4 numbers a, b, c and d. Ram figures that the smallest average of some three of these four numbers is 30 and the largest average of some three of these 4 is 40. What is the range of values the average of all 4 numbers can take?

We can assume a, b,c d are in ascending order (with the caveat that numbers can be equal to each other)

a + b + c = 90
b + c + d = 120

We need to find the maximum and minimum value of a + b + c + d.

a + b + c + d = 120 + a. So, this will be minimum when a is minimum. Given a + b + c = 90. a is minimum when b + c is maximum. If b + c is maximum, d should be minimum. Given that b + c + d = 120, the minimum value d can take is 40 as d cannot be less than b or c. The highest value b +c can take is 80, when b = c= d = 40. When b = c = d = 40, a = 10. a + b + c + d = 130. Average = 32.5

Similarly, a + b + c + d = 90 + d. So, this will be maximum when d is maximum. Given b + c + d = 120. d is maximum when b + c is minimum. If b + c is minimum, a should be maximum. Given that a + b + c = 90, the maximum value a can take is 30 as a cannot be greater than b or c. The lowest value b +c can take is 60, when a = b= c = 30. When a = b = c = 30, d = 60. a + b + c + d = 150. Average = 37.5

So, the average has to range from 32.5 to 37.5

Q. The average of 5 distinct positive integers if 33. What are the maximum and minimum possible values of the median of the 5 numbers if the average of the three largest numbers within this set is 39?

Let the numbers be a, b, c, d, e in ascending order. a + b + c + d + e = 165. Average of the three largest numbers is 39, so c + d + e = 117, or a + b = 48.

We need to find the maximum and minimum possible values of c.

For minimum value, a and b have to be minimum. a +b = 48. Let us assume a =23, b =25, ca can be as low as 26.
23, 25, 26, 45, 46 is a possible sequence that satisfies the conditions.

For maximum value, we need d + e to be minimum as c + d + e = 117. we can have c = 38, d =39 and e =40. Or, the maximum value c can take = 38

23, 25, 38, 39, 40 is a possible sequence that satisfies the conditions specified


Q. Consider 5 distinct positive numbers a, b, c, d, and e. The average of these numbers is k. If we remove b from this set, the average drops to m (m is less than k). Average of c, b, d and e is K. We also know that c is less than d and e is less than k. The difference between c and b is equal to the difference between e and d. Average of a, b, c and e is greater than m. Write down a, d, c, d and e in ascending order.

Average of a, b, c, d and e is k, Average of b, c, d and e is also k, this implies that a = k
If we remove b, the average drops, this implies that b is higher than the average
e is less than k, or e is less than a. c is less than d.

e < a < b, c < d

Average of a, b, c and e is greater than average a, c, d and e. This tells us that d < b

From this, we get that b is the largest number

b - c = d - e
b + e = c + d
a + b + c +d + e = 5a
Or, b + c + d + e = 4a

b + e = 2a, c + d = 2a. Or, e, a, b is an AP, c, a, d is an AP.

e is the smallest number (as b is the largest number)

Or, the ascending order should be e c a d b

Mean Median - Solutions to Basic Questionsn

Have given below the solutions to the basic questions in Mean Median .

Averages online

View more presentations from rajeshb1980.

Averages - Questions

Have given below few questions from averages. These are slightly more challenging than the basic questions

1. Consider 4 numbers a, b, c and d. Ram figures that the smallest average of some three of these four numbers is 30 and the largest average of some three of these 4 is 40. What is the range of values the average of all 4 numbers can take?
2. The average of 5 distinct positive integers if 33. What are the maximum and minimum possible values of the median of the 5 numbers if the average of the three largest numbers within this set is 39?
3. Consider 5 distinct positive numbers a, b, c, d, and e. The average of these numbers is k. If we remove b from this set, the average drops to m (m is less than k). Average of c, b, d and e is K. We also know that c is less than d and e is less than k. The difference between c and b is equal to the difference between e and d. Average of a, b, c and e is greater than m. Write down a, d, c, d and e in ascending order.

Mean Median Basic questions

Have given a few very simple questions in mean, median. Will post more challenging questions through the week.

1. The average revenue of company A in a year was Rs.30,000 a month. During the first 5 months, the company saw an average revenue of Rs. 25,000 per month and in the last 6 months, the company saw an average of Rs.33,000 per month. What was the revenue level in the 6th month?

2. The captain of a cricket squad of 14 players is 28 years old. The wicket-keeper is 32 years old. If these two players are replaced by two younger players, the average age of the team drops by 1 year. What is the average age of the two new players.?

3. The average age of the students of class A comprising 25 students is 19 years and that of the students of class B comprising 15 students is 11 years. What is the average age of the students of the two classes taken together?

4. The average weight of a class comprising 29 students is 40 kgs. If the weight of the teacher is also included, the average weight increases by 0.5 kgs. What is the weight of the teacher?

5. 30 friends go out for dinner. 12 of them spend Rs.32.5 each for the dinner while the rest of them spend Rs.3 more than the average expense of all 30. What was the total money spent?

6. The range of the height of male members in a team is 9 cms, and that of the female members in the team is 8 cms. What is the range of the height of the team if the shortest man in the team is 3 cms taller than the shortest woman in the team?

7. The median of 68, 93, 109, x, y, and 97 is 97. What is the least possible average of x and y?

Happy cracking