*Thursday, May 31st, 2018*

Average is mainly defined as the sum of observations divided by the number of observations. Keeping in view the Quantitative Aptitude section of any competitive exam and especially CAT, 1-2 questions are asked from this chapter. It is not a difficulty chapter if you know the tricks to solve the question in very less time.

**Average= (Sum of observations)/(Number of observations)**

**Important Points: **In past year Cat questions, a number of times concepts of age/height/weight and average has been mixed together. Here are some important points to remember,

- When a person replaces another person then:
- If the average is increased, then

Age of new person= Age of person who left + (Increase in average * total number of persons) - If the average is decreased, then

Age of new person= Age of person who left – (Decrease in average * total number of persons)

- If the average is increased, then
- When a person joins the group:
- In case of an increase in average,

Age of new member= Previous average+ (Increase in average * Number of members including new member) - In case of decrease in average,

Age of new member= Previous average- (Decrease in average * Number of members including new member)

- In case of an increase in average,
- In the Arithmetic Progression there are two cases:
- When the number of terms is odd – the average will be the middle term.
- When number of terms is even – the average will be the average of two middle terms.

- Sum of 1
^{st}n consecutive natural numbers = [n(n+1)]/2

Average of 1^{st} n consecutive natural numbers = (n+1)/2

To understand the concept properly, let’s try doing some questions.

**Example 1:** The average weight of a class of 24 students is 36 years. When the weight of the teacher is also included, the average weight increases by 1kg. What is the weight of the teacher?

**Solution:** We know that, in case of increase in average,

Age of new member= Previous average+ (Increase in average X Number of members including new member)

Therefore,

Age of new member = 36 + (1 × 25)

Age of new member = 36 + 25 = 61

**Example 2: **Find the average of first 30 natural numbers.

**Solution: **Average of first n natural numbers = [(n+1)/2]

Therefore, Average=((30+1))/(2)

= (31)/(2) = 15.5

**Example 3**: ** **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 to 60 – 3 students, 61 to 70 – 8 students, 71 to 80 – 7 students, 81 to 90 – 4 students, 91 to 100 – 3 students.

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

A. 72 B. 71.2 C. 70.6 D. 69.2

**Solution: **Let us look at the constraints and some hints here.

- There would be total of 25 students in the class. (3+8+7+4+3)
- Number of students who scored less than 75 is less than total number of students who scored more than 75.
- We have to make the average minimum. This means that if it is given, 3 students between 51-60, we should assume there score to be 51 only.

Based on these, let’s see how we can find the average to be minimum.

- There are three students in 51-60 range. To make the average minimum, we assume that all the three have got 51 marks.
- There are eight students in 61-70 range. To make the average minimum, we assume that all the three have got 61 marks.
- We have to make sure that the average is the least. This means that the numbers 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 we ensure that the average is least, i.e. how do we 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 12th 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

**Example 4: **Natural numbers 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The medians of these 5 groups are A, B, C, D and E. If the average of these medians is m, what are the smallest and the largest values m can take?

**Solution: **Let us try to construct a scenario for getting the smallest value of the average. We need to have the minimum value for each of the 5 medians.

Now, what is the lowest value the median of any of the 5 groups can take?

The median is the middle term among the 5 terms. So, it cannot be 1 or 2, but it can be 3. If the set were 1, 2, 3, 4, 5, the median would be 3. If we choose a sub-group such as this, the smallest median from the next group will be 8. The next group could be 6, 7, 8, 9, 10. The idea is to create sub-groups in such a way so that the second group can have a smaller median than 8.

Now, even a sub-group that has the numbers 1, 2, 3, 24, 25 would have the median as 3. So, if we want to keep the medians as small as possible, we should choose sub-groups that have small numbers till the median, and then very large numbers. Because, these large numbers do not affect the median, we will still have small numbers to choose from for the next group. So, choose sub-groups such that the first 3 numbers are small, the final two are as large as possible.

1, 2, 3, 24, 25

4, 5, 6, 22, 23

7, 8, 9, 20, 21

10, 11, 12, 18, 19

13, 14, 15, 16, 17

would be an ideal set of groups. The medians would be 3, 6, 9, 12, 15, and the average of the medians would be 9.

Similar set of groups can be found to find the highest value of the average. The medians would be 23, 20, 17, 14, 11 and the average would be 17. Groups would be as below:

1, 2, 23, 24, 25

3, 4, 20, 21, 22

5, 6, 17, 18, 19

7, 8, 14, 15, 16

9, 10, 11, 12, 13

Correct Answer: Smallest value: 9, Highest Value: 17

**Example 5:** The average score in an examination of 10 students of a class is 60. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 40 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is……

**A. 99 B. 100 C. 87 D. 95**

**Solution: **

- Since the average is 60, total marks scored is 600
- None of the students have scored less than 40 or higher than 100.
- New average for remaining 5 students is 55. Therefore, total marks for remaining 5 students would be (55*5) = 275
- Thus, Sum of marks for top 5 students = (600-275) = 325
- Let’s call them a, b, c, d and e. Now, in order to maximize what the top scorer “e” gets, all the others have to get the least possible scores (and at the same time, they should also get distinct integers.)
- The least possible score of the top 5 should be at least equal to the highest of the bottom 5.
- Now we want to make sure that the highest of the bottom 5 is the least possible. This can be done by making all scores equal to 55. If some scores are less than 55, some other scores have to be higher than 55 to compensate and make the average 55. Thus the highest score is the least only when all of the scores in bottom 5 is 55.
- So now, we have the lowest value that the top 5 can score, which is 55. The others have to get distinct integer scores, and as few marks as possible, so that “e” gets the maximum.
- So, 55 + 56 + 57 + 58 + e = 325
- e = 99 marks.

**Example 6: **Consider a sequence of seven consecutive integers. The average of first five integers is n. The average of all seven integers is:

- a) n
- b) n + 1
- c) k x n, where k is a function of n
- d) n +(2/7).

** Solution: **Let’s solve this by assumption. Let the 7 consecutive integers be 1,2,3,4,5,6,7

Average of first five integers = 3

Average of first seven integers = 4.

∴ Answer is option (b)

**Alternative Method:**

Let a set of any 5 integers be a, a+1, a+2, a+3, a+4

Here, the average will be the middle term i.e. a+2.

It is given that a+2= n.

Now, similarly, a set of 7 integers can be expressed as a, a+1, a+2, a+3, a+4, a+5, a+6.

In this set, the average will be the middle term i.e. a+3.

The term a+3 can be written as; a+2+1.

As a+2= n,

∴a+3= n+1.

So, the answer is option (b).

**Example 7:** A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

- 240 meters
- 360 meters
- 420 meters
- 600 meters

**Solution: **When the train crosses a man standing on a platform, the distance covered by the train is equal to the length of the train.

However, when the same train crosses a platform, the distance covered by the train is equal to the length of the train plus the length of the platform.

The extra time that the train takes when crossing the platform is on account of the extra distance that it has to cover = length of the platform.

Therefore, length of the platform = speed of train * extra time taken to cross the platform

Length of platform = 72 kmph * (30-18) seconds = 72 kmph * 12 seconds

Converting 72 kmph into m/sec, we get 72 kmph = (5/18) * 72 m/sec = 20 m/sec

Therefore, length of the platform = 20 * 12 = 240 meters.

**Conclusion: **The concept of averages has various applications while solving different quantitative aptitude question. Some of the important application is mentioned above along with examples. Concept of average is often combined with various other concepts such as age, speed and distance, natural numbers etc. If we know the basics and have solved some of the CAT questions on these topics, it is not difficult to solve average related problems.

You can also see: Divisibility Rules for CAT Quantitative Aptitude Preparation

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