PROBLEMS ON AVERAGES

Question 1:
15 numbers are arranged in random order. The average of 15 numbers is 54.  The average of first 8 numbers is 64 and the average of last 8 numbers is 60. Find the 8th number.
Solution:
Average of 15 numbers = 64 => Sum of 15 numbers = 15*54 = 810
Sum of 15 numbers = Sum of first 7 numbers + 8th number (x) + Sum of last 7 numbers = 810
(The framing of above equation is important. Because, we are asked to find the 8th number. So, the sum of 15 numbers can be split up like above)
Average of first 8 numbers = 64 => Sum of first 7 numbers + x (8th number) = 64*8 = 512 ---- (1)
Average of last 8 numbers = 60 => Sum of last 7 numbers + x (8th numbers)= 60*8 = 480 ---- (2)
Adding 1 and 2, Sum of first 7 nos.+ x+ Sum of last 7 nos.+x = 512+480 = 992
=> 810 + x = 992 => x = 992-810 = 182
Question 2:
The average age of 12 players in the Indian Hockey team is the same as it was 18 months ago because an old player has been replaced by a younger player now. Find the difference in the age in the older and the younger player.
Solution:
Sum of ages of 12 players before 18 months (1.5 years) = 12x (Assuming x as the average)
Sum of ages of 12 players after 18 months ( now with the old player) = 12x+ 12*1.5 = 12x+18
Given that the sum of the ages 12 players now with replacement of young player = sum of the ages of 12 players 18 months ago = 12x
Sum of the ages of 12 players with old player - Sum of the ages of 12 players with the young player = Difference between the ages of the players
=> 12x+18 - 12x = 18 which is the required difference.
Question 3:
The average of 13 consecutive numbers is 324. If there is another series of consecutive numbers which starts with the same number as the previous series, then what is the average of first 20 numbers of the series?
Solution:
Note: The average of first n consecutive numbers is given by median
So, n=13 13+1/2 = 7th number (The 7th value is 324)
Since the first 20 numbers of the series also starts with the same number, the median will be the average of 10th and 11th value (The 10th value will be 324+3=327, 11th value will be 328)
Average = (327+328)/2 = 327.5
Question 4:
The average age of 20 women is 49. The age of the youngest woman is 32 years and that of the oldest woman is 56. If two women with ages 45 and 50 leave the group and three women join the group, the average group remains unchanged. What is the average of three women who joined the group later?
Solution:
Average of 20 women = 49 => sum of ages of 20 women = 20*49 = 980
980-45-50+3*a = 49*21 =>885+3*a = 1029 => 3a = 144 => a = 48 years
Question 5:
The arithmetic mean of two positive numbers exceeds their geometric mean by 5. The geometric mean exceeds the harmonic mean by 4. Find a-b.
Solution:
(a+b)/2 - sqrt (ab) =5  => (a+b)/2 = 5+ Sqrt(ab) --- (1)
sqrt(ab) -2ab/ (a+b) =4 --- (2)
Sub 1 in 2
sqrt(ab) - 2/(5+sqrt(ab)) =4 --- (3)
Solving 3 we get, ab=400, sqrt(ab)= 20
Substituting these values in 1, a+b = 50, From the values of ab and a+b, we can see that the values of a and b are 40 and 10. So, the difference a-b = 40-10= 30 (Ans)

CONCEPTS ON AVERAGES FOR CAT- II

MEASURES OF CENTRAL TENDENCY:
These are used to measure the 'central' or 'expected' value of the data set.
Standard deviation and Variance are measures of spread. That is, they indicate how spread the data is.
Five most common measures of central tendency:
1. Arithmetic mean
2. Geometric mean
3. Harmonic mean
4. Median
5. Mode


CONCEPT 4
ARITHMETIC MEAN:
It is the method of calculating standard average, also called as 'mean'. It is used to find the central tendency of any random distribution of numbers.
A.M = (x1+x2+x3+.... +xn) / n
Example:
Company A has shown profit of Rs.30 lakhs, 42, 45, 48 and 52 for rhe last five years
Company B has shown profit of Rs. 28 lakhs, 46, 50, and 57 for the last four years
Which company has shown better profits?
This can be answered by calculating the arithmetic mean of both companies A and B
A.M of A = 43.4 lakhs A.M of B = 45. 25 lakhs => Obviously, Company B had higher profits.
NOTE
1. AM is not a very good measure of central tendency.
Why? the mean tends to be influenced by highest value of the set. Example: Average marks of a set of 10, 12, 12, 12, 13, and 25=> AM= 14  But, 5 out of 6 scores are below than this. Median should be adopted in such cases.


CONCEPT 5
GEOMETRIC MEAN:
G.M of n positive numbers is the mean calculated by taking nth root of the product of these numbers.
GM = (x1*x2*x3*.....*xn)^ (1/n) 
Geometric mean is used to calculate the average growth rates or interest rates.
Example 1:
A certain amount is deposited in a bank. For four years, the bank compounds the interest rate of 10%, 12%, 14% and 15% respectively. Compounded net interest rate = ?
Soln:
If the principal is Rs.100, the amount at the end of first year, P1 = 100* 1.1
Likewise, After 4 years, P4 = 100*1.1*1.12*1.14*1.15 ---- (1)
Another way of expressing P4 is by compound interest formula
P4 = 100 *(1+(r/100))^4--- (2) Equating 1 and 2, we get r = 12.7 %
Example 2:
If the population of a village increases by 10 % in one year and then decreases by 20 % in next year,
The average rate of increase/decrease =? This type of problem can be solved by GM of the percentages.
Soln:
First year increase of 10% =>The value will be 100+10 = 110%
Next year decrease of 20% => The value will be 100-20 = 80%
So, the average rate will be (110%*80*)^(1/2) which is the GM  = 93.8%
Therefore, the average rate of decrease per year = 6.2%
Check: Assume a population of 100 people. In first year, 10% increase will give 110. In second year, 20% decrease of 110 will give 88.  => So, there has been 12% decrease in two years.
Approximately, 6% decrease per year.
Example 3:
You need to take vehicle loan from a bank. The fixed rate is 12% for the last 3 years. The floating rates are 10%,12% and 13% for the last 3 years. Which one is best to adopt.
Soln:
Fixed rate is same = 12%
GM of floating rate = (1.10*1.12*1.13)^(1/3) = 1.116 => Avg floating interest for 3 years = 11.6%
So, adoption of floating rate of interest is best.


CONCEPT 6
HARMONIC MEAN:
Harmonic mean of a set of numbers is given by the following formula.
HM = n / ((1/x1)+(1/x2)+(1/x3)+....(1/xn))
For any two quantities a and b, HM is given by the formula,
HM = 2ab / a+b
NOTE:
1. Harmonic mean is the least of the three Pythagorean means (AM, GM and HM). Why? The HM tends towards the smaller values which is opposite to AM which tends towards larger values
2. HM is used in many situations involving rates and ratios. In such cases, HM gives the best average.
3. Also, HM is used to find average speed when equal distances are covered at different speeds
Example 1:
A person travels from city P to city Q of distance 300 km at an average speed 60 km/hr.
On the way back, from Q to P, he covers the same distance at 50 km/hr. Average speed for entire journey=?
Soln:
Average speed is the HM of the two numbers,
Average speed = (2*60*50)/(50+60) = (6000/110) = 54.54 km/hr
ALTERNATE METHOD TO CALCULATE AVERAGE SPEED
A person plans a trip from city A to city B
Decides to travel half  the distance by flight at speed of 336 km/hr
And half of the remaining distance by train at 60 km/hr
And the rest by car at 40 km/hr  What is the estimated average speed?
Soln:
Notice here, the distances travelled by train and car are equal, now first calculate the average speed of train and car. Step 1=> Ratio of smaller speed to larger speed => 40:60= 2:3 (x:y)
Step 2=> Difference between two speeds |40-60|= 20 kmph
Step 3=> Difference/ (x+y) => 20/5 = 4
Step 4 =>Value obtained by step 3 *x => 4*2 = 8
Step 5=>Add the value obtained in step 4 to smaller speed => 40+8 = 48 kmph
To find the average speed of the trip, 48:336 => 1: 7
Difference = 336-48= 288, Average speed = 48+(288/8)*1 = 84 kmph


CONCEPT 6
RELATION BETWEEN MEANS:
For two numbers a and b,
1. AM = (a+b)/2
2. GM= (a*b)^(1/2)
3. HM = 2ab/(a+b)
4. GM = sqrt (AM*HM)
5. AM, GM and HM will be in geometric progression,  AM>=GM>=HM. This will be useful in inequalities.


CONCEPT 7
WEIGHTED AVERAGES:
The term 'weight' stands for the relative importance that is attached to the values For example, consider the scenario that the average ages of different departments in a company are known, and there is a need to calcuate the combined average. In such case, a simple mean would not be correct because the 'weight' (no of persons) in different departments will vary. So, the weighted average is given by
Weighted average = (w1x1 +w2x2 +w3x3+...+wnxn) / (x1+x2+x3+...+xn) where w1, w2 and w3 are the weights of the respected values.
Example 1:
Weighted average of 1L milk @ Rs. 30, 4L milk @ Rs. 22 , 2L milk @ Rs.25 =?
Weighted average = ((1*30)+(4*22)+(2*25))/ (1+4+2) = Rs. 24
Example 2:
Three math classes X, Y and Z take an algebra test.
Average score of X is 83. Avg score of class Y is 76. Average score of class Z is 85
Average score of class X and class Y is 79. Average score of class Y and class Z is 81.
What is the average score of all the three classes? (CAT 2001)
Soln:
Let no of students in classes X,Y and Z be x, y and z respectively
=> 63x+76y = 79( x+y) --- (1)  ; 76y+85z = 81(y+z) ---- (2)
=>  4x= 3y                                ;  4z= 5y
Combining,  4x= 3y  ;  4z= 5y  by the rule of proportions 
( given that a:b and b:c, To find a:b:c => a*b: b*b: b*c), we get 20x = 15y = 12 z
which is, x:y:z = 3:4:5
The average of three classes= (83*3)+(76*4)+(85*5) /(3+4+5) = 81.5
Example 3:
A final year MBA student gets 50% in exam and 80% in assignments. If the exam should count for 70% of final result and 30% for assignment, what will be final score if the college decide to use weighted harmonic mean to uneven performances? (FMS 2007)
Weighted harmonic mean = (W1+W2)/((W1/X1)+(W2/X2))
                                       = (70+30)/((70/50)+(30/80)) = 56.34% (app)


CONCEPT 7
MEDIAN:
Median is the middle value of the group of numbers arranged in an ascending order or descending order.
1.If the number of values in a given set is odd, then the median will be (n+1)/2 value
Median of 31, 43, 32, 34, 36, 42, 33, 39 and 40.
First arrange in ascending/descending  order => 31, 32, 33, 36, 39, 40, 42, 43 and 45.
No of values = 9 (odd), median = (9+1)/2 = 5th value => 39
2. If the number of values in a given set is even, then there will be two middle values say x and y. Median will be (x+y)//2.
Median of 102, 99, 111, 101, 98, 87, 105, 100.
First arrange in ascending/descending order => 111, 105, 102, 101, 100, 99, 98, 87
No of values = 8(even), median = (8/2) =  average of 4th and 5th value => (101+100)/2 = 100.5
NOTE:
The median of a set of values may or may not be equal to, less than or greater than the mean set of values.
Example 1:
In the data set {2,5,7,8,X}, the arithmetic mean is the same as median. Determine the value of X. Assume X>=8 (JMET 2009)
Soln:
Given that the median is same as Arithmetic mean. So, calculate the median first.
n= 5. Median = (5+1)/2 = 3rd value = 7   =>  Therefore,Arithmetic mean =7 
That is (2+5+7+8+X)/5 = 7 => X =13


CONCEPT 8
MODE:
1. Mode is the number that occurs most frequently in a given set of numbers.
2. If two or more values appear the same number of times, then the data set does not have a unique mode.
3. The mode of a set may or may not be equal to the median or mean of the set of values.
Example 1:
Find the mode of the set 1,2,3,7,9,2,12,2,13,4,10,2,8,2,6 ?
Arranging in ascending order => 1,2,2,2,2,2,3,4,5,6,7,8,9,10,12,13
2 occurs maximum number of times ( 5 times) So, the mode is 2

CONCEPTS ON AVERAGES FOR CAT- I

CONCEPT 1
An average is the most likely middle value of a set of data. It is the mean value of a data set around which the numbers are clustered and hence it is a representative of the set.
Average = Sum of all the values of the set of numbers / Total number of values
Example (JMET 2009)
A student in a business school calculates his cumulative average.
QT and QB were his last two tests.
83 marks in QT increases average by +2 marks
75 marks in QB increases average by +1
Next test is reasoning. If he gets 51 in reasoning, his average will be ---?
Soln:
Average -- a number of tests -- n (Before QT)
After QT
a+2 = (an + 83)/ (n+1) => a+2n = 81 ---- (1)
After OB
a+3= (an + 83 +75)/ (n+2) => 2a+3n = 152 ---- (2)
Solving 1 and 2 we get, n= 10, a = 61
After Reasoning,
Average = (an+83+75+51)/ (n+3) Sub values of a and n,
we get average = 63
NOTE:
1. If the value of each item of the group ----> increases or decreases by the same value x
    The average of the group                  ----> increases or decreases by x
2. This is useful in dealing with ages.  If average age of a group of people is x years
                                                        After n years , the average will be x+n
Why? With each passing year, the age of each person increases by 1
3. If the value of each item of the group ---> multiplied/ divided by the same value x (x not equal to 0)
    average of the group                        ----> multiplied/ divided by x
4. Net deficit due to the numbers below average = Net surplus of the numbers above average.


CONCEPT 2
ALTERNATE METHOD TO CALCULATE AVERAGES:
When it is used? If we have large group of numbers that may lead to tedious calculations.
Another way to call the method is assumed average method.
A simple example: Calculate the average of 102, 105, 92, 103, 96 and 98.
Assume any arbitrary average between 92 and 105 (between smallest value and largest value)
Let us take arbitrary average = 100. Now, find the difference of each number
102- 100 = 2, 105-100= 5. 92-100= -8, 103-100 = 3, 96-100= -4, 98-100= -2.
2. 5. -8, 3, -4, -2 are all called as deviations. Sum of all deviations = -4
Actual average = 100-(4 /6) = 99.33


CONCEPT 3
CALCULATION OF AVERAGES WHEN NEW VALUES ARE ADDED:
When all the values of a set is replaced by the average, the sum will not change.
Consider set of 5 numbers; 43, 51, 74, 60 and 22.
Average = (22+43+51+60+74)/5 = (250/5) = 50
The concept is, if we replace all the 5 values of the set to 50, the sum will still remain 250. (5*50 = 250)
If a new item 68 gets added, then the new average will be (250+68)/(5+1) = 318/6 = 53
A more simple way:  If a new number 68 is added, it is 18 more than the original average.
So, the number should distribute 18 equally to all the 6 numbers including itself ==> 18/6 = 3 to all the values
The new average is 50 +3 = 53
Thus the formula is,
New average after addition = >
Original avg + (Value of newly added item (V) - Original average (O)) / The number of items after addition
If V-O is negative, the the average will reduce.
A converse approach:
Given that the average of 5 numbers is 50 and when another number is added, the average becomes 53.
Find the newly added number.
Ans: (53*6) - (50*5) = 68
or Easy way --> Original average (5 nos) = 50, New average (6 nos) = 53
So, 3 is distributed equally to all the 6 numbers by the new number, (3*6)= 18 --> New no. = 50+18= 68
Example 1:
The average of 4 numbers is 12, 13, 17, and 18 is (12+13+17+!8)/4 = 60/4 = 15
Suppose if a 5th number is added, the average becomes 16.  Find the newly added number.
To solve this, visualise addition of a fifth number, which increases average by 1.
15+ 1 = 16
15+ 1 = 16
15+ 1 = 16
15+ 1 = 16
Thus +1 appearing 4 times is due to the fifth number, which is able to maintain the average as 16 first and then give 'one' to each of the first 4. So, the fifth number is 16+4 = 20.
Example 2:
Average wt of players = 60 kg
If a player of 68 kg joins the group, average weight = 61 kg
No. of players at initial stage = ?
Soln:
Assume, each player weight = 60 kg.
Another player joins, the average = 61 kg (This means he brings extra 8 kg which has to be distrubuted equally.) Since, the average increases by 1, 8 kg would be distributed to 8 players
So, initially there will be 7 players.
Example 3:
Average weight of a school football team ( 22 members including the goal keeper) decreases by half a kilogram when the goalie is not included . Average age initially = 60 kg Goal keeper's weight?
Soln:
After goalie removal, the average will be 60 - 0.5 = 59.5 kg
So, the goalie age will be 0.5*21 = 10.5 kg in excess of the initial average => 10.5+ 60 = 70.5 kg.
Example 4:
The average age of class of 30 students and a teacher reduces by 0.5 years, if we exclude the teacher. If the initial average is 14 years, find the age of the class teacher.
Soln:
The teacher after fulfilling the average of 14 years is also able to give 0.5 years to each of 30 students 
(30*0.5 =15) ==> he has 15 years to give over and maintain his own average of 14 years.
Age of teacher = 14 + (30 * 0.5) = 29 years.
(Note: The problem should be viewed as change in average from 13.5 to 14 when teacher is included)
Example 5:
The average marks of a group of 20 students on a test is reduced by 4 when the topper who scored 90 marks is replaced by a new student. How many marks did the new student have?
Soln:
The new student should have scored (20 *4) 80 marks less than the topper.So, the mark of new student = 90-80 = 10 marks. 
Example 6:
The average age of 3 students A, B and C is 48 marks. 
Another student D joins the group => the new average = 44
Another student E who has 3 more marks than D joins the group => the average = x
Average of 4 students B,C,D and E becomes 43 marks. How much did A get in the exam?
Soln:
Solve while reading
By first sentence, Total marks of 3 students(A+B+C) = 48*3 = 144 marks.
If D, joins  the total marks of 4 students(A+B+C+D) = 44*4 = 176 marks. ---- (1)
176-144= 32 marks is what D has got.
E's mark will be 32+3 = 35 marks.
Total marks of 4 students (B+C+D+E) = 43*4 = 172 marks ----1(2)
Subtracting 1 and 2 we get, A-E = 4 marks. E's mark is 35 => A's mark is 35+4 = 39 marks.

Verbal Ability and Logical Reasoning of CAT – Tips and Techniques

Verbal Ability and Logical Reasoning is the second section in CAT 2011. I would like to share some tips and facts about this section of CAT. These are purely my approaches which I employ in section 2 and this article is based on the different strategies I use. I hope that it would be useful to everyone.

SECTION 2 : (VERBAL ABILITY AND LOGICAL REASONING)

An easy way to score good in verbal part and clear the cut offs is to practice more Reading Comprehension passages. This is because you’ll get more questions on RC rather than any other sub section in Verbal Ability. In Verbal Ability, I give importance to the following sub sections in the given order.

1. Reading comprehension
2. Sentence correction
3. Para jumbles
4. Fill in the blanks
5. Para completion and Critical reasoning

1) Reading Comprehension

As I said, I will tell you the strategies used by me that I find useful.

The first step is to skim through the RC passage to get an idea about the field of the passage. It may be from philosophy, medicine, conversations, technology, etc. If I feel that the topic is completely new to me and difficult to interpret, I’ll skim the questions following the Reading Comprehension Passage. It makes it easier for me to find out the answers when I go back to the RC passage.

If you do not understand any passage, answer the factual answers (if there are any).

Inferential questions such as those that ask the central idea, style and tone of the passage should be answered when you completely understand the passage and it’s theme. If you try to attempt it without understanding the passage, you’ll most probably get it wrong.

You have to work out such techniques and see whether it suits you. And for doing that, you need to practice a lot. Do atleast 3-4 RCs a day and then compare your answers with the official answers and analyse why you have answered incorrectly. Analysing is the most important part. Only by doing a proper analysis can you improve upon your errors.

2) Sentence correction:

I believe that for questions on Sentence Correction, you will need to have a strong foundation with proper grammatical concepts. To crack any type of Sentence Correction, you need to learn the following:

1. Proper usage of verb (tense forms) according to the context.

2. Subject-verb agreement.

3. Spelling mistakes and confusing words.

4. Articles and prepositions.

5. Phrasal verbs and idiomatic usage.

6. Punctuation.

If someone feels that he/she is too weak in cracking Sentence Correction Questions, the reason may lie in the lack of fundamental knowlege of Grammar. So, the need is to work on grammar concepts and learn them.

3) Parajumbles:

For solving Parajumble questions, you need to practice a lot. While solving, try to find the links between the sentences. Thereby, eliminate the options which are too close.

The first step should be to guess the opening sentence. The opening sentence gives you a general idea about the passage. And then try to find out by the links as suggested by the options after eliminating the irrelavant ones. Be active on ParaJumble thread at pagalguy. This is the best source for finding ParaJumble questions.

4) Fill in the blanks:

You need a good vocabulary to crack difficult sets from this section. It cannot be improved much from now if you are preparing for CAT 2011. If you find a Fill in the blank question in which the words are totally new, just leave it. Guessing will mostly get you a wrong answer and fetch you negative marks in this sub section.

5) Para Completion and Critical reasoning:

These are a bit rare type of question these days. But indeed, you can expect 1-2 questions in CAT too. As they say, Expect the Unexpected. So, practice them through PG threads. There is a thread for GMAT critical reasoning discussions in PG.

It is the best one for CR. Both para completion and CR are modified forms of RCs. Once you get good hold in RCs, you’ll definitely be able to answer these type of questions too..

6) Logical reasoning:

I found testfunda books and TIME trishna book for DI and LR to be good on LR. And another best source is the past years’ original CAT papers and AIMCATs, CL mock CATS with explanations for the solutions. Practice is the key to LR. It is like playing a new game in lap/mobile. The more you get addicted to play, the more you’ll unlock the levels and become an expert.

I hope that everyone will be able to increase their score in the Verbal ability and Logical Reasoning section in a month’s time if they follow the aforesaid suggestions properly.