AVERAGES - Averages and Mixtures - How to Prepare for Quantitative Aptitude for CAT

How to Prepare for Quantitative Aptitude for CAT (2014)

Block II: Averages and Mixtures

...BACK TO SCHOOL

  • The Relevance of Average

Average is one of the most important mathematical concepts that we use in our day-to-day life. In fact, even the most non-mathematical individuals regularly utilize the concept of averages on a day-to-day basis.

So, we use averages in all the following and many more instances.

  • How a class of students fared in an exam is assessed by looking at the average score.
  • What is the average price of items purchased by an individual.
  • A person might be interested in knowing his average telephone expenditure, electricity expenditure, petrol expenditure, etc.
  • A manager might be interested in finding out the average sales per territory or even the average growth rate month to month.
  • Clearly there can be immense application of averages that you might be able to visualise on your own.
  • The Meaning of an Average

An average is best seen as a representative value which can be used to represent the value of the general term in a group of values.

For instance, suppose that a cricket team had 10 partnerships as follows:

1st wicket 28

2nd wicket 42

3rd wicket 112

4th wicket 52

5th wicket 0

6th wicket 23

7th wicket 41

8th wicket 18

9th wicket 9

10th wicket 15

On adding the ten values above, we get a total of 340—which gives an average of 34 runs per wicket, i.e. the average partnership of the team was 34 runs.

In other words, if we were to replace the value of all the ten partnerships by 34 runs, we would get the same total score. Hence, 34 represents the average partnership value for the team.

Suppose, in a cricket series of 5 matches between 2 countries, you are given that Team A had an average partnership of 58 Runs per wicket while Team B had an average partnership of 34 runs per wicket. What conclusion can you draw about the performance of the two teams, given that both the teams played 5 complete test matches?

Obviously, Team B would have performed much worse than Team A: For that matter, if I tell you that the average daytime high temperature of Lucknow was 18° C for a particular month, you can easily draw some kind of conclusion in your mind about the month we could possibly be talking about.

Thus, you should realize that the beauty of averages lies in the fact that it is one single number that tells you a lot about the group of numbers—hence, it is one number that represents an entire group of numbers.

But one of the key concepts that you need to understand before you move into the chapters of this block is the concept of WEIGHTED AVERAGES.

As always, the concept is best explained through a concrete example.

Suppose I had to buy a shirt and a trousers and let us say that the average cost of a shirt was ` 1200 while that of a trousers was ` 900.

In such a case, the average cost of a shirt and a trousers would be given by (1200 + 900)/2 = 1050.

This can be visualised on the number line as:

(midpoint) = answer

As you can easily see in the figure, the average occurs at the midpoint of the two numbers.

Now, let us try to modify the situation:

Suppose, I were to buy 2 trousers and 1 shirt. In such a case I would end up spending (900 + 900 + 1200) = ` 3000 in buying a total of 3 items. What would be my average in this case?

Obviously, 3000/3 = ` 1000!!

Clearly, the average has shifted!!

On the number line we could visualise this as follows:

(Answer)  (midpoint)

It is clearly visible that the average has shifted towards 900 (which was the cost price of the trousers—the larger purchased item.)

In a way, this shift is similar to the way a two pan weighing balance shifts on weights being put on it. The balance shifts towards the pan containing the larger weight.

Similarly, in this case, the correct average (1000) is closer to 900 than it is to 1200. This has happened because the number of elements in the group of average 900 is greater than the number of elements in the group having average 1200. Since, this is very similar to the system of weights, we call this as a weighted average situation.

At this stage, you should realize that weighted averages are not solely restricted to two groups. We can also come up with a weighted average situation for three groups (although in such a case the representation of the weighted average on the number line might not be so easily possible.) In fact, it is the number line representation of a weighted average situation that is defined as alligation (when 2 groups are involved).

Pre-assessment Test

1.

X’s age is 1/10th of Y’s present age. Y’s age will be thrice of Z’s age after 10 years. If Z’s eighth birthday was celebrated two years ago, then the present age of X must be

(a) 5 years

(b) 10 years

(c) 15 years

(d) 20 years

2.

Dravid was twice as old as Rahul 10 years back. How old is Rahul today if Dravid will be 45 years old 15 years hence?

(a) 20 years

(b) 10 years

(c) 30 years

(d) None of these

3.

A demographic survey of 100 families in which two parents were present revealed that the average age A, of the oldest child, is 15 years less than ½ the sum of the ages of the two parents. If X represents the age of one parent and Y the age of the other parent, then which of the following is equivalent to A?

(a) – 15

(b) + 15

(c) – 15

(d) X + Y – 7.5

4.

If 10 years are subtracted from the present age of Randy and the remainder divided by 12, then you would get the present age of his grandson Sandy. If Sandy is 19 years younger to Sundy whose age is 24, then what is the present age of Randy?

(a) 80 years

(b) 70 years

(c) 60 years

(d) None of these

5.

Two groups of students, whose average ages are 15 years and 25 years, combine to form a third group whose average age is 23 years. What is the ratio of the number of students in the first group to the number of students in the second group?

(a) 8 : 2

(b) 2 : 8

(c) 4 : 6

(d) None of these

6.

A year ago, Mohit was four times his son’s age. In six years, his age will be 9 more than twice his son’s age. What is the present age of the son?

(a) 10 years

(b) 9 years

(c) 20 years

(d) None of these

7.

In 1952, I was as old as the number formed by the last two digits of my birth year. When I mentioned this interesting coincidence to my grandfather, he surprised me by saying that the same applied to him also.

The difference in our ages is:

(a) 40 years

(b) 50 years

(c) 60 years

(d) None of these

8.

The average age of three boys is 18 years. If their ages are in the ratio 4:5:9, then the age of the youngest boy is

(a) 8 years

(b) 9 years

(c) 12 years

(d) 16 years

9.

“I am eight times as old as you were when I was as old as you are”, said a man to his son. Find out their present ages if the sum of their ages is 75 years.

(a) 40 years and 35 years

(b) 56 years and 19 years

(c) 48 years and 27 years

(d) None of these

10.

My brother was 3 years of age when my sister was born, while my mother was 26 years of age when I was born. If my sister was 4 years of age when I was born, then what was the age of my father and mother respectively when my brother was born?

(a) 35 years, 33 years

(b) 35 years, 29 years

(c) 32 years, 23 years

(d) None of these

11.

Namrata’s father is now four times her age. In five years, he will be three times her age. In how many years, will he be twice her age?

(a) 5

(b) 20

(c) 25

(d) 15

12.

A father is twice as old as his daughter. 20 years back, he was seven times as old as the daughter. What are their present ages?

(a) 24, 12

(b) 44, 22

(c) 48, 24

(d) none of these

13.

The present ages of three persons are in the proportion of 5:8:7. Eight years ago, the sum of their ages was 76. Find the present age of the youngest person.

(a) 20

(b) 25

(c) 30

(d) None of these

14.

The average age of a class is 14.8 years. The average age of the boys in the class is 15.4 years and that of girls is 14.4 years. What is the ratio of boys to girls in the class?

(a) 1 : 2

(b) 3 : 2

(c) 2 : 3

(d) None of these

15.

In an organisation, the daily average wages of 20 illiterate employees is decreased from ` 25 to ` 10, thus the average salary of all the literate (educated) and illiterate employees is decreased by ` 10 per day. The number of educated employees working in the organisation are:

(a) 15

(b) 20

(c) 10

(d) 25

16.

Mr.Akhilesh Bajpai while going from Lucknow to Jamshedpur covered half the distance by train at the speed of 96 km/hr, half the rest of the distance by his scooter at the speed of 60 km/hr and the remaining distance at the speed of 40 km/hr by car. The average speed at which he completed his journey is:

(a) 64 km/hr

(b) 56 km/hr

(c) 60 km/hr

(d) 36 km/hr

17.

There are four types of candidates in AMS Learning Systems preparing for the CAT. The number of students of Engineering, Science, Commerce and Humanities is 400, 600, 500 and 300 respectively and the respective percentage of students who qualified the CAT is 80%, 75%, 60% and 50% respectively the overall percentage of successful candidates in our institute is:

(a) 67.77%

(b) 66.66%

(c) 68.5%

(d) None of these

18.

Mr. Jagmohan calculated the average of 10 ‘Three digit numbers’. But due to mistake he reversed the digits of a number and thus his average increased by 29.7. The difference between the unit digit and hundreds digit of that number is:

(a) 4

(b) 3

(c) 2

(d) can’t be determined

Directions for Questions 19 and 20: Answer the questions based on the following Information.

Production pattern for number of units (in cubic feet) per day.

Days

1

2

3

4

5

6

7

Numbers of units

150

180

120

250

160

120

150

For a truck that can carry 2,000 cubic feet, hiring cost per day is `1,000. Storing cost per cubic feet is `5 per day. Any residual material left at the end of the seventh day has to be transferred.

19.

If all the units should be sent to the market, then on which days should the trucks be hired to minimize the cost:

(a) 2nd , 4th , 6th , 7th

(b) 7th

(c) 2nd , 4th , 5th , 7th

(d) None of these

20.

If the storage cost is reduced to ` 0.9 per cubic feet per day, then on which day/days, should the truck be hired?

(a) 4th

(b) 7th

(c) 4th and 7th

(d) None of these

ANSWER KEY

1. (a)

2. (a)

3. (a)

4. (b)

5. (b)

6. (b)

7. (b)

8. (c)

9. (c)

10. (d)

11. (b)

12. (c)

13. (b)

14. (c)

15. (c)

16. (a)

17. (a)

18. (b)

19. (a)

20. (b)

SCORE INTERPRETATION ALGORITHM FOR PRE-ASSESSMENT TEST OF BLOCK II

If You Scored: < 12: (In Unlimited Time)

Step One

Go through the first chapter of the block, viz., Averages. When you do so, concentrate on clearly understanding each of the concepts explained in the chapter theory.

Then move onto the LOD 1 exercises. These exercises will provide you with the first level of challenge. Try to solve each and every question provided under LOD 1. While doing so do not think about the time requirement .Once you finish solving LOD 1, revise the questions and their solution processes.

Also at this stage study the concept of averages from your school text books (Class 8, 9 & 10) and solve all the questions which are available to you in those books.

Step Two

After finishing LOD 1 of Averages, move into LOD 2 and then LOD 3 of this chapter.

Step Three

Go to the chapter of alligations and study the shortcuts provided carefully. Understand the use of the alligation process as clearly as possible.

Then move to the LOD 1 exercise of the same. (Note: The chapter of alligations does not have an LOD2 & LOD 3 exercise)

Step Four

Go to the practice test given at the end of the block and solve it. While doing so, first look at the score you get within the time limit mentioned. Then continue to solve the test further without a time limit and try to evaluate the improvement in your unlimited time score.

In case the growth in your score is not significant, go back to the theory of each chapter and review each of the questions you have solved for both the chapters.

If You Scored: > 12 (In Unlimited Time)

Follow the same process as above. The only difference is that the school book work is optional – do it only if you feel you need to. However, your concentration during the solving of the two chapters has to be on developing your speed at solving questions on this chapter.

Chapter 3. AVERAGES

THEORY

The average of a number is a measure of the central tendency of a set of numbers. In other words, it is an estimate of where the center point of a set of numbers lies.

The basic formula for the average of n numbers x1, x2, x3, … xn is

An = (x1 + x2 + x3 + … + xn)/n = (Total of set of n numbers)/n

This also means An × n = total of the set of numbers.

The average is always calculated for a set of numbers.

Concept of weighted average: When we have two or more groups whose individual averages are known, then to find the combined average of all the elements of all the groups we use weighted average. Thus, if we have kgroups with averages A1, A2 ... Ak and having n1, n2 ... nk elements then the weighted average is given by the formula:

Aw =

Another meaning of average The average [also known as arithmetic mean (AM)] of a set of numbers can also be defined as the number by which we can replace each and every number of the set without changing the total of the set of numbers.

Properties of average (AM) The properties of averages [arithmetic mean] can be elucidated by the following examples:

Example 1: The average of 4 numbers 12, 13, 17 and 18 is:

Solution: Required average = (12 + 13 + 17 + 18)/4 = 60/4 = 15

This means that if each of the 4 numbers of the set were replaced by 15 each, there would be no change in the total.

This is an important way to look at averages. In fact, whenever you come across any situation where the average of a group of ‘n’ numbers is given, you should visualise that there are ‘n’ numbers, each of whose value is the average of the group. This view is a very important way to visualise averages.

This can be visualised as

12 Æ +3 Æ 15

13 Æ +2 Æ 15

17 Æ –2 Æ 15

18 Æ –3 Æ 15

60 Æ +0 Æ 60

Example 2: In Example 1, visualise addition of a fifth number, which increases the average by 1.

15 + 1 = 16

15 + 1 = 16

15 + 1 = 16

15 + 1 = 16

The +1 appearing 4 times is due to the fifth number, which is able to maintain the average of 16 first and then ‘give one’ to each of the first 4.

Hence, the fifth number in this case is 20

Example 3: The average always lies above the lowest number of the set and below the highest number of the set.

Example 4: The net deficit due to the numbers below the average always equals the net surplus due to the numbers above the average.

Example 5: Ages and averages: If the average age of a group of persons is x years today then after n years their average age will be (x + n).

Also, n years ago their average age would have been (x – n). This happens due to the fact that for a group of people, 1 year is added to each person’s age every year.

Example 6: A man travels at 60 kmph on the journey from A to B and returns at 100 kmph. Find his average speed for the journey.

Solution:

Average speed = (total distance)/(total time)

If we assume distance between 2 points to be d

Then

Average speed = 2d/[(d/60) + (d/100)] = (2 × 60 × 100)/ (60 + 100) = (2 × 60 × 100)/160 = 75

Average speed = (2S1 ◊ S2)/(S1 + S2) [S1 and S2 are speeds]

of going and coming back respectively.

Short Cut The average speed will always come out by the following process:

The ratio of speeds is 60 : 100 = 3 : 5 (say r1 : r2)

Then, divide the difference of speeds (40 in this case) by r1 + r2 (3 + 5 = 8, in this case) to get one part. (40/8 = 5, in this case)

The required answer will be three parts away (i.e. r1 parts away) from the lower speed.

Check out how this works with the following speeds:

S1 = 20 and S2 = 40

Step 1: Ratio of speeds = 20 : 40 = 1 : 2

Step 2: Divide difference of 20 into 3 parts (r1 + r2) Æ = 20/3 = 6.66

Required average speed = 20 + 1 × 6.66

Note: This process is essentially based on alligations and we shall see it again in the next chapter.

Exercise for Self-practice

Find the average speed for the above problem if

(1)S1 = 20  S2 = 200

(2)S1 = 60  S2 = 120

(3)S1 = 100  S2 = 50

(4)S1 = 60  S2 = 180

WORKED-OUT PROBLEMS

Problem 3.1 The average of a batsman after 25 innings was 56 runs per innings. If after the 26th inning his average increased by 2 runs, then what was his score in the 26th inning?

Solution Normal process:

Runs in 26th inning = Runs total after 26 innings – Runs total after 25 innings

= 26 × 58 – 25 × 56

For mental calculation use:

(56 + 2) × 26 – 56 × 25

= 2 × 26 + (56 × 26 – 56 × 25)

= 52 + 56 = 108

Short Cut Since the average increases by 2 runs per innings it is equivalent to 2 runs being added to each score in the first 25 innings. Now, since these runs can only be added by the runs scored in the 26th inning, the score in the 26th inning must be 25 × 2 = 50 runs higher than the average after 26 innings (i.e. new average = 58).

Hence, runs scored in 26th inning = New Average + Old innings × Change in average

= 58 + 25 × 2 = 108

Visualise this as

Average in first 25 innings

Average after 26 innings

56

58

56

58

56

58

…

…

…

…

25 times…

26 times…

Difference in total is two, 25 times and 58 once, that is, 58 + 25 × 2.

Problem 3.2 The average age of a 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.

Solution Normal process:

Age of teacher = Total age of (students + teacher)

–Total age of students

= 31 × 14 – 30 × 13.5

= 434 – 405

= 29 years

Short Cut The teacher after fulfilling the average of 14 (for the group to which he belonged) is also able to give 0.5 years to the age of each of the 30 students. Hence, he has 30 × 0.5 Æ 15 years to give over and above maintaining his own average age of 14 years.

Age of teacher = 14 + 30 × 0.5 = 29 years

(Note: This problem should be viewed as change of average from 13.5 to 14 when teacher is included.)

Problem 3.3 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?

Solution Normal process:

Let initial average be x.

Then the initial total is 20x

New average will be (x – 4) and the new total will be 20 (x – 4) = 20x – 80.

The reduction of 80 is created by the replacement.

Hence, the new student has 80 marks less than the student he replaces. Hence, he must have scored 10 marks.

Short Cut The replacement has the effect of reducing the average marks for each of the 20 students by 4. Hence, the replacement must be 20 × 4 = 80 marks below the original.

Hence, answer = 10 marks.

Problem 3.4 The average age of 3 students A, B and C is 48 marks. Another student D joins the group and the new average becomes 44 marks. If another student E, who has 3 marks more than D, joins the group, the average of the 4 students B, C, D and E becomes 43 marks. Find how many marks A got in the exam.

Solution Solve while reading. The first sentence gives you a total of 144 for A, B and C’s marks. Second sentence: When D joins the group, the total becomes 44 × 4 = 176. Hence D must get 32 marks.

Alternatively, you can reach this point by considering the first 2 statements together as:

D’s joining the group reduces the average from 48 to 44 marks (i.e. 4 marks).

This means that to maintain the average of 44 marks, D has to take 4 marks from A, 4 from B and 4 from C Æ A total of 12 marks. Hence, he must have got 32 marks.

From here:

The first part of the third sentence gives us information about E getting 3 marks more than 32 Æ Hence, E gets 35 marks.

Now, it is further stated that when A is replaced by E, the average marks of the students reduces by 1 to 43.

Mathematically this can be shown as

A + B + C + D = 44 × 4 = 176 while, B + C + D + E

= 43 × 4 = 172

Subtracting the two equations, we get A – E = 4 marks.

Hence, A would have got 39 marks.

Alternatively, you can think of this as:

The replacement of A with E results in the reduction of 1 mark from each of the 4 people who belong to the group. Hence, the difference is 4 marks. Hence, A would get 4 marks more than E i.e. A gets 39 marks.

Problem 3.5 The mean temperature of Monday to Wednesday was 27 °C and of Tuesday to Thursday was 24 °C. If the temperature on Thursday was 2/3rd of the temperature on Monday, what was the temperature on Thursday?

Solution From the first sentence, we get that the total from Monday to Wednesday was 81 while from Tuesday to Thursday was 72. The difference is arising out of the replacement of Monday by Thursday.

This can be mathematically written as

Mon + Tue + Wed = 81(1)

Tue + Wed + Thu = 72(2)

Hence, Mon – Thu = 9

We have two unknown variables in the above equation. To solve for 2 unknowns, we need a new equation. Looking back at the problem we get the equation:

Thu = (2/3) × Mon

Solving the two equations we get: Thursday = 18 °C.

However, in the exam, you should avoid using equation-solving as much as possible. You should, ideally, be able to reach half way through the solution during the first reading of the question, and then meet the gap through the use of options.

The answer to this problem should be got by the time you finish reading the question for the first time.

Thus suppose we have the equations:

M – T = 9 and T = 2M/3 or T/M = 2/3 and have the options for T as

(a)12    (b) 15

(c)18    (d) 27

To check which of these options is the appropriate value, we need to check one by one.

Option (a) gives T = 12, then we have M = 21. But 12/21 π 2/3. Hence, this is not the correct option.

Option (b) gives T = 15, then M = 24. But again 15/24 π 2/3. Hence, this is not the correct option.

Option (c) gives T = 18, then M = 27. Now 18/27 = 2/3. Hence, this is the correct option.

So we no longer need to check for option (d).

However, if we had checked for option (d) then T = 27, so M = 36. But again 27/36 π 2/3. Hence, this is not the correct option.

In the above, we used ‘solving-while-reading’ and ‘option-based’ approaches.

These two approaches are very important and by combining the two, you can reach amazing speeds in solving the question.

You are advised to practice both these approaches while solving questions, which will surely improve your efficiency and speed. You will see that, with practice, you will be able to arrive at the solution to most of the LOD I problems (given later in this chapter) even as you finish reading the questions. And since it is the LOD I level problems that appear in most examinations (like CMAT, Bank PO, MAT, Indo MAT, NMIMS, NIFT, NLS and most other aptitude exams) you will gain a significant advantage in solving these problems.

On LOD II, LOD III and CAT type problems, you will find that using solving-while-reading and option-based approaches together would take you through anywhere between 30–70% of the question by the time you finish reading the question for the first time.

This will give you a tremendous time advantage over the other students appearing in the examination.

Problem 3.6 A person covers half his journey by train at 60 kmph, the remainder half by bus at 30 kmph and the rest by cycle at 10 kmph. Find his average speed during the entire journey.

Solution Recognise that the journey by bus and that by cycle are of equal distance. Hence, we can use the short cut illustrated earlier to solve this part of the problem.

Using the process explained above, we get average speed of the second half of the journey as

10 + 1 × 5 = 15 kmph

Then we employ the same technique for the first part and get

15 + 1 × 9 = 24 kmph (Answer)

Problem 3.7 A school has only 3 classes that contain 10, 20 and 30 students respectively. The pass percentage of these classes are 20% , 30% and 40% respectively. Find the pass percentage of the entire school.

Solution

Using weighted average: = = 33.33%

Alternatively, we can also use solving-while-reading as

Recognize that the pass percentage would be given by

As soon as you get into the second line of the question get back to the first sentence and get the total number of passed students = 2 + 6 + 12 and you are through with the problem.

LEVEL OF DIFFICULTY (I)

1.

The average age of 24 students and the principal is 15 years. When the principal’s age is excluded, the average age decreases by 1 year. What is the age of the principal?

(a) 38

(b) 40

(c) 39

(d) 37

2.

The average weight of 3 men A, B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg more than that of D, replaces A then the average weight of B, C, D and E becomes 78 kg. The weight of A is

(a) 70 kg

(b) 72 kg

(c) 79 kg

(d) 78 kg

3.

The mean temperature of Monday to Wednesday was 37 °C and of Tuesday to Thursday was 34 °C. If the temperature on Thursday was 4/5 that of Monday, the temperature on Thursday was

(a) 38 °C

(b) 36 °C

(c) 40 °C

(d) 39 °C

4.

Three years ago, the average age of A, B and C was 27 years and that of B and C 5 years ago was 20 years. A’s present age is

(a) 30 years

(b) 35 years

(c) 40 years

(d) 48 years

5.

Ajit Tendulkar has a certain average for 9 innings. In the tenth inning, he scores 100 runs thereby increasing his average by 8 runs. His new average is

(a) 20

(b) 24

(c) 28

(d) 32

6.

The average of the first five multiples of 7 is

(a) 20

(b) 21

(c) 28

(d) 30

7.

There are three fractions A, B and C. If A = and B = 1/6 and the average of A, B and C is 1/12. What is the value of C?

(a) –1/2

(b) – 1/6

(c) –1/3

(d) – 1/4

8.

The marks obtained by Hare Rama in Mathematics, English and Biology are respectively 93 out of 100, 78 out of 150 and 177 out of 200. Find his average score in percent.

(a) 87.83

(b) 86.83

(c) 76.33

(d) 77.33

9.

The average monthly expenditure of a family was ` 2750 for the first 3 months, ` 3150 for the next three months and ` 6750 for the next three months. Find the average income of the family for the 9 months, if they save `650 per month.

(a) 4866.66

(b) 5123.33

(c) 4666.66

(d) 4216.66

10.

The average age of a family of 6 members is 22 years. If the age of the youngest member be 7 years, what was the average age of the family at the birth of the youngest member?

(a) 15

(b) 18

(c) 21

(d) 12

11.

The average age of 8 persons in a committee is increased by 2 years when two men aged 35 years and 45 years are substituted by two women. Find the average age of the two women.

(a) 48

(b) 45

(c) 51

(d) 42

12.

The average temperature for Wednesday, Thursday and Friday was 40 °C. The average for Thursday, Friday and Saturday was 41 °C. If the temperature on Saturday was 42 °C, what was the temperature on Wednesday?

(a) 39 °C

(b) 44 °C

(c) 38 °C

(d) 41 °C

13.

The speed of the train in going from Nagpur to Allahabad is 100 km/hr while when coming back from Allahabad to Nagpur, its speed is 150 km/hr. Find the average speed during the whole journey.

(a) 125

(b) 75

(c) 135

(d) 120

14.

The average weight of a class of 29 students is 40 kg. If the weight of the teacher be included, the average rises by 500 gm. What is the weight of the teacher?

(a) 40.5 kg

(b) 50.5 kg

(c) 45 kg

(d) 55 kg

15.

The average of 3 numbers is 17 and that of the first two is 16. Find the third number.

(a) 15

(b) 16

(c) 17

(d) 19

16.

The average weight of 19 men in a ship is increased by 3.5 kg when one of the men, who weighs 79 kg, is replaced by a new man. Find the weight of the new man upto 2 decimal places

(a) 105.75

(b) 107.55

(c) 145.50

(d) 140.50

17.

The age of Shaurya and Kauravki is in the ratio 2 : 6. After 5 years, the ratio of their ages will become 6 : 8. Find the average of their ages after 10 years.

(a) 12

(b) 13

(c) 17

(d) 24

18.

Find the average of the first 97 natural numbers.

(a) 47

(b) 37

(c) 48

(d) 49

19.

Find the average of all prime numbers between 30 and 50.

(a) 39.8

(b) 38.8

(c) 37.8

(d) 41.8

20.

If we take four numbers, the average of the first three is 16 and that of the last three is 15. If the last number is 18, the first number is

(a) 20

(b) 21

(c) 23

(d) 25

21.

The average of 5 consecutive numbers is n. If the next two numbers are also included, the average will.

(a) increase by 1

(b) remain the same

(c) increase by 1.4

(d) increase by 2

22.

The average of 50 numbers is 38. If two numbers, namely, 45 and 55 are discarded, the average of the remaining numbers is

(a) 36.5

(b) 37

(c) 37.6

(d) 37.5

23.

The average of ten numbers is 7. If each number is multiplied by 12, then the average of the new set of numbers is

(a) 7

(b) 19

(c) 82

(d) 84

24.

In a family of 8 males and a few ladies, the average monthly consumption of grain per head is 10.8 kg. If the average monthly consumption per head be 15 kg in the case of males and 6 kg in the case of females, find the number of females in the family.

(a) 8

(b) 7

(c) 9

(d) 15

25.

Average marks obtained by a student in 3 papers is 52 and in the fourth paper he obtains 60 marks. Find his new average.

(a) 54

(b) 52

(c) 55

(d) 53.5

26.

The average earning of Shambhu Nath Pandey for the initial three months of the calendar year 2002 is ` 1200. If his average earning for the second and third month is ` 1300 find his earning in the first month?

(a) 900

(b) 1100

(c) 1000

(d) 1200

27.

In a hotel where rooms are numbered from 101 to 130, each room gives an earning of ` 3000 for the first fifteen days of a month and for the latter half, ` 2000 per room. Find the average earning per room per day over the month. (Assume 30 day month)

(a) 2250

(b) 2500

(c) 2750

(d) 2466.66

28.

The average weight of 5 men is decreased by 3 kg when one of them weighing 150 kg is replaced by another person. Find the weight of the new person.

(a) 165 kg

(b) 135 kg

(c) 138 kg

(d) 162 kg

29.

The average age of a group of men is increased by 5 years when a person aged 18 years is replaced by a new person of aged 38 years. How many men are there in the group?

(a) 3

(b) 4

(c) 5

(d) 6

30.

The average score of a cricketer in three matches is 22 runs and in two other matches, it is 17 runs. Find the average in all the five matches.

(a) 20

(b) 19.6

(c) 21

(d) 19.5

31.

The average of 13 papers is 40. The average of the first 7 papers is 42 and of the last seven papers is 35. Find the marks obtained in the 7th paper.

(a) 23

(b) 38

(c) 19

(d) 39

32.

The average age of the Indian cricket team playing the Nagpur test is 30. The average age of 5 of the players is 27 and that of another set of 5 players, totally different from the first five, is 29. If it is the captain who was not included in either of these two groups, then find the age of the captain.

(a) 75

(b) 55

(c) 50

(d) 58

33.

Siddhartha has earned an average of 4200 dollars for the first eleven months of the year. If he justifies his staying on in the US on the basis of his ability to earn at least 5000 dollars per month for the entire year, how much should he earn (in dollars) in the last month to achieve his required average for the whole year?

(a) 14,600

(b) 5,800

(c) 12,800

(d) 13,800

34.

A bus goes to Ranchi from Patna at the rate of 60 km per hour. Another bus leaves Ranchi for Patna at the same time as the first bus at the rate of 70 km per hour. Find the average speed for the journeys of the two buses combined if it is known that the distance from Ranchi to Patna is 420 kilometers.

(a) 64.615 kmph

(b) 64.5 kmph

(c) 63.823 kmph

(d) 64.82 kmph

35.

A train travels 8 km in the first quarter of an hour, 6 km in the second quarter and 40 km in the third quarter. Find the average speed of the train per hour over the entire journey.

(a) 72 km/h

(b) 18 km/h

(c) 77.33 km/h

(d) 78.5 km/h

36.

The average weight of 6 men is 68.5 kg. If it is known that Ram and Tram weigh 60 kg each, find the average weight of the others.

(a) 72.75 kg

(b) 75 kg

(c) 78 kg

(d) 80 kg

37.

The average score of a class of 40 students is 52. What will be the average score of the rest of the students if the average score of 10 of the students is 61.

(a) 50

(b) 47

(c) 48

(d) 49

38.

The average age of 80 students of IIM, Bangalore of the 1995 batch is 22 years. What will be the new average if we include the 20 faculty members whose average age is 37 years?

(a) 32 years

(b) 24 years

(c) 25 years

(d) 26 years

39.

Out of three numbers, the first is twice the second and three times the third. The average of the three numbers is 88. The smallest number is

(a) 72

(b) 36

(c) 42

(d) 48

40.

The sum of three numbers is 98. If the ratio between the first and second is 2 : 3 and that between the second and the third is 5 : 8, then the second number is

(a) 30

(b) 20

(c) 58

(d) 48

41.

The average height of 30 girls out of a class of 40 is 160 cm and that of the remaining girls is 156 cm. The average height of the whole class is

(a) 158 cm

(b) 158.5 cm

(c) 159 cm

(d) 157 cm

42.

The average weight of 6 persons is increased by 2.5 kg when one of them whose weight is 50 kg is replaced by a new man. The weight of the new man is

(a) 65 kg

(b) 75 kg

(c) 76 kg

(d) 60 kg

43.

The average age of three boys is 15 years. If their ages are in the ratio 3 : 5 : 7, the age of the youngest boy is

(a) 21 years

(b) 18 years

(c) 15 years

(d) 9 years

44.

The average age of A, B, C and D five years ago was 45 years. By including X, the present average age of all the five is 49 years. The present age of X is

(a) 64 years

(b) 48 years

(c) 45 years

(d) 40 years

45.

The average salary of 20 workers in an office is ` 1900 per month. If the manager’s salary is added, the average salary becomes ` 2000 per month. What is the manager’s annual salary?

(a) ` 24,000

(b) ` 25,200

(c) ` 45,600

(d) None of these

46.

If a, b, c, d and e are five consecutive odd numbers, then their average is

(a) 5(a + b)

(b) (a b c d e)/5

(c) 5(a + b + c + d + e)

(d) None of these

47.

The average of first five multiples of 3 is

(a) 3

(b) 9

(c) 12

(d) 15

48.

The average weight of a class of 40 students is 40 kg. If the weight of the teacher be included, the average weight increases by 500 gm. The weight of the teacher is

(a) 40.5 kg

(b) 60 kg

(c) 62 kg

(d) 60.5 kg

49.

In a management entrance test, a student scores 2 marks for every correct answer and loses 0.5 marks for every wrong answer. A student attempts all the 100 questions and scores 120 marks. The number of questions he answered correctly was

(a) 50

(b) 45

(c) 60

(d) 68

50.

The average age of four children is 8 years, which is increased by 4 years when the age of the father is included. Find the age of the father.

(a) 32

(b) 28

(c) 16

(d) 24

51.

The average of the first ten natural numbers is

(a) 5

(b) 5.5

(c) 6.5

(d) 6

52.

The average of the first ten whole numbers is

(a) 4.5

(b) 5

(c) 5.5

(d) 4

53.

The average of the first ten even numbers is

(a) 18

(b) 22

(c) 9

(d) 11

54.

The average of the first ten odd numbers is

(a) 11

(b) 10

(c) 17

(d) 9

55.

The average of the first ten prime numbers is

(a) 15.5

(b) 12.5

(c) 10

(d) 12.9

56.

The average of the first ten composite numbers is

(a) 12.9

(b) 11

(c) 11.2

(d) 10

57.

The average of the first ten prime numbers, which are odd, is

(a) 12.9

(b) 13.8

(c) 17

(d) 15.8

58.

The average weight of a class of 30 students is 40 kg. If, however, the weight of the teacher is included, the average becomes 41 kg. The weight of the teacher is

(a) 31 kg

(b) 62 kg

(c) 71 kg

(d) 70 kg

59.

Ram bought 2 toys for ` 5.50 each, 3 toys for ` 3.66 each and 6 toys for ` 1.833 each. The average price per toy is

(a) ` 3

(b) ` 10

(c) ` 5

(d) ` 9

60.

30 oranges and 75 apples were purchased for ` 510. If the price per apple was ` 2, then the average price of oranges was

(a) ` 12

(b) ` 14

(c) ` 10

(d) ` 15

61.

The average income of Sambhu and Ganesh is ` 3,000 and that of Arun and Vinay is ` 500. What is the average income of Sambhu, Ganesh, Arun and Vinay?

(a) ` 1750

(b) ` 1850

(c) ` 1000

(d) ` 2500

62.

A batsman made an average of 40 runs in 4 innings, but in the fifth inning, he was out on zero. What is the average after fifth inning?

(a) 32

(b) 22

(c) 38

(d) 49

63.

The average weight of 40 teachers of a school is 80 kg. If, however, the weight of the principal be included, the average decreases by 1 kg. What is the weight of the principal?

(a) 109 kg

(b) 29 kg

(c) 39 kg

(d) None of these

64.

The average temperature of 1st, 2nd and 3rd December was 24.4 °C. The average temperature of the first two days was 24 °C. The temperature on the 3rd of December was

(a) 20 °C

(b) 25 °C

(c) 25.2 °C

(d) None of these

65.

The average age of Ram and Shyam is 20 years. Their average age 5 years hence will be

(a) 25 years

(b) 22 years

(c) 21 years

(d) 20 years

66.

The average of 20 results is 30 and that of 30 more results is 20. For all the results taken together, the average is

(a) 25

(b) 50

(c) 12

(d) 24

67.

The average of 5 consecutive numbers is 18. The highest of these numbers will be

(a) 24

(b) 18

(c) 20

(d) 22

68.

The average of 6 students is 11 years. If 2 more students of age 14 and 16 years join, their average will become

(a) 12 years

(b) 13 years

(c) 21 years

(d) 19 years

69.

The average of 8 numbers is 12. If each number is increased by 2, the new average will be

(a) 12

(b) 14

(c) 13

(d) 15

70.

Three years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average of the family is the same today. What is the age of the baby?

(a) 1 year

(b) 2 years

(c) 6 months

(d) 9 months

71.

Sambhu’s average daily expenditure is ` 10 during May, ` 14 during June and ` 15 during July. His approximate daily expenditure for the 3 months is

(a) ` 13 approximately

(b) ` 12

(c) ` 12 approximately

(d) ` 10

72.

A ship sails out to a mark at the rate of 15 km per hour and sails back at the rate of 20 km/h. What is its average rate of sailing?

(a) 16.85 km

(b) 17.14 km

(c) 17.85 km

(d) 18 km

73.

The average temperature on Monday, Tuesday and Wednesday was 41 °C and on Tuesday, Wednesday and Thursday it was 40 °C. If on Thursday it was exactly 39 °C, then on Monday, the temperature was

(a) 42 °C

(b) 46 °C

(c) 23 °C

(d) 26 °C

74.

The average of 20 results is 30 out of which the first 10 results are having an average of 10. The average of the rest 10 results is

(a) 50

(b) 40

(c) 20

(d) 25

75.

A man had seven children. When their average age was 12 years a child aged 6 years died. The average age of the remaining 6 children is

(a) 6 years

(b) 13 years

(c) 17 years

(d) 15 years

76.

The average income of Ram and Shyam is ` 200. The average income of Rahul and Rohit is ` 250. The average income of Ram, Shyam, Rahul and Rohit is

(a) ` 275

(b) ` 225

(c) ` 450

(d) ` 250

77.

The average weight of 35 students is 35 kg. If the teacher is also included, the average weight increases to 36 kg. The weight of the teacher is

(a) 36 kg

(b) 71 kg

(c) 70 kg

(d) 45 kg

78.

The average of x, y and z is 45. x is as much more than the average as y is less than the average. Find the value of z.

(a) 45

(b) 25

(c) 35

(d) 15

79.

Find the average of four numbers 2, 5, 4, 8.

(a) 5

(b) 3

(c) 16

(d) 3

80.

The average salary per head of all the workers in a company is ` 95. The average salary of 15 officers is ` 525 and the average salary per head of the rest is ` 85. Find the total number of workers in the workshop.

(a) 660

(b) 580

(c) 650

(d) 460

81.

The average age of 8 men is increased by 2 years when one of them whose age is 24 years is replaced by a woman. What is the age of the woman?

(a) 35 years

(b) 28 years

(c) 32 years

(d) 40 years

82.

The average monthly expenditure of Ravi was ` 1100 during the first 3 months, ` 2200 during the next 4 months and ` 4620 during the subsequent five months of the year. If the total saving during the year was ` 2100, find Ravi’s average monthly income.

(a) ` 1858

(b) ` 3108.33

(c) ` 3100

(d) None of these

83.

Shyam bought 2 articles for ` 5.50 each, and 3 articles for ` 3.50 each, and 3 articles for ` 5.50 each and 5 articles for ` 1.50 each. The average price for one article is

(a) ` 3

(b) ` 3.10

(c) ` 3.50

(d) ` 2

84.

In a bag, there are 150 coins of Re. 1, 50 p and 25 p denominations. If the total value of coins is ` 150, then find how many rupees can be constituted by 50 p coins.

(a) 16

(b) 20

(c) 28

(d) None of these

LEVEL OF DIFFICULTY (II)

1.

With an average speed of 40 km/h, a train reaches its destination in time. If it goes with an average speed of 35 km/h, it is late by 15 minutes. The length of the total journey is:

(a) 40 km

(b) 70 km

(c) 30 km

(d) 80 km

2.

In the month of July of a certain year, the average daily expenditure of an organisation was ` 68. For the first 15 days of the month, the average daily expenditure was ` 85 and for the last 17 days, ` 51. Find the amount spent by the organisation on the 15th of the month.

(a) ` 42

(b) ` 36

(c) ` 34

(d) ` 52

3.

In 1919, W. Rhodes, the Yorkshire cricketer, scored 891 runs for his county at an average of 34.27; in 1920, he scored 949 runs at an average of 28.75; in 1921, 1329 runs at an average of 42.87 and in 1922, 1101 runs at an average of 36.70. What was his county batting average for the four years?

(a) 36.23

(b) 37.81

(c) 35.88

(d) 28.72

4.

A train travels with a speed of 20 m/s in the first 10 minutes, goes 8.5 km in the next 10 minutes, 11 km in the next 10, 8.5 km in the next 10 and 6 km in the next 10 minutes. What is the average speed of the train in kilometer per hour for the journey described?

(a) 42 kmph

(b) 35.8 kmph

(c) 55.2 kmph

(d) 46 kmph

5.

One-fourth of a certain journey is covered at the rate of 25 km/h, one-third at the rate of 30 km/h and the rest at 50 km/h. Find the average speed for the whole journey.

(a) 600/53 km/h

(b) 1200/53 km/h

(c) 1800/53 km/h

(d) 1600/53

6.

Typist A can type a sheet in 6 minutes, typist B in 7 minutes and typist C in 9 minutes. The average number of sheets typed per hour per typist for all three typists is

(a) 265/33

(b) 530/63

(c) 655/93

(d) 530/33

7.

Find the average increase rate if increase in the population in the first year is 30% and that in the second year is 40%.

(a) 41

(b) 56

(c) 40

(d) 38

8.

The average income of a person for the first 6 days is ` 29, for the next 6 days it is ` 24, for the next 10 days it is ` 32 and for the remaining days of the month it is ` 30. Find the average income per day.

(a) ` 31.64

(b) ` 30.64

(c) ` 29.26

(d) Cannot be determined

9.

In hotel Jaysarmin, the rooms are numbered from 101 to 130 on the first floor, 221 to 260 on the second floor and 306 to 345 on the third floor. In the month of June 2012, the room occupancy was 60% on the first floor, 40% on the second floor and 75% on the third floor. If it is also known that the room charges are ` 200, ` 100 and ` 150 on each of the floors, then find the average income per room for the month of June 2012.

(a) ` 151.5

(b) ` 88.18

(c) ` 78.3

(d) ` 65.7

10.

A salesman gets a bonus according to the following structure: If he sells articles worth ` x then he gets a bonus of ` (x/100 – 1). In the month of January, his sales value was ` 100, in February it was ` 200, from March to November it was ` 300 for every month and in December it was ` 1200. Apart from this, he also receives a basic salary of ` 30 per month from his employer. Find his average income per month during the year.

(a) ` 31.25

(b) ` 30.34

(c) ` 32.5

(d) ` 34.5

11.

A man covers half of his journey by train at 60 km/h, half of the remainder by bus at 30 km/h and the rest by cycle at 10 km/h. Find his average speed during the entire journey.

(a) 36 kmph

(b) 30 kmph

(c) 24 kmph

(d) 18 kmph

12.

The average weight of 5 men is decreased by 3 kg when one of them weighing 150 kg is replaced by another person. This new person is again replaced by another person whose weight is 30 kg lower than the person he replaced. What is the overall change in the average due to this dual change?

(a) 6 kg

(b) 9 kg

(c) 12 kg

(d) 15 kg

13.

Find the average weight of four containers, if it is known that the weight of the first container is 100 kg and the total of the second, third and fourth containers’ weight is defined by f(x) = x2 – 3/4 (x2) where x = 100

(a) 650 kg

(b) 900 kg

(c) 750 kg

(d) 450 kg

14.

There are five boxes in a cargo hold. The weight of the first box is 200 kg and the weight of the second box is 20% higher than the weight of the third box, whose weight is 25% higher than the first box’s weight. The fourth box at 350 kg is 30% lighter than the fifth box. Find the difference in the average weight of the four heaviest boxes and the four lightest boxes.

(a) 51.5 kg

(b) 75 kg

(c) 37.5 kg

(d) 112.5 kg

15.

For Question 14, find the difference in the average weight of the heaviest three and the lightest three.

(a) 116.66 kg

(b) 125 kg

(c) 150 kg

(d) 112.5 kg

16.

A batsman makes a score of 270 runs in the 87th inning and thus increases his average by a certain number of runs that is a whole number. Find the possible values of the new average.

(a) 98

(b) 184

(c) 12

(d) All of these

17.

19 persons went to a hotel for a combined dinner party. 13 of them spent ` 79 each on their dinner and the rest spent ` 4 more than the average expenditure of all the 19. What was the total money spent by them?

(a) 1628.4

(b) 1534

(c) 1492

(d) None of these

18.

There were 42 students in a hostel. Due to the admission of 13 new students, the expenses of the mess increase by ` 31 per day while the average expenditure per head diminished by ` 3. What was the original expenditure of the mess?

(a) ` 633.23

(b) ` 583.3

(c) ` 623.3

(d) ` 632

19.

The average price of 3 precious diamond studded platinum thrones is ` 97610498312 if their prices are in the ratio 4:7:9. The price of the cheapest is:

(a) 5, 65, 66, 298.972

(b) 5, 85, 66, 29, 897.2

(c) 58, 56, 62, 889.72

(d) None of these

20.

The average weight of 47 balls is 4 gm. If the weight of the bag (in which the balls are kept) be included, the calculated average weight per ball increases by 0.3 gm. What is the weight of the bag?

(a) 14.8 gm

(b) 15.0 gm

(c) 18.6 gm

(d) None of these

21.

The average of 71 results is 48. If the average of the first 59 results is 46 and that of the last 11 is 52. Find the 60th result.

(a) 132

(b) 122

(c) 134

(d) 128

22.

A man covers 1/3rd of his journey by cycle at 50 km/h, the next 1/3 by car at 30 km/h, and the rest by walking at 7 km/h. Find his average speed during the whole journey.

(a) 14.2 kmph

(b) 15.3 kmph

(c) 18.2 kmph

(d) 12.8 kmph

23.

The average age of a group of 14 persons is 27 years and 9 months. Two persons, each 42 years old, left the group. What will be the average age of the remaining persons in the group?

(a) 26.875 years

(b) 26.25 years

(c) 25.375 years

(d) 25 years

24.

In an exam, the average was found to be x marks. After deducting computational error, the average marks of 94 candidates got reduced from 84 to 64. The average thus came down by 18.8 marks. The numbers of candidates who took the exam were:

(a) 100

(b) 90

(c) 110

(d) 105

25.

The average salary of the entire staff in an office is ` 3200 per month. The average salary of officers is ` 6800 and that of non-officers is ` 2000. If the number of officers is 5, then find the number of non-officers in the office?

(a) 8

(b) 12

(c) 15

(d) 5

26.

A person travels three equal distances at a speed of x km/h, y km/h and z km/h respectively. What will be the average speed during the whole journey?

(a) xyz/(xy + yz + zx)

(b) (xy + yz + zx)/xyz

(c) 3xyz/(xy + yz + xz)

(d) None of these

Directions for Questions 27 to 30: Read the following passage and answer the questions that follow.

In a family of five persons A, B, C, D and E, each and everyone loves one another very much. Their birthdays are in different months and on different dates. A remembers that his birthday is between 25th and 30th, of B it is between 20th and 25th, of C it is between 10th and 20th, of D it is between 5th and 10th and of E it is between 1st to 5th of the month. The sum of the date of birth is defined as the addition of the date and the month, for example 12th January will be written as 12/1 and will add to a sum of the date of 13. (Between 25th and 30th includes both 25 and 30).

27.

What may be the maximum average of their sum of the dates of birth?

(a) 24.6

(b) 15.2

(c) 28

(d) 32

28.

What may be the minimum average of their sum of the dates of births?

(a) 24.6

(b) 15.2

(c) 28

(d) 32

29.

If it is known that the dates of birth of three of them are even numbers then find maximum average of their sum of the dates of birth.

(a) 24.6

(b) 15.2

(c) 27.6

(d) 28

30.

If the date of birth of four of them are prime numbers, then find the maximum average of the sum of their dates of birth.

(a) 27.2

(b) 26.4

(c) 28

(d) None of these

31.

The average age of a group of persons going for a picnic is 16.75 years. 20 new persons with an average age of 13.25 years join the group on the spot due to which the average of the group becomes 15 years. Find the number of persons initially going for the picnic.

(a) 24

(b) 20

(c) 15

(d) 18

32.

A school has only four classes that contain 10, 20, 30 and 40 students respectively. The pass percentage of these classes are 20%, 30%, 60% and 100% respectively. Find the pass % of the entire school.

(a) 56%

(b) 76%

(c) 34%

(d) 66%

33.

Find the average of f(x), g(x), h(x), d(x) at x = 10. f(x) is equal to x2 + 2, g(x) = 5x2 – 3, h(x) = log x2 and d(x) = (4/5)x2

(a) 170

(b) 170.25

(c) 70.25

(d) 70

34.

Find the average of f(x) – g(x), g(x) – h(x), h(x) – d(x), d(x) – f(x)

(a) 0

(b) –2.25

(c) 4.5

(d) 2.25

35.

r where r = n.

(a)

(b)

(c)

(d)

36.

The average of ‘n’ numbers is z. If the number x is replaced by the number x1, then the average becomes z1. Find the relation between n, z, z1, x and x1.

(a)

(b)

(c)

(d)

37.

The average salary of workers in Mindworkzz is ` 2,000, the average salary of faculty being ` 4,000 and the management trainees being ` 1,250. The total number of workers could be

(a) 450

(b) 300

(c) 110

(d) 500

Directions for Questions 38 to 41: Read the following and answer the questions that follows.

During a criket match, India playing against NZ scored in the following manner:

Partnership

Runs scored

1st wicket

112

2nd wicket

58

3rd wicket

72

4th wicket

92

5th wicket

46

6th wicket

23

38.

Find the average runs scored by the first four batsmen.

(a) 83.5

(b) 60.5

(c) 66.8

(d) Cannot be determined

39.

The maximum average runs scored by the first five batsmen could be

(a) 80.6

(b) 66.8

(c) 76

(d) Cannot be determined.

40.

The minimum average runs scored by the last five batsmen to get out could be

(a) 53.6

(b) 44.4

(c) 66.8

(d) 0

41.

If the fifth down batsman gets out for a duck, then find the average runs scored by the first six batsmen.

(a) 67.1

(b) 63.3

(c) 48.5

(d) Cannot be determined

42.

The weight of a body as calculated by the average of 7 different experiments is 53.735 gm. The average of the first three experiments is 54.005 gm, of the fourth is 0.004 gm greater than the fifth, while the average of the sixth and seventh experiment was 0.010 gm less than the average of the first three. Find the weight of the body obtained by the fourth experiment.

(a) 49.353 gm

(b) 51.712 gm

(c) 53.072 gm

(d) 54.512 gm

43.

A man’s average expenditure for the first 4 months of the year was ` 251.25. For the next 5 months the average monthly expenditure was ` 26.27 more than what it was during the first 4 months. If the person spent ` 760 in all during the remaining 3 months of the year, find what percentage of his annual income of ` 3000 he saved in the year.

(a) 14%

(b) –5.0866%

(c) 12.5%

(d) None of these

44.

A certain number of trucks were required to transport 60 tons of steel wire from the TISCO factory in Jamshedpur. However, it was found that since each truck could take 0.5 tons of cargo less, another 4 trucks were needed. How many trucks were initially planned to be used?

(a) 10

(b) 15

(c) 20

(d) 25

45.

One collective farm got an average harvest of 21 tons of wheat and another collective farm that had 12 acres of land less given to wheat, got 25 tons from a hectare. As a result, the second farm harvested 300 tons of wheat more than the first. How many tons of wheat did each farm harvest?

(a) 3150, 3450

(b) 3250, 3550

(c) 2150, 2450

(d) None of these

LEVEL OF DIFFICULTY (III)

Directions for Questions 1 to 8: Read the following:

There are 3 classes having 20, 25 and 30 students respectively having average marks in an examination as 20, 25 and 30 respectively. If the three classes are represented by A, B and C and you have the following information about the three classes, answer the questions that follow:

A Æ Highest score 22, Lowest score 18

B Æ Highest score 31, Lowest score 23

C Æ Highest score 33, Lowest score 26

If five students are transferred from A to B.

1.

What will happen to the average score of B?

(a) Definitely increase

(b) Definitely decrease

(c) Remain constant

(d) Cannot say

2.

What will happen to the average score of A?

(a) Definitely increase

(b) Definitely decrease

(c) Remain constant

(d) Cannot say

In a transfer of 5 students from A to C

3.

What will happen to the average score of C?

(a) Definitely increase

(b) Definitely decrease

(c) Remain constant

(d) Cannot say

4.

What will happen to the average score of A?

(a) Definitely increase

(b) Definitely decrease

(c) Remain constant

(d) Cannot say

In a transfer of 5 students from B to C (Questions 5–6)

5.

What will happen to the average score of C?

(a) Definitely increase

(b) Definitely decrease

(c) Remain constant

(d) Cannot say

6.

Which of these can be said about the average score of B?

(a) Increases if C decreases

(b) Decreases if C increases

(c) Increases if C decreases

(d) Decreases if C decreases

7.

In a transfer of 5 students from A to B, the maximum possible average achievable for group B is

(a) 25

(b) 24.5

(c) 25.5

(d) 24

8.

For the above case, the maximum possible average achieved for group A will be

(a) 20.66

(b) 21.5

(c) 20.75

(d) 20.5

9.

What will be the minimum possible average of Group A if 5 students are transferred from A to B?

(a) 19.55

(b) 21.5

(c) 19.33

(d) 20.5

10.

If 5 students are transferred from B to A, what will be the minimum possible average of A?

(a) 20.69

(b) 21

(c) 20.75

(d) 20.6

11.

For question 10, what will be the maximum average of A?

(a) 23.2

(b) 22.2

(c) 18.75

(d) 19

Directions for Questions 12 to 17: Read the following and answer the questions that follow.

If 5 people are transferred from A to B and another independent set of 5 people are transferred back from B to A, then after this operation (Assume that the set transferred from B to A contains none from the set of students that came to B from A)

12.

What will happen to B’s average?

(a) Increase if A’s average decreases

(b) Decrease always

(c) Cannot be said

(d) Decrease if A’s average decreases

13.

What can be said about A’s average?

(a) Will decrease

(b) Will always increase if B’s average changes

(c) May increase or decrease

(d) Will increase only if B’s average decreases

14.

At the end of the 2 steps mentioned above (in the direction) what could be the maximum value of the average of class B?

(a) 25.4

(b) 25

(c) 24.8

(d) 24.6

15.

For question 14, what could be the minimum value of the average of class B?

(a) 22.4

(b) 24.2

(c) 25

(d) 23

16.

What could be the maximum possible average achieved by class A at the end of the operation?

(a) 25.2

(b) 26

(c) 23.25

(d) 23.75

17.

What could be the minimum possible average of class A at the end of the operation?

(a) 21.4

(b) 19.2

(c) 28.5

(d) 20.25

Directions for Questions 18 to 23: Read the following and answer the questions that follow.

If 5 people are transferred from C to B, further, 5 more people are transferred from B to A, then 5 are transferred from A to B and finally, 5 more are transferred from B to C.

18.

What is the maximum possible average achieved by class C?

(a) 30.833

(b) 30

(c) 29.66

(d) 30.66

19.

What is the maximum possible average of class B?

(a) 26

(b) 27

(c) 25

(d) 28

20.

What is the maximum possible average value attained by class A?

(a) 22.75

(b) 23.75

(c) 23.5

(d) 24

21.

The minimum possible value of the average of group C is:

(a) 26.3

(b) 27.5

(c) 29.6

(d) 28

22.

The minimum possible average of group B after this set of operation is

(a) 21.6

(b) 21.4

(c) 21.8

(d) 21.2

23.

The minimum possible average of group A after the set of 3 operation is

(a) 20

(b) 20.3

(c) 20.4

(d) 19.8

24.

Which of these will definitely not constitute an operation for getting the minimum possible average value for group A:

(a) Transfer of five 31s from B to A

(b) Transfer of five 26s from C to B

(c) Transfer of five 22s from A to B

(d) Transfer of five 33s from C to B

25.

For getting the lowest possible value of C’s average, the sequence of operations could be

(a)Transfer five 33s from C to B, five 23s from B to A, five 18s from A to B, five 18s from B to C

(b)Transfer five 33s from C to B, 31s from B to A,

(c)Both a and b

(d)None of the above

26.

If we set the highest possible average of class C as the primary objective and want to achieve the highest possible value for class B as the secondary objective, what is the maximum value of class B’s average that is attainable?

(a) 27

(b) 26

(c) 25

(d) 24

27.

For Question 26, if the secondary objective is changed to achieving the minimum possible average value of class B’s average, the lowest value of class B’s average that could be attained is

(a) 22.2

(b) 23

(c) 22.6

(d) 22

28.

For question 27, what can be said about class A’s average?

(a) Will be determined automatically at 22.25

(b) Will have a maximum possible value of 22.25

(c) Will have a minimum possible value of 22.25

(d) Will be determined automatically at 22.5

29.

A team of miners planned to mine 1800 tons of ore during a certain number of days. Due to technical difficulties in one-third of the planned number of days, the team was able to achieve an output of 20 tons of ore less than the planned output. To make up for this, the team overachieved for the rest of the days by 20 tons. The end result was that the team completed the task one day ahead of time. How many tons of ore did the team initially plan to ore per day?

(a) 50 tons

(b) 100 tons

(c) 150 tons

(d) 200 tons

30.

According to a plan, a team of woodcutters decided to harvest 216 m3 of wheat in several days. In the first three days, the team fulfilled the daily assignment, and then it harvested 8 m3 of wheat over and above the plan everyday. Therefore, a day before the planned date, they had already harvested 232 m3 of wheat. How many cubic metres of wheat a day did the team have to cut according to the plan?

(a) 12

(b) 13

(c) 24

(d) 25

31.

On an average, two liters of milk and one liter of water are needed to be mixed to make 1 kg of sudha shrikhand of type A, and 3 liters of milk and 2 liters of water are needed to be mixed to make 1 kg of sudha shrikhand of type B. How many kilograms of each type of shrikhand was manufactured if it is known that 130 liters of milk and 80 liters of water were used?

(a) 20 of type A and 30 of type B

(b) 30 of type A and 20 of type B

(c) 15 of type A and 30 of type B

(d) 30 of type A and 15 of type B

32.

There are 500 seats in Minerva Cinema, Mumbai, placed in similar rows. After the reconstruction of the hall, the total number of seats became 10% less. The number of rows was reduced by 5 but each row contained 5 seats more than before. How many rows and how many seats in a row were there initially in the hall?

(a) 20 rows and 25 seats

(b) 20 rows and 20 seats

(c) 10 rows and 50 seats

(d) 50 rows and 10 seats

33.

One fashion house has to make 810 dresses and another one 900 dresses during the same period of time. In the first house, the order was ready 3 days ahead of time and in the second house, 6 days ahead of time. How many dresses did each fashion house make a day if the second house made 21 dresses more a day than the first?

(a) 54 and 75

(b) 24 and 48

(c) 44 and 68

(d) 04 and 25

34.

A shop sold 64 kettles of two different capacities. The smaller kettle cost a rupee less than the larger one. The shop made 100 rupees from the sale of large kettles and 36 rupees from the sale of small ones. How many kettles of either capacity did the shop sell and what was the price of each kettle?

(a) 20 kettles for 2.5 rupees each and 14 kettles for 1.5 rupees each

(b) 40 kettles for 4.5 rupees each and 24 kettles for 2.5 rupees each

(c) 40 kettles for 2.5 rupees each and 24 kettles for 1.5 rupees each

(d) either a or b

35.

An enterprise got a bonus and decided to share it in equal parts between the exemplary workers. It turned out, however, that there were 3 more exemplary workers than it had been assumed. In that case, each of them would have got 4 rupees less. The administration had found the possibility to increase the total sum of the bonus by 90 rupees and as a result each exemplary worker got 25 rupees. How many people got the bonus?

(a) 9

(b) 18

(c) 8

(d) 16

Directions for Questions 36 to 39: Read the following and answer the questions that follows.

In the island of Hoola Boola Moola, the inhabitants have a strange process of calculating their average incomes and expenditures. According to an old legend prevalent on that island, the average monthly income had to be calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 9 months per year. This would lead to people having an underestimation of their savings since there would be an underestimation of the income and an overestimation of the expenditure per month.

36.

If the minister for economic affairs decided to reverse the process of calculation of average income and average expenditure, what will happen to the estimated savings of a person living on Hoola Boola Moola island?

(a) It will increase

(b) It will decrease

(c) It will remain constant

(d) Will depend on the value

37.

If it is known that Mr. Magoo Hoola Boola estimates his savings at 10 Moolahs and if it is further known that his actual expenditure is 288 Moolahs in an year (Moolahs, for those who are not aware, is the official currency of Hoola Boola Moola), then what will happen to his estimated savings if he suddenly calculates on the basis of a 12 month calendar year?

(a) Will increase by 5

(b) Will increase by 15

(c) Will increase by 10

(d) Will triple

38.

Mr. Boogie Woogie comes back from the USA to Hoola Boola Moola and convinces his community comprising 546 families to start calculating the average income and average expenditure on the basis of 12 months per calendar year. Now if it is known that the average estimated income on the island is (according to the old system) 87 Moolahs per month, then what will be the change in the average estimated savings for the island of Hoola Boola Moola. (Assume that there is no other change).

(a) 251.60 Moolahs

(b) 565.5 Moolahs

(c) 625.5 Moolahs

(d) Cannot be determined

39.

Mr. Boogle Woogle comes back from the USSR and convinces his community comprising 273 families to start calculating the average income on the basis of 12 months per calendar year. Now if it is known that the average estimated income in his community is (according to the old system) 87 Moolahs per month, then what will be the change in the average estimated savings for the island of Hoola Boola Moola. (Assume that there is no other change).

(a) 251.60 Moolahs

(b) 282.75 Moolahs

(c) 312.75 Moolahs

(d) Cannot be determined

Directions for Questions 40 to 44: Read the following and answer the questions that follows.

The Indian cricket team has to score 360 runs on the last day of a test match in 90 overs, to win the test match. This is the target set by the opposing captain Brian Lara after he declared his innings closed at the overnight score of 411 for 7.

The Indian team coach has the following information about the batting rates (in terms of runs per over) of the different batsmen:

Assume that the run rate of a partnership is the weighted average of the individual batting rates of the batsmen involved in the partnership (on the basis of the ratio of the strike each batsman gets, i.e. the run rate of a partnership is defined as the weighted average of the run rates of the two batsmen involved weighted by the ratio of the number of balls faced by each batsman).

Since decimal fractions of runs are not possible for any batsman, assume that the estimated runs scored by a batsman in an inning (on the basis of his run rate and the number of overs faced by him) is rounded off to the next higher integer immediately above the estimated value of the runs scored during the innings.

For example, if a batsman scores at an average of 3 runs per over for 2.1666 overs, then he will be estimated to have scored 2.1666 × 3 = 6.5 runs in his innings, but since this is not possible, the actual number of runs scored by the batsman will be taken as 7 (the next higher integer above 6.5).

Runs scored per over in different batting styles

Name of Batsman

Defensive

Normal

Aggressive

Das

3

4

5

Dasgupta

2

3

4

Dravid

2

3

4

Tendulkar

4

6

8

Laxman

4

5

6

Sehwag

4

5

6

Ganguly

3

4

5

Kumble

2

3

4

Harbhajan

3

4

5

Srinath

3

4

5

Yohannan

2

3

4

Also, this rounding off can take place only once for one innings of a batsman.

Assume no extras unless otherwise stated.

Assume that the strike is equally shared unless otherwise stated.

40.

If the first wicket pair of Das and Dasgupta bats for 22 overs and during this partnership Das has started batting normally and turned aggressive after 15 overs while Dasgupta started off defensively but shifted gears to bat normally after batting for 20 overs, find the expected score after 22 overs.

(a) 65

(b) 71

(c) 82

(d) 58

41.

Of the first-wicket partnership between Das and Dasgupta as per the previous question, the ratio of the number of runs scored by Das to those scored by Dasgupta is:

(a) 46 : 25

(b) 96 : 46

(c) 41 : 32

(d) Cannot be determined

42.

The latest time by which Tendulkar can come to bat and still win the game, assuming that the run rate at the time of his walking the wicket is into 2.5 runs per over, is (assuming he shares strike equally with his partner and that he gets the maximum possible support at the other end from his batting partner and both play till the last ball).

(a) After 50 overs

(b) After 55 overs

(c) After 60 overs

(d) Cannot be determined

43.

For question 42, where Tendulkar batted aggressively and assuming that it is the Tendulkar-Laxman pair that wins the game for India (after Tendulkar walks into bat with the current run rate at 2.5 per over, and at the latest possible time for him to win the game with maximum possible support from the opposite end), what will be Tendulkar’s score for the innings (assume equal strike)?

(a) 105

(b) 120

(c) 135

(d) None of these

44.

For questions 42 and 43, if it was Laxman who batted with Tendulkar for his entire innings, then how many runs would Laxman score in the innings?

(a) 105

(b) 75

(c) 90

(d) Cannot be determined

Directions for Questions 45 to 49: Read the following and answer the questions that follow.

If Sachin Tendulkar walks into bat after the fall of the fifth wicket and has to share partnerships with Ganguly, Kumble, Harbhajan, Srinath and Yohannan, who have batted normally, defensively, defensively, defensively and defensively respectively while Tendulkar has batted normally, aggressively, aggressively, aggressively and aggressively respectively in each of the five partnerships that lasted for 12, 10, 8, 5 and 10 overs respectively, sharing strike equally with Ganguly and keeping two-thirds of the strike in his other four partnerships, then answer the following questions:

45.

How many runs did Sachin score during his innings?

(a) 128

(b) 212

(c) 176

(d) None of these

46.

The highest partnership that Tendulkar shared in was worth:

(a) 60

(b) 61

(c) 62

(d) 58

47.

The above partnership was shared with:

(a) Ganguly

(b) Yohannan

(c) Kumble

(d) All three

48.

If India proceeded to win the match based on the runs scored by these last five partnerships (assuming the last wicket pair remained unbeaten), what could be the maximum score at which Tendulkar could have come into bat:

(a) 103 for 5

(b) 97 for 5

(c) 100 for 5

(d) 104 for 5

49.

For Question 48, what could be the minimum score at which Tendulkar could have come to bat:

(a) 103 for 5

(b) 97 for 5

(c) 104 for 5

(d) 98 for 5

ANSWER KEY

Level of Difficulty (I)

1. (c)

2. (c)

3. (b)

4. (c)

5. (c)

6. (b)

7. (b)

8. (d)

9. (a)

10. (b)

11. (a)

12. (a)

13. (d)

14. (d)

15. (d)

16. (c)

17. (a)

18. (d)

19. (a)

20. (b)

21. (a)

22. (d)

23. (d)

24. (b)

25. (a)

26. (c)

27. (b)

28. (b)

29. (b)

30. (a)

31. (c)

32. (c)

33. (d)

34. (a)

35. (a)

36. (a)

37. (d)

38. (c)

39. (d)

40. (a)

41. (c)

42. (a)

43. (d)

44. (c)

45. (d)

46. (d)

47. (b)

48. (d)

49. (d)

50. (b)

51. (b)

52. (a)

53. (d)

54. (b)

55. (d)

56. (c)

57. (d)

58. (c)

59. (a)

60. (a)

61. (a)

62. (a)

63. (c)

64. (c)

65. (a)

66. (d)

67. (c)

68. (a)

69. (b)

70. (b)

71. (a)

72. (b)

73. (a)

74. (a)

75. (b)

76. (b)

77. (b)

78. (a)

79. (a)

80. (a)

81. (d)

82. (b)

83. (c)

84. (d)

Level of Difficulty (II)

1. (b)

2. (c)

3. (c)

4. (c)

5. (c)

6. (b)

7. (a)

8. (d)

9. (a)

10. (c)

11. (c)

12. (a)

13. (a)

14. (b)

15. (a)

16. (d)

17. (d)

18. (a)

19. (d)

20. (d)

21. (b)

22. (b)

23. (c)

24. (a)

25. (c)

26. (c)

27. (c)

28. (b)

29. (d)

30. (a)

31. (b)

32. (d)

33. (b)

34. (a)

35. (b)

36. (c)

37. (c)

38. (d)

39. (a)

40. (d)

41. (d)

42. (c)

43. (b)

44. (c)

45. (a)

Level of Difficulty (III)

1. (b)

2. (d)

3. (b)

4. (d)

5. (d)

6. (b)

7. (b)

8. (a)

9. (c)

10. (d)

11. (b)

12. (b)

13. (b)

14. (c)

15. (a)

16. (c)

17. (d)

18. (a)

19. (b)

20. (b)

21. (b)

22. (b)

23. (a)

24. (a)

25. (a)

26. (d)

27. (c)

28. (a)

29. (b)

30. (c)

31. (a)

32. (a)

33. (a)

34. (c)

35. (b)

36. (a)

37. (b)

38. (d)

39. (b)

40. (b)

41. (b)

42. (c)

43. (b)

44. (d)

45. (b)

46. (b)

47. (b)

48. (d)

49. (b)

Solutions and Shortcuts

Level of Difficulty (I)

1.P = 25 × 15 – 24 × 14 = 375 – 336 = 39

2.D’s weight = 4 × 80 – 3 × 84 = 320 – 252 = 68. E’s weight = 68 + 3 = 71.

Now, we know that A + B + C + D = 4 × 80 = 320 and B + C + D + E = 78 × 4 = 312. Hence, A’s weight is 8 kg more than E ’s weight. A = 71 + 8 = 79.

3.Monday + Tuesday + Wednesday = 3 × 37 = 111;

Tuesday + Wednesday + Thursday = 3 × 34 = 102. Thus, Monday – Thursday = 9 and

Thursday = 4 × Monday/5 Æ Thursday = 36 and Monday = 45.

4.Today’s total age of A, B and C = 30 × 3 = 90.

Today’s total age for B and C = 25 × 2 = 50.

C ’s age = 90 – 50 = 40.

5.9x + 100 = 10(x + 8) Æ x = 20 (average after 9 innings). Hence, new average = 20 + 8 = 28.

6.7 × 3 = 21.

7.1/4 + 1/6 + C = 3 × 1/12 Æ C = –1/6.

8.His total score is 93 + 78 + 177 = 348 out of 450. % score = 77.33

9.Average income over 9 months = [3 × (2750 + 650) + 3 × (3150 + 650) + 3 × (6750+650)]/9

= [3 × 3400 + 3 × 3800 + 3 × 7400]/9 = 4866.66

10.Today’s total age = 6 × 22 = 132 years. Total age of the family excluding the youngest member (for the remaining 5 people) = 132 – 7 = 125. Average age of the other 5 people in the family = 25 years.

7 years ago their average age = 25 – 7 =18 years.

11.If the average age of 8 people has gone up by 2 years it means the total age has gone up by 16 years. Thus the total age of the two women would be: 35 + 45 + 16 = 96. Hence, their average age = 48.

12.W + T + F = 120; T + F + S = 123 Æ S – W = 3. Hence temperature on Wednesday = 42 – 3 = 39.

13.The average speed can be calculated by assuming a distance of 300 km (LCM of 100 and 150). Then time taken @ 100 kmph = 3 hours and time taken @ 150kmph = 2 hours. Average speed = Total distance/total time = 600/5 = 120 kmph.

14.Teacher’s weight = 40.5 × 30 – 40 × 29 = 1215 – 1160 = 55.

15.3 × 17 – 2 × 16 = 51 – 32 = 19.

16.The weight of the new man would be 19 × 3.5 kgs more than the weight of the man he replaces. New man’s weight = 79 + 19 × 3.5 = 145.5 kgs.

17.Let their current ages be x and 3x (ratio of 2:6). Then their ages after 5 years would be x + 5 and 3x + 5. Now it is given that (x + 5)/(3x + 5) = ¾ Æ x = 1 and hence their current ages are 1 years and 3 years respectively. After 10 years their average age would be 12 years.

18.The average would be given by the average of the first and last numbers (since the series 1, 2, 3, 4…97 is an Arithmetic Progression).

Hence, the average = (1 + 97)/2 = 49

19.We need the average of the numbers: 31, 37, 41, 43 and 47

Average = Total/number of numbers Æ 199/5 = 39.8

20.Let the numbers be a, b, c and d respectively. a + b + c = 16 × 3 = 48 and b + c + d = 15 × 3 = 45.

Also, since d = 18, we have b + c = 45 – 18 = 27. Hence, a = 48 – (b + c) Æ a = 21.

21.If the numbers are a + 1, a + 2, a + 3, a + 4 and a + 5 the average would be a + 3. If we take 7 numbers as:

a + 1, a + 2, a + 3, a + 4, a + 5, a + 6 and a + 7 their average would be a+4. Hence, the average increases by 1.

22.Total of 48 numbers = 50 × 38 – 45 – 55 = 1800. Average of 48 numbers = 1800/48 = 37.5.

23.When we multiply each number by 12, the average would also get multiplied by 12. Hence, the new average = 7 × 12 = 84.

24.Let the number of ladies be n. Then we have 8 × 15 + n × 6 = (8 + n) × (10.8) Æ 120 + 6n = 86.4 + 10.8n Æ 4.8n = 33.6 Æ n = 7.

25.(3 × 52 + 60)/4 = 216/4 = 54.

26.1200 × 3 – 1300 × 2 = 1000.

27.(2000 × 15 + 3000 × 15)/30 = (2000 + 3000)/2 = 2500.

28.The decrease in weight would be 15 kgs (5 people’s average weight drops by 3 kgs). Hence, the new person’s weight = 150 × 15 = 135.

29.When a person aged 18 years is replaced by a person aged 38 years, the total age of the group goes up by 20 years. Since this leads to an increase in the average by 5 years, it means that there are 20/5 = 4 persons in the group.

30.(22 × 3 + 17 × 2)/5 = 100/5 = 20.

31.Let the number of marks in the 7th paper be M. Then the total of the first seven papers = 7 × 42 while the total of the last 7 (i.e. 7th to 13th papers) would be 7 × 35.

Total of 1st 7 + total of 7th to 13th = total of all 13 + marks in the 7th paper Æ

7 × 42 + 7 × 35 = 13 × 40 + M

539 = 520 + M Æ M = 19.

(Note: We write this equation since marks in the seventh paper is counted in both the first 7 and the last 7)

32.Let the captain’s age by C. Then: 11 × 30 = 27 × 5 + 29 × 5 + C Æ 330 = 135 + 145 + C Æ C = 50.

33.His earning in the 12th month should be: 5000 × 12 – 4200 × 11 = 60000 – 46200 = 13800.

34.Total distance by total time = 840/13 = 64.615.

35.In three quarters of an hour the train has traveled 54 km. Thus, in a full hour the train would have traveled 1/3rd more (as it gets 1/3rd time more). Thus, the speed of the train = 54 + 1 × 54/3 = 54 + 18 = 72.

36.Total weight of all 6 = 68.5 × 6. Total weight of Ram and Tram = 60 × 2 =120. Average weight of the 4 people excluding Ram and Tram = (68.5 × 6 – 120)/4 = 72.75 kg.

37.10 × 61 + 30 × A = 40 × 52 Æ A = (2080 – 610)/30 = 1470/30 = 49.

38.(80 × 22 + 20 × 37)/100 = 2500/100 = 25

39.If we take the first number as 6n, the second number would be 3n and the third would be 2n. Sum of the three numbers = 6n + 3n + 2n = 11n = 88 × 3 Æ n = 24. The smallest number would be 2n = 48.

40.The ratio between the first , second and third would be: 10:15:24. Since their total is 98, the numbers would be 20, 30 and 48 respectively. The second number is 30.

41.(30 × 160 + 10 × 156)/40 = 159. (note this question can also be solved using the alligation method explained in the next chapter.

42.The total weight of the six people goes up by 15 kgs (when the average for 6 persons goes up by 2.5 kg). Thus, the new person must be 15 kgs more than the person who he replaces.Hence, the new person’s weight = 50 + 15 = 65 kg.

43.Total age = 3 × 15 = 45. Individual ages being in the ratio 3:5:7 their ages would be 9, 15 and 21 years respectively. The youngest boy would be 9 years.

44.50 × 4 + X = 49 × 5 Æ X = 45.

45.(20 × 1900 + M) = 21 × 2000 Æ M = 4000. Hence, the salary is ` 4000 per month which also means ` 48,000 per year.

46.5 consecutive odd numbers would always be in an Arithmetic progression and their average would be the middle number. The average would be ‘c’ in this case.

47.The average of 3, 6, 9, 12 and 15 would be 9.

48.40 × 40 + T = 41 × 40.5 Æ T = 1660.5 – 1600 = 60.5 kgs.

49.If the number of questions correct is N, then the number of wrong answers is 100 – N. Using this we get:

N × 2 – (100 – N) × 0.5 = 120 Æ 2.5 N = 170 Æ N = 68.

50.Required age of the father will be given by the equation: 5 × 12 = 4 × 8 + F Æ F = 28.

51.Required average = (1 + 2 + 3 + … + 10)/10 = 55/10 = 5.5. Alternately you could use the formula for sum of the first n natural numbers as n(n + 1)/2 with n as 10. Then average = Sum/10 = 10 × 11/2 × 10 = 5.5

52.Required average = (0 + 1 + 2 + … + 9)/10 = 45/10 = 4.5

53.Required average = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)/10 = 110/10 = 11. Alternately you could use the formula for sum of the first n even natural numbers as n(n + 1) with n as 10. Then average = Sum/10 = 10 ×11/10 = 11.

54.The sum of the first n odd numbers = n2. In this case n = 10 Æ Sum = 102 = 100. Required average = 100/10 = 10.

55.Required average = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10 = 129/10 = 12.9

56.Required average = (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18)/10 = 112/10 = 11.2

57.Required average = (3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31)/10 = 158/10 = 15.8

58.Teacher’s weight = 31 × 41 – 30 × 40 = 1271 – 1200 = 71.

59.Required average = (2 × 5.5 + 3 × 3.666 + 6 × 1.8333)/11 = (11 + 11 + 11)/11 = 3

60.30 × P + 75 × 2 = 510 Æ P = (510 – 150)/30 = 12

61.Required average = (2 × 3000 + 2 × 500)/4 = 7000/4 = 1750.

62.Required average = Total runs/ total innings = (40 × 4 + 0)/5= 160/5 = 32.

63.Principal’s weight = 41 × 79 – 40 × 80 = 3239 – 3200 = 39.

64.Temperature on 3rd December = 24.4 × 3 – 24 × 2 = 73.2 – 48 = 25.2

65.Average age 5 years hence would be 5 years more than the current average age. Hence, 20 + 5 = 25.

66.Required average = (20 × 30 + 30 × 20)/50 = 1200/50 = 24.

67.The numbers would form an AP with common difference 1 and the middle term (also the 3rd term) as 18. Thus, the numbers would be 16, 17, 18, 19 and 20. The highest of these numbers would be 20.

68.Required average = (6 × 11 + 14 + 16)/8 = 96/8 = 12

69.The new average would also go up by 2. Hence, 12 + 2 = 14.

70.Total age 3 years ago for 5 people = 17 × 5 = 85. Today, the family’s total age = 17 × 6 = 102. The age of the 5 older people would be 85 + 3 × 5 = 100. Hence, the baby’s age is 2 years.

71.Required average = (10 × 31 + 14 × 30 + 15 × 31)/92 = (310 + 420 + 465)/92 = 12.98 (which is closest to 13.

72.Assume a distance of 60 km. In such a case, the Required average = Total distance/Total time = (60 + 60)/(4 + 3) = 120/7 = 17.14

73.(Mon + Tue + wed) = 41 × 3 = 123. (Tue + Wed + Thu) = 40 × 3 = 120.

Mon – Thu = 123 – 120 = 3. Since Thursday’s temperature is given as 39, Monday’s temperature would be 39 + 3 = 42.

74.Required average = (30 × 20 – 10 × 10)/10 = 500/10 = 50.

75.Total age of 7 children = 12 × 7 = 84 years. When the 6 year old child dies, the total age of the remaining 6 children would be 84 – 6 = 78. Required average = 78/6 = 13 years.

76.Required average = (2 × 200 + 2 × 250)/4 = 900/4 = 225.

77.Teacher’s weight = 36 × 36 – 35 × 35 = 1296 – 1225 = 71 kgs.

78.The statement ‘x is as much more than the average as y is less than the average signifies that the numbers x, y, z form an Arithmetic Progression with z as the middle term. z’s value would then be equal to the average of the three numbers. This average is given as 45. Hence, the correct answer is z = 45.

79.The sum of the given 4 numbers is 20.75. The required average = 20.75/4. Option (a) is correct.

80.Let the number of non officer workers in the company be W. Then we will have the following equation: (15 × 525 + W × 85) = (15 + W) × 95 Æ 10 W = 15 × 430 Æ W = 645. Thus, the total number of workers in the company would be 645 + 15 = 660.

81.The woman’s age would be 8 × 2 = 16 years more than the age of the man she replaces. Age of the woman = 24 + 2 × 8 = 40 years.

82.Required average income = (Total expenditure + total savings]/12

= [(1100 × 3 + 2200 × 4 + 4620 × 5) + 2100]/12 = 37300/12 = 3108.333

83.Required average = (2 × 5.5 + 3 × 3.5 + 3 × 5.5 + 5 × 1.5)/13 = 45.5/13 = 3.5

84.For 150 coins to be of a value of ` 150, using only 25 paise, 50 paise and 1 Re coins, we cannot have any coins lower than ther value of 1 `. Thus, the number of 50 paise coins would be 0. Option (d) is correct.

Level of Difficulty (II)

1.The train needs to travel 15 minutes extra @35 kmph. Hence, it is behind by 8.75 kms. The rate of losing distance is 5 kmph. Hence, the train must have travelled for 8.75/5 = 1 hour 45 minutes. @ 40 kmph Æ 70 km.

Alternatively, you can also see that 12.5% drop in speed results in 14.28% increase in time. Hence, total time required is 105 minutes @ 40 kmph Æ 70 kilometers.

Alternatively, solve through options.

2.Standard question requiring good calculation speed. Obviously, the 15th day is being double counted. Calculations can be reduced by thinking as:

Surplus in first 15 days – Deficit in last 17 days = 255 – 289 Æ Net deficit of 34. This means that the average is reducing by 34 due to the double counting of the 15th day. This can only mean that the 15th day’s expenditure is ` 68 – 34 = 34.

(Lengthy calculations would have yielded the following calculations:

85*15 + 51*17 – 68*31 = 34)

3.Find out the number of innings in each year. Then the answer will be given by:

(4270/119 = 35.88)

4.Find the total distance covered in each segment of 10 minutes. You will get total distance = 46 kilometers in 50 mins.

5.Assume that the distance is 120 km. Hence, 30 km is covered @ 25 kmph, 40 @ 30 kmph and so on.

Then average speed is 120/total time

6.In three hours the total number of sheets typed will be: 60/6 + 60/7 + 60/9 = 10 + 8.57 + 6.66.

Hence the number of sheets/hour is 25.23/3 = 8.41 is equivalent to 530/63.

7.100 Æ 130 Æ 182. Hence, 82/2 = 41.

8.You do not know the number of days in the month. Hence, the question cannot be answered.

9.The number of rooms is 18 + 16 + 30 on the three floors respectively.

Total revenues are: 18*200 + 16*100 + 30*150 = 9700 required average = 9700/110 = 88.18.

Note here that if you could visualize here that since the number of rooms is 110 the decimal values cannot be .3 or .7 which effectively means that options 3 and 4 are rejected.

10.Replace x with the sales value to calculate the bonus in a month.

11.Use the same process as Q. No. 5 above.

12.The weight of the second man is 135 and that of the third is 105. Hence, net result is a drop of 45 for 5 people. Hence, 9 kg is the drop.

13.Put x = 100 to get the weight of the containers. Use these weights to find average weight as 2600/4 = 650.

14.The weight of the boxes are 1st box Æ 200, 3rd box Æ 250 kg, 2nd box Æ 300 kg, 4th box Æ 350 and 5th box Æ 500 kg. Hence difference between the heavier 4 and the lighter 4 is 300. Hence, difference in the averages is 75.

15.Difference between heaviest three and lightest three totals is: (350 + 500) – (300 + 200) = 350

Difference in average weights is 350/3 = 116.66.

16.Part of the runs scored in the 87th innings will go towards increasing the average of the first 86 innings to the new average and the remaining part of the runs will go towards maintaining the new average for the 87th innings. The only constraint in this problem is that there is an increase in the average by a whole number of runs. This is possible for all three options.

17.Assume x is the average expenditure of 19 people. Then, 19x = 13*79 + 6(x + 4).

18.42 A + 31 = 55 (A – 3) Æ 13 A = 196 Æ A = 196/13 = 15.07. Total expenditure original = 15.07 × 42 = 633.23

19.The total price of the three stones would be 97610498312 × 3 = 292831494936. Since, this price is divided into the three stones in the ratio of 4 : 7 : 9, the price of the cheapest one would be = (4 × 2928314936/20) = 58566298987.2

20.The average weight per ball is asked. Hence, the bag does not have to be counted as the 48th item.

21.71*48 = 59*46 + x + 11*52 Æ x = 72.

Alternately, this can be solved by using the concepts of surpluses and deficits as:

2*59 (deficit) – 4 * 11(surplus) + 48 (average to be maintained by the 60th number) = 118-44 + 48 = 122.

22.Solve through the same process as the Q. No. 5 of this chapter.

23.(14 * 333 – 2 * 504)/12.

24..

25.Use alligation to solve. 20———32———68. Thus, 5 corresponds to 12, hence for 36 the answer will be 15.

26.Let the equal distances be ‘d’ each. Then 3d/(d/x + d/y + d/z) = 3xyz/(x + y + z).

27–30.You have to take between 25th and 30th to mean that both these dates are also included.

27.The maximum average will occur when the maximum possible values are used. Thus:

A should have been born on 30th, B on 25th, C on 20th, D on 10th and E on 5th. Further, the months of births in random order will have to be between August to December to maximize the average.

Hence the total will be 30 + 25 + 20 + 10 + 5 + 12 + 11 + 10 + 9 + 8 = 140. Hence average is 28.

28.The minimum average will be when we have 1 + 5 + 10 + 20 + 25 + 1 + 2 + 3 + 4 + 5 = 76. Hence, average is 15.2.

29.This does not change anything. Hence the answer is the same as Q. 27.

30.The prime dates must be 29th, 23rd, 19th and 5th. Hence, the maximum possible average will reduce by 4/5 = 0.8. Hence, answer will be 27.2.

31.Solve using alligation. Since 15 is the mid-point of 13.25 and 16.75, the ratio is 1:1 and hence there are 20 people who were going for the picnic initially.

32.The number of pass candidates are 2 + 6 + 18 + 40 = 66 out of a total of 100. Hence, 66%.

33&34. Put x = 10 in the given equations and find the average of the resultant values.

35.Solve through options.

36.nz–x + x1 = nz1 Æ Simplify to get Option (c) correct.

37.By alligation the ratio is 3:8. Hence, only 110 is possible.

Q38-41:

38.You don’t know who got out when. Hence, cannot be determined.

39.Since possibilities are asked about, you will have to consider all possibilities. Assume, the sixth and seventh batsmen have scored zero. Only then will the possibility of the first 5 batsmen scoring the highest possible average arise. In this case the maximum possible average for the first 5 batsmen could be 403/5 = 80.6.

40.Again it is possible that only the first batsman has scored runs.

41.We cannot find out the number of runs scored by the 7th batsman. Hence answer is (d).

42.You can take 53 as the base to reduce your calculations. Otherwise the question will become highly calculation intensive.

43.251.25*4 + 277.52 * 5 + 760 = 3152.6

44.Solve using options. 20 is the only possible value.

45.Check through options to solve.

Level of Difficulty (III)

1.Definitely decrease, since the highest marks in Class A is less than the lowest marks in Class B.

2.Cannot say since there is no indication of the values of the numbers which are transferred.

3.It will definitely decrease since the highest possible transfer is lower than the lowest value in C.

4.The effect on A will depend on the profile of the people who are transferred. Hence, anything can happen.

5.Cannot say since there is a possibility that the numbers transferred are such that the average can either increase, decrease or remain constant.

6.If C increases, then the average of C goes up from 30. For this to happen it is definite that the average of B should drop.

7.The maximum possible average for B will occur if all the 5 transferees from A have 22 marks.

8.The average of Group A after the transfer in Q. 7 above is:

(400 – 18*5)/15 = 310/15 = 20.66

9.(400 – 22*5)/15 = 19.33

10.400 + 23*5 = 515. Average = 515/25 = 20.6

11.400 + 31*5 = 555. Average = 555/25 = 22.2

12.Will always decrease since the net value transferred from B to A will be higher than the net value transferred from A to B.

13.Since the lowest score in Class B is 23 which is more than the highest score of any student in Class A. Hence, A’s average will always increase.

14.The maximum possible value for B will happen when the A to B transfer has the maximum possible value and the reverse transfer has the minimum possible value.

15.For the minimum possible value of B we will need the A to B transfer to be the lowest possible value while the B to A transfer must have the highest possible value. Thus, A to B transfer Æ 18* 5 while B to A transfer will be 31*5. Hence answer is 22.4.

16.The maximum value for A will happen in the case of Q. 15. Then the increment for group A is:

31*5 – 18*5 = 5*(31 – 18) = 65.

Thus maximum possible value is 465/20 = 23.25.

17.Minimum possible average will happen for the transfer we saw in Q. 14. Thus the answer will be 405/20 = 20.25.

18.The maximum possible value for C will be achieved when the transfer from C is of five 26’s and the transfer back from B is of five 31’s. Hence, difference is totals will be +25. Hence, max. average = (900 + 25)/30 = 30.833.

[Note here that 900 has come by 30*30]

19.For the maximum possible value of Class B the following set of operations will have to hold:

Five 33’s are transferred from C to B, whatever goes from B to A comes back from A to B, then five 23’s are transferred from B to C. This leaves us with:

Increase of 50 marks Æ average increases by 2 to 27.

20.A will attain maximum value if five 33’s come to A from C through B and five 18’s leave A. In such a case the net result is going to be a change of +75. Thus the average will go up by 75/20 = 3.75 to 23.75.

21–23.Will be solved by the same pattern as the above questions.

24.Only option A will give us the required situation since the transfer of five 31’s increases the value of the average of group A.

25–28.Will be solved by the same pattern as above questions.

29–35.These are standard questions using the concept of averages. Hence, analyse each and every sentence by itself and link the interpretations. If you are getting stuck, the only reason is that you have not used the information in the questions fully.

36.Monthly estimates of income is reduced as the denominator is increased from 12 to 14 at the same time the monthly estimate of expenditure is increased as the denominator is reduced from 12 to 9. Hence, the savings will be underestimated.

37–39.Use the averages formulae and common sense to answer.

40–49.The questions are commonsensical with a lot of calculations and assumptions involved. You have to solve these using all the information provided.

40.Das’s score = 15*2 + 7*2.5 = 47.5 Æ 48.

Dasgupta’s score = 20*1 + 2*1.5 = 23

41.From the above the answer is 48:23 = 96:46.

42–44.By maximum possible support from the other end, you have to assume that he has Laxman or Sehwag batting aggressively for the entire tenure at the crease. Strike has to be shared equally.

42.Through options, After 60 overs, score would be 150. Then Tendulkar can score @ 4 runs per over (sharing the strike and batting aggressively) and get maximum support @ 3 runs per over. Thus in 30 overs left the target will be achieved.

43.Tendulkar’s score for the innings will be 30*4 = 120.

44.We do not know when Laxman would have come into bat. Hence this cannot be determined.

45–49.Build in each of the conditions in the problem to form a table like:

Partnership PartnerOvers faced Tendulkar’s scorePartner’s score

6th wicket Ganguly 12 6 overs × 6 6 overs × 4

7th wicket and so on

8th wicket

9th wicket

10th wicket