MATHS TRICKS: AVERAGES
Maths Tricks : To find out the averages we use the below given formaula
Average of ‘n’ numbers = (Sum of the numbers) / n
Eg: Average of 40, 60, 25, 75 = (40+60+25+75)/4 = 200/4 = 50
This is known to us all..
In competitive exams like Kerala PSC, state PSC, UPSC, SSC, Banks etc..,., the question asked in averages is as follows (these are just a few examples….)

CLASSROOM/
When teachers age added,subtracted what is new average or
If teachers age added,subtracted what will be the change in average .. or
If a new boy comes, goes then what will be change in average or find the age of new boy etc..

MARKS
Increase or decrease in average when marks of a new student added or vice versa

CRICKETER and INNINGS
Increase or decrease in average score of a batsman, or individual score of a cricketer when we are given averages

LET US LEARN THROUGH EXAMPLES……
Q) The average age of 30 students in a class is 10. When teachers age is added the average age becomes 11. Find out the age of the teacher?
Solution)

CONVENTIONAL METHOD TO SOLVE AVERAGES
Total number of students = 30
Average age of students  = 10
Therefore the sum of ages of all 30 students = 30 x 10 = 300
When teachers age is added the total number becomes = 30 + 1 = 31
The new average = 11
Therefore the sum of ages of students and the teacher = 31 x 11 = 341
Teachers age = (Sum of ages of teacher + students) – (Sum of ages of students)
= 341- 300
= 41

VERY SIMPLE METHOD – USE YOUR COMMONSENSE:
Let us assume that the teachers age was also 10. In such case the average of all 31 people ( 30 students + 1 teacher) remains same which is 10.
But this is not the case. On arrival of teacher the average has increased by 1. So 1 has been added to all the 31 people of the class.
It means teachers age should be more than 10. How much more ? It is 31 more …
10 + 31 = 41

FORMULA METHOD
TEACHERS AGE = NEW AVERAGE + (NUMBER OF STUDENTS X DIFFERENCE OF AVERAGES)
In above case
TEACHERS AGE = 11 + ( 30 X 1) = 11+ 30 = 41

GENERAL FORMULA
 

NEW PERSON’S AGE OR WEIGHT = NEW AVERAGE + ( NUMBER OF PEOPLE BEFORE NEW PERSON CAME X DIFFERENCE BETWEEN THE NEW AND OLD AVERAGES)
TYPE II
Q) The average  marks obtained by 10 students in an exam is 55. When a boy who scored 65 marks left the class and another one joined the average reduced by 3. What was the marks of new boy?
Solution)
CONVENTIONAL METHOD
Average marks of 10 students = 55
Total marks = 55 x 10 = 550
Marks of 9 student after boy scoring 65 left = 550-65 = 485
New average for 10 students = 55 – 3 = 52
Total marks of 10 students now = 52 x 10 = 520
Therefore the marks of new boy = 520 – 485 = 35
COMMONSENSE
If the boy who came also scored 65 then no problem at all, the average would have remained same that is 55. But the average reduced by 3. So from each person he reduced 3.
Total persons  are 10
So total marks he reduced is 10 x 3 = 30
So from 65 we can reduce this 30. We get 65-30= 35
FORMULA METHOD
NEW PERSON’S AGE/WEIGHT/MARKS = AGE/WEIGHT/MARKS OF THE PERSON WHO LEFT + (TOTAL PERSONS  X  DIFFERENCE IN AGE/MARKS/WEIGHT)
NEW PERSON’S MARKS = 65 + (10 X -3) = 65 -30 = 35

Questions on clocks (or even calendars) are not really frequent in CAT these days. They used to be really popular few years ago. Having said that, it is always better to understand some of the basic principles and the types of problems that get asked. They might come in handy in case of other exams like CMAT, MAT, SNAP, etc.
Clock problems can be broadly classified in two categories:

a) Problems on angles
b) Problems on incorrect clocks

Problems on angles

Before we actually start solving problems on angles, we need to get couple of basic facts clear:

· Speed of the hour hand = 0.5 degrees per minute (dpm) {The hour hand completes a full circle or 360 degrees in 12 hours or 720 minutes}

· Speed of the minute hand = 6 dpm {The minute hand completes a full circle in 60 minutes}

· At ‘n’ o’ clock, the angle of the hour hand from the vertical is 30n
The questions based upon these could be of the following types
Example 1: What is the angle between the hands of the clock at 7:20
At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In 20 minutes,
Hour hand = 210 + 20*(0.5) = 210 + 10 = 220 {The hour hand moves at 0.5 dpm}
Minute hand = 20*(6) = 120 {The minute hand moves at 6 dpm}
Difference or angle between the hands = 220 – 120 = 100 degrees
Example 2: At what time do the hands of the clock meet between 7:00 and 8:00
Ans: At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In ‘t’ minutes
Hour hand = 210 + 0.5t
Minute hand = 6t
They should be meeting each other, so
210 + 0.5t = 6t

=> t = 210/5.5 = 420/11= 38 minutes 2/11th minute
Hands of the clock meet at 7 : 38 : 2/11th
Example 3: At what time do the hands of a clock between 7:00 and 8:00 form 90 degrees?
Ans: At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In ‘t’ minutes
Hour hand = 210 + 0.5t
Minute hand = 6t
The difference between them should be 90 degrees. Please note that it can be both before the meeting or after the meeting. You will get two answers in this case, one when hour hand is ahead and the other one when the minute hand is ahead.
Case 1: 210 + 0.5t – 6t = 90

=> 5.5t = 120

=> t = 240/11 = 21 minutes 9/11th of a minute
Case 2: 6t – (210 + 0.5t) = 90

=> 5.5t = 300

=> t = 600/11 = 54 minutes 6/11th of a minute
So, the hands of the clock are at 90 degrees at the following timings:

7 : 21 : 9/11th and 7 : 54 : 6/11th
Some other results which might be useful:

· Hands of a clock meet at a gap of 65 5/11 minutes.

· The meetings take place at 12:00:00, 1:05:5/11, 2:10:10/11 … and so on.

· Hands of a clock meet 11 times in 12 hours and 22 times in a day.

· Hands of a clock are perfectly opposite to each other (i.e. 180 degrees) 11 times in 12 hours and 22 times a day. {Same as above}

· Any other angle is made 22 times in 12 hours and 44 times in a day
Problems on incorrect clocks

Such sort of problems arise when a clock runs faster or slower than expected pace. When solving these problems it is best to keep track of the correct clock.

Example 4: A watch gains 5 seconds in 3 minutes and was set right at 8 AM. What time will it show at 10 PM on the same day?
Ans: The watch gains 5 seconds in 3 minutes => 100 seconds in 1 hour.
From 8 AM to 10 PM on the same day, time passed is 14 hours.
In 14 hours, the watch would have gained 1400 seconds or 23 minutes 20 seconds.
So, when the correct time is 10 PM, the watch would show 10 : 23 : 20 PM

Example 5: A watch gains 5 seconds in 3 minutes and was set right at 8 AM. If it shows 5:15 in the afternoon on the same day, what is the correct time?

Ans: The watch gains 5 seconds in 3 minutes => 1 minute in 36 minutes
From 8 AM to 5:15, the incorrect watch has moved 9 hours and 15 minutes = 555 minutes.
When the incorrect watch moves for 37 minutes, correct watch moves for 36 minutes.

=> When the incorrect watch moves for 1 minute, correct watch moves for 36/37 minutes
=> When the incorrect watch moves for 555 minutes, correct watch moves for (36/37)*555 = 36*15 minutes = 9 hours
=> 9 hours from 8 AM is 5 PM.
=> The correct time is 5 PM.

I am sure you would have heard the proverb that even a broken clock is right twice a day. However, a clock which gains or loses a few minutes might not be right twice a day or even once a day. It would be right when it had gained / lost exactly 12 hours.

Example 6: A watch loses 5 minutes every hour and was set right at 8 AM on a Monday. When will it show the correct time again?

Ans: For the watch to show the correct time again, it should lose 12 hours.
It loses 5 minutes in 1 hour
=> It loses 1 minute in 12 minutes
=> It will lose 12 hours (or 720 minutes) in 720*12 minutes = 144 hours = 6 days
=> It will show the correct time again at 8 AM on Sunday.

 

 1.The scientific study of beauty is called?

Answer: Aesthetics 

2.The seasonal winds of south Asia is known as .............?

 Answer: Monsoons 

3.International womens day is observed on -

Answer : March 

8 4.Pick out the incorrect singular - plural group from among the following?

     (a)Child - Children

     (b)story - series

     (c)Mouse - Mice

     (d)Wolf - wolves

Answer: Story - Series 

5.There is ............hourly bus from here to the capital ,

Answer: an

If the world inspector is coded as 123456789, what is the code of inspection?

Answer: 1234567182 

6.Who is the wrong member in the following group?

     (a) Kissy

    (b)Lissy

    (c)Dizzy

    (d)Missy

Answer:Dizzy 

7. IAS=538,

IAS+IFS=1136. what is IAS x IFS?

Answer: 321724 

 8.Find the missing member in the series WXYZ, VWXY...........TUVW :-

Answer: UVMX 

9.If X:Y = 5:3 What is (x+y) :(x-y)?

 Answer: 4:1

10Which number will complete the given series: 1,9,19,31,45,.....................?

       Answer: 61 11.(1.2)2 + (1.5)2=...............? 

       Answer: 0.369 12.(-1)000 =? 

       Answer: -1 13.10[2+(20-.5)x3]-10=? 

       Answer: 130 

14.A train of 100 Meters length is running at a speed of 72 km per hour. What time it take to pass a man standing at the railway track?

Answer:5 seconds 

15. Express 30 paise as the percentage of Rs.6 -

 Answer:5%