Most of the points that I have listed below are very basic.
In a Arithmetic Progression a,b,c.
The arithmetic mean b = (a+c)/2.
If we insert 'n' arithmetic means between 2 terms a and b,
we will get an AP with n+2 terms.
Take 2 numbers 10 and 20.
If we are inserting 1 Arithmetic mean between these 2 terms,
it will be 15.
Now 15 is the Arithmetic mean of 10 and 20.
If we are inserting 3 Arithmetic means between 10 and 20.
The AP becomes
10,12.5,15,17.5,20
Now the Arithmetic mean of the 3 inserted numbers 12.5,15,17.5 is
15 which is the arithmetic mean of the original 2 numbers.
Here in the AP 10,12.5,15,17.5,20 we have 5 terms,
To determine the arithmetic means to be inserted,
we can use the formula
(Difference between the 2 terms)/Number of means to be inserted+1
Lets see the case explained above.
We need to insert 1 AM between 10 and 20.
So 20-10/(1+1)
We get 5.
We need add this to the smaller number.
We get 15.
We see that in an AP with odd number of terms,
first term+last term = second term + second last term = 2*middle term
Some properties that will of help while solving problems.
If Sum of 'P' terms = Sum of 'Q' terms then
Sum of 'P+Q' terms = 0
Consider the AP 2,1,0,-1,-2
Here Sum upto 1 term is 2
Sum of 4 terms is also 2.
Sum of the 5 terms is 0.
If 'P'th term=Q and 'Q'th term=P
'P+Q'th term = 0
Consider 4,3,2,1,0
4th term is 1
1st term is 5
and 5th term is 0
Wednesday, June 18, 2008
Speed Maths 3
Yet another method to find Squares faster.
I will start with an example.
Say We need to find the square of 16.
The nearest 10 is 20.
To get 20, we add 4 to 16,
i.e 16+4=20
Since we have added 4, to get the nearest ten.
We subtract 4 from 16..
16-4=12
To get the square of 16, 20 * 12 + (4 ^ 2)
=240 + 16 = 256.
For a 2 digit number we go for the nearest 10.
If it is a 3 digit number we should go for the nearest 100.
Few more examples
1) 24
Nearest 10 is 20.
24- 4=20
So to get the square, 20 * 28 + (4 ^ 2 )
The answer is 576.
2) 111
Nearest 100 is 100
111-11 = 100
So to get the square 100 * 122 + (11^ 2)
=12200+121
=12321
Will post the next one on progressions.
But new concepts in progressions are limited.
So that will be a small post .
I will start with an example.
Say We need to find the square of 16.
The nearest 10 is 20.
To get 20, we add 4 to 16,
i.e 16+4=20
Since we have added 4, to get the nearest ten.
We subtract 4 from 16..
16-4=12
To get the square of 16, 20 * 12 + (4 ^ 2)
=240 + 16 = 256.
For a 2 digit number we go for the nearest 10.
If it is a 3 digit number we should go for the nearest 100.
Few more examples
1) 24
Nearest 10 is 20.
24- 4=20
So to get the square, 20 * 28 + (4 ^ 2 )
The answer is 576.
2) 111
Nearest 100 is 100
111-11 = 100
So to get the square 100 * 122 + (11^ 2)
=12200+121
=12321
Will post the next one on progressions.
But new concepts in progressions are limited.
So that will be a small post .
Friday, June 13, 2008
Is High reading speed necessary to do well in the Verbal Section
This is something that goes through everybody's mind.
There are all kinds of views on this.
In my opinion "A high reading speed is not necessary to do well
in the verbal section". It is certainly not sufficient to do well
in the verbal section.
Off late , CAT has moved towards smaller passages.
So Numbe of words in a RC Passage is typically 800-900
as opposed to 1200+.Also passages are very abstract and difficult.
So you may have to read the passage , and then come back to the passage
after taking a look at the questions.Considering the difficulty level of the CAT
passages these days, I think regression is unavoidable.
CAT has increased the time available and simultaneously reduced the
number of questions.Even if you have a very high reading speed , assuming
that you get 3 RC passages in CAT , the time you save will be 2-3 minutes.
Now IMHO this is negligible.
What is required is a moderate reading speed and good comprehension.
Also an ability to read vague and abstract passages.There are times when I
attempt a very difficult passage but the effort of reading drains me
completely and I have no enthusiasm to attend the next section.
One must ideally read a lot and read from a cross section of subjects.
Increased reading is the only way to do well in RCs.If you do have a
great reading speed, feel happy even though it may not give you a
great advantage in CAT.Everybody who talks about reading speed,
talks about the case studies and the amount of reading required while
doing MBA.So maybe your reading speed will keep you in good stead there.
What is a high reading speed ? 500+
Moderate- 300+
Even with a speed of 250+, one should be able to RCs well.
But a speed less than that means no regular reading or more likely
an aversion to reading.Now that's not a good thing.Read a few books,
read 2 papers regularly,read editorials regularly.Read atleast 1 weekly
magazine.6 months of this should definitely bring the reading speed to
250+. That should be enough to keep your heads above the water.
There are all kinds of views on this.
In my opinion "A high reading speed is not necessary to do well
in the verbal section". It is certainly not sufficient to do well
in the verbal section.
Off late , CAT has moved towards smaller passages.
So Numbe of words in a RC Passage is typically 800-900
as opposed to 1200+.Also passages are very abstract and difficult.
So you may have to read the passage , and then come back to the passage
after taking a look at the questions.Considering the difficulty level of the CAT
passages these days, I think regression is unavoidable.
CAT has increased the time available and simultaneously reduced the
number of questions.Even if you have a very high reading speed , assuming
that you get 3 RC passages in CAT , the time you save will be 2-3 minutes.
Now IMHO this is negligible.
What is required is a moderate reading speed and good comprehension.
Also an ability to read vague and abstract passages.There are times when I
attempt a very difficult passage but the effort of reading drains me
completely and I have no enthusiasm to attend the next section.
One must ideally read a lot and read from a cross section of subjects.
Increased reading is the only way to do well in RCs.If you do have a
great reading speed, feel happy even though it may not give you a
great advantage in CAT.Everybody who talks about reading speed,
talks about the case studies and the amount of reading required while
doing MBA.So maybe your reading speed will keep you in good stead there.
What is a high reading speed ? 500+
Moderate- 300+
Even with a speed of 250+, one should be able to RCs well.
But a speed less than that means no regular reading or more likely
an aversion to reading.Now that's not a good thing.Read a few books,
read 2 papers regularly,read editorials regularly.Read atleast 1 weekly
magazine.6 months of this should definitely bring the reading speed to
250+. That should be enough to keep your heads above the water.
Thursday, June 12, 2008
TIME AIMCAT 0919
The results of the next AIMCAT is out.
Compared to 920, This went a lot better for me and that
shows in the result.I have 2 ver good sections -DI and Verbal.
Quant was horrible.
Made many stupid mistakes in Quant.Need to improve the accuracy
and be careful while doing Quant.
Positives :
.For the first time, very good accuracy rate in Verbal and DI
.A good DI section after a long time.
.After a poor verbal in CAT 07, Good verbal scores in the mocks
give me some confidence
Negatives:
No improvement in Quant despite the effort put in.
Need to do practice more sums ....
Compared to 920, This went a lot better for me and that
shows in the result.I have 2 ver good sections -DI and Verbal.
Quant was horrible.
Made many stupid mistakes in Quant.Need to improve the accuracy
and be careful while doing Quant.
Positives :
.For the first time, very good accuracy rate in Verbal and DI
.A good DI section after a long time.
.After a poor verbal in CAT 07, Good verbal scores in the mocks
give me some confidence
Negatives:
No improvement in Quant despite the effort put in.
Need to do practice more sums ....
Wednesday, June 11, 2008
Last 2 digits of an Expression
Last 2 digits is the remainder when Divided by 100.
This can be found usind Binomial theorem.
The method described below is quick and is
an extension of the Binomial Theorem.
For expressions ending in '5'
Last 2 digits will always be 25.
For '1'
I will first so some examples and then follow them with an
explanation.
Take 41 ^ 73
The last digit/ the right most digit will always be 1
For the last but one digit take the ten's digit of the base i.e 4 in the above
example and multiply it will the unit's digit of the power i.e 3 in the example.
So 2.
The last 2 digits will be 21.
Some more examples.
31 ^ 123 =
Right most is one.
Last but one will be 3 (ten's digit of the base) * 3 (unit's digit of the power)=9
So last 2 digits will be 91
81 ^ 1423 = 41
Last but one here will be 8*3
For Numbers ending in 3,7,9
From the power cycles,
3 ^ 4k/4k+4 and 7 ^ 4k/4k+4 , 9 ^ 2k will end in 1.
So using this we can convert them into expressions ending in 1.
233 ^ 56
The last 2 digits will depend on 33 ^ 56
This becomes (33 ^ 4 ) ^ 14
33 ^ 4 is (33 ^ 2 ) * (33 ^ 2)
I have already written about quicker methods to get squares.
33 ^ 2 will end in 89
Finally we get 21 ^ 14
This can be found using the method explained for 1.
For expressions ending in even numbers
Some pointers to remember
1024 ^ even number will always end in 76
1024 ^ odd number will always end in 24
1024 is 2 ^ 10
For most expressions ending in an even number,
we can take 2 ^ x as common and express that as 10k+
Say we have 74 ^ 288
74 is 2 * 37
so we get (2 ^ 288)*(37^288)
2 ^ 288 will be (2 ^ 10)^28 * (2 ^ 8)
We can proceed with this
and use earlier methods to get 37 ^ 288.
Agreed it's not an easy method by any means.
This can be found usind Binomial theorem.
The method described below is quick and is
an extension of the Binomial Theorem.
For expressions ending in '5'
Last 2 digits will always be 25.
For '1'
I will first so some examples and then follow them with an
explanation.
Take 41 ^ 73
The last digit/ the right most digit will always be 1
For the last but one digit take the ten's digit of the base i.e 4 in the above
example and multiply it will the unit's digit of the power i.e 3 in the example.
So 2.
The last 2 digits will be 21.
Some more examples.
31 ^ 123 =
Right most is one.
Last but one will be 3 (ten's digit of the base) * 3 (unit's digit of the power)=9
So last 2 digits will be 91
81 ^ 1423 = 41
Last but one here will be 8*3
For Numbers ending in 3,7,9
From the power cycles,
3 ^ 4k/4k+4 and 7 ^ 4k/4k+4 , 9 ^ 2k will end in 1.
So using this we can convert them into expressions ending in 1.
233 ^ 56
The last 2 digits will depend on 33 ^ 56
This becomes (33 ^ 4 ) ^ 14
33 ^ 4 is (33 ^ 2 ) * (33 ^ 2)
I have already written about quicker methods to get squares.
33 ^ 2 will end in 89
Finally we get 21 ^ 14
This can be found using the method explained for 1.
For expressions ending in even numbers
Some pointers to remember
1024 ^ even number will always end in 76
1024 ^ odd number will always end in 24
1024 is 2 ^ 10
For most expressions ending in an even number,
we can take 2 ^ x as common and express that as 10k+
Say we have 74 ^ 288
74 is 2 * 37
so we get (2 ^ 288)*(37^288)
2 ^ 288 will be (2 ^ 10)^28 * (2 ^ 8)
We can proceed with this
and use earlier methods to get 37 ^ 288.
Agreed it's not an easy method by any means.
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