Q 1) Total number of Integral solutions of x^2 + 4 * y^2 = 100
a. 6
b. 8
c. 10
d. 12
e. 14
OA : 12
Q 1) Total number of Integral solutions of x^2 + 4 * y^2 = 100
a. 6
b. 8
c. 10
d. 12
e. 14
OA : 12
So one fine day you had this epiphany that a management degree from a premium B school would solve your current problems (ranging from impressing your crush to improving your salary) and will change your life for good. You decided to give CAT and thought what a funny name that is for an exam. You googled for the syllabus and found the most comforting sentences you ever read in your life – High school Math and English. You are happy as you always scored well in Math and you loved reading Harry potter series. You joined some forums where you met your fellow aspirants - Clark Kent, Bruce Wayne, Peter Parker and Pandit Gangadhar Vidyadhar Mayadhar Omkarnath Shastri. Everything was peaceful until one day you saw some weird numbers like 123456…. 1000 digits or 9999^55555. You were bemused thinking why on earth these kind of questions appearing in a forum that should be focusing on high school math. To add up to your pain, all your above said mortal friends turned in to Superman, Batman, Spiderman, Shaktiman and solved them in a blink of an eye. Some even complained why such super easy questions are being posted.
When I started seeing all such numbers, my expression was almost like this and the case would be similar for most of the mortal newbies.
Don’t worry, it is just a very common initial crisis most of the CAT aspirants face (even for people from math background) and with very minimal effort, you can also put a cape and come up with some fancy names for yourself.
In this article we will learn some simple and interesting concepts to solve 10 such "Monster" number waala questions. If you know how to solve all the given questions without taking more than a minute per question, you need not spend time on this page and can explore other interesting articles we have in MBAtious. For others, let’s have some fun!
Soltuions:
Please let me know in case of errors or better methods :)
Happy learning!
We will now learn another important concept, divisibility check of numbers. These concepts will be used extensively during prime factorization, while calculating HCF/LCM and also in finding remainders. Here we will just focus on how we can easily check whether a given number is divisible by some common divisors or not.
We will start with the most important one in the list
Any digit repeated ( P - 1 ) times is divisible by P, where P is a prime > 5 |
Find the remainder when 7777.... (100 digits) is divided by 13
We know Remainder[777... 96 digits ( 12 * 8 ) / 13 ] = 0
so Remainder [ 777... 100 digits / 13 ] = Remainder [ 7777 ( remaining 4 digits) / 13 ] = 3
Now we will see some other rules which most of you are already famililar with
Divisible by 2: If the last digit is divisible by 2.
12, 142, 68…
Divisible by 4: If the last 2 digits are divisible by 4.
724, Last 2 digits (24) gives a number divisible by 4.
Divisible by 8: If the last 3 digits are divisible by 8.
1040, Last 3 digits (040) gives a number divisible by 8.
Got the pattern ?
A number is divisible by 2^{n }if the last n digits are divisible by 2^{n} |
Divisible by 3: Sum of digits of the number is divisible by 3.
15672, sum of digits = 1+5+6+7+2 = 21 = 3 * 7, hence divisible by 3.
Divisible by 9: If the sum of the digits is divisible by 9
972036, sum of the digits = 9 + 7 + 2 + 0 + 3 + 6 = 27, divisible by 9.
Why this is true ? Let the number be ab, where a and b are the digits. We know ab = 10a + b = 9a + (a+b). So ab is divisible by 3 (or 9) if (a+b) is divisible by 3 (or 9) :)
Divisible by 33, 333, 3333… & 99, 999, 9999…
To check a given number is divisible by 333…3 (n digits) just see whether the sum of digits taken n at a time from right to left is divisible by 333…3 (n digits). If yes then the original number is also divisible by 33…3 (n digits) Same can be applied for checking the divisibility of a given number with 99…9 (n digits). Check if the sum of digits taken n at a time from right is divisible by 999…9 (n digits). If yes then the original number is also divisible by 99…9 (n digits) |
It is easy to understand through examples.
Is 627 divisible by 33 ?
Take 2 digits from right at a time, and get the sum.
27 + 06 = 33, hence divisible by 33
Is 22977 divisible by 333 ?
Take 3 digits from right at a time and find the sum.
977 + 022 = 999, hence divisible by 333
Is 6435 divisible by 99 ?
35 + 64 = 99. As per the above rule, 6435 is divisible by 99.
How much time you need to tell whether the number 1000000998 is divisible by 999 ?
Divisible by 5: If the last digit is 5 or 0.
E.g. 625, 310 etc…
Divisible by 25: If the last two digits are divisible by 25
Eg: 125, 50 etc..
Divisible by 125: If the last three digits are divisible by 125
Eg: 1250, 3500 etc..
A number is divisible by 5^{n }if the last n digits are divisible by 5^{n} |
Divisible by 7: Subtract twice the unit digit from the remaining number.If the result is divisible by 7, the original number is.
14238, 1423 - 2 * 8 = 1407 = 201 * 7, hence divisible by 7
Divisible by 11: If the difference between the sum of digits at the odd place and the sum of digits at the even place is zero or divisible by 11.
1639, (9+6) - (3+1) = 11, divisible by 11
What is the remainder when 10^{5 }- 560 is divided by 11 ?
10^{5 }- 560 = 99440Divisible by 101: Mark off the number in groups of two digits starting from the right, and add the two-digit groups together with alternating signs. If the sum is divisible by 101 then the original number is also divisible by 101.
Eg: 4512276, (76 + 51) - (22 + 4) = 101, hence divisible by 101.
Divisible by 1001: Mark off the number in groups of three digits starting from the right, and add the three-digit groups together with alternating signs. If the sum is divisible by 1001 then the original number is also divisible by 1001.
Eg: 9533524, (524 + 9 ) - 533 = 0, hence divisible by 1001.
Divisible by 10^{n }+ 1 : Mark off the number in groups of n digits starting from the right, and add the n-digit groups together with alternating signs. If the sum is divisible by 10^{n} + 1 then the original number is also divisible by 10^{n} + 1. |
Divisible by 13: If the difference of the number of its thousands and the remainder of its division by 1000 is divisible by 13.
2184, 2 - 184 = -182, so divisible by 13.
Divisible by 111: Add the digits in block of 3 from right to left. The number is divisible by 101 if the sum is a multiple of 111 or is zero.
12659328 = 328 + 659 + 12 = 999 = 111 * 9, divisible by 111
To check divisibility by a number, check divisibility by highest power of each of its prime factors. |
Eg: To check for the divisibility by 24, check for divisibility by 2^{3} and 3 as 24 = 2^{3} * 3
72 is divisible by 24 as 72 is divisible by 8 and 3.
Another important one for you is a generic method
To test for divisibility by a number ( say D ), where D ends in 1, 3, 7, or 9 : |
Eg: Find the remainder when 1054 is divided by 17
Step 1 : 17 * 7 = 119
Step 2 : m = (119 + 1) / 10 = 12
1054 = 10 * 105 + 4, t = 105 and q =4
mq + t = 48 + 105 = 153 = 17 * 9, Remainder[1054/17] = 0
OK! So we got some neat tricks. But how are we going to engage them to find the remainders. Again, we will learn from some examples.
What is the remainder when 2111756 is divided by 8 ?
We know the divisibility check for 8. Just find the remainder of the number formed from the last three digits with 8. That is our answer.
Remainder [ 2111756 / 8 ] = Remainder [ 756 / 8 ] = 4
What is the remainder when 2345987572219134 is divided by 9 ?
We know the sum of the digits should be a multiple of 9. Here just find the remainder of the sum of the digits with 9.
Remainder [ 2345987572219136 / 9 ] = Remainder [ sum of the digits / 9 ] = 2
Question asked by @deepalis727
Find the maximum value of (x - 6)^2 (11 - x)^3 for 6 < x < 11.
Data Interpretation accounts for a major chunk of questions in common aptitude test format. Mastering this area is comparatively less tiresome than the other two major sections, Quantitative Aptitude and Verbal Ability. Data Interpretation tests our ability to understand and apply data. Usually DI won’t challenge us with complicated stuffs. Given enough time we can solve most of the Data Interpretation questions with good accuracy. But Data Interpretation lures us in to its web by giving seemingly harmless questions with hidden traps such as unwanted data and unnecessary calculations. If we do not approach DI smartly, we will waste our time.
Some best practices while dealing with Data interpretation are
Rather than going on and on with theory we will try to understand these concepts by solving some Data Interpretation questions. Here we go.
The table below reports annual statistics related to rice production in selected states of India for a particular year. (CAT 2005)
1.Which two states account for the highest productivity of rice (tons produced per hectare of rice cultivation)?
(1) Haryana and Punjab (2) Punjab and Andhra Pradesh
(3) Andhra Pradesh and Haryana (4) Uttar Pradesh and Haryana
2. How many states have a per capita production of rice (defined as total rice production divided by its population) greater than Gujarat?
(1) 3 (2) 4 (3) 5 (4) 6
3. An intensive rice producing state is defined as one whose annual rice production per million of population is at least 400,000 tons. How many states are intensive rice producing states?
(1) 5 (2) 6 (3) 7 (4) 8
What is given?
Data is represented using a quantitative method (Table).
Given Data are:
Total area available
Area used for rice cultivation (as % of total area)
Rice production
Population
What is asked?
Rice productivity = tons produced per hectare of rice cultivation
Per capita production of rice = total rice production divided by its population
Intensive rice producing states = annual rice production per million of population is at least 400,000 tons (0.4 million tons)
What is the effort required?
Question#1
We need to find out the area under rice cultivation in hectares as the given data is as % of the total area. Options have only 4 States in them (Haryana, Punjab, Andhra Pradesh and Uttar Pradesh). Hence, we need to find area under rice cultivation in hectares ONLY for four states.
Question#2
We need to find the number of states Per capita production of rice is greater than Gujarat.We may not have to calculate the actual value as we are asked to find out whether the value is higher or lower than a given reference point (of Gujarat’s).
Question #3
Here also we may not have to find the actual values as we are asked to find out the values higher than a reference point (400,000 tons).
Question#1:
Total Area under rice cultivation (in million hectares) for the required 4 states
Haryana: 4 * 80/100 = 3.2
Punjab: 5 * 80/100 = 4
A.P: 28 * 80/100 = 22.4
U.P: 24 * 70/100 = 16.8
Rice production in ton per hectare for the required four states
Haryana: 19.2/3.2 = 6
Punjab: 24/4 = 6
A.P: 112/22.4 = 5
U.P: 67.2/16.8 = 4
Hence Haryana & Punjab account for the highest productivity of rice.
Here trap that is we may start finding out the area under rice cultivation for all the states rather than the required four. Never jump to solution before understanding what is really required to solve the questions.
Question#2:
We are asked to find the number of states with Per capita production of rice higher than Gujarat.
Per capita production of rice for Gujarat:
X = 24 / 51 = (20.4 + 2.04 + 1.02 + 0.51 + …)/51 ≈ 0.4 + 0.04 + 0.02 + 0.01 ≈ 0.47
This is a neat trick while solving fractions. Concept is simple. Write numerator as a sum or difference of numbers which can be written in terms of denominator. In the above expression, 24 is written as sum of 20.4 (51 * 0.4), 2.04 (51 * 0.04), 1.02 (51 * 0.02) and 0.51 (51 * 0.01).
Coming back, we got Per capita production of rice for Gujarat, say X = 0.47 ≈ 0.5
We need to find states where
Production / Population > 0.5
= > Production > 0.5 * Population
= > 2 * Production > Population
If the Population is less than or equal to double the Production, per capita production of rice for that state will be higher than X.
If the Population is considerably more than double the Production, Per capita production of rice for that state will be lesser than X.
If Population is only little more than double the Production we may need to solve for actual value (as we approximated)
We can easily find out that 3 states (Haryana, Punjab and A.P) have its Per capita production of rice higher than Gujarat. For Maharashtra we may not be able to easily compare with the reference value. So we need to solve only that one.
Per capita production of rice for Maharashtra
48/97 ≈ (38.8 + 4.85 + 3.88 + …)/97
≈ 0.4 + 0.05 + 0.04 ≈ 0.49 which is greater than X (0.47)
Hence four states (Haryana, Punjab, A.P and Maharashtra) have Per capita production higher than Gujarat.
We didn’t find the actual per capita production for all states. We just checked whether the values are above or below the reference point. Only in case of closer values we solved using approximation. Here if we spend time to find all the actual values it would have been a terrible waste of time, which was the hidden trap.
Question #3
Here we need to find states whose annual rice production per million population is 400,000 tons (0.4 million tons) or more. In this case also we are asked to find out relative values based on a reference value (greater than 0.4 million tons/million population). We don’t have to solve for actual values.
Production/Population > 0.4
= > Production/0.4 > Population
= > Production * 2.5 > Population
we need to identify states where Production * 2.5 > Population.
We can get the status for all the states without doing any complex calculation.
May be we need to do some quick check for U.P
67.2 * 2.5 = 60 * 2.5 + 7 * 2.5 + 0.2 * 2.5
= 150 + 17.5 + 0.5 = 168 > 166, Hence YES.
We have eight states which qualify as intensive rice producing states.
By converting division by 0.4 to comparatively easier multiplication with 2.5 we were able to find the relative values much quicker. If we go with the conventional approach of solving each fractions and calculating the actual values we will end up spending lot of time and chances are more that we mess up with our calculations, another trap.
Just imagine that we had one more question to find the state ranked at number five among the top intensive rice producing states with one among the option as ‘None of these’. Here, we need actual values and we cannot limit the calculations based on options as one option is ‘none of these’. We know how to solve this question but it is better to leave it and come back if we are left with some time and no better question remaining.
Another important aspect in Data Interpretation is qualitative analysis. We will solve one from that bucket too.
Answer the questions on the basis of the data presented in the figure below.
(CAT 2003 Retest)
1. During 1996-2002, the number of commodities that exhibited a net overall increase and a net overall decrease, respectively, were
(1) 3 and 3 (2) 2 and 4 (3) 4 and 2 (4) 5 and 1
2. The number of commodities that experienced a price decline for two or more consecutive years is
(1) 2 (2) 3 (3) 4 (4) 5
3. For which commodities did a price increase immediately follow a price decline only once in this period?
(1) Rice, Edible oil, and Dal
(2) Egg and Dal
(3) Onion only
(4) Egg and Onion
What is given?
Data is represented using a Qualitative method (Line graph). Given data are
Price of commodities from year 1996 to 2002
Trend of commodity price during the tenure 1996 – 2006
What is asked?
Net overall increase/decrease of commodities during the period
Trend in price fluctuations
Effort required:
Question 2 & 3 are purely qualitative in nature. Visual inspection can yield the answer. Both are very much doable and can fetch us some easy marks.
Question 1, we have to find the net increase/decrease of price for the period. We can either approach it using conventional quantitative methods or can use a qualitative approach. If the qualitative approach crossed your mind by now, it’s all yours... else don’t waste time in this one unless you don’t have an easier question left to do.
As a matter of fact, you don’t have to touch your pen to solve all the three questions.
Question #2
By visual inspection, answer is five. Except for edible oil all other commodity price decline for two or more consecutive years during the tenure. We need to find the pattern of 2 continuous dips in the graph for the commodity during the period.
Question #3
This question needs us to find the commodities which has a price decline followed by a price increase and only once. We need to find a particular pattern in the plot which is a dip and rise coming together and only once. A Dip-Rise together makes a V shaped plot. Hence we are asked to identify those commodities whose plot has one and only one V in it.
Before start solving, we will see another important aspect here. Options are there for helping us. Not just the right one, but all options! Build the habit to bring down the calculations to the minimum and do only what is really required. Less calculations means lesser time required and lesser chance for committing mistakes :)
Carefully observe the options to see whether there is any chance to simplify our efforts. Here Option 3 needs us to find the pattern only for one commodity (Onion). The trick here is that if Onion satisfies the condition, then we can eliminate options 1 and 2 as Onion is not there in both of them. Otherwise if Onion does not satisfy the condition then we can remove option 3 and 4 as Onion is included in them.
Onion has one V and only one V in its plot. Option 1 and 2 out from our attention. :)
Now to mark between option 3 and 4 we need to take the plot of Egg. We have one and only one V in egg plot too... hence option 4 is the right one.
Question #1
Many mistakes we commit during our aptitude exam is not mathematical, it is psychological. By using best method or not, it is relatively easy to solve the 2^{nd} and 3^{rd} questions in this set. Once we get those two questions into our basket it is natural to urge to ‘Conquer’ the whole set by solving the first question. First question is also easy if we got the correct approach. Otherwise we are in loss in terms of time.
We are asked to find the net increase/decrease in the overall price in the period. There is no optimisation available as we need to find the status of all the commodities to get the answer.
We can go with the quantitative method, like
Rice price fluctuation during the period: 10 - > 14 - > 13 - > 10 - > 11 - > 12 - > 14
Net price difference = +4 – 1 – 3 + 1 + 2 + 2 = +4. (Hence net is positive)
If we do this for all the commodities and for all the years, we will find 42 values from the graph, calculate the difference between previous data and then find the net value... ouch!
While discussing number line we shared a concept. If a we take 100 steps forward and then 99 steps backwards in a number line it is effectively taking only one step forward. Same applies here. We are asked to get the net value of price difference during the period. Just take the difference between initial and final point of the plot for each commodity. If final point is above the initial point, net is increase. If final point is below the initial point, net is decrease. Visual inspection will suffice. Edible oil and Dal has a net decrease and other four show a net increase. Hence option 3.
Data Interpretation is an approach based section than a syllabus based one. If we don’t treat Data Interpretation carefully, DI will treat us bad, waste our time and trick us into an awful situation where we have doable questions remaining with no time left to solve them. There is not much theory to learn but mastering DI comes through practice. It is perfectly OK to take some time to pick up the DI skills. The focus while building your Data Interpretation skills should be on your ability to spot the right questions and pick the best approach quickly.
Happy learning :)
Remainder theorem is a very important topic in number system and can be learnt easily. We will try to learn some interesting concepts regarding remainders with examples. Here we go!
Definition of remainder
If a and d are natural numbers, with d # 0, it can be proved that there exist unique integers q and r, such that a = qd + r and 0 ≤ r < d. The number q is called the quotient, while r is called the remainder. Dividend = Divisor × Quotient + Remainder. |
if r = 0 then we say that a is perfectly divisible by d or d is a factor of a. For example, we say 8 is a factor of 40 because 40 leaves a remainder 0 with 8.
By definition remainder cannot be negative. |
Now just to give an example, 17 = 3 * 5 + 2, which means 17 when divided by 5 will give 2 as remainder. Well that was simple!
Find Remainder[ (12 * 13 * 14) / 5 ]
Remainder [ (12 * 13 * 14) / 5 ]
= Remainder [2184/5] = 4. But this method is not the right one for us :)
In order to find the remainder of an expression find the individual remainder and replace each term with the respective remainders. Eg: Remainder[(100 + 30 * 4 - 8 ) / 7] = Remainder[(Remainder[100/7] + Remainder[30/7] * Remainder[4/7] - Remainder[8/7] ) / 7 ] = Remainder[(2 + 2 * 4 - 1)/7] = Remainder[9/7] = 2 |
In the above case 12, 13 and 14 will give remainders 2, 3 and 4 respectively when divided by 5. So replace them with the respective remainders in the expression and find the remainder again.
Remainder [ (12 * 13 * 14) / 5 ] = Remainder[(2 * 3 * 4) / 5] = Remainder[ 24 / 5 ] = 4.
Note:
One common mistake while dealing with remainders is when we have common factors in both dividend and divisor. Example, what is the remainder when 15 is divided by 9
15 / 9 is same as 5 / 3, remainder 2. Correct? No 15/9 will give a remainder of 6.
Where we slipped?
Always remember that if we find remainder after cancelling common terms make sure we multiply the remainder obtained with the common factors we removed.
In previous case we will get correct answer (6) when we multiply the remainder obtained (2) with the common factor we removed (3).
What is the remainder of 1421 * 1423 * 1425 when divided by 12 ? ( CAT 2000 )
1421, 1423 and 1425 gives 5, 7 and 9 as remainders respectively when divided by 12.
Remainder [ (1421 * 1423 * 1425 ) / 12 ] = Remainder [ (5 * 7 * 9) ] / 12, gives a remainder of 3.
Find the reminder when 1! + 2! + 3! + . . . . .. . .. 99! + 100! is divided by the product of first 7 natural numbers
From 7! the remainder will be zero. Why ? because 7! is nothing but product of first 7 natural numbers and all factorial after that will have 7! as one of the factor. so we are concerned only factorials till 7!, i.e, 1! + 2! + 3! + 4! + 5! + 6!
1! + 2! + 3! + 4! + 5! + 6! = 873 and as 7! > 873 our remainder will be 873
What is the remainder when 64^{999 }is divided by 7? (GMAT Type Question)
Many of us get intimidated with such numbers, always remember that the key to crack quant is a strong hold of basic concepts.
Remainder [64^{999 }/ 7 ] = Remainder[ 64 * 64 * …. 64 (999 times) / 7 ]
Remainder[64/7] = 1, hence Remainder [64^{999 }/ 7 ] = Remainder [ 1 ^{999 }/ 7 ] = 1
What is the remainder when 444^{444}^{ ^ 444} is divided by 7 ? (GMAT Type Question)
Remainder[444/7] = 3
Remainder[ 444 ^{444 ^ 444 }/ 7 ] = Remainder [ 3 ^{444 ^ 444 }/ 7 ]
= Remainder [ ( 3^{2 }) ^{222 ^ 444 }/ 7 ] = Remainder [ 2^{222 ^ 444 }/ 7 ] ( As Remainder [ 3^{2 }/ 7 ] = 2 )
= Remainder [ ( 2^{3} ) ^{74 ^ 444 }/ 7] = Remainder [ 1 ^{74 ^ 444 }/ 7 ] = 1 ( As Remainder [ 2^{3 }/ 7 ] = 1 )
Concept of negative remainder
We saw earlier that by definition remainder cannot be negative. But considering negative remainder is a very useful exam trick.
For example,
What is the remainder when 2^{11} divided by 3?
The easiest method for this one will be using the concept of negative remainders.
Here 2 when divided by 3 gives a remainder of -1. (Say)
2 = 3 * 1 + (-1), remainder is -1, which is theoretically incorrect but let’s cheat!
So we are asked to find (-1) * (-1) * … 11 times divided by 3.
Which is Remainder[-1/3] = -1.
Whenever you are getting negative number as a remainder, make it positive by adding the divisor to the negative remainder. |
Here required answer is 3 + (-1) = 2.
Remainder when (41 * 42) is divided by 43
Use negative remainder concept,
Remainder [ 41 * 42 / 13 ] = Remainder[(-2) * (-1) / 43 ]( as 41 = 43 * 1 – 2 and 42 = 43 * 1 – 1)
= Remainder [ 2 /43 ] = 2 (here we got a positive remainder itself, so no need of correction)
Some useful concepts while dealing with remainder are given below.
Remainder[(ax + 1)^{n }/ a] = 1 for all values of n. |
Find the remainder when 100^{99 }is divided by 11
Remainder[100^{99 }/ 11 ] = Remainder[(11* 9 + 1)^{99 }/ 11] = 1.
Remainder[(ax - 1)^{n }/ a ] = 1 when n is even Remainder[(ax - 1)^{n }/ a ] = (a-1) when n is odd. |
Find the remainder when 21^{875 }is divided by 17.
Remainder[21 / 17] = 4, so we need to find Remainder[4^{875}/ 17]
4^{2 }= 16 = (17 – 1), we can write the expression as Remainder[(4^{2})^{437 }* 4 / 17]
= Remainder[(17 – 1)^{437 }* 4 / 17] = Remainder[(17-1) * 4 / 17] = Remainder[64 / 17] = 13.
Remainder[(a^{n }+ b^{n}) / (a + b) ] = 0 when n is odd. |
Remainder[(2^{101 }+ 3^{101}) / 5 ] = 0
What is the remainder when 15^{23 }+ 23^{23 }is divided by 19 ? ( CAT 2004 )
15^{23 }+ 23^{23 }is divisible by 15 + 23 = 38 ( as 23 is odd).
So Rem[(15^{23 }+ 23^{23})/19] = 0
Remainder[(a^{n }+ b^{n} + c^{n} + ...) / (a + b + c + ...) ] = 0 if ( a + b + c + ... are in Arithmetic progression and n is odd |
What is the remainder when 16^{3 }+ 17^{3 }+ 18^{3 }+ 19^{3} is divided by 70 ? ( CAT 2005 )
Apply the above funda. Here n =3 ( odd ), 16 + 17 + 18 + 19 = 70 and 16,17,18 and 19 are in AP. Remainder is 0
Remainder[(a^{n }- b^{n}) / (a + b) ] = 0 when n is even. |
Remainder[(5^{100 }– 2^{100}) / 7] = 0
Remainder [(a^{n }- b^{n}) / (a - b) ] = 0 |
Now we can say Remainder[(101^{75 }– 76^{75}) / 25] = 0 in no time..!
The remainder when f(x) = a + bx + cx^{2}+ dx^{3}+ … is divided by (x-a) is f(a) |
Rem [(3x^{2 }+ 4x + 1) / (x-2)] = f(2) = 3 * 2^{2 }+ 4 * 2 + 1 = 21
Cyclic property of remainders
Sometimes it is easy to find the remainder by using the cyclic property of remainders, remainders forming a pattern.
As a thumb rule if we divide p^{n }with q, the remainder will follow a pattern. |
For example,
Remainder [ 2^{1 }/ 3 ] = 2, Remainder [ 2^{2} / 3 ] = 1, Remainder [ 2^{3 }/ 3 ] = 2, Remainder [ 2^{4 }/ 3 ] = 1 and so on.
Pattern repeats in cycles of 2. Remainder [ 2^{n }/ 3 ] = 2 when n is odd and 1 when n is even.
With this information, we can find Remainder [ 2^{3276 }/ 3 ] = 1 very quickly.
One more, Remainder [ 9^{1} / 11 ] = 9, Remainder [ 9^{2} / 11 ] = 4, Remainder [ 9^{3} / 11 ] = 3 , Remainder [ 9^{4} / 11 ] = 5
Remainder [ 9^{5} / 11 ] = 1, Remainder [ 9^{6} / 11 ] = 9, Remainder [ 9^{7} / 11 ] = 4, Remainder [ 9^{8} / 11 ] = 3
Pattern repeats in cycles of 5. So if we are asked to find Remainder [ 9^{100 }/ 11 ]， we know it is 1. (100 is in the form 5n and we know remainder for 5 is 1.. cool right ?)
Note:
Remainder [9^{3}/11] = Remainder [Remainder [9^{2}/11] * Remainder [9^{2}/11] ) / 11] = Remainder [ 4 * 9 / 11] = 3.
This funda comes very handy in scenarios like this. Like we dont have to solve Rem [9^{8 }/11] because we already know Rem[9^{4 }/11] as 5..
Rem [9^{8 }/11] = Remainder [ 9^{4 }* 9^{4 }/11 ] = Remainder [ (5 * 5) / 11 ] = 3
Also, Remainder [9^{7 }/11] = Remainder [9^{3 }* 9^{4 }/ 11 ] = Remainder [ ( 3 * 5 ) / 11 ] = 4 ( as Rem [9^{3 }/11] = 3 and Rem [9^{4 }/11] = 5 )
What is the remainder when 7^{100 }is divided by 4?
Remainder[ 7^{1 }/ 4 ] = 3, Remainder[ 7^{2 }/ 4 ] = 1, Remainder[ 7^{3 }/ 4 ] = 3, Remainder[ 7^{4 }/ 4 ] = 1 and so on...
Pattern repeats in cycles of 2. Remainder [ 7^{n }/ 4 ] is 3 when n is odd and is 1 when n is even.
7^{100 }when divided by 4 gives a remainder of 1.
(Same can be solved using other methods also)
Find the remainder when 3^{99^99} is divided by 7
Find the pattern of remainder when 3^{n }is divided by 7.
Remainder [ 3^{1 }/ 7 ] = 3, Remainder [ 3^{2 }/ 7 ] = 2
Remainder [ 3^{3 }/ 7 ] = 6 (Don’t calculate Rem[3^{3}/7] we already have Rem[3^{1}/7] & Rem[3^{2}/7])
Remainder [ 3^{4 }/ 7 ] = 4 (using Rem[3^{2}/7])
Remainder [ 3^{5 }/ 7 ] = 5 (using Rem[3^{3}/7] and Rem[3^{2}/7] )
Remainder [ 3^{6 }/ 7 ] = 1 (using Rem[3^{3}/7])
Remainder [ 3^{7 }/ 7 ] = 3 (using Rem[3^{3}/7] and Rem[3^{4}/7] )
Remainder [ 3^{8 }/ 7 ] = 2 (using Rem[3^{4}/7])
Pattern repeats in cycles of 6. (We can do this easily from Euler’s theorem, as φ(7) = 6, hence Remainder [3^{6}/7] = 1. Explained just to get the idea of patterns in remainders)
Now our task is to find Remainder [ 99^{99}/ 6 ]
Remainder [99^{99}/6] = Remainder [3^{99}/6]
Remainder [ 3^{1} / 6 ] = 3, Remainder [ 3^{2} /6 ] = 3, Remainder [ 3^{3} / 6 ] = 3 and so on..!
Hence 99^{99 }can be written as 6n + 3.
Remainder [ 3^{99^99 }/ 7 ] = Remainder [ 3^{(6n + 3)}/ 7 ]
we have found out the pattern of 3 divided by 7 repeats in cycles of 6. So we need to find the Remainder [ 3^{3} / 7 ] to get the answer which is equal to 6
Euler’s Remainder Theorem
We say two numbers ( say a and b ) are co-prime to each other when HCF(a,b) = 1, i.e, no divisor divide both of them completely at the same time.
Eg: 21 and 8 are co primes because they don't have any common factors except 1
In number theory, Euler's totient or phi function, φ(n) is an arithmetic function that counts the number of positive integers less than or equal to n that are relatively prime to n. |
Take n = 9, then 1, 2, 4, 5, 7 and 8, are relatively prime to 9. Therefore, φ(9) = 6.
How to find Euler’s totient?
Say n = P_{1}^{a }x P_{2}^{b }x P_{3}^{c }x ... ( where P_{1} P_{2}, P_{3} ... are prime factors of n ) φ(n) = n x (1 - 1/P_{1}) x (1 - 1/P_{2}) x (1 - 1/P_{3}) x .... If n is prime then φ(n) = n - 1 |
φ(100) = 100 x (1-1 / 2) x (1- 1 / 5) = 100 x 1 / 2 x 4 / 5 = 40
φ(9) = 9 x (1 – 1/3) = 6
Euler’s Remainder theorem states that for co prime numbers M and N, Remainder [M^{φ}^{(N) }/ N] = 1 Always check whether the numbers are co primes are not. Euler’s theorem is applicable only for co prime numbers |
What is the remainder when 21^{865 }is divided by 17
Remainder[21/17] = 4
Remainder [ 21^{865}/ 17 ] = Remainder [ 4^{865}/ 17 ]
4 and 17 are co prime numbers. ( A prime number is always coprime to any other number)
φ(17) = 17 x ( 1 – 1 / 17) = 16.
So Euler’s theorem says Remainder [ 4^{16}/ 17 ] = 1
To use this result in the given problem we need to write 865 in 16n + r form.
865 = 16 * 54 + 1, so 4^{865 }can be written as 4^{16 * 54 }x 4
Remainder[4^{865}/17] = Remainder[ 4^{16*54}/17] * Remainder[4/17] = 1 * 4 = 4
What is the remainder when 99^{999999 }is divided by 23
Remainder[99/23] = 7
Remainder[99^{999999}/23] = Remainder[7^{999999}/23]
7 and 23 are co prime numbers
Here 23 is prime, so φ(23) = 22
So by applying Euler's theorem we can say that Remainder[7^{22}/23] = 1
In order to use this result in our problem we need to write 999999 in 22n + r form. Before rushing into dividing 999999 by 22, just think whether we have any better way to do that. We know, 999999 = 22n + r , 0 ≤ r < 22. 999999 is divisible by 11 and so is 22. which means r is also a multiple of 11. Only numbers which are less than 22 and is a multiple of 11 is 11 and 0. But as 999999 is odd and 22n is even, r should be odd. so r = 11. Saved our time right ? ;)
999999 = 22n + 11
Remainder[7^{999999}/23] = Remainder[7^{22n}/23] * Remainder[7^{11}/23] = 1 * Remainder[7^{11}/23] = Remainder[7^{2 }* ^{5 }* 7/23]
= Remainder[ 3^{5} * 7/23] ( as Remainder[7^{2}/23] = 3 )
= Remainder[3^{3 }* 3^{2 }* 7/23] = Remainder[4 * 9 * 7/23] = Remainder[28 * 9/23] = Remainder[45/23] = 22
Find the remainder when 97 ^{97 ^ 97 }is divided by 11
Remainder [97/11] = 9
So, Remainder [97^{97^97}/11] = Remainder [9^{ 97^97}/11]
From Euler’s theorem, Remainder [9^{10}/11] = 1
97 and 10 are again co primes. So φ(10) = 10 (1-1/2) (1-1/5) = 4
Remainder [97^{4}/11] = 1
97 = 4n + 1, So Remainder [97^{97}/11] = Remainder [97/10] = 7
Means 97^{97 }can be written as 10n + 7
Now our original question,
Remainder [9^{97^97}/11] = Remainder [9^{10n+7}/11] = Remainder [9^{7}/11] = 4 :)
Fermat’s little theorem
Euler’s theorem says that if p is a prime numberand a and p are co-primes then a^{φ}^{(p) }/ p always gives a remainder of 1.
Now we know for any prime number p, φ(p) = p - 1
Remainder of a^{( p – 1 )}/ p is 1, which is Fermat’s little theorem |
We can derive other useful results like
Wilson’s theorem
Remainder[(p-1)! / p ] = (p-1), if p is a prime number. |
We can also derive some useful results like
Example:
Remainder of [22!/23] = 22
Remainder of [21!/23] = 1
I hope the explanations are clear are correct. Please let me know if any concepts regarding remainders are missed out or incase of any errors.. Happy learning :)
@hemant_malhotra
f(-1) < 0
f(0) < 0
f(1) > 0
f(2) > 0
So one real solution between 0 and 1
I often get confused when it comes to placing the comma, using a semi colon etc. Hopefully life would be a little better after this article. Punctuations are grammatical tools that allow you to keep your intended meaning clear. When used correctly, they help in understanding the message properly and aid easy reading with necessary pauses and stops. The improper usage of punctuations, either an overdose or under utilisation, introduces the risk of confusing and misinforming the reader. When the punctuation is missing, even the most well-intentioned of statements can turn into a minefield of unintended meaning.
Example 1
Let’s eat, Grandma.
Let’s eat Grandma.
Example 2
I love cooking, my dogs, and my family.
I love cooking my dogs and my family.
Comma is important! Seems like it will take a lifetime for me to understand all the contexts of using a comma. So, I decided to focus on some common errors and simple rules that will help me look less stupid while using a comma. Commas are often used to separate small meaningful statements. Comma provides clarity in a sentence like a pause while we actually say the sentence. If you cannot justify a pause at that part of the sentence, a comma may not be required. Say the above example sentences aloud and you can see how a pause and a comma is related.
Check whether we need a comma in the examples below.
The book on the table, is mine.
We don’t need a comma here. Because the clause after comma (is mine) doesn’t make sense on its own and needs the clause before comma for completion. We don’t have to put a comma in between them. Let them be together and live happily ever after. So correct version would be - The book on the table is mine.
College students, who are absent, ought to be questioned.
Do we need a comma here? No! Why? Because if you remove the clause within coma (who are absent) the remaining sentence – College students ought to be questioned - is incomplete/incorrect. The sentence needs the clause –who are absent - to be meaningful. So let’s not put poor college kids into trouble and remove the comma. Correct version – College students who are absent ought to be questioned.
What about College students, who are young, enjoy their lives.
If we remove the clause within the comma (who are young) the remaining clause is still meaningful. As –who are young – clause doesn’t add anything to the meaning of the sentence, comma is required.
If the identification comes after the name, it should always be surrounded by commas.
The boy answered quickly, what his mentor asked.
Here also comma is not required as the ‘his’ in the clause after comma cannot be meaningful without the first clause. Correct version – The boy answered quickly what his mentor asked
If the dependent clause comes in the beginning of the sentence, comma is required.
Whatever his mentor asked, the boy answered quickly. (Correct)
Rakesh wanted to be either a teacher, or an engineer.
Pairs like either-or, not only – but also etc. cannot be separated by comma. So correct version – Rakesh wanted to be either a teacher or an engineer.
John’s eldest son Robert is a doctor.
When the identifier makes sense in the sentence by itself, then the name is non-essential and you use a comma before it. Otherwise, no comma. Correct version – John’s eldest son, Robert is a doctor. (As eldest son make sense by itself). If John has more than one son then John’s son Robert is a doctor is correct
Dedicated to my parents, Ayn Rand and God.
Three or more items in a series should be separated by commas, including the final item, which comes after the conjunction (almost always AND or OR). This final comma is known as a serial comma (also called Oxford comma and Harvard comma)
Dedicated to my parents, Ayn Rand and God. (Wrong – Author’s parents can’t be Ayn Rand and God!)
Dedicated to my parents, Ayn Rand, and God (Correct)
As a parting shot, check whether the below question uses comma properly and then solve it! (share your answers (both verbal and quant) as comment :D
Two tanks, T1 and T2, of equal capacities, are provided with inlet taps, A and B, of different flow rates, respectively. Tap A begins to fill the first tank. Seven minutes later, tap B is opened. Nine minutes after that, the total quantity of water in the two tanks is just enough to fill one tank completely. Exactly t minutes after that both the tanks are full. What is the value of t? ( Answer is 12 )
credits : @shashank_prabhu
Let the circumference be 120 units and speed of A, B, C be 5 units, 1 unit and 1 unit respectively. So, A and B will meet once every 20 seconds and A and C will meet once every 30 seconds. So, in 1 minute, A gives B 3 flags and gives C 2 flags. So, we can safely say that in the first 30 distributions, B gets 18 flags and C gets 12 flags. Post that, after 20 seconds, B gets a flag, at the 30th second C gets a flag and at the 40th second, B gets another flag. So, in total, B gets 20 flags and C gets 13 flags. So, difference is 7.
Let the distance PQ be d and speed of river be R and speed of boat be B
d/(B + R) = 4
d/(B - R) = 6
Solve for B.
4(B + R) = 6(B - R)
4B + 4R = 6B - 6R
10R = 2B
R = B/5
d/(B + B/5) = d/(6B/5) = 5d/6B = 4
d = 24B/5
If we put d = 24 units, then B = 5 and R = 1
So first hour boat will cover 6 units, second hour boat will cover 5 * 1.4 + 1 = 8 units and third hour 10 units.
So 24 units will be covered in 3 hours
Statutory warning : It's always best to sketch the curves than mugging up the formulae.
Some useful formulae:
Area bounded by the curves |ax +/- m | = p and |by +/- n| = q is 4pq/ab sq units
Area bounded by |ax +/- m| + |by +/- n| = k is 2k^2/(ab)
Area bounded by |ax + by| = k and |ax - by| = k is 2k^2/ab
Area bounded by |ax + by| + |ax - by| = k is k^2/(ab)
We will see in detail how we got these fancy formulas and how easy it is to derive them whenever you need it!
Area of a square with diagonal d = d^2/2
Area of a rhombus with diagonal d1 and d2 = (d1 x d2)/2
Area of parallelogram = base x height
Some important graphs
|x| = k (say k = 2) will give 2 lines parallel to y axis. One of which represents x = 2 and the other x = - 2
Similarly |y| = k will give 2 lines parallel to x axis. One of which represents y = 2 and the other y = - 2
So |x| = 2 AND |y| = 2 we will give us a square with side = 2k as shown below.
Now what about |x + 3| = 2 AND |y + 1| = 2 ?
|x + 3| = 2 => x = -1 or x = -5 (4 units)
|y + 1| = 2 => y = 1 or y = -3 ( 4 units)
So this should also give a square of side 4 units
Graph is as below
One interesting thing here is even if we plot |x + 6| = 2 and |y + 8| = 2
|x + 6| = 2 => x = -8 or x = -4 (4 units)
|y + 8| = 2 => y = -6 or y = -10 (4 units)
It will still give a square of side 4 units.
Or say if we plot |x - 3| = 2 and |y + 4| = 2
|x - 3| = 2 => x = 1 or x = 5 (4 units)
|y + 8| = 2 => y = -6 or y = -10 (4 units)
again, a square of side 4 units.
So all the values of |x +/- a| = k and |y +/- b| = k will yield a square of side 2k units. (definitely in different coordinates depending on a and b)
What if the value of k is different. Like |x + 3| = 1 and |y - 1| = 2 ?
|x + 3| = 1 => x = -2 or x = -4 (2 units)
|y - 1| = 2 => y = 3 or y = -1 (4 units)
A rectangle with sides 2 and 4.
We get a rectangle instead of square. That's it. Graph will be like
We can say in general |ax +/- m | = p and |by +/- n| = q will plot a rectangle with side 2p/a and 2q/b.
Hence area = 4pq/ab.
Now we will see another important type.
|x| + |y| = 2 will give a square with diagonal = 2k as shown below.
Here also if you plot any |x +/- a| + |y +/- b| = 2 it will still yield a square with diagonal = 2k.
So in general, |x +/- a| + |y +/- b| = k will yield a square of diagonal 2k.
What about |3x| + |4y| = 12 ?
Graph is as below
here it is a rhombus with diagonals 6 and 8 (which is nothing but 12 x 2/3 and 12 x 2/4, where 3 and 4 are our coefficients). Area here is (2 x 12^2)/(3x4)
Here also if you try something like |3x + 6| + |4y - 4| = 12, it will plot the same shape (a rhombus with diagonals as 6 and 8 and the only change will be the position of the rhombus in the xy plane which won't alter the area)
So in general, Area of |ax +/- m| + |by +/- n| = k is 2k^2/ab
We saw the graph of |x| + |y| = k, what about the graph of |x| - |y| = k ?
For example graph of |x| - |y| = 3 is as below
means it won't bound any region in the xy plane.
what about |x - y| = k type ?
For example, graph of |x - y| = 3 looks like
so this one also won't bound any region in the xy plane (good, lesser formulas!)
|x + y| = k case is also same. for example |x + y| = 3 won't bound any region in the xy plane. Graph is as below
What if we combine both ? i.e |x + y| = 3 and |x - y| = 3. Can you guess from the above graphs how it would turn out ?
Bounded region is as below
A square with diagonal as 6.
For example, the graph of |3x + 2y| = 6 and |3x - 2y| = 6 is shown as below
So in general, if we plot |ax + by| = k and |ax - by| = k, we will get a rhombus with diagonals 2k/a and 2k/b.
Okay. so now we can solve the graph for |x + y| = 4 and |x - y| = 4. But what about |x + y| + |x - y| = 4 ?
Graph is as below
We can see that the graph will give a square of side = k
so area = k^2 = 16
what would be the graph of |2x + y| + |2x - y| = 8 ?
Graph will be like
Rectangle with side 8 and 4. Area = 8 x 4 = 32
If you see this, the sides are nothing but 8/2 and 8/1. Where 2 and 1 are nothing by coefficients of x and y
So we can say that the area covered by the graph |ax + by| + |ax - by| = k is k/a * k/b = k^2/ab
How to plot the graph of a line, say 3x + 2y = 6 ?
when x = 0, y = 3
when y = 0, x = 2
Graph is as below
x^2 + y^2 = r^2 is the equation of a circle with radius = r and center at origin.
For example, x^2 + y^2 = 9 will plot a circle as below
Share the formulas/concepts which we missed out and point out errors (if any).
Reading comprehension is often considered as an “Open book” test. Where we have the complete data in front of us and we just need to refer and answer. This is not exactly what a passage question does. Let’s take a real life scenario. Your Boss asks you to represent your department in an organizational level summit. You are given a report which is agreed upon by the department and this should be your reference while taking any decision in the meeting.
Now, what will be your strategy? Will you take a print and start turning pages throughout the meeting…? Doesn’t sound promising right… Of course we will carry a printout but that will be for a ‘Just in case’ scenario. We will do our homework and ensure that we understand the intent of the report along with the content and whatever decision we take will be based on that. Better our homework is, better our participation will be. For reading comprehension, context is not much different than this meeting room. We are given the ‘report’ in the form of a passage, and we need to solve the questions based on our understanding of the passage.
We know that majority of aptitude questions checks the quality of our approach than our knowledge. While checking the approach, question maker wants to see our ability to decide on which question to choose and which question to leave, which method to employ and which method to avoid and so on. In this list one very important aspect is our ability to identify useful information from a chunk of data. This is a key aspect that is tested in passage questions.
Any meaningful language construct can be divided into two main parts, Intent and Content. Intent deals with what is that to be conveyed and content deals with how the intent is conveyed.
Consider a paragraph from a CAT 2008 passage,
“Language is not a cultural artifact that we learn the way we learn to tell time or how the federal government works. Instead, it is a distinct piece of the biological makeup of our brains. Language is a complex, specialized skill, which develops in the child spontaneously, without conscious effort or formal instruction, is deployed without awareness of its underlying logic, is qualitatively the same in every individual, and is distinct from more general abilities to process information or behave intelligently. For these reasons some cognitive scientists have described language as a psychological faculty, a mental organ, a neural system, and a computational module. But I prefer the admittedly quaint term “instinct.” It conveys the idea that people know how to talk in more or less the sense that spiders know how to spin webs. Web-spinning was not invented by some unsung spider genius and does not depend on having had the right education or on having an aptitude for architecture or the construction trades. Rather, spiders spin spider webs because they have spider brains, which give them the urge to spin and the competence to succeed. Although there are differences between webs and words, I will encourage you to see language in this way, for it helps to make sense of the phenomena we will explore.”
While trying to comprehend a passage we usually get caught in content and start spending unnecessary attention to details. Like in the above passage, intent of the author is to establish that language is a biological instinct rather than learned skill. Even if we don’t know who qualifies as a cognitive scientist or the meaning of ‘quaint’, we will be still able to comprehend the intent. Also we don’t need to spend any time on the working of federal government or the process of web spinning… they are all illustrations and details which support the core idea, intent. Until there is a question which specifically focus on content (which are not that common), we can win a passage by comprehending the intent.
There are umpteen methodologies to tackle a passage. Best method is the one that suits YOU the best. In this article I will share my favourite RC strategy, Prick & Pick. Like our famous potato chips pack, which look promising at the beginning, passages looks bulky. But if we prick, then the chips pack becomes disappointingly empty and our passage becomes relaxingly simple :)
Process is simple,
PRICK: We need a needle to prick the passage. Prepare the needle by understanding what is needed from the passage. What we NEED is our ‘NEED’le. Glance through questions and understand the keywords that appear in them. These keywords shall give us our needle. While doing so, don’t read the options as it fails the purpose of having keywords. Reading options will create unnecessary confusions and diversions even before we start reading the passage.
PICK: We will read the passage and focus only on data which are related to what we need. PICK the intent of the passage and if you come across the keywords from the NEEDle, pay extra heed even if it is a content based portion.
Here one important aspect is that we are NOT trying to ‘locate’ the keywords in the passage. Sometimes test takers will read the questions and start fanatically searching the keywords in the passage. There is no guarantee that those keywords will appear in the passage. Even if they appear it is not necessary that the question can be answered just by reading that particular portion alone. Always remember, Reading comprehension is not an open book test. (Well… people who have experienced open book tests will disagree with me stating them as simple. We know what the exam is going to be like when professor says bring any reference you need, it is an open book test! ;) )
While reading for the first time (Picking) we stay within the framework we formed from our required information (while Pricking). As we read, ensure that we mark the important statements. In a paper based test, use your pen and in case of online test, use text highlighter. Also don’t try to mug up the passage. Even though sounds silly, many of us read and re read the passage to an extent that we know them by heart once we comeback from exam hall :)
Enough theory… lets Prick & Pick…! Below passage is from CAT 2007
The difficulties historians face in establishing cause-and-effect relations in the history of human societies are broadly similar to the difficulties facing astronomers, climatologists, ecologists, evolutionary biologists, geologists, and palaeontologists. To varying degrees each of these fields is plagued by the impossibility of performing replicated, controlled experimental interventions, the complexity arising from enormous numbers of variables, the resulting uniqueness of each system, the consequent impossibility of formulating universal laws, and the difficulties of predicting emergent properties and future behaviour. Prediction in history, as in other historical sciences, is most feasible on large spatial scales and over long times, when the unique features of millions of small-scale brief events become averaged out. Just as I could predict the sex ratio of the next 1,000 new born but not the sexes of my own two children, the historian can recognize factors that made inevitable the broad outcome of the collision between American and Eurasian societies after 13,000 years of separate developments, but not the outcome of the 1960 U.S. presidential election. The details of which candidate said what during a single televised debate in October 1960 could have given the electoral victory to Nixon instead of to Kennedy, but no details of who said what could have blocked the European conquest of Native Americans.
How can students of human history profit from the experience of scientists in other historical sciences? A methodology that has proved useful involves the comparative method and so-called natural experiments. While neither astronomers studying galaxy formation nor human historians can manipulate their systems in controlled laboratory experiments, they both can take advantage of natural experiments, by comparing systems differing in the presence or absence (or in the strong or weak effect) of some putative causative factor. For example, epidemiologists, forbidden to feed large amounts of salt to people experimentally, have still been able to identify effects of high salt intake by comparing groups of humans who already differ greatly in their salt intake; and cultural anthropologists, unable to provide human groups experimentally with varying resource abundances for many centuries, still study long-term effects of resource abundance on human societies by comparing recent Polynesian populations living on islands differing naturally in resource abundance. The student of human history can draw on many more natural experiments than just comparisons among the five inhabited continents. Comparisons can also utilize large islands that have developed complex societies in a considerable degree of isolation (such as Japan, Madagascar, Native American Hispaniola, New Guinea, Hawaii, and many others), as well as societies on hundreds of smaller islands and regional societies within each of the continents. Natural experiments in any field, whether in ecology or human history, are inherently open to potential methodological criticisms. Those include confounding effects of natural variation in additional variables besides the one of interest, as well as problems in inferring chains of causation from observed correlations between variables. Such methodological problems have been discussed in great detail for some of the historical sciences. In particular, epidemiology, the science of drawing inferences about human diseases by comparing groups of people (often by retrospective historical studies), has for a long time successfully employed formalized procedures for dealing with problems similar to those facing historians of human societies. In short, I acknowledge that it is much more difficult to understand human history than to understand problems in fields of science where history is unimportant and where fewer individual variables operate. Nevertheless, successful methodologies for analysing historical problems have been worked out in several fields. As a result, the histories of dinosaurs, nebulae, and glaciers are generally acknowledged to belong to fields of science rather than to the humanities.
1. Why do islands with considerable degree of isolation provide valuable insights into human history?
(1) Isolated islands may evolve differently and this difference is of interest to us.
(2) Isolated islands increase the number of observations available to historians.
(3) Isolated islands, differing in their endowments and size may evolve differently and this difference can be attributed to their endowments and size.
(4) Isolated islands, differing in their endowments and size, provide a good comparison to large islands such as Eurasia, Africa, Americas and Australia.
(5) Isolated islands, in so far as they are inhabited, arouse curiosity about how human beings evolved there.
2. According to the author, why is prediction difficult in history?
(1) Historical explanations are usually broad so that no prediction is possible.
(2) Historical out comes depend upon a large number of factors and hence prediction is difficult for each case.
(3) Historical sciences, by their very nature, are not interested in a multitude of minor factors, which might be important in a specific historical outcome.
(4) Historians are interested in evolution of human history and hence are only interested in long term predictions.
(5) Historical sciences suffer from the inability to conduct controlled experiments and therefore have explanations based on a few long-term factors.
3. According to the author, which of the following statements would be true?
(1) Students of history are missing significant opportunities by not conducting any natural experiments.
(2) Complex societies inhabiting large islands provide great opportunities for natural experiments.
(3) Students of history are missing significant opportunities by not studying an adequate variety of natural experiments.
(4) A unique problem faced by historians is their inability to establish cause and effect relationships.
(5) Cultural anthropologists have overcome the problem of confounding variables through natural experiments.
We need our NEEDle... Just glance through the questions (remember, just questions, not options). Keywords we got from questions are ‘Isolated islands’ and ‘Difficulty in prediction’.
How easily and quickly we spot and comprehend relevant portions depends on our comfort level with lengthy passages. Build the habit of reading two editorials daily. While doing so pick from different genres, like pick one from general news and another from business. This will not only help in improving reading speed and comprehension skills but also will be of great support during GD and interview preparations.
Let’s PICK now… This is our first read and the most important part. We will try to comprehend the intent and also will pay extra heed if we come across any keywords. I am highlighting portions which I felt relevant with GREEN (intent) & GREY (NEEDle). Other portions (illustrations and fillers) are not highlighted. It should be possible to answer most of the questions if we are able to get a good understanding of GREEN and GREY portions.
The difficulties historians face in establishing cause-and-effect relations in the history of human societies are broadly similar to the difficulties facing astronomers, climatologists, ecologists, evolutionary biologists, geologists, and palaeontologists. To varying degrees each of these fields is plagued by the impossibility of performing replicated, controlled experimental interventions, the complexity arising from enormous numbers of variables, the resulting uniqueness of each system, the consequent impossibility of formulating universal laws, and the difficulties of predicting emergent properties and future behaviour.[Explains the difficulty]. Prediction in history, as in other historical sciences, is most feasible on large spatial scales and over long times, when the unique features of millions of small-scale brief events become averaged out [Due to the keyword Difficulty in prediction]. Just as I could predict the sex ratio of the next 1,000 new born but not the sexes of my own two children, the historian can recognize factors that made inevitable the broad outcome of the collision between American and Eurasian societies after 13,000 years of separate developments, but not the outcome of the 1960 U.S. presidential election. The details of which candidate said what during a single televised debate in October 1960 could have given the electoral victory to Nixon instead of to Kennedy, but no details of who said what could have blocked the European conquest of Native Americans.
How can students of human history profit from the experience of scientists in other historical sciences? A methodology that has proved useful involves the comparative method and so-called natural experiments. While neither astronomers studying galaxy formation nor human historians can manipulate their systems in controlled laboratory experiments, they both can take advantage of natural experiments, by comparing systems differing in the presence or absence (or in the strong or weak effect) of some putative causative factor [Suggests the solution to the difficulty mentioned in first paragraph]. For example, epidemiologists, forbidden to feed large amounts of salt to people experimentally, have still been able to identify effects of high salt intake by comparing groups of humans who already differ greatly in their salt intake; and cultural anthropologists, unable to provide human groups experimentally with varying resource abundances for many centuries, still study long-term effects of resource abundance on human societies by comparing recent Polynesian populations living on islands differing naturally in resource abundance. The student of human history can draw on many more natural experiments than just comparisons among the five inhabited continents. Comparisons can also utilize large islands that have developed complex societies in a considerable degree of isolation [Due to the keyword Isolated islands] (such as Japan, Madagascar, Native American Hispaniola, New Guinea, Hawaii, and many others), as well as societies on hundreds of smaller islands and regional societies within each of the continents. Natural experiments in any field, whether in ecology or human history, are inherently open to potential methodological criticisms. Those include confounding effects of natural variation in additional variables besides the one of interest, as well as problems in inferring chains of causation from observed correlations between variables [Explains the challenges in natural experiments]. Such methodological problems have been discussed in great detail for some of the historical sciences. In particular, epidemiology, the science of drawing inferences about human diseases by comparing groups of people (often by retrospective historical studies), has for a long time successfully employed formalized procedures for dealing with problems similar to those facing historians of human societies. In short, I acknowledge that it is much more difficult to understand human history than to understand problems in fields of science where history is unimportant and where fewer individual variables operate. Nevertheless, successful methodologies for analysing historical problems have been worked out in several fields. As a result, the histories of dinosaurs, nebulae, and glaciers are generally acknowledged to belong to fields of science rather than to the humanities.
Before solving the questions I will share the intent which I got from the passage. This is just my understanding which may be incomplete or can be wrong… there are verbal experts in our gang and I am sure they will help us in improving our RC skills… :)
While analysing and formulating a pattern in historical sciences, the sample data is spread across a large span of time period and multitude of occurrences. It is not easy to reproduce them under controlled conditions as the outcome is dependent on lot of factors. Historical science enthusiasts observe the cause/effects absence/presence of various natural/social attributes for a long period of time and then predict what can be the trend for another long period of time. While doing so they get a high level picture which may not consider all the factors that can alter the outcome. Even though learning from nature has challenges associated with them, we are able to deal with them to a certain extent.
With this understanding let’s try to solve the questions…
1. Why do islands with considerable degree of isolation provide valuable insights into human history?
(1) Isolated islands may evolve differently and this difference is of interest to us.
(2) Isolated islands increase the number of observations available to historians.
(3) Isolated islands, differing in their endowments and size may evolve differently and this difference can be attributed to their endowments and size.
(4) Isolated islands, differing in their endowments and size, provide a good comparison to large islands such as Eurasia, Africa, Americas and Australia.
(5) Isolated islands, in so far as they are inhabited, arouse curiosity about how human beings evolved there.
Option1: True
Option2: True
Option3: Not only True, but also explains why Option 1 & 2 are True
Option4: True. But More than just comparing two different systems a historian will be more interested in knowing HOW the differences got shaped by the absence/presence of various attributes in each systems (endowments and size). This is explained neatly in option 3.
Option 5: ‘Arising curiosity’ is not a valid reason from our passage.
It is safe to go with Option 3. What say?
2. According to the author, why is prediction difficult in history?
(1) Historical explanations are usually broad so that no prediction is possible.
(2) Historical out comes depend upon a large number of factors and hence prediction is difficult for each case.
(3) Historical sciences, by their very nature, are not interested in a multitude of minor factors, which might be important in a specific historical outcome.
(4) Historians are interested in evolution of human history and hence are only interested in long term predictions.
(5) Historical sciences suffer from the inability to conduct controlled experiments and therefore have explanations based on a few long-term factors.
Option 1: ‘No prediction is possible’ cannot be concluded from the passage. It says prediction is difficult and is more feasible in longer time periods…
Option 2: True
Option 3: ‘not interested in a multitude of minor factors’ is not the intent of the passage. Historians are interested but it is difficult to replicate those factors in a controlled environment.
Option 4: ‘only interested in long term predictions’ is not the intent of the passage. Short term predictions will be dependent on lot of factors hence is difficult. Predictions are more feasible over a period of time where the outcomes can be correlated with fewer attributes.
Option 5: True
Between Option 2 and Option 5, Option 2 also explains why Option 5 is true. Historical sciences suffer from the inability to conduct controlled experiments (because Historical out comes depend upon a large number of factors) and have explanations based on a few long-term factors (because prediction is difficult for each case).
3. According to the author, which of the following statements would be true?
(1) Students of history are missing significant opportunities by not conducting any natural experiments.
(2) Complex societies inhabiting large islands provide great opportunities for natural experiments.
(3) Students of history are missing significant opportunities by not studying an adequate variety of natural experiments.
(4) A unique problem faced by historians is their inability to establish cause and effect relationships.
(5) Cultural anthropologists have overcome the problem of confounding variables through natural experiments.
Option 1: Not true, they are conducting natural experiments.
Option 2: True
Option 3: True
Option 4: Not true, it is not unique just to historians. Passage talks about historical sciences in general.
Option 5: Not true, they have not overcome the problem.
Between Option 2 and Option 3, Option 3 explains why Option 2 is True. As there are not many options to conduct constructive natural experiments, each possibility is an opportunity.
Our approaches in verbal are evolved over a period of time. Here we will deal with situations like Option A is ‘more correct’ than Option B. Someone’s “Obviously Correct” can be a “Definitely Wrong” for someone else. Every option, correct or incorrect, is a learning opportunity in verbal section. Along with understanding why a particular option is correct it is important that we have clarity on why other options are incorrect. We need to harness the maximum potential of collective wisdom.
Happy Learning :)
@Rohit-Rathore
Q3) Let say we had x coins in the beginning
After 1st loot : x - 1 - (x-1)/3 = 2/3(x-1) = (2x-2)/3
After 2nd loot : (2x - 2)/3 - 1 - [(2x - 2)/3 - 1]/3 = 2/3[(2x-2)/3 - 1] = (4x - 10)/9
After 3rd loot : (4x - 10)/9 - 1 - [((4x - 10)/9 - 1]/3 = 2/3 [(4x - 10)/9 - 1] = (8x - 38)/27
Now the final value (after throwing away the fake coin) = (8x - 38)/27 - 1 = (8x - 65)/27 is a multiple of 3.
So we can write, (8x - 65)/27 = 3k
x = (81k + 65)/8
x is an integer, 81K + 65 should be a multiple of 8, means (81K + 65) Mod 8 = 0
We know 81 Mod 8 = 1 and 65 mod 8 = 1. So to get remainder 0, we need to adjust K in a way that 81K mod 8 = 7.
Least value would be K = 7
For K = 7, x = (81 * 7 + 65)/8 = 79, which should (could) be our answer.
Let me know in case of any Logical/Calculation errors.
It's June already and with 6 months left for the D-day, if you are still struggling to finalize your preparation strategy or to enjoy your preparation hours, it is high time to sort out the gaps. In this note, I will try to share some of my thoughts in this regard. This won't be about How to prepare for CAT in 6 months, but mostly about why you didn't prepare for CAT till now. (As usual, ignore the grammar mistakes :) )
Don't get addicted to planning, you already know what needs to be done
There is a lot of demand for CAT preparation strategies that our prep forums are flooded with them, guiding and misguiding aspirants (some with very useful tips and many of them with just toppers boasting). Many CAT aspirants use these articles to reassure themselves to continue their evil plan to loiter time like - I agree I did nothing useful till now but look at this IIM A guy. He cracked CAT in just 3 months.. means I can also do it in 3 months. Wrong! The person who wrote the article will have a completely different set of strengths and weakness compared to yours. Sooner or later, you will find other articles which says CAT syllabus can be done in 2 months, 50 days, 1 month, 1 week and finally from someone who cracked it without any preparation at all. Most of you, who are preparing for CAT, already knows what needs to be done to crack the exam and trust me when I say, that's all it needs. There is no magic tricks here. Finish your theory, Solve good number of questions and take the mocks. Done!
Build a purpose (No, not the philosophical one)
Most of the CAT aspirants are not motivated enough to enjoy their preparation. They still can't convince their heart and brain to read a newspaper editorial everyday or to attend the awesome sessions happening in various forums. They get exhausted/bored soon and the main culprit is usually not the lack of a plan but the lack of a purpose. Why you need to read editorials, why you need to solve those intimidating quant problems or why you need to spend your time solving lengthy DILR sets ? I will try to explain this with another question!
Why you want to spend 2 precious years and 2 million INR into an MBA program ?
You might say - for a higher pay, to extent your academic pursuit, to earn some social currency, to enter a challenging work environment or to enjoy a high profile lifestyle (you know all those suits, foreign travels, star hotel stays... which makes your FB profile looks like a Karan Johar Movie ;) ). All are good reasons. But have you ever thought what's happening inside a B school that makes its students worth of all this good stuffs ? Again, there is no magic here. I started my career a decade back and I had the good fortunate to work closely with people who were graduated from top business schools. All these folks were so wonderful that it was a treat watching them work. In my observation, the common trait of these people were their ability to handle pressure and to get things done. They go through various hurdles every single day and somehow they managed to build an attitude like "Hey look, another challenge! Let's get this done!". What is the role of a B-School in this ? Did they transformed them ? I feel No. They didn't transformed them, but the grilling MBA program polished the skills they already had. The Life inside a B-School is tough and many a times, even getting a decent sleep itself becomes a challenge. Once you managed the hell, fire walk becomes a cake walk, right ?
A B-school do not CREATE this attitude, they polish it through their rigorous curriculum. You have to create this attitude yourself. Whatever challenges you face in your preparation, which stops you from doing what is necessary to crack CAT, is your opportunity to prove that you deserve to be in a good Business School.. To prove that you are a person who is worth that fat pay check.. that you are a person who can handle challenging business roles and can get things done.. If you are not building this kind of attitude and just keep dreaming about that sweet pay-check and Euro trip, it is not going to workout. So If a question is posing an evil smile, just say, "I don't know how to solve you yet, But I will learn it and then I will solve you!"
Check out our collection of 10,000+ quant questions and 150+ concept notes - Mission IIMpossible 2019 - (Quant)
All the best folks!