Before we talk of number system, lets discuss the question which I asked in comment #10 on article 3.

The answer is 10%. Let’s see how to get the answer. After explaining how to get the answer, I will give you the logic. 100-85 = 15, 100-80 = 20, 100-75 = 25, 100-70 = 30. Now 15 + 20 + 25 + 30 = 90. This 90 when subtracted from 100 is 10 which is the answer.
Logic: ————–|————————|———————————|————————————–|————–
Say this line above represents 100% of the girls. The Red ones(upto first ‘|” mark) are the ones who are not fair and therefore the balance i.e. the black ones are fair. Similarly the blue ones(from first ‘|’ mark and upto 2nd ‘|” mark) on the line do not have blue eyes. Therefore the rest i.e. red ones and black ones do have blue eyes. And similarly the yellow ones(from 2nd ‘|’ mark and upto 3rd ‘|” mark) are not tall and therefore the rest i.e. red, blue and black are tall. And similarly the green ones(from 3rd ‘|’ mark and upto 4th ‘|” mark) are not rich and therefore the rest i.e. red, blue and yellow and black, are. See carefully now, these are the black ones who have all the four characteristics. And this will be 10%.

I do not have a solution better than this. Here note that I have kept the ‘Don’t have’ disjoint to arrive at ‘have ‘at least’’. The method is useful only in this type of question. And note that this ques(with diff data) has been asked many times in last 12 CATs and many other exams. Ques: Is this Maths? No. But its common sense.

Number system comprises of rational and irrational numbers, positive and negative numbers, even and odd numbers, prime and composite numbers. It also has Natural numbers, whole numbers, Integers, Perfect numbers, co-prime numbers etc. I am assuming that all of you know all this. If not, then please read any book. Please note you will, almost always find a ques on these in CAT.

So here come a few questions for you. Real CAT type questions.

Q1: 3ab59 = x + y. x and y are natural numbers and x < y. Both x and y are prime. What is x?

Q2: 2^10 = 1024. 1024 = p + q. p and q both are integers. p < q. Both p and q are not even. Both p and q are not odd. None of the two is negative. What is p? What is q?

Q3: How many rational numbers exists between two rational numbers?

Q4: Give one rational number between 1.3333……. and 1.4444…….(Both are recurring decimal numbers).

Q5: How many perfect numbers are there which are greater than 100?

Divisibility rules: If you want to understand divisibility rules of a number, I suggest instead of mugging the things, look into the multiples of a number. You will be able to see a trend/pattern almost always. If you don’t, then keep looking for the pattern, you will for sure find a pattern, almost always. Caution: Not all the numbers have divisibility rules. But these are only few. So few that u could be effected practically. I will talk more about divisibility rules with passage of time and with your feedback.

Q6: If a^b+b^a=100, where a and b are natural numbers, find the number of pairs of (a,b) such that the product a.b is maximum.(Ques by Courtesy of Mr.Rajneesh Tripathi)

Q7. Is X = 42.317317317…….(recurring), a rational number or irrational? If it’s rational, what is it’s p/q form. (Ques by Courtesy of Mr.Rajneesh Tripathi)(I suggest in this ques note the time in which u solved this question).

Q8. X = 0.ababab………(recurring). X is multiplied by Y and the result is an integer. What can Y be?
98 b) 198 c)298 d)398 e) None of these

Q9: 1001 ^ 1001 is expanded and sum of the digits is taken. Now sum of the digits of ‘sum of the digits from previous calculation’ is taken. And so on till we get a single digit. What will be this digit?

Q10. abcde x 4 = edcba. What are a, b, c, d and e respectively. a, b, c, d and e are all distinct.

Only this much for now. I am sure all of you will be able to post the answers within 24 hours. I think that should be enough time for you for you are preparing for a 2.5 hour exam having about 75 questions.

All d best to all of you.

VM (& Rajneesh)

P.S.: Mr.Rajneesh Tripathi is a seasoned CAT trainer, teaching for last 6+ years. He is joining our team now on and will be an invaluable resource to all of you and to us as well.Similar Posts:

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Answers

Hi All.

Good news for all of you is that I am back in India and the bad news associated with it is, BSNL is taking hell lot of time in activating broadband at my home.

Anyways. Here are the long awaited answers. I am not answering individual comments as that will be too long a post for all to read.

Here the answers go.

1. Both x and y are prime. All primes except 2 are odd. 3ab59 is odd. So 1 of x and y has to be even and other odd. The only even prime we know is 2 and thats smallest as well. Since x < y, x has to be 2.

2. Here the ‘click’ point should be ‘None of the 2 is -ve’. If you start from that you can easily figure out that p and q both are non -ve i.e. 0 and +ve. Hence p = 0 and q = 1024.

3. Answer is infinite. Most of you are correct.

4. I am asking what’s your dad’s name. You are telling me how many uncles you have. I am asking you to tell me 1 rational no. between the 2 given rational no. and am not asking how many between the 2. Simplest answer as per me is (1.3333….. + 1.444…..)/2. Yes answers of many are correct. I better suggest revise 9th class NCERT funda of inserting n rational numbers between 2 given rational numbers.
Say 1.333… = a and 1.444… = b and I ask you to tell me 2 rationals between a and b. Then let d = (b-a)/3. 1st rational number will be a + d and 2nd will be a + 2d. I hope this is enough for you to recollect the basics.

5. opinionated and all others, please note only a finite number of perfect numbers are known so far. If i can recall from my ‘getting older’ memory, there are 46 perfect numbers known and 2 of them are less then 100. so the answer should be 44. Howzz that??

6. Its the coolest one. Answer is 2 such pair and those are (1,99) and (99,1). Do u still need an explaination??

7. Ans is (42317-42)/99= 42275/99. Max 3 seconds job after u have read the ques. Please note in short i have also explained the process.

8. Ans is b. See the explaination of 7.

9. Ans is 5. This was a 15 sec question. Logic is 1001 ^ 1001 divided by 9, whatever is the remainder is the answer. Why – because thats the concept. To know more of the concept I need a face to face class cant explain by writing.

10. a= 2, b= 1, c = 9, d = 7 and e = 8. This involves more of logic and less of NUMBERS. All u guys do a little brain storm. Its difficult for me to explain this by writing. By the way this was an old CAT actual question. and there were 5 ques. I solved this in 3 minutes 15 sec. which I think is worth 5 ques.

Here comes a ques based on the last set of questions. Lets see how fast I get a reply.

Q1. 3p5q7 = x + y. x and y are natural numbers and x < y. Both x and y are prime. What is x?

Q2. Give one rational number between 1.9999……. and 2.0.

All d best.
VMSimilar Posts:

None Found

Hi everyone,
Yesterday I have posted the article on last two digits concept. I am still awaiting your invaluable comments & suggestions for that post. Now as i have been to classroom teaching for a good period of time I have faced maximum students complaining that they can’t remember or say summarise the important facts behind any topic in quantitative area. Whether its an approach or problem solving, You always require a good back up of those important facts. Here I have tried to put some of those useful points which must be considered as your back up tool while you approach any relevant problem.

You can have a look one by one.

1. If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x^3+3x^2+2x+6=0 has no positive roots

2. For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.

3. For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)

4. Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.

5.
For a cubic equation ax^3+bx^2+cx+d=o
• Sum of the roots = – b/a
• Sum of the product of the roots taken two at a time = c/a
• Product of the roots = -d/a

For a bi-quadratic equation ax^4+bx^3+cx^2+dx+e = 0
• Sum of the roots = – b/a
• Sum of the product of the roots taken three at a time = c/a
• Sum of the product of the roots taken two at a time = -d/a
• Product of the roots = e/a

6. If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)

7. Consider the two equations

a1x+b1y=c1
a2x+b2y=c2

Then,
If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
If a1/a2 = b1/b2 c1/c2, then we have no solution.
If a1/a2 b1/b2, then we have a unique solution.

8. Roots of x^2 + x + 1=0 are 1, w, w^2 where 1 + w + w^2=0 and w^3=1

9. |a| + |b| = |a + b| if a*b>=0
else, |a| + |b| >= |a + b|

10. The equation ax2+bx+c=0 will have max. value when a0. The max. or min. value is given by (4ac-b^2)/4a and will occur at x = -b/2a

11.
If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.

If for two numbers x*y=k (a constant), then their SUM is MINIMUM if
x=y (=root(k)). The minimum sum is then 2*root (k).

12. Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.

13. For any 2 numbers a, b where a>b

a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)

(GM)^2 = AM * HM

14. For three positive numbers a, b, c

(a + b + c) * (1/a + 1/b + 1/c)>=9

15. For any positive integer n

2<= (1 + 1/n)^n = ab + bc + ca
If a=b=c, then the case of equality holds good.

17. a^4 + b^4 + c^4 + d^4 >= 4abcd (Equality arises when a=b=c=d=1)

18. (n!)^2 > nn

19. If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s

20. If n is even, n(n+1)(n+2) is divisible by 24

21. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + …….+ a^(n-1) ) ……Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 – 14^3)

22. e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ……..to infinity
Note: 2 < e =k/2, then, a is the only prime factor for that no.

35. To find out the sum of 3-digit nos. formed with a set of given digits

This is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits)

Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8.
Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there)
= 25 * 24 * 11111
=6666600

36. Consider the equation x^n + y^n = z^n

As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.

37. Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then,
N^(p-1) – 1 is always divisible by p.

38. 145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e.

145 = 1! + 4! + 5!

39.
Where a no. is of the form a^n – b^n, then,
• The no. is always divisible by a – b
• Further, the no. is divisible by a + b when n is even and not divisible by
a + b when n is odd

Where a no. is of the form a^n + b^n, then,
• The no. is usually not divisible by a – b
• However, the no. is divisible by a + b when n is odd and not divisible by
a + b when n is even
40. The relationship between base 10 and base ‘e’ in log is given by
log10N = 0.434 logeN

41. WINE and WATER formula

Let Q – volume of a vessel, q – qty of a mixture of water and wine be removed each time from a mixture, n – number of times this operation is done and A – final qty of wine in the mixture, then,

A/Q = (1-q / Q)^n

42. Pascal’s Triangle for computing Compound Interest (CI)

The traditional formula for computing CI is
CI = P*(1+R/100)^N – P

Using Pascal’s Triangle,

Number of Years (N)
——————-
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
… 1 …. …. … … ..1

Eg: P = 1000, R=10 %, and N=3 years. What is CI & Amount?

Step 1:
Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331

The coefficients – 1,3,3,1 are lifted from the Pascal’s triangle above.

Step 2:
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331 (leaving out first term in step 1)

If N =2, we would have had,
Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs.1210
CI = 2 * 100 + 1* 10 = Rs.210

43. Suppose the price of a product is first increased by X% and then decreased by Y% , then, the final change % in the price is given by:
Final Difference% = X – Y – XY/100

Eg) The price of a T.V set is increased by 40 % of the cost price and then is decreased by 25% of the new price. On selling, the profit made by the dealer was Rs.1000. At what price was the T.V sold?

Applying the formula,
Final difference% = 40 – 25 – (40*25/100) = 5 %.

So if 5 % = 1,000
Then, 100 % = 20,000.
Hence, C.P = 20,000
& S.P = 20,000+ 1000= 21,000

44. Where the cost price of 2 articles is same and the mark up % is same, then, marked price and NOT cost price should be assumed as 100.

45.
Where ‘P’ represents principal and ‘R’ represents the rate of interest, then, the difference between 2 years’ simple interest and compound interest is given by P * (R/100)^2

The difference between 3 years’ simple interest and compound interest is given by (P * R^2 *(300+R))/100^3

46.If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.

If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.

If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time

If A can finish a work in X time and B in Y time and A, B & C together in S time then
• C can finish that work alone in (XYS)/ (XY-SX-SY)
• B+C can finish in (SX)/(X-S); and
• A+C can finish in (SY)/(Y-S)

47. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n^5

48.
When an unbiased coin is tossed odd no. (n) of times, then, the no. of heads can never be equal to the no. of tails i.e. P (no. of heads=no. of tails) = 0

When an unbiased coin is tossed even no. (2n) of times, then,
P (no. of heads=no. of tails) = 1-(2nCn/2^2n)

49. Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m!

Eg)1. Suppose there are 10 girls dancing one after the other. What is the probability of A dancing before B dancing before C?

Here n=10, m=3 (i.e. A, B, C)

Hence, P (A>B>C) = 1/3!
= 1/6

Eg)2. Consider the word ‘METHODS’. What is the probability that the letter ‘M’ comes before ‘S’ when all the letters of the given word are used for forming words, with or without meaning?

P (M>S) = 1/2!
= 1/2

50. CALENDAR

Calendar repeats after every 400 years.

Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.

Century has 5 odd days and leap century has 6 odd days.

In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.

January 1, 1901 was a Tuesday.

51.
For any regular polygon, the sum of the exterior angles is equal to 360 degrees, hence measure of any external angle is equal to 360/n (where n is the number of sides)

For any regular polygon, the sum of interior angles =(n-2)*180 degrees
So measure of one angle is (n-2)/n *180

If any parallelogram can be inscribed in a circle, it must be a rectangle.

If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal).

52. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order)
53.
For any quadrilateral whose diagonals intersect at right angles, the area of the quadrilateral is
0.5*d1*d2, where d1, d2 are the length of the diagonals.

For a cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), where
s=(a + b + c + d)/2
Further, for a cyclic quadrilateral, the measure of an external angle is equal to the measure of the interior opposite angle.

Area of a Rhombus = Product of Diagonals/2

54. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]

55. Area of a triangle

1/2*base*altitude
1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B
root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2
a*b*c/(4*R) where R is the circumradius of the triangle
r*s ,where r is the inradius of the triangle

56. In any triangle
a=b*cos C + c*cos B
b=c*cos A + a*cos C
c=a*cos B + b*cos A
a/sin A=b/sin B=c/sin C=2R, where R is the circumradius
cos C = (a^2 + b^2 – c^2)/2ab
sin 2A = 2 sin A * cos A
cos 2A = cos^2 (A) – sin^2 (A)

57. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1

58. Appollonius Theorem

In a triangle ABC, if AD is the median to side BC, then
AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2)

59.
In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.

60. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.

61. Let W be any point inside a rectangle ABCD, then,
WD^2 + WB^2 = WC^2 + WA^2

62. Let a be the side of an equilateral triangle, then, if three circles are drawn inside this triangle such that they touch each other, then each circle’s radius is given by a/(2*(root(3)+1))

63.
Distance between a point (x1, y1) and a line represented by the equation
ax + by + c=0 is given by |ax1+by1+c|/Sq(a^2+b^2)
Distance between 2 points (x1, y1) and (x2, y2) is given by
Sq((x1-x2)^2+ (y1-y2)^2)

64. Where a rectangle is inscribed in an isosceles right angled triangle, then, the length of the rectangle is twice its breadth and the ratio of area of rectangle to area of triangle is 1:2.

Now I sincerely hope that this will be of your great help.

All the very Best

Rajneesh TripathiSimilar Posts:

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Concept of Last Two Digits

Hello All,
First of all let me appreciate the opportunity of being a part of this team. This is my first article to this project.
Before I start the concept of last two digits lets have a quick look at the concept behind unit’s digit. I know most of you might be aware of this concept called cyclicity. All you need to do, is just catch the last digit of the base and apply the rule of cyclicity. You can follow the cyclicity table give below for your convenience.
Base Index-1 Index-2 Index-3 Index-4
Type 4K+1 4K+2 4K+3 4K
1 1 1 1 1
2 2 4 8 6
3 3 9 7 1
4 4 6 4 6
5 5 5 5 5
6 6 6 6 6
7 7 9 3 1
8 8 4 2 6
9 9 1 9 1
0 0 0 0 0

e.g Last digit or unit digit of 127^127.(Please note that this expression is to be read as 127 to the power 127. Also all other similar expressions are also to be followed in the same pattern.)
Just pick up last digit of the base i.e 7 and divide the given power by 4 & check out the remainder.
The remainder when 127 is divided by 4 is 3. Hence the power of the given expansion is of the type 4K+3. Therefore the unit’s digit will be 3.
How simple is that.
Now lets move on to the concept of last two digits. The fundamental behind the last two gits is just like unit’s digit concept.
Here you just need to catch the last two digits.
e.g Find the last two digits of the product of 1023 X 2027 X 5096 X 5979.
Just pick up the last two digits of all the numbers.
i.e 23, 27, 96 & 79
Multiply any pair using crisscross method of multiplication and note down the last two digits only.
i.e 23 X 27 gives you 21 as last two digits similarly 96 x 79 gives you 84 as last two digits.
Now again multiply the obtained results to get the answer.
i.e 21 x 84 which gives you 64 as last two digits.
Hope you find it doable comfortably.
Now lets move on to the cyclicity concept of last two digits.
1. If you have 1 as your unit’s digit in any base then whatever be the ten’s place, you just need to write 1 as your unit’s place as simple as that & just multiply the unit digit of index/power to the ten’s place of the base and note down the unit digit of multiplication as your ten’s digit of expansion.
e.g Last two digits of 21^144.
You have 1 as unit’s place and 2 as ten’s place in your base in the given example.
You know that unit digit of this expansion has to be 1 always.
Now for ten’s digit of this expansion you pick up the unit digit of the power 144, which is 4 & multiply it to the ten’s digit of the base which is 2.
i.e 4 X 2 which will be 8 & this will be the ten’s place of the expansion.
Hence last two digits come out to be 81.
Similarly 3, 7 & 9 can be converted to such an extent so that you can get unit digit 1 and after that you can easily do the process of 1 as unit’s place.
One more very interesting and very important result is the cyclicity of 24 as last two digits of the base.
Please follow if 24^even power it will always end by 76 as last two digits
& if 24^odd power, it will always end by 24 as last two digits.
Also 2, 4 & 6 can be broken into factors to get the last two digits.
e.g Last two digits of 196^224.
The base 196 can be converted to small factors like 196 = 4 X 49
Now the problem becomes simple 4^224 x 49^224

Or (4^5)^44 X 4^4 X 2401^112

which gives you 76, 56 & 01 as last two digits respectively.
Multiply step by step as the process discussed earlier and get the last two digits.
The answer will be 56 for this example.
Concept of compliment of the base
e.g Last two digits of 39^44 .
Compliment of the base will be 61, hence lets take out last two digits of (-61)^44 which results to -41 as last two digits. But this has to be positive and to make it positive just add the complimentary base i.e 100 to the last result -41. Hence the answer comes out to be 59.
And lastly when 5 is your unit’s place in the base. It can be of two forms.
Either ten’s place is odd or ten’s place is even.
i.e either of the type 15, 35, 55, 75, 95, (Lets say Case-1)
or of the type 25, 45, 65, 85. (Lets say Case-2)
Case-1 last two digits will be 75 for any odd power .
And for any even power last two digits will be 25.
Case-2 Last two digits will be 25 for any power .
Now you can have any example of the types discussed in this article and easily take out the last two digits of the same.
Thats all for now. I believe that you all will be able to understand the concept of last two digits from this article. I will be waiting for your responses and valuable suggestions.
All the best
Rajneesh TripathiSimilar Posts:

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Today I got a mail from TCY online and they are giving some 11th hour strategies with lectures, tests and some download material. I have gone through some of them and they are good; as they are for free, just kidding; they are really good. You don’t have to pay to listen to the lectures. Yes, lectures are in audio-visual format, so they are more interactive and helpful. Go & have a look for yourself. Click the image to go the page where TCY has put all the stuff.Similar Posts:

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• Vedic Math for Quant CAT 2008

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Found one more collection of tips on vedic maths for solving quant questions fast.

Note: These will be more helpful to those who are preparing for next year’s CAT as if you are aiming this years CAT then now its not a good time to learn a new method of solving a question, instead revise what you already know. Waise, this is my suggestion.Similar Posts:

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Please click the image to read some of the tips provided by arun sharma himself. Arun Sharma is an IIM, Bangalore pass out and now a days give coaching to CAT aspirants. His book on quant is very beneficial for CAT aspirants. I too refer his book. For me seriously his tips are  of great help. If you don’t believe give it atleast 5 min and you will got to know about his mettle.

Try it.Similar Posts:

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List of Topics

Are you still searching the source where you can find tests to practice? If yes, then you have once again come to the right place. Yesterday, I came across the site of TCYonline where you can register it for free and chose the topic of test which you want to give. All you have to do is register for free. List of topics is beside this text.

Problem: You can select only two topics from the list.

Solution: You can make multiple accounts. I haven’t tried it but I’m sure they won’t caught you. Only thing is you would require more than one email account and with so many email providers; I don’t think it’s not an issue.

You can also refer this to your younger cousins who are in school or preparing for any other competitive exams as “Nishkaam Seva” is what our “Bhagwat Gita” teaches. It will also help you improving your PR (public relation) skills.

MBA Section

Here is the screen shot of MBA Section which is divided in various sections. Note: Many tests are uploaded by users of this site but I don’t think they made it on their own. It’s just that they have it and are helping other users.

JUST CLICK ON ANY IMAGE TO ENTER INTO THEIR HOME PAGE.

We apologise  for the blurr image as we had to do it in order to fit it in available space.Similar Posts:

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50 Keys to CAT

Today I found a collection of 50 important formulae for Quantitative Ability Section of CAT – Arithmetic, Algebra, Geometry and Modern Mathematics. These are very good for quick revision. You should take their print outs or save it on your PC. Just click the image to open the file.

Thanks: TestFunda.com

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A simple thanks will suffice. Waiting for your thanks.Similar Posts:

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CLOCKS.
• Here “#” means degrees.
• Normal clock has 60 divisions and each division =6 #
• In 60 minutes , minute hand moves 60 divisions whereas hour hand covers 5 divisions whereas hour hand covers 5 divisions, therefore minute hand overtakes 55 divisions in 60 minutes, so to overtake 1 division it needs 12/11 minutes.
• Minute hand covers 12 times the no. of divisions covered by hour hand in same time.
• Whenever u need to imagine; imagine the position of hand at exact hour, because u know the gap between the hour hand and minute hand=( hour*5) and minute hand is behind
• Angle between hands – (11m-60h)/2, If the angle calculated does not exists in the ans. options, then subtract your ans. from 360#
• Whatever position comes 1 in an hour, it takes place 11 times in 12 hrs.
• When u get ans. in decimals then remember base is 60.
• For every hour the gap is 60/13
• I f two watches move at different speeds, then they show the same time when the gap is 12hrs. or 720 minutes.
• We use the concept of ratio to find time.
In a normal watch hands coincide every 65 5/11 minutes. If it is written that hands are coinciding every 65 minutes. it means, true time is 65 minutes but watch is giving 65 5/11 minutes, so it is gaining 5/11 minutes every 65 minutes, therefore to calculate the variance , we find the number of time periods of 65 minutes and multiply it by 5/11

Note: If you want to practice more related to this topic, then you can download the “Neha’s Funda” in the right section of this page. It contain lot of fundaes and practice questions. Enjoy. Thanks appreciated.

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A number is divisible by ……..By 7, 11, and 13 if it consists of but four places, the first and fourth being occupied by the same
significant figures, and the second and third by ciphers.”
(A cipher is a 0.) It continues:
“Thus, 2002, 3003, and 5005 are divisible by 7, 11, and 13.”

Proof:  7×11x13 = 1001Similar Posts:

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Remember: This method doesn’t give all the combinations of other sides. For e.g If the given side is 25cm then this method doesn’t give other two sides as 20cm and 15cm. So always keep open your mind. Kewal Dikhaawe pe mat jaao, apni akal lagao. 

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1. 1st January 0001 was a Monday.
2. Calendar repeats after every 400 years.
3. Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
4. Odd days- remainder obtained when no. of days is divided by 7. Normal year has 1 odd day and leap year has 2 odd days.
5. Calendar moves ahead by number of odd days.
6. While checking leap year just analyze whether February falls in that period or not.
7. Century has 5 odd days and leap century has 6 odd days.
8. Take out net odd days.( add all the odd days and again divide by 7)
9. In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
10. 1st January 1901 was Tuesday.
11. We calculate odd days on the basis of the previous month.

Example – what day is it on 29th August 1982?
Method- As we know 1/1/1901 was a Tuesday now we take 1982 and 1901
- 1982-1901=81 years.
- 81/4=20…. ( disregard decimal)
- 81+20=101
- 101/7- remainder is 3, so 3 days from Tuesday is Friday.
-Now check whether it is a leap year or not. In this case it is not a leap year. Therefore 2nd July will be Friday
- Now we have to go month wise . 2nd august = 3odd days= so from Friday 3 odd days will be Monday so 29th August will be a Sunday.

Here is the link to check the day on a specific date. Keep Practicing, and you will find this method at your fingertips.Similar Posts:

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Example- A horse and a carriage were bought for Rs. 12000. the carriage was sold at a loss of 10% , horse was sold at a profit of 20%. Together I received Rs.13500. what is the C.P. and S.P. of each.

Method-Assume everything to be a horse, so I shud have sold everything at 20% profit.20% of 12000+12000= 14400.But I received Rs. 13500 only that makes a difference of 900 or Rs. 900 are less. This also makes a difference of 30% coz we calculated 40% profit (20+20%), but we had [+20%+(-10%)]=10%
I calculated 30% more on carriage.
30% or carriage=900
therefore 100%=3000
so now we have the individual cost of the horse, the carriage is for 9000/-and the S.P. can b calculated now.

We have similar questions in many forms which can b done by applying the above concept

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(1) Subtract the ones digit from N: Subtract 3 from 273: 273
– 3
270
(2) Dividing the result by 10: Divide 270 by 10: 270 ÷ 10 = 27
(3) Subtract—from that result—twice
the original ones digit: Subtract, from 27, twice 3:

27

– 6

21:  a multiple of 7
Since this procedure leads to a multiple of 7, it must be that the original number, 273, is a
multiple of 7, as shown in this long division:
A much quicker way to use the procedure is shown here
2 7 3
– 6
2 1
x 2
Since 21 is a multiple of 7, it must
be that 273 is a multiple of 7.

Let’s apply this more direct procedure to a larger multiple of 7: 7 x 568 = 3,976:
3 9 7 6
– 1 2
3 8 5
x 2
We don’t know if 385 is a multiple of 7
or not, so we continue the procedure.
3 8 5
– 1 0
2 8

Since 28 is a multiple of 7, it must be that
both 385 and 3,976 are multiples of 7.
What is the proof behind this procedure?
Consider, without loss of generality, a three digit number, N = 100a + 10b + c, where a, b, and c are
single digits, a ≠ 0.
Applying the Divisibility Rule for 7 to 100a + 10b + c, we get:
(1) Subtract c: 100a + 10b
(2) Divide by 10: 10a + b
(3) Subtract twice c: 10a + b – 2c
Claim: 100a + 10b + c is a multiple of 7 if and only if 10a + b – 2c is a multiple of 7
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Lets take a number: 89

Only thing you should remember is square of numbers from 1 to 9.

1) Write the Square of first digit i.e 8^2 = 64

2) Find 2xfirst digit x second digit i.e = 2×8x9=144

3) Write the Square of second digit i.e 9^2 = 81

Now write them in sequence as shown in picture. { Result = Square of 89 }

Last digit of in sequence will become the last digit of Result. After this take the remaining digits i.e 8 (in this case) and add it to second number of sequence. Last digit of addition will become the second last digit of Result. Now take the remaining digits of this number i.e 15(in this case) and add it to the first number of sequence i.e 64. Now this addition has given us the final digits of Result. Therefore, Square of 89 = 7921.

Steps will remain the same only digits will change. May be here it seems that its a difficult method but I can assure that once you expertise this method, you won’t feel the same.  I hope I have been able to explain this method in best possible way. If you still have some problem in this method, you can mail it to me at gaurav@mbatutes.com.Similar Posts:

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1) Figure formed by joining the mid points of a Square:

Remember: Figure formed will be Square. And the area of this Square is half the area of the enclosing square. For eg. If the side of enclosing square is : a cms, then

side of the Square formed by joining mid points = b= a/(sqrt2).

We got this result using Pythagoras Theorum.

Area of this Square : b^2 = a^2/2 = Half of the enclosing Square.

2. When enclosing figure is a Rectangle.

Remember: Figure formed by joinging the mid points of this figure won’t be rectangle or square. It will be a Rhombus (whose diagnols bisect each other at 90deg and all sides are equal)

Area of this rhombus too will be half of the area of the enclosing rectangle. Let the sides of rectangle be p cms and q cms (p<q)

Then Length of Rhombus = sqrt(p^2 + q^2)/2 { Using Pythagoras Theorum}

Area of Rhombus = half the product of diagnols.

= (1/2)* pq {As length of diagnols of rhombus will be same as sides of rectangle enclosing it}

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None Found

First of all, What is Negative Remainder:

eg. When 14 is divided by 8

Then, 14= 8×1+6

or

14=8×2-2

Therefore, We have two remainder and they are +6 and -2 where (-2) is the Negative Remainder. I hope I am able to make it clear.

Now, I make a small assumption that you know the remainder theorum and you make use of it while calculating the remainder. Ok, Now lets take an example

Q. Find the remainder when 14×15 is divided by 8

Ans. Traditional method: Find remainder after dividing 14 and 15 by 8, multiply them and find the final answer i.e

a)14/8= 8×1+6

b)15/8=8×1+7

c)find remainder after dividing 6×7 by 8 = 42/8=8×5+2

Therefore, Remainder is 2

New method,

a)14=8×2-2 (minus 2 is the negative remainder)

b)15=8×2-1 ( minus 1 is the negative remainder)

Final Remainder is = (-2)x(-1) divided by 8 =2

Therefore Remainder is 2.

Though both the method looks same in calculation level but when you some difficult problem then this negative remainder method will help a lot and will save some time.s

Remember: If at the end remainder is negative then substract that number from the divisor for eg. If the answer was (-3) then remainder would be = 8-3=5.Similar Posts:

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To form a square we need to select same number of vertical and horizontal lines. Therefore,

No of Squares grid has = 1×1 + 2×2 + 3×3 + 4×4 …………. + nxn

= (nx(n+1))/2

To form a rectangle we can select any number of horizontal and vertical lines. Therefore,

No of rectangles grid has = (1+2+3+…….n)x(1+2+3+……..+n)

= (1+2+3+……………n)^2

= ((nx(n+1)/2))^2

= Square of sum of first n natural numbers = Sum of cube of first n natural numbers

Main Confusion: Sometimes students take n as number of lines presents instead of number of rows and column. Always take n as number of rows or column present in grid.Similar Posts:

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A) 7 ( not 6 which lot of students do on the D day.) So remember this, it will help you for sure.Similar Posts:

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A,H,M,I,O,    T,U,V,W,X,Y

Remember it as it will save time in the questions which deal with mirror image of letters. If they are not directly used in the question even then this knowledge will help as you are now aware that english alphabets contain eleven letters which are mirror image of self.Similar Posts:

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Q. The question is to find the number of ways for reaching B from A. Also you cannot move upwards or towards left.

A. Never start counting the ways even if the grid is small. When you see the figure you will notice that In Order to move from A to B you have to cross x (3 in figure. 1) horizontal steps and y (3 in figure.1) vertical steps.

LET,

Horizontal steps be marked as 1

Vertical steps be marked as 0

Therefore, you need to find different combinations of 111……(x times) 00000….. (y times) binary code.  Now this problem has become of Permutation and Combination in which you have to arrange the digits of a number and tell how many numbers are possible with the digit of this number. So all those people who have basics of Permutation and Combination will be able to solve this question very easily.

Solution is = (x+y)!/(x!*y!)

= (3+3)!/(3!+3!)

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1 mile = 1.609 km

8 kmph = 5 mile per hour  (approx)

18 kmph = 5 m/s­

15 mile per hour = 22 ft/sec

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Any six-digit, or twelve-digit, or eighteen-digit, or any such number with number of digits equal to multiple of 6, is divisible by EACH of 7, 11 and 13 if all of its digits are same .

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