The CAT Community

2009-02-14T02:39:00.000-08:00

GEOMETRY – Important Notes/ Formulae

(1) Property of tangents - 1

(2) Property of tangents – 2

The point joining the centre of the circle (O) and the point (P) bisects the angle BPA

(3) Property of tangents and secants

2009-02-14T01:41:00.000-08:00

GEOMETRY – Important Notes/ Formulae

(1) The area of shaded in green can calculated as A = 2 (Area of sector 120 deg) – Area of rhombus AO1BO2.

Hence the green shaded area A is

(2) Property of Chords of a circle

(3) External angle bisector property

2009-02-14T01:16:00.000-08:00

GEOMETRY - Important notes/Formlae

(1) Length of angular bisector of a triangle

where AD is the bisector of angle A

(2) Angular bisector Theorem

where AD is the bisector of angle A

(3) Medians of a triangle divides it into 6 equal parts

2009-02-03T02:59:00.000-08:00

PROGRESSIONS - Important Notes/Formulae

(1) Progression formulae

(2) Sum of 1st ‘n’ natural nos = ∑n = n(n+1)/2
Sum of the squares of 1st ‘n’ natural nos = ∑(n^2) = n (n+1) (2n+1) /6
Sum of the cubes of 1st ‘n’ natural nos = ∑(n^3) = [∑n]^2 = (n^2) [(2n+1)^2] / 4

(3) a,b,c are in arithmetic progression if b= (a+b)/2

(4) a,b,c are in harmonic progression if b= 2ac/ (a+c)

(5) If the terms of any progression is specified as a fuction f(x) then sum fo the progression series is given as
Sn = ∑(Tn) = ∑ [f(x)]
Eg: If terms of a progression series is given by the function f(x)

2009-02-03T01:48:00.000-08:00

TIME AND DISTANCE - Important Notes/Formulae

(1) Circular track motion – Time taken to meet for the first time is given below

(a) If 2 persons run in the SAME direction

(b) If 2 persons run in OPPOSITE direction

(2) Clocks - Angle between the Hr hand and Min hand at H : M
α = 30H – (11m/2)

(3) Clocks – At what time between ‘T’ and ‘T+1’ hrs, does the angle between the Hr hand and min hand is ‘β’ ?
β = Difference between (6x) and [30T + (x/2)] , Form this eqn find ‘x’.
Therfore at T : x the angle between the hands will be β.

(4) If a person travels distance of ‘m’ at a speed of ‘a’ and a distance of ‘n’ at a speed of ‘b’, then
Avg speed = ab(m+n) / (ma+nb)

(5) 2 persons a and b start from A and B towards each other, they meet after ‘t’ secs. a takes ‘t1’secs to meet the point B after meeting b. Similarly b takes ‘t2’ secs to meet A, after meeting a. Then,
t = √(t1*t2)

(6) If a person travels to a Distance ‘D’ at ‘u’ Kmph, he is late by ‘p’ hrs.
If he travels at ‘v’ Kmph, he is late by ‘q’ hrs. Then,
D = uv(p+q) / (u+v)

(7) If speedof a boat is upstream is ‘u’ and downstream is ‘v’, then
Speed of boat in still water = (u+v)/2
Speed of stream = (u-v)/2

2009-02-02T23:53:00.000-08:00

NUMBERS – Important Notes/Formulae

(1) Sample Q: Find the min no. of identical square tiles required to cover up a rectangular area of with dimensions ‘x’ and ‘y’ units
Solution: Ans = (Total rectangular area / Area of the square)

(2) Find the number of 5’s present when the numbers from 1-500 are written down on a sheet of paper.
Solution: Instead of counting the number of 5’s totally, spit the no. of 5’s by the place of the digits.
No. of 5’s in unit digit = 5*10 = 50
No. of 5’s in ten’s digit = 5*10 = 50
No. of 5’s in hundred’s digit = 1 = 1 → Ans = 101

(3) Find the minimum no. of standard weights required for measuring any whole number weight upto 38Kg?
This question can also be asked as : Find the minimum no. of positive integers which can produce all numbers between 1-38 by either addition or subtraction.
Solution:
For this type of question add up the power of 3, as shown below
(3^0) + (3^1) + (3^2) + (3^3) = 1+3+9+27=40 (>38)
Hence using 1kg, 3kg, 9kg, 27kg weights alone we can measure any weight between 1-38. → Ans = 4 weights

(4) There are ‘a’ apples, ‘b’ bananas and ‘m’ mangoes, Find the maximum no. of boxes required to arrange the fruits so that no. of fruits in each box is the same and the no. of fruits of each variety in each box is the same.
Ans = HCF (a,b,m)

2009-02-02T23:30:00.000-08:00

NUMBERS – Important Notes/Formulae

(1) If a number “N” when divided by a,b,c leaves the same remainder ‘r’, then
“N” can be expressed in the below formats
N = LCM(a,b,c) + r
and
N = HCF [ (c-a), (c-b) ]

(2) If “N” is divides a,b,c and leaves same remainder, then
N=k(a-b) (k can take fractional values as well)

(3) If a number “N” when divided by a,b,c leaves the remainders (a-x), (b-y), (c-z), then
N = LCM(a,b,c) – x

(4) LCM and HCF of Fractions
LCM [ (a/x), (b/y), (c/z) ] = LCM (a,b,c) / HCF (x,y,z)
HCF [ (a/x), (b/y), (c/z) ] = HCF (a,b,c) / LCM (x,y,z)

(5) a*b = LCM(a,b) * HCF(a,b)

2009-02-02T23:03:00.000-08:00

NUMBERS - Impotant notes/Formulae

(1) Sample Q:
(i) Find the remainder when (2222...upto...1000digits) is divided by 3125?
(ii) Find the remainder when (2222...upto...1000digits) is divided by 40?

Solution (i): Rem (Any number/5n) = Rem(Last 5 digits of the number/5n)
315 = 5n (Multiple of 5)
Therefore Ans = Rem (22222/3125)=347
Solution (ii): Rem (Any number/2) = Rem(Last 2 digits of the number/2)
40=2n (Multiple of 2)
Therefore Ans = Rem (22/40)=22

(2) Sample Q: How many perfect squares are there in the first 50 whole numbers?
Solution: The trap here is the word ewhole numbers which includes '0' as well.
Please note that the number '0' is also a perfect square (Since 0 = 0*0).

(3) Sample Q (Successive division): Find the number "N" which when divided successively by a,b,c leaves remainder x,y,z.
Solution: The general form of such a number "N" --> N = (abc)k + [(zb+y)a + x]

(4) The values of Rem [(a/N), (a^2/N), (a^3/N), (a^4/N).....] (where a,N are positive integers), follow a specific repetative pattern.
Eg: Rem [(4/9), (4^2/9), (4^3/9), (4^4/9)....] = Rem [(4/9), (16/9), (64/9), (256/9)....] = 4,7,1,4,7,1,....

2009-02-01T03:57:00.000-08:00

NUMBERS - Important Formulae/Notes

(1) Product of any 3 consecutive natural numbers is divisible by 6
Eg: 4*5*6=300 (div by 6)

(2) Any prime no. That is greater than 3 can be expressed in the following form
N = (6k + 1) or (6k – 1)

(3) Find the highest power of “a” that divides N!

Now the highest power of "a" that divides N! = (x+y+z)

(4) If a number “N” is divided by a,b,c and leaves reminders a,b,c, then,
N = LCM(a,b,c) + A
A = (bz+y)a + x

(5) A perfect number is a number that can be expressed in the below format,
where p = any prime no

(6) If
(a-1)n = ka + 1 → (a is even)
(a-1)n = ka - 1 → (a is odd)
(a+1)n = ka + 1 → (a is even or odd)

(7) Sample Q: Find the greatest power of 30 in 200!
Solution: First divide 30 its atomic factors 30 = 5*3*2
Now greatest power of 30 in 200! = greatest power of 5 in 200!
To find the greatest power of 5 in 200!,

Now ans = Greatest power of 30 in 200! = 40+8+1 = 49

2009-01-31T22:19:00.000-08:00

NUMBERS - Important Formulae/Notes

(1) No. of Factors / Co-primes

If any integer is be expressed in the below format,

then,

Where,
Nf → No. of Factors of N
Sf → Sum of all the Factors of N
Nc → No. of co-primes to N (that are lesser than N)
Sc → Sum of all co-primes to N (that are lesser than N)

(2) Divisibility Rule

3 - Sum of the digits should be divisible by 3
Eg: 36 → 3+6=9 (divisible by 3)

4 - Last 2 digits of the number should be divisible by 4
Eg: 2248 → last 2 digits = 48 (divisible by 4) → 2248 is divisible by 4

6 - Number should be divisible by both 2 & 3
Eg; 5124 is divisible by 2. Also 5+1+2+4=12 (divisible by 3) → 5124 is divisible by 6

7 - If Number is of the form (N = abc), then (ab - 2c) should be divisible by 7
Eg: 322 → 32-2(2)=28 (divisible by 7) → 322 is divisible by 7

8 - Last 3 digits of the number should be divisible by 8
Eg: 36184 → Last 3 digits = 184 (divisible by 8) → 36184 is divisible by 8

9 - Sum of the digits should be divisible by 9
Eg: 4302 → 4+3+0+2=9 (divisible by9) → 4302 is is divisible by 9

11 - (sum of even digits) - (sum of odd digits) should be a multiple of 11
Eg: 6182 → (6+8)-(1+2)=11 (divisible by 11) → 6182 is divisible by 11

19 - If Number is of the form (N = abc), then (ab + 2c) should be divisible by 19
Eg: 798 → 79+2(8)=95 (19*5=95, divisible by 19) → 798 is divisible by 19