S.
Danesh Kumar
Tnpsc - Aptitude
APTITUDE FORMULAS
NUMBER SYSTEM
Integers :
Z = {….,-3, -2, -1, 0, 1, 2, 3,….}
Natural Numbers : ( Positive integers)
N = {1, 2, 3, …..}
Whole Numbers : ( Non-negative integers )
W = {0, 1, 2, 3,…..}
I Every natural number is a whole number.
II 0 is a whole number which is not a natural number.
Negative integers : {-1, -2, -3,….}
Non-positive integers : { 0, -1, -2 ,-3, …..}
Even number : {2,4,6,…..}
Odd number : {1,3,5,….}
Prime number :
If it has exactly two factors , namely itself and 1.
Co-primes :
Two natural number A and B are said to be co-primes if their HCF is 1.
Ex: (2,3) , (7,9)
S. Danesh Kumar
Tnpsc - Aptitude
Fibonacci sequence :
F1 = F2 = 1 and Fn = Fn-1 + Fn-2 >, n = 3, 4, …..
Fibonacci sequence = 1, 1, 2, 3, 5, 8, 13, 21, 34,…..
Special Series
1. Sum of first n natural numbers :
1 + 2 + 3 + 4 + …… + n =
2. Sum of first n natural odd numbers :
1 + 3 + 5 + ……. upto n terms = n2
1 + 3 + 5 + …….. + l = [ if last number ‘l’ given]
3. Sum of first n natural even numbers :
2 + 4 + 6 +…….. upto n terms = n2 + n
4. Sum of first n natural square numbers :
12 + 22 + 32 + ………. + n2 =
12 - 22 + 32 - 42 ……. upto n terms
If n is even number then ;>
If n is odd number then ;>
5. Sum of first n natural cube numbers :
13 + 23 + 33 + ………. + n =
S. Danesh Kumar
Tnpsc - Aptitude
A.P
Terms;
tn = a + (n-1) d = L
sum:
Sn = [2a + (n - 1) d];
Sn = (a + L)
G.P
Terms:
tn = arn-1 = L
sum:
Sn =
na , r=1
Infinity
-1 < r < 1;>
a + ar + ar2 +……. =
S. Danesh Kumar
Tnpsc - Aptitude
Simplification
1. (a + b)2 = a2 + 2ab + b2
2. (a - b)2 = a2 - 2ab + b2
3. (a + b)2 = (a - b)2 +4ab
4. (a-b)2 = (a+b)2 - 4ab
5. a2 - b2 = (a +b) (a - b)
6. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
7. (a + b)3 = a3 + b3 +3ab (a + b)
8. (a - b)3 = a3 - b3 - 3ab (a - b)
9. a3 + b3 = (a + b) (a2 - ab + b2)
10. a3 - b3 = (a - b) (a2 + ab + b2)
11. a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)
If a + b + c = 0, then a3 + b3 + c3 = 3abc
12. xm x xn = xm + n
13.
14.
15.
16. (xy)m = xm ym
17.
18.
VBODMAS
V - Bar v.fh :
B - Bracket (milg;Gf;Fwp ) v.fh : [ ] { } ( )
O - Orders v.fh : powers, square root etc..,
D - Division - tFj;jy;
M - Multiplication - ngUf;fy; A - Addition - $l;ly; S - Subtraction - fopj;jy;
S. Danesh Kumar
Tnpsc - Aptitude
L.C.M & H.C.F
1) L.C.M X H.C.F = product of numbers.
2) Ratio of numbers X H.C.F = numbers
3) = H.C.F
4) L.C.M =
5) H.C.F =
PERCENTAGE
1) % or % = x 100
2) % & % = I1 + I2 +
3) % & % = D1 + D2 -
4) % & % = I-D-
5) same (%) =
6) after = present x
7) before = present x
8) n( A U B ) = n(A) + n (B) - n (AnB)
9) error% = x 100
S. Danesh Kumar
Tnpsc - Aptitude
RATIO & PROPORTION
1) a : b :: c : d => ( bx c) x (axd)
2) fourth proportion = a : b :: c : d
3) third proportion = a : b :: b : c
4) mean proportion ( a,b) =
5) compounded proportion (a:b),(c:d),(e:f) = ( ace : bdf )
6) duplicate ratio (a:b) =
7) sub duplicate ratio(a:b) =
8) triplicate ratio (a:b) =
9) sub triplicate ratio (a:b) =
PROFIT OR LOSS
1) profit(P) = selling price(S.P) - cost price(C.P) ; loss(L) = cost price(C.P) - selling price(S.P)
2) P% = X 100 ; L% = X 100
3) S.P = C.P ; S.P = C.P
4) market price ( M.P) = C.P
5) C.P of (x) article = S.P of (y) article
i) P% = x 100
ii) L% = x 100
S. Danesh Kumar
Tnpsc - Aptitude
SIMPLE INTEREST & COMPOUND INTEREST
S.I ( simple interest)
S.I =
P = principal
N = no of years
R = rate of interest (%)
Amount (A) = S.I + P
C.I ( compound interest)
P =A
C.I = A - P
S.I & C.I
1year = 0 ( same)
2years P = diference x
3years P=
INSTALLMENT
S.I => Pn +
C.I => present worth =
RECURRING OR CUMMULTIVE DEPOSIT
Amount maturity = total money deposit + interest
Total money deposit = monthly deposit + n ( in months)
I=
S. Danesh Kumar
Tnpsc - Aptitude
TIME & WORK
1) A = given , B = given, total ( A&B) = ?
Total =
2) Total (A&B) = given, A= given, B= ?
Total =
3) A = given, B = given, C = given, Total = ?
4) A&B = given, B&C = given, C&A = given ; Total = ?
Total = 2 x Answer
5) M1 X D1 = M2 X D2
6) M1 X D1 X H1 = M2 X D2 X H2
7)
8)
PIPES AND CISTERN
1) A = fill, B = fill, Total = ?
TOTAL =
2) A = fill, B = empty, Total = ?
TOTAL =
3) A= fill, B = fill, C = fill, Total = ?
4) A = fill, B = fill, C = empty, Total = ?
=
S. Danesh Kumar
Tnpsc - Aptitude
MENSURATION - AREA
SQUARE rJuk;
1. Area gug;gsT = a2
2. Perimeter Rw;wsT = 4a
3. Diagonal d given then Area %iytpl;lk; d vdpy; mjd; gug;gsT =
4. Diagonal %iytpl;lk;
RECTANGLE nrt;tfk; :
1. Area gug;gsT = l b
2. Perimeter Rw;wsT = 2(l + b)
3. Diagonal %iytpl;lk;
RHOMBUS rha;rJuk; :
1. Area gug;gsT
2. Perimeter Rw;wsT = 4a
PARALLOGRAM ,izfuk;
1. Area gug;gsT = b × h ( base x height ) (mbg;gf;fk; × cauk;)
2. Perimeter Rw;wsT = 2(a + b)
CIRCLE tl;lk;
1. Area (gug;gsT) = πr2
2. Perimeter (Rw;wsT) = 2πr
3. Diameter = 2r
S. Danesh Kumar
Tnpsc - Aptitude
SEMI CIRCLE miutl;lk;
1. Area (gug;gsT) =
2. Perimeter (Rw;wsT) = (π+2) r or
QUADRANT fhy;tl;lk;
1. Area (gug;gsT) =
2. Perimeter (Rw;wsT) = r
SECTOR
1. Length of Arc (l) tl;lf;Nfhzj;jpd; tpy;ypd; ePsk; = π
2. Area of a sector (tl;lf; Nfhzj;jpd; gug;gsT) = π
3. Perimeter of a sector (tl;lf;Nfhzj;jpd; Rw;wsT) = l + 2r
EQUILATERAL TRIANGLE
1. Area (gug;gsT) =
2. Perimeter (Rw;wsT) = 3a
3. Height cauk; =
4. Radius of circumcircle =
SCALAR TRIANGLE
1. Area (gug;gsT) =
2. Perimeter (Rw;wsT) = a + b + c
3. Semi perimeter = 4. Radius of incircle =
S. Danesh Kumar
Tnpsc - Aptitude
ISOSCELES TRIANGLE
1. Area (gug;gsT) = or
2. Perimeter (Rw;wsT) = 2a + b or 2a + 2
3. Height cauk; = or
RIGHT ANGLED TRIANGLE
1. Area (gug;gsT) = ( base x height )
2. Perimeter (Rw;wsT) = AB + BC + CA or Base + Height + Hypotenus
3. Largest side < Sum of other two sides
TRIANGLE WITH VERTICES:
(x1,y1), (x2,y2),(x3,y3)
Area=
TRAPEZIUM rhptfk;
1. Area (gug;gsT) = Height x sum of parallel sides or
= cauk; × (,izg; gf;fq;fspd; $Ljy;)
2. Perimeter (Rw;wsT) = AB + BC + CD + DA or a + b + c + d
QUADILATERAL ehw;fuk;
1. Area (gug;gsT) = d x ( h1 + h2)
2. Perimeter (Rw;wsT) = AB + BC + CD + DA
Area of 4 walls of a room = 2 ( l + b ) h
S. Danesh Kumar
Tnpsc - Aptitude
Length of canvas =
MENSURATION - VOLUME
CUBE fdr;rJuk; :
1. VOLUME fd msT = a3
2. L.S.A gf;fg; gug;G = 4a2
3. T.S.A nkhj;jg; gug;G = 6a2
4. Diagonal %iytpl;lk;
CUBOID fdr; nrt;tfk;
1. VOLUME fd msT = l × b × h
2. L.S.A gf;fg; gug;G = 2 (l + b)h
3. T.S.A nkhj;jg; gug;G = 2 (lb + bh + hl)
4. Diagonal %iytpl;lk; (d)
SPHERE Nfhsk;
1. VOLUME fd msT = π
2. C.S.A tisg;gug;G = 4πr2
HOLLOW SPHERE cs;sPlw;w Nfhsk;
1. VOLUME fd msT = π
2. Area -motor cyclist circus = 4πr2
HEMISPHERE miuf;Nfhsk;
1. VOLUME fd msT = π
2. C.S.A gf;fg; gug;G = 2πr2
3. T.S.A nkhj;jg; gug;G = 3πr2
S. Danesh Kumar
Tnpsc - Aptitude
HOLLOW HEMISPHERE cs;sPlw;w miuf;Nfhsk;
1. VOLUME fd msT = π
2. C.S.A gf;fg; gug;G = 2π(R2 + r2)
3. T.S.A nkhj;jg; gug;G = π(3R2 + r2)
CONE $k;G
1. VOLUME fd msT = πr2h
2. C.S.A gf;fg; gug;G = πrl
3. T.S.A nkhj;jg; gug;G = πr(l + r)
4. Slant height (l ) =
5. Base Area = πr2
CYLINDER cUis
1. VOLUME fd msT = πr2h
2. C.S.A gf;fg; gug;G = 2πrh
3. T.S.A nkhj;jg; gug;G = 2πr(h + r)
4. Base Area = πr2
HOLLOW CYLINDER cs;sPlw;w cUis
1. VOLUME fd msT = πh (R+r) (R-r)
2. C.S.A gf;fg; gug;G = 2π (R+r) (R+r-h)
3. T.S.A nkhj;jg; gug;G = 2πh (R+r)
FRUSTUM $kgpd; ,ilf;fz;lk
1. VOLUME fd msT = π
2. C.S.A gf;fg; gug;G =
3. T.S.A nkhj;jg; gug;G = π π π
4. L =
S. Danesh Kumar
Tnpsc - Aptitude
SECTOR CONVERTED INTO A CONE
1. CSA of a cone = Area of the sector
Πrl = π
2. Length of the sector = Base circumference of the cone
Volume of water flows out through a pipe = Cross section area x speed x time
No. of new solids obtained by recasting =
CUBE PAINTED
0 side painted =
1 side painted = 6
2 side painted = 12
3 side painted = 8( always)
4 side painted = 0
STATISTICS
RANGE
Range = L - S
Co efficient of Range =
Variance =
S.D =
S.D of natural number =
Co efficient of variance =
Mode = 3Median - 2Mean
S. Danesh Kumar
Tnpsc - Aptitude
TIME, SPEED, DISTANCE:
1. Distance = Speed x Time D=SxT m (or) Km
2. Speed = S= (or)
3. Time = T= s (or) hr
4. to = …….. x
5. to = …….x
6. Average Speed = ; if Distance is same
7. Speed = a : b , then Time = b : a (or) :
UNITS CONVERSION:
1. 1 m = 10 dm
2. 1 m = 100 cm
3. 1 cm = 10 mm
4. 1 Km = 1000 m
5. 1 Hectare = 10,000
6. 1 Litre = 1000
7. 1 Kg = 1000 g
8. 1 hr = 3600 s
9. 1 day = 1440 minutes
10. 1 Rs = 100 Paise
