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Tamil Model Test Question paper 16 Sets of collection Prepared by Target TNPSC FB Group

Tamil Model Test Question paper

16 Sets of collection

Prepared by Target TNPSC FB Group

சிறப்புமிக்க மாதிரித் தேர்வு 50-ஐ கீழ்க்கண்ட இணைப்பின் மூலம் பெற்றுக் கொள்ளலாம். இதன் உள்ளடக்கம் கீழே விரிவாக அளிக்கப்பட்டுள்ளது. சுமார் 154 பக்கங்களை கொண்டு 1600 பொதுத்தமிழ் கேள்விகள் மற்றும் விடைகளை கொண்டது. இது 6-12 வரை உள்ள பாடத்திட்டத்தை அடிப்படையாகக் கொண்டதாகும்.

உள்ளடக்கம்: 

1. 6-ம் வகுப்பு தமிழ் - 1 SET (MOCK TEST 40)

2. 6 & 7ம் வகுப்பு தமிழ் - 1 SET (MOCK TEST 10)

3. 7-ம் வகுப்பு தமிழ் -1 SET (MOCK TEST 41)

4. 8-ம் வகுப்பு தமிழ் – 2 SETS (MOCK TEST 11, 43)

5. 9-ம் வகுப்பு தமிழ் – 2 SETS (MOCK TEST 12, 44)

6. 10-ம் வகுப்பு தமிழ் – 2 SETS (MOCK TEST 13, 46)

6. 11-ம் வகுப்பு தமிழ் – 1 SET (MOCK TEST 15)

8. 12-ம் வகுப்பு தமிழ் – 1 SET (MOCK TEST 16)

9. பொதுத்தமிழ் பகுதி அ &  1 SET (MOCK TEST 02)

10. பொதுத்தமிழ் பகுதி இ 1 SET (MOCK TEST 01)

11. பொதுத்தமிழ் 3 SETS (MOCK TEST 19, 35, 42)


[Total 154 Pages & 1600 Questions] -11.2 MB

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{Flash News} TRB Government Polytechnic Recruitment Notification 2017 - Canceled : High Court Order

Flash News: அரசு பாலிடெக்னிக் ஆசிரியர் தேர்வு அறிவிப்பு ரத்து: ஐகோர்ட் உத்தரவு.
அரசு பாலிடெக்னிக்கில் காலியாக உள்ள1058 ஆசிரியர் பணியிடங்களை நிரப்ப அரசு வெளியிட்ட ஆணை ரத்து செய்யப்பட்டுள்ளது.
எம்.இ. பட்டதாரிகள் விண்ணப்பங்கள் ஏற்க மறுக்கப்படுவதாக தொடரப்பட்ட வழக்கில் உயர்நீதிமன்றம் ஆணை பிறப்பித்துள்ளது. பி.இ. படித்தவர்கள் மட்டுமே விண்ணப்பிக்க வேண்டும் என அரசு அறிவித்திருந்தது.


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TNPSC Aptitude Tips & Tricks, Model Question Paper


Quantitative Aptitude Tips & Tricks

With examples


Finding number of Factors 
To find the number of factors of a given number, express the number as a product of powers of prime numbers

In this case, 48 can be written as 16 * 3 = (24 * 3)
Now, increment the power of each of the prime numbers by 1 and multiply the result.
In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)
Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.



Sum of n natural numbers
-> The sum of first n natural numbers = n (n+1)/2
-> The sum of squares of first n natural numbers is n (n+1)(2n+1)/6
-> The sum of first n even numbers= n (n+1)
-> The sum of first n odd numbers= n^2

Finding Squares of numbers
To find the squares of numbers near numbers of which squares are known
To find 41^2, Add 40+41 to 1600 =1681
To find 59^2, Subtract 60^2-(60+59) =3481

Finding number of Positive Roots
If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.
Eg: x^4+3x^2+2x+6=0 has no positive roots . 

Finding number of Imaginary Roots
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) .
Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)

Reciprocal Roots
The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a

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

Finding Sum of the roots
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 biquadratic 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

Maximum/Minimum
-> If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if x=y(=k/2). The maximum product is then (k^2)/4
-> If for two numbers x*y=k (=constant), then their SUM is MINIMUM if x=y(=root(k)). The minimum sum is then 2*root (k). 
--> 

Inequalties
-> x + y >= x+y (stands for absolute value or modulus) (Useful in solving some inequations)
-> a+b=a+b if a*b>=0 else a+b >= a+b
-> 2<= (1+1/n)^n <=3 -> (1+x)^n ~ (1+nx) if x<<<1> When you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.

Product Vs HCF-LCM 
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

AM GM HM
For any 2 numbers a>b a>AM>GM>HM>b (where AM, GM, HM stand for arithmetic, geometric, harmonic menasa respectively) (GM)^2 = AM * HM

Sum of Exterior Angles
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 in
Square-----=90
Pentagon--=108
Hexagon---=120
Heptagon--=128.5
Octagon---=135
Nonagon--=140
Decagon--=144

Problems on clocks
Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other. That is, the minute hand describes 6 degrees /minute the hour hand describes 1/2 degrees / minute. Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute.
The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight. (This can be derived from the above).

Co-ordinates
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]

Ratio
If a1/b1 = a2/b2 = a3/b3 =.............., then each ratio is equal to (k1*a1+ k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is also equal to (a1+a2+a3+............./b1+b2+b3+..........)

Finding multiples
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)

Exponents
e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity 2 <>GP
-> In a GP the product of any two terms equidistant from a term is always constant.
-> The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .

Mixtures
If Q be the volume of a vessel q qty of a mixture of water and wine be removed each time from a mixture n be the number of times this operation be done and A be the final qty of wine in the mixture then,
A/Q = (1-q/Q)^n
--> 

Some Pythagorean triplets:
3,4,5----------(3^2=4+5)
5,12,13--------(5^2=12+13)
7,24,25--------(7^2=24+25)
8,15,17--------(8^2 / 2 = 15+17 )
9,40,41--------(9^2=40+41)
11,60,61-------(11^2=60+61)
12,35,37-------(12^2 / 2 = 35+37)
16,63,65-------(16^2 /2 = 63+65)
20,21,29-------(EXCEPTION)

Apollonius theorem
Apollonius theorem could be applied to the 4 triangles formed in a parallelogram.

Function
Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

Finding Squares
To find the squares of numbers from 50 to 59
For 5X^2 use the formulae
(5X)^2 = 5^2 +X / X^2
Eg ; (55^2) = 25+5 /25 =3025
(56)^2 = 25+6/36 =3136
(59)^2 = 25+9/81 =3481
--> 

Successive Discounts
Formula for successive discounts
a+b+(ab/100)
This is used for successive discounts types of sums. Like 1999 population increases by 10% and then in 2000 by 5% so the population in 2000 now is 10+5+ (50/100) =+15.5% more that was in 1999 and if there is a decrease then it will be proceeded by a -ve sign and likewise.

Rules of Logarithms:
-> loga(M)=y if and only if M=ay
-> loga(MN)=loga(M)+loga(N)
-> loga(M/N)=loga(M)-loga(N)
-> loga(Mp)=p*loga(M)
-> loga(1)=0-> loga(ap)=p
-> log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [ Note the alternating sign . .Also note that the ogarithm is with respect to base e ]


Read more tips & tricks visit  http://www.ias.org.in

General Aptitude model question paper: click here and download
 

All the Best.!!! @ tnpscnet.com
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1058 PGTRB Polytechnic - Study Material | TRB Notification Syllabus and Study Materials all in one Page | www.trb.nic.in last date: 07.07.2017

Teacher's Recruitment Board
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Post Graduate Lecturers in Polytechnique 

Notification - Syllabus - Study Materials - Hall Ticket - Result
all in one

Applications are invited only through online mode from eligible candidates upto 11.59 P.M on 07.07.2017 for the Direct Recruitment of Lecturers (Engineering / Non- Engineering) in Government Polytechnic Colleges for the year 2017-2018.

No. of Vacancies – 1058 Posts
Name of the Post: Lecturers (Engineering / Non Engineering) in Govt. Polytechnic Colleges
Salary: Scale of Pay Rs.15600-39100 + AGP Rs.5400/-

Important Dates:
Date of Notification: 16.06.2017
Application through online from: 17.06.2017
Last date to apply online: 07.07.2017
Date of Written Examination: 13.08.2017

Online Apply: Click here

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