TNPSC Maths Shortcuts
Are there any shortcut methods to solve Maths Question? This page provides you TNPSC Maths Shortcuts In Tamil. By referring shortcuts, you may save time for another section like English, General Studies & others. TNPSC stands for Tamil Nadu Public Service Commission. It conducts recruitment exam in which many applicants apply for getting placed for Group 1, Group 2, Group 3, and Group 4 positions. Now by applying this TNPSC Maths Shortcut, you will be able to optimize your time & score well in TNPSC Maths paper. In addition to this, we are also listing TNPSC Maths Study Material PDF which you may download.
Students, preparing for TNPSC exam may refer these shortcut techniques as time is the main factor which helps you to crack paper. We are placing video of TNPSC Maths Shortcuts In Tamil which will help you a lot. There are many shortcuts for topics including percentage, simplification, ratio & proportion, height & distance, average in Maths. To know more about TNPSC Maths Shortcuts In Tamil, go through this page customized by team of recruitmentresult.com
TNPSC Maths Shortcuts
TNPSC Maths Questions With Explanations
Ques 1: Which one of the following is not a prime number?
 31
 61
 71
 91
Ans: 4
Explanation:
91 is divisible by 7. So, it is not a prime number.
Ques 2: (935421 x 625) = ?
 575648125
 584638125
 584649125
 585628125
Ans: 2
Explanation:
935421 x 625 = 935421 x 5^{4}= 935421 x {10/2}^{4}
=935421 x 10^{4}/2^{4}=9354210000/16
= 584638125
Ques 3: Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
 39, 30
 41, 32
 42, 33
 43, 34
Ans: 3
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Explanation:
Let their marks be (x + 9) and x.
Then, x + 9 = 56/100(x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33
Ques 4: In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is of the number of students of 8 years of age which is 48. What is the total number of students in the school?
 72
 80
 120
 100
Ans: 4
Explanation:
Let the number of students be x. Then,
Number of students above 8 years of age = (100 – 20)% of x = 80% of x.
80% of x = 48 +2/3 of 48
80/100 of x = 80
x = 100.
Ques 5: A student multiplied a number by 3/5instead of 5/3. What is the percentage error in the calculation?
 34%
 44%
 54%
 64%
Ans: 4
Explanation:
Let the number be x.
Then, error = 5/3 of x3/5of x=16/15 of x
Error% = {16x/15 of 3/5x of 100} % = 64%.
Ques 6: In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
 2700
 2900
 3000
 3100
Ans: 1
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Explanation:
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
{45/100x 6000}= 2700.
Ques 7: Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
 2 : 5
 3 : 5
 4 : 5
 6 : 7
Ans: 3
Explanation:
Let the third number be x.
Then, first number = 120% of x = 120x/100=6x/5
Second number = 150% of x =150x/100= 3x/2
Ratio of first two numbers = {6x/5: 3x/2} = 12x: 15x = 4: 5
Ques 8: A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?
 500
 1500
 2000
 None of these
Ans: 3
Explanation:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x – 3x = 1000
x = 1000.
B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
Ques 9: In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:
 20 litres
 30 litres
 40 litres
 60 litres
Ans: 4
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Explanation:
Quantity of milk =(60 x2/3)litres = 40 litres.
Quantity of water in it = (60 40) litres = 20 litres.
New ratio = 1: 2
Let quantity of water to be added further be x litres.
Then, milk: water = (40/20+x)
Now, (40/20 +x) = ½
20 + x = 80
x = 60.
Quantity of water to be added = 60 litres.
Ques 10: The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
 4 years
 8 years
 10 years
 None of these
Ans: 1
Explanation:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years.
Ques 11: A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
 7
 8
 9
 10
Ans: 4
Explanation:
Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, B’s age = 2x = 10 years.
Ques 12: A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
 14 years
 18 years
 20 years
 22 years
Ans: 4
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Explanation:
Let the son’s present age be x years. Then, man’s present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.
Ques 13: The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:
 12 years
 14 years
 18 years
 20 years
Ans: 4
Explanation:
Let the present ages of son and father be x and (60 x) years respectively.
Then, (60 – x) – 6 = 5(x – 6)
54 – x = 5x – 30
6x = 84
x = 14.
Son’s age after 6 years = (x+ 6) = 20 years
Ques 14: There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
 20
 80
 100
 200
Ans: 3
Explanation:
Let the number of students in rooms A and B be x and y respectively.
Then, x – 10 = y + 10 =x y = 20…. (i)
and x + 20 = 2(y – 20) =x – 2y = 60 …. (ii)
Solving (i) and (ii) we get: x = 100, y = 80.
The required answer A = 100.
Ques 15: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
 4
 7
 9
 13
Ans: 1
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Explanation:
Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)
H.C.F. of 48, 92 and 140 = 4.
Ques 16: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
 4
 10
 15
 16
Ans: 4
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
In 30 minutes, they will toll together 30/2 + 1 = 16 times.
Ques 17: The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
 9000
 9400
 9600
 9800
Ans: 3
Explanation:
Greatest number of 4digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 – 399) = 9600.
Ques 18: Three numbers are in the ratio of 3: 4: 5 and their L.C.M. is 2400. Their H.C.F. is:
 40
 80
 120
 200
Ans: 1
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Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Ques 19: The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
 74
 94
 184
 364
Ans: 4
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Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Ques 20: In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
 360
 480
 720
 5040
Ans: 3
Explanation:
The word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
TNPSC Maths Study Material
TNPSC Maths Preparation Book  Author name 
Quantitative Aptitude for Competitive Examinations  R S Aggarwal

S. Chand’s Mental Mathematics  R.S. Aggarwal and Vikas Aggarwal 
Fast Track Objective Arithmetic  Arihant 
Objective Arithmetic  R.S.Aggarwal – 1990 
Contenders who are seeking for Best books of TNPSC Maths may refer books listed above. By going through these books, you will also know about TNPSC Maths Shortcuts. Make a proper study plan & allot time to topics added in TNPSC Syllabus as well in these books. Simplification, ratio & proportion, height & distance, average, data interpretation is among the key topics from which questions are asked in Maths exam.
Final words:
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