UPSC CSAT Formula PDF

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UPSC CSAT Formula

UPSC CSAT Formula PDF read online or download for free from the official website link given at the bottom of this article.

UPSC CSAT (Civil Services Aptitude Test) was introduced in the year 2011 as a part of the UPSC Civil Services Exam (Preliminary) to test the analytical skills, reasoning ability and aptitude of IAS aspirants. It is a part of the UPSC Prelims (Civil Services Exam – Preliminary). However, the Union Public Service Commission (UPSC) refers to the exam as General Studies (GS) Paper – II.

The total marks for the CSAT exam is 200, and there are 80 questions. Each question carries 2.5 marks. Candidates need to score at least 33% or at least 66 marks to qualify for this paper.

CSAT Exam Pattern:

CSAT or Civil Services Aptitude Test was introduced in 2011 to test candidates’ analytical skills. It is the second paper in the UPSC Prelims. Officially, it is known as General Studies Paper II.
The CSAT exam pattern for UPSC:

  • Number of questions: 80 Objective-Type (MCQ) questions
  • Negative Marking: Yes (1/3rd of the maximum marks for the question)
  • Time: 2 hours
  • Type of Exam: Offline exam
  • Date of CSAT exam: 10th October 2021
  • Language of CSAT exam paper: English/Hindi
  • Maximum Marks: 200
  • CSAT qualifying marks: 66 marks (33% qualifying criteria)

UPSC CSAT Formula

Algebraic Identities

  •  n(n + l)(2n + 1) is always divisible by 6.
  •  32n leaves remainder = 1 when divided by 8
  • n3 + (n + 1 )3 + (n + 2 )3 is always divisible by 9
  • 102n + 1 + 1 is always divisible by 11
  • n(n2- 1) is always divisible by 6
  • n2+ n is always even
  • 23n-1 is always divisible by 7
  • 152n-1 +l is always divisible by 16
  • n3 + 2n is always divisible by 3
  • 34n – 4 3n is always divisible by 17
  • n! + 1 is not divisible by any number between 2 and n(where n! = n (n – l)(n – 2)(n – 3)…….3.2.1)
  • for eg 5! = 5.4.3.2.1 = 120 and similarly 10! = 10.9.8…….2.1= 3628800
  • The product of n consecutive numbers is always divisible by n!.
  • If n is a positive integer and p is a prime, then np – n is divisible by p.
  • |x| = x if x ≥ 0 and |x| = – x if x ≤ 0.
  • Minimum value of a2.sec2Ɵ + b2.cosec2Ɵ is (a + b)2; (0° < Ɵ < 90°)for eg. minimum value of 49 sec2Ɵ + 64.cosec2Ɵ is (7 +2 = 225.
  • Among all shapes with the same perimeter a circle has the largest area.
  • If one diagonal of a quadrilateral bisects the other, then it also bisects the quadrilateral.
  • sum of all the angles of a convex quadrilateral = (n – 2)180°
  • Number of diagonals in a convex quadrilateral = 0.5n(n – 3)
  • let P, and Q are the midpoints of the nonparallel sides BC and AD of a trapezium ABCD. Then,
    • ΔAPD = ΔCQB.
  • a2 – b2 = (a + b)(a – b)
  • a2 + b2 + c2 + 2(ab + bc + ca) = (a + b + c)2
  • (a ± b)2 = a2 + b2± 2ab
  • (a + b + c + d)2 = a2 + b2 + c2 + d2 + 2(ab + ac + ad + bc + bd + cd)
  • (a ± b)3 = a3 ± b3 ± 3ab(a ± b)
  • (a ± b)(a2 + b2 m ab) = a3 ± b3
  • (a + b + c)(a2 + b2 + c2 -ab – bc – ca) = a3 + b3 + c3 – 3abc =
  • 1/2 (a + b + c)[(a – b)2 + (b – c)2 + (c – a)2]
    •  when a + b + c = 0, a3 + b3 + c3 = 3abc
  • (x + a)(x + b) (x + c) = x3 + (a + b + c) x2 + (ab + bc + ac)x + abc
  • (x – a)(x – b) (x – c) = x3 – (a + b + c) x2 + (ab + bc + ac)x – abc
  • a4 + a2b2 + b4 = (a2 + ab + b2)( a2 – ab + b2)
  • a4 + b4 = (a2 – √2ab + b2)( a2 + √2ab + b2)
  • an + bn = (a + b) (a n-1 – a n-2 b +  a n-3 b2 – a n-4 b3 +…….. + b n-1)(valid only if n is odd)
  • an – bn = (a – b) (a n-1 + a n-2 b +  a n-3 b2 + a n-4 b3 +……… + b n-1){were n ϵ N)
  • (a ± b)2n is always positive while -(a ± b)2n is always negative, for any real values of a and b
  • (a – b)2n = (b – a)2” and (a – b)2n+1 = – (b – a)2n+1
  • if α and β are the roots of equation ax2 + bx + c = 0, roots of cx” + bx + a = 0 are 1/α and 1/β. if α and β are the roots of equation ax2 + bx + c = 0, roots of ax2 – bx + c = 0 are -α and -β.

Divisibility Rule

  • An even number or a number with an even last digit, such as 0, 2, 4, 6, and 8.
  • The number should be divisible by 3 when the sum of all its digits is considered.
  • The result of the number’s last two digits must be either 00 or a multiple of 4.
  • Numbers with the digits 0 or 5 in the ones place.
  • A number that can be divided by both 2 and 3
  • Multiple of 7 is obtained by deducting twice the number’s last digit from the other   digits.
  • The sum of a number’s last three digits must be divisible by 8 or equal to 000.
  • The number should be divisible by 9 when the sum of all its digits is considered.
  • Any number with the digit 0 in the ones position.
  •  A number is divisible by 11 when the sums of its alternative digits differ from each other.
  •  A number that can be divided by 3 and 4.

Profit Loss: Formulas

  • Cost Price = Selling Price ( No profit No loss)
  • Gain Percentage = (Gain × 100)/(C.P.)
  • Loss Percentage = (Loss × 100)/(C.P.)
  • Profit, P = SP – CP; SP>CP
  • Loss, L = CP – SP; CP>SP
  • P% = (P/CP) x 100
  • L% = (L/CP) x 100
  • SP = {(100 + P%)/100} x CP
  • SP = {(100 – L%)/100} x CP
  • CP = {100/(100 + P%)} x SP
  • CP = {100/(100 – L%)} x SP
  • Discount = MP – SP
  • SP = MP -Discount
  • For false weight, profit percentage will be P% = [(True weight – false weight)/ false weight] x 100.
  • When there are two successful profits, say m% and n%, then the net percentage profit equals to [m+n+(mn/100)]
  • When the profit is m%, and loss is n%, then the net % profit or loss will be: [m-n-(mn/100)]
  • If a product is sold at m% profit and then again sold at n% profit then the actual cost price of the product will be: CP = [100 x 100 x P/(100+m)(100+n)]. In case of loss, CP = [100 x 100 x L/(100-m)(100-n)]
  • If P% and L% are equal then, P = L and %loss = P2/100

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UPSC CSAT Formula

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