GRE Quantitative Puzzles Powers and Logarithms

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TODAY, I CONTINUE TO GIVE TO YOU SOME BASIC CONCEPTS, BEFORE WE PLUNGE OURSELVES TO SOLVING GRE Quantitative Puzzles Powers and Logarithms
A^M = X => M = LOGAX
EXAMPLE 10^2 = 100. THUS, LOG10100 = 2
RULES AND PROPERTIES OF LOGARITHMS:
LOGA(M*N) = LOGAM + LOGAN
LOGA(M/N) = LOGAM – LOGAN
LOGAA = 1
LOGA1 = 0
LOGA(MN) = N* LOGAM
LOGAM = 1/(LOGMA)
LOGAM = LOGBM / LOGBA
PROFIT & LOSS:
PROFIT = SELLING PRICE – COST PRICE (SP – CP)
PROFIT % = PROFIT*100/(COST PRICE)
VENN DIAGRAMS:
N(A) = N(A-B) + N(A INTERSECTION B)
N(B) = N(B-A) + N(A INTERSECTION B)
N(AUB) = N(A-B) + N(A INTERSECTION B) + N(B-A)
N(AUB) = N(A) + N(B) – N(A INTERSECTION B)
SIMPLE & COMPOUND INTEREST:
SIMPLE INTEREST, SI = P*R*T/100, WHERE P = PRINCIPAL, R = INTEREST RATE %AGE PER ANY UNIT OF TIME, T = NUMBER OF TIME UNITS LAPSED
COMPOUND INTEREST : AMOUNT AFTER COMPOUNDING, A = P*(1+R/100)^T
WHEN THE RATES OF INTEREST VARY OVER A PERIOD OF TIME, THE AMOUNT BECOMES, UPON COMPOUNDING, A = P*(1+R1/100)^T1*(1+R2/100)^T2*(1+R3/100)^T3*…… WHERE T1, T2, AND T3 ARE THE TIME UNITS ON WHICH THE INTEREST RATES R1, R2, AND R3 RESPECTIVELY APPLY
VOLUMES, SURFACE AREAS, DIAGONALS, HEIGHTS ETC.:
CUBOID
VOLUME = L*B*H CUBIC UNITS
SURFACE AREA = 2*(L*B+B*H+H*L) SQUARE UNITS
DIAGONAL = (L^2+B^2+H^2)^0.5 UNITS
CUBE
VOLUME = A^3 CUBIC UNITS
SURFACE AREA = 6*A^2 SQUARE UNITS
DIAGONAL = (3^0.5)*A UNITS
CYLINDER
VOLUME = PI*(R^2)*H CUBIC UNITS
CURVED SURFACE AREA = 2*PI*R*H SQUARE UNITS
TOTAL SURFACE AREA = 2*PI*R*(H+R) SQUARE UNITS
CONE
SLANT HEIGHT = (H^2+R^2)^0.5 UNITS
VOLUME = (1/3)*PI*(R^2)*H CUBIC UNITS
CURVED SURFACE AREA = PI*R*L SQUARE UNITS
TOTAL SURFACE AREA = PI*R(R+L) SQUARE UNITS
FRUSTUM OF CONE
VOLUME = (PI*H/3)*(R^2+r^2+R*r) CUBIC UNITS
SLANT HEIGHT = L = [(R-r)^2+H^2]^0.5
CURVED SURFACE AREA = PI*L*(R+r) SQUARE UNITS
TOTAL SURFACE AREA = PI*[R^2+r^2+L*(R+r)] SQUARE UNITS
SPHERE
VOLUME = (4/3)*PI*R^3 CUBIC UNITS
SURFACE AREA = 4*PI*R^2 SQUARE UNITS
HEMISPHERE
VOLUME = (2/3)*PI*R^3 CUBIC UNITS
CURVED SURFACE AREA = 2*PI*R^2 SQUARE UNITS
TOTAL SURFACE AREA = 3*PI*R^2 SQUARE UNITS
PYRAMID
VOLUME = (1/3)*AREA OF BASE*HEIGHT SQUARE UNITS
WHOLE SURFACE AREA = AREA OF BASE + SUM OF THE AREAS OF LATERAL FACES SQUARE UNITS
TRIGONOMETRY:
SINE (THETA) = SIN (THETA) = AB/OB
COSINE (THETA) = COS (THETA) = OA/OB
TANGENT (THETA) = TAN (THETA) = AB/OA
COSECANT (THETA) = COSEC (THETA) = 1 / [ SINE (THETA) ]
SECANT (THETA) = SEC (THETA) = 1 / [ COS (THETA) ]
COT (THETA) = 1 / [ TAN (THETA) ]
SIN (-THETA) = – SIN (THETA)
COS (-THETA) = COS (THETA)
TAN (-THETA) = – TAN (THETA)
COSEC (-THETA) = – COSEC (THETA)
SEC (-THETA) = SEC (THETA)
COT (-THETA) = – COT (THETA)
SIN^2 (THETA) + COS^2 (THETA) = 1
SIN (2*THETA) = 2 *SIN (THETA) * COS (THETA)
COS (2*THETA) = COS^2 (THETA) – SIN^2(THETA)
TAN (THETA + BETA) = [TAN (THETA) + TAN (BETA)] / [1 – TAN (THETA) * TAN (BETA)]
TAN (THETA – BETA) = [TAN (THETA) – TAN (BETA)] / [1 + TAN (THETA) * TAN (BETA)]
1 + TAN^2 (THETA) = SEC^2 (THETA)
1 + COT^2 (THETA) = COSEC^2 (THETA)
TRIGONOMETRIC RATIOS OF CERTAIN ANGLE VALUES:
0 DEGREES
30 DEGREES
45 DEGREES
60 DEGREES
90 DEGREES
SIN (THETA)
0
1/2
1/(2^0.5)
(3^0.5)/2
1
COS (THETA)
1
(3^0.5)/2
1/(2^0.5)
1/2
0
TAN (THETA)
0
1/(3^0.5)
1
(3^0.5)
INDEFINITE
COSEC (THETA)
INFINITE
2
(2^0.5)
2/(3^0.5)
1
SEC (THETA)
1
2/(3^0.5)
(2^0.5)
2
INDEFINITE
COT (THETA)
INFINITE
(3^0.5)
1
1/(3^0.5)
0

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