SSC CGL Discussion - 2015 tips and tricks

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Quick read of current question before your exam -RBI Assistant 2015 all Questions with solutions 





SOME QUICK MATHS FORMULAS
============================


1. Sum of first n natural numbers = n(n+1)/2
2. Sum of the squares of first n natural numbers = n
(n+1)(2n+1)/6
3. Sum of the cubes of first n natural numbers = [n(n
+1)/2]^2
4. Sum of first n natural odd numbers = n^2
5. Average = (Sum of items)/Number of items

------>>>Arithmetic Progression (A.P.):
An A.P. is of the form a, a+d, a+2d, a+3d,
where a is called the ‘first term’ and d is called the
‘common difference’
1. nth term of an A.P. tn = a + (n-1)d
2. Sum of the first n terms of an A.P. Sn = n/2[2a
+(n-1)d] or Sn = n/2(first term + last term)

----->>>Geometrical Progression (G.P.):
A G.P. is of the form a, ar, ar2, ar3, …
where a is called the ‘first term’ and r is called the
‘common ratio’.
1. nth term of a G.P. tn = arn-1
2. Sum of the first n terms in a G.P. Sn = a|1-rn|/|1-
r|

---->>>>Permutations and Combinations :
nPr = n!/(n-r)!
nPn = n!
nP1 = n
nCr = n!/(r! (n-r)!)
nC1 = n
nC0 = 1 = nCn
nCr = nCn-r
nCr = nPr/r!
Number of diagonals in a geometric figure of n sides =
nC2-n

------>>>>Tests of Divisibility :
A number is divisible by 2 if it is an even number.
A number is divisible by 3 if the sum of the digits is
divisible by 3.
A number is divisible by 4 if the number formed by
the last two digits is divisible by 4.
A number is divisible by 5 if the units digit is either 5
or 0.
A number is divisible by 6 if the number is divisible
by both 2 and 3.
A number is divisible by 8 if the number formed by
the last three digits is divisible by 8.
A number is divisible by 9 if the sum of the digits is
divisible by 9.
A number is divisible by 10 if the units digit is 0.
A number is divisible by 11 if the difference of the
sum of its digits at odd places and the sum of its
digits at even places, is divisible by 11.
H.C.F and L.C.M :
H.C.F stands for Highest Common Factor. The other
names for H.C.F are Greatest Common Divisor (G.C.D)
and Greatest Common Measure (G.C.M).
The H.C.F. of two or more numbers is the greatest
number that divides each one of them exactly.
The least number which is exactly divisible by each
one of the given numbers is called their L.C.M.
Two numbers are said to be co-prime if their H.C.F.
is 1.
H.C.F. of fractions = H.C.F. of numerators/L.C.M of
denominators
L.C.M. of fractions = G.C.D. of numerators/H.C.F of
denominators
Product of two numbers = Product of their H.C.F. and
L.C.M.
------>>>>PERCENTAGES :
If A is R% more than B, then B is less than A by R /
(100+R) * 100
If A is R% less than B, then B is more than A by R /
(100-R) * 100
If the price of a commodity increases by R%, then
reduction in consumption, not to increase the
expenditure is : R/(100+R)*100
If the price of a commodity decreases by R%, then the
increase in consumption, not to decrease the
expenditure is : R/(100-R)*100

---->>>PROFIT & LOSS :
Gain = Selling Price(S.P.) – Cost Price(C.P)
Loss = C.P. – S.P.
Gain % = Gain * 100 / C.P.
Loss % = Loss * 100 / C.P.
S.P. = (100+Gain%)/100*C.P.
S.P. = (100-Loss%)/100*C.P.
If CP(x), Gain(y), Gain%(z). Then y = x*z/100. [Same
in case of Loss]

------->>>>RATIO & PROPORTIONS:
The ratio a : b represents a fraction a/b. a is called
antecedent and b is called consequent.
The equality of two different ratios is called
proportion.
If a : b = c : d then a, b, c, d are in proportion. This is
represented by a : b :: c : d.
In a : b = c : d, then we have a* d = b * c.
If a/b = c/d then ( a + b ) / ( a – b ) = ( c + d ) / ( c
– d ).

------->>>TIME & WORK :
If A can do a piece of work in n days, then A’s 1 day’s
work = 1/n
If A and B work together for n days, then (A+B)’s 1
days’s work = 1/n
If A is twice as good workman as B, then ratio of work
done by A and B = 2:1
------>>>PIPES & CISTERNS :
If a pipe can fill a tank in x hours, then part of tank
filled in one hour = 1/x
If a pipe can empty a full tank in y hours, then part
emptied in one hour = 1/y
If a pipe can fill a tank in x hours, and another pipe
can empty the full tank in y hours, then on opening
both the pipes,
the net part filled in 1 hour = (1/x-1/y) if y>x
the net part emptied in 1 hour = (1/y-1/x) if x>y
------->>>TIME & DISTANCE :
Distance = Speed * Time
1 km/hr = 5/18 m/sec
1 m/sec = 18/5 km/hr
Suppose a man covers a certain distance at x kmph
and an equal distance at y kmph. Then, the average
speed during the whole journey is 2xy/(x+y) kmph.
--------->>>PROBLEMS ON TRAINS :
Time taken by a train x metres long in passing a
signal post or a pole or a standing man is equal to
the time taken by the train to cover x metres.
Time taken by a train x metres long in passing a
stationary object of length y metres is equal to the
time taken by the train to cover x+y metres.
Suppose two trains are moving in the same direction
at u kmph and v kmph such that u>v, then their
relative speed = u-v kmph.
If two trains of length x km and y km are moving in
the same direction at u kmph and v kmph, where u>v,
then time taken by the faster train to cross the slower
train = (x+y)/(u-v) hours.
Suppose two trains are moving in opposite directions
at u kmph and v kmph. Then, their relative speed = (u
+v) kmph.
If two trains of length x km and y km are moving in
the opposite directions at u kmph and v kmph, then
time taken by the trains to cross each other = (x+y)/
(u+v)hours.
If two trains start at the same time from two points A
and B towards each other and after crossing they take
a and b hours in reaching B and A respectively, then
A’s speed : B’s speed = (√b : √a)
-------->>>>SIMPLE & COMPOUND INTERESTS :
Let P be the principal, R be the interest rate percent
per annum, and N be the time period.
Simple Interest = (P*N*R)/100
Compound Interest = P(1 + R/100)^N – P
Amount = Principal + Interest
when rate of interest time n principal are constant den
principal=(C.I.-S.I.)*(100/R)^N
------>>>LOGARITHM:
If a^m = x , then m = loga(x).
Properties :
logx(x) = 1
logx(1) = 0
loga(x*y) = loga(x) + loga(y)
loga(x/y) = log ax – log ay
loga(x) = 1/logx(a)
loga(x^p) = p(loga(x))
loga(x) = logb(x)/logb(a)
Note : Logarithms for base 1 does not exist.
----->>>AREA & PERIMETER :
Shape Area Perimeter
Circle ∏ (Radius)2 2∏(Radius)
Square (side)2 4(side)
Rectangle length*breadth 2(length+breadth)
Area of a triangle = 1/2*Base*Height or
Area of a triangle = √ (s(s-(s-b)(s-c)) where a,b,c are
the lengths of the sides and s = (a+b+c)/2
Area of a parallelogram = Base * Height
Area of a rhombus = 1/2(Product of diagonals)
Area of a trapezium = 1/2(Sum of parallel sides)
(distance between the parallel sides)
Area of a quadrilateral = 1/2(diagonal)(Sum of sides)
Area of a regular hexagon = 6(√3/4)(side)2
Area of a ring = ∏(R2-r2) where R and r are the outer
and inner radii of the ring.
Area of a circle=πr^2 or πd^2/4
Area of semi-circle=πr^2/2
Area of a quadrant of a circle=πr^2/4
Area enclosed by two concentric circles=π(R^2-r^2)
Area of a sector=Ɵ/180 degree *πr
No of revolutions completed by a rotating wheel in 1
minute=distance moved in 1 minute/circumference
------>>>VOLUME & SURFACE AREA :
Cube :
Let a be the length of each edge. Then,
Volume of the cube = a3 cubic units
Surface Area = 6a2 square units
Diagonal = √ 3 a units
Cuboid :
Let l be the length, b be the breadth and h be the
height of a cuboid. Then
Volume = lbh cu units
Surface Area = 2(lb+bh+lh) sq units
Diagonal = √ (l2+b2+h2)
Cylinder :
Let radius of the base be r and height of the cylinder
be h. Then,
Volume = ∏r2h cu units
Curved Surface Area = 2∏rh sq units
Total Surface Area = 2∏rh + 2∏r2 sq units
Cone :
Let r be the radius of base, h be the height, and l be
the slant height of the cone. Then,
l2 = h2 + r2
Volume = 1/3(∏r2h) cu units
Curved Surface Area = ∏rl sq units
Total Surface Area = ∏rl + ∏r2 sq units
Sphere :
Let r be the radius of the sphere. Then,
Volume = (4/3)∏r3 cu units
Surface Area = 4∏r2 sq units
Hemi-sphere :
Let r be the radius of the hemi-sphere. Then,
Volume = (2/3)∏r3 cu units
Curved Surface Area = 2∏r2 sq units
Total Surface Area = 3∏r2 sq units
Prism :
Volume = (Area of base)(Height)
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