# Railway Exam Aptitude โ Simplification

Railway Exam Aptitude โ Simplification Set-12:
Dear Readers, Here we have given from Aptitude Questions for Railway Exam 2016. Candidates those who are all preparing for these exams can use this material.

Introduction:
For simplifying an expression containing various types of operation, the order of operating various operations, should be strictly followed as accordance with.

VBODMAS Rule
This rule gives the correct order in which various operations regarding simplification are to be performed, so as to find out the values of given expressions in simple ways. Let us see what these letters mean. Clearly, the order will be as follows

FirstVinculum bracket is solved, [Remember โ 5 โ 10 = – 15 but โ 5 โ 10 = -(-5) = 5]

SecondBrackets are to be solved in order given above, [first, then second, then third, and then fourth]

ThirdOperation of โOf โ is done.

FourthOperation of โDivisionโ is done,

FifthOperation of โMultiplicationโ is done,

SixthOperation of โAdditionโ is done and

SeventhOperation of โSubtractionโ is done.

Important Formulae
ยท       (a + b)2 = a2+ 2ab + b2
ยท       (a โ b) 2 = a2– 2ab + b2
ยท       (a + b) (a โ b) = a2–  b2
ยท       (x + a) (x + b) = x2 + (a + b) x + ab
ยท       (a + b + c)2 = a2+ b2+ c2 + 2ab + 2bc + 2ca
ยท       (a + b )3 = a3+ b3+ 3ab (a + b)
ยท       (a – b )3 = a3– b3– 3ab (a + b)
ยท       a3 โ b3 = (a  – b)3 + 3ab (a + b)
ยท       a3 + b3 = (a + b) (a2 – ab + b2 )
ยท       a3 – b3 = (a – b) (a2 + ab + b2 )

1).Simplify the expression [ 11 / 2(1 / 5) ] รท (11 / 5) of 2(1 / 2) – 2
a)  3 (7 / 22)
b)  3 (15 / 22)
c)  3 (13 / 22)
d)  3 (21 / 22)

2).(14 ร 14  – 46) / (11 ร 6  – 42)  = ?
a)  10
b)  15
c)  3
d)  8

3).11(1 / 3) ร 4(8 / 10) รท ? = 22(2 / 3)
a)  3.6
b)  2.4
c)  1.2
d)  4.8

1). b) 2). c) 3). c)

Solution:

1.Applying VBODMAS rule,
[11 ร (1 / 2(1 / 5) ) รท (11 / 5) of 2(1 / 2) โ 2 = 5 / (11 / 5) ร 2(1 / 2) โ 2
= 5 / (11 / 5) ร (5 / 2) – 2
= 5 ร (5 / 11) ร (5 / 2) – 2
= (125 / 22) โ 2 = 3(15 / 22)

2.Putting x for (?) and applying VBODMAS rule, we get
x = [(196 โ 46) / (66 โ 16)]
x = (150 / 50)
x = 3

3.Putting x for (?) and solving it gives
11(1 / 3) ร 4(8 / 10) รท x = 22(2 / 3)
11(1 / 3) ร 4(8 / 10) = 22(2 / 3) ร x
x = (1 / 2) ร 4(8 / 10)
x = 2.4

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