# Simplification Questions Quiz 8 for SBI Clerk Pre, IBPS Clerk Pre, IBPS RRB

Direction : Study the following questions carefully and choose the right answer.
Important for :
1
What will come in place of question mark (?) in the following question:

333.44 + 33.4444 + 3.444 + 3333.4 = ?

» Explain it
C
333.44 + 33.4444 + 3.444 + 3333.4 = ?

3 + 33 + 333 + 3333 + .4 + .44 + .444 + .4444 = ?

3 (1234) + 4(.4321) = ?

3702 + 1.7284 = ?

? = 3703.7284

Hence, option C is correct.
2
What will come in place of question mark (?) in the following question:

 1 13 – 2 7 + 5 4 = ? + 7 6 18 9 3
» Explain it
C
 1 13 – 2 7 + 5 4 = ? + 7 6 18 9 3

 1 13 – 2 7 + 5 4 – 7 = ? 6 18 9 3

 1 – 2 + 5 + ( 13 – 7 + 4 – 7 ) = ? 6 18 9 3

 6 – 2 + ( 39 – 7 + 8 – 42 ) = ? 18

 4 + ( – 2 ) = ? 18

 3 + ( 1 – 1 ) = ? 9

 ? = 3 + 8 9

 ? = 3 8 9

Hence, option C is correct.

3
What will come in place of question mark (?) in the following question:

?2 – 18.75 – 11.25 = 9.09% of 396 + 295
» Explain it
D
?2 – 18.75 – 11.25 = 9.09% of 396 + 295

?2 – 30 = 396 ÷ 11 + 295

?2 = 36 + 295 + 30

?2 = 361

? = 19

Hence, option D is correct.
4
What will come in place of question mark (?) in the following question:

(37.5% of 1160) ÷ (441256) = ? + 23

» Explain it
D
 37.5% of 1160 = ? + 23 441 – 256

 ( 1160 × 3 ) 8

= ? + 23
21 – 16

 435 = ? + 23 5

87 – 23 = ?

? = 64

Hence, option D is correct.
5
What will come in the place of question mark (?) in the following question:

(11.275 + 101.237 + 13.5) ÷  64 – (17.657 + 18.976) ÷ 22 = ?2 + 8.379 ÷ 4
» Explain it
C
(11.275 + 101.237 + 13.5) ÷ 64 – (17.657 + 18.976) ÷ 22 = ?2 + 8.379 ÷ 4

(126.012) ÷ 4 – (36.633) ÷ 4 = ?2 + 8.379 ÷ 4

(126.012) ÷ 4 – (36.633) ÷ 4 – (8.379) ÷ 4 = ?2

(126.012 - 45.012) ÷ 4 = ?2

?2 = 81 ÷ 4

? = 9/2 = 4.5

Hence, option C is correct.
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Introduction:

Simplification Questions frequently asked in Bank, LIC and Railways Exams, Here we are going to discuss about simplification rules, formulas, and practice simplification problems with detailed solution.

To solve complex arithmetical expression, we have to solve addition (+), subtraction (–), multiplication (×), division (÷), opening of brackets, etc systematically.

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.
V = Vinculum Bar
B = Brackets

Order of removing brackets

1st Vinculum Brackets (Bar Brackets) ‘–’
2nd Small Brackets (Circular Brackets) ‘( )’
3rd Middle Brackets (Curly Brackets) ‘{ }’
4th Square Brackets (Big Brackets) ‘[ ]’
 O = Of D = Division M = Multiplication A = Addition S = Subtraction

Order of above mentioned operations is as same as the order of letters in the ‘VBODMAS’ from the left to right as

 V → B → O → D → M → A → S Left to Right

Clearly, the order will be as follows:

First Vinculum bracket is solved,

Remember
[ – 5 – 10 = – 15 but – 5 – 10 = – ( – 5 ) = 5],

Second  Brackets are to be solved in order given above,
[First, then second, then third and then fourth]

Third    Operation of ‘Of’ is done,

Fourth  Operation of division is done,

Fifth     Operation of multiplication is done,

Sixth    Operation of addition is done, and

Seventh  Operation of subtraction is done

Absolute value of a Real Number:

If  m is a real number, then its absolute value is defined as

 |m| = { m, if m > 0 –m, if m < 0

|3| = 3 and |– 3|= – (–3)3

Ex. Simplify 4 – [6 – {12 – (10 – 8 – 6)}]

Given expression:

= 4 – [6 – {12 – (10 – 8 + 6)}]            Remove Vinculum

= 4 – [6 – {12 – 8}]                              Remove ( )

= 4 – [6 – 4]                                          Remove { }

= 4 – 2 = 4                                            Remove [ ]

Example 2. Simplify

 1 ÷ 3 of (6 + 8 × 3 – 2) + [ 1 ÷ 7 – { 3 + 8 }] . 7 5 25 7 14

 Soln.  1 ÷ 3 of (6 + 8 × 1 ) + [ 1 ÷ 7 – 14 ] 7 5 25 14

 = 1 ÷ 3 of (6 + 8) + [ 1 × 25 – 1 ] 7 5 7

 = 1 ÷ 3 of 14 + [ 5 – 1 ] 7 7

 = 1 ÷ 6 + [ – 2 ] = 1 – 2 = 7 – 12 = – 5 . 7 6 7 42 42

Example 3.

IF               4x   = 2, then x = ?

1         1
 1 +
x
1– x

Soln. we have                4x   = 2

1 +          1
 1 +
x
1– x

⇒                                      4x   = 2

1 +       1
 1 (1– x)

⇒                                     4x   = 2

 1 +    (1 – x)

⇒                         4x = 2[ 1 + (1 – x)]

⇒                        4x = 2(2 – x)

⇒                        4x = 4 – 2x

⇒                        4x + 2x = 4 ⇒ 6x = 4

 ∴                      x = 4 = 2 6 3

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