Model 1 Simplification using VBODMAS Practice Questions Answers Test with Solutions & More Shortcuts
simplification PRACTICE TEST [4 - EXERCISES]
Model 1 Simplification using VBODMAS
Model 2 Simplification based on square & square root
Model 3 Simplification based on cube & cube root
Model 4 Simplification with continued fraction
Question : 21
5 - [4 - {3 - (3 - 3 - 6)}] is equal to :
a) 0
b) 10
c) 4
d) 6
Answer »Answer: (b)
Using Rule 1,
An expression must be simplified by following defined order/sequence known as VBODMAS, which is given by:
1st step, | V | - | Vineculum (line brackets)/Bar |
B | - | Brackets | |
O | - | Of | |
D | - | Division | |
M | - | Multiplication | |
A | - | Addition | |
Last step, | S | - | Subtraction |
There are four types of brackets given below.
- – → Line/Bar
- ( ) → Simple or Small Bracket/open brackets
- { } → Curly Brackets/Braces
- [ ] → Square Brackets/Closed brackets
These brackets must be solved in given order only.
? = 5 - [4 - {3 - (3– 3 - 6)}]
= 5 - [4 - { 3 - (– 6)}]
= 5 - [4 - {3 + 6}]
= 5 - [4 - 9]
= 5 + 5 = 10
Question : 22 [SSC CGL Prelim 2005]
The value of $2/3 × 3/{5/6 ÷ {2/3} of 1{1/4}}$ is :
a) $2/3$
b) 2
c) $1/2$
d) 1
Answer »Answer: (b)
Using Rule 1,
$2/3 × 3/{5/6 ÷ {2/3} of 1{1/4}}$
$2/3 × 3/{5/6 ÷ {2/3} of {5/4}}$
$2/3 × 3/{{5/6} ÷ {10/12}}$
= $2/3 × 3/{5/6 × 12/10} = 2/3 × 3/1 = 2$
Question : 23
If ${1120}/{√{P}}$ = 80, then P is equal to
a) 225
b) 14
c) 196
d) 140
Answer »Answer: (c)
80 × $√{P}$ = 1120
$√{P} = 1120/80 = 14$
P = $(14)^2 = 196$
Question : 24 [SSC CGL Prelim 2003]
Simplify : ${2{3/4}}/{1{5/6}} ÷ {7/8} × (1/3 + 1/4) + 5/7 ÷ {3/4} of {3/7}$
a) 3$2/9$
b) $56/77$
c) $2/3$
d) $49/80$
Answer »Answer: (a)
Using Rule 1,
The given expression
${2{3/4}}/{1{5/6}} ÷ {7/8} × (1/3 + 1/4) + 5/7 ÷ {3/4} of {3/7}$
= ${11/4}/{11/6} ÷ {7/8} × ({4+3}/12) + 5/7 ÷ {3/4} of {3/7}$
= $(11/4 × 6/11) ÷ {7/8} × 7/12 + 5/7 ÷ (3/4 × 3/7)$
= ${3/2} ÷ {7/8} × 7/12 + {5/7} ÷ {9/28}$
= ${3/2} × {8/7} × 7/12 + {5/7} ×{28/9}$
= $1 + 20/9 = {9+20}/9 = 29/9 = 3{2}/9$
Question : 25 [SSC CGL Prelim 1999]
1– [5 - {2 + (– 5 + 6 - 2) 2}] is equal to :
a) –2
b) –4
c) 0
d) 2
Answer »Answer: (b)
Using Rule 1,
? = 1 - [5 - {2 + (–1)2}]
= 1 - [5 - {2 - 2}]
= 1 - [5 - 0]
= 1 - 5 = - 4
IMPORTANT quantitative aptitude EXERCISES
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