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
Model 1 Simplification using VBODMAS Shortcuts »
Click to Read...Model 1 Simplification using VBODMAS Online Quiz
Click to Start..simplification Shortcuts and Techniques with Examples
quantitative aptitude CATEGORIES
RECENT BLOGS
![SSC CHSL 2023 Tier 2: Syllabus PDF & Exam Pattern [Changed]](https://careericons.com/adminicon/bunch/images/ssc-chsl-tier-2-2023-new-revised-syllabus-pdf-s.jpg "SSC CHSL 2023 Tier 2: Syllabus PDF & Exam Pattern [Changed]")
Recently Added Subject & Categories For All Competitive Exams
New 100+ Compound Interest MCQ with Answers PDF for IBPS
Compound Interest verbal ability questions and answers solutions with PDF for IBPS RRB PO. Aptitude Objective MCQ Practice Exercises all competitive exams
Continue Reading »
100+ Mixture and Alligation MCQ Questions PDF for IBPS
Most importantly Mixture and Alligation multiple choice questions and answers with PDF for IBPS RRB PO. Aptitude MCQ Practice Exercises all Bank Exams
Continue Reading »
IBPS Profit and Loss Questions Solved Problems with PDF
Most important Profit and Loss multiple choice questions and answers with PDF for IBPS RRB PO. 100+ Aptitude MCQ Practice Exercises all competitive exams
Continue Reading »
100+ Average Aptitude Questions Answers solutions MCQ PDF
New Average multiple choice questions and answers with PDF for IBPS RRB PO. 100+ Quantitative Aptitude MCQ Practice Exercises all competitive exams
Continue Reading »