Model 1 Simplification using VBODMAS Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 4 EXERCISES
The following question based on simplification topic of quantitative aptitude
(a) $6/55$
(b) $√{2}2/27$
(c) $5/27$
(d) $1/9$
The correct answers to the above question in:
Answer: (a)
Using Rule 2,
$1/{5×6} + 1/{6×7} + 1/{7×8} + 1/{8×9} + 1/{9×10} + 1/{10×11}$
= $1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 +1/9 - 1/10 +1/10 - 1/11 +1/11$
= $1/5 - 1/11 ={11 - 5}/55 = 6/55$
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Read more problems based on vbodmas Based Quantitative Aptitude Questions and Answers
Question : 1
$9 - 1{2/9} of 3{3/11} ÷ 5{1/7} of {7/9}$ is equal to :
a) $3/4$
b) 8
c) 8$32/81$
d) 9
Answer »Answer: (b)
Using Rule 1,
Expression
= $9 - 11/9 of {36/11} ÷ {36/7} of 7/9$
= $9 - 11/9 × {36/11} ÷ {36/7} × 7/9$
= 9 - 4 ÷ 4
= $9 - 4 × 1/4$ = 9 - 1 = 8
Question : 2
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 : 3
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 : 4
If I = $3/4 ÷ {5/6}$, II = 3 ÷ [(4 ÷ 5) ÷ 6], III = [3 ÷ (4 ÷ 5)] ÷ 6, IV = 3 ÷ 4 (5 ÷ 6) then
a) All are equal
b) I and II are equal
c) I and III are equal
d) I and IV are equal
Answer »Answer: (d)
Using Rule 1,
I. = $3/4 × 6/5 = 9/10$
II. = 3 ÷ $[4/5 × 1/6] = 3 ÷ {2/15} = 45/2$
III. = $[3 ÷ {4/5}] ÷ 6 = {15/4} ÷ 6 = 5/8$
IV. =$3 ÷ {4} × 5/6 = 3 ÷ {10/3} = 9/10$
Obviously, (I) and (IV) are equal
Question : 5
The value of ${0.1 × 0.1 × 0.1 + 0.2 × 0.2 × 0.2 + 0.3 × 0.3 × 0.3 - 3 × 0.1 × 0.2 × 0.3}/{0.1 × 0.1 + 0.2 × 0.2 + 0.3 × 0.3 - 0.1 × 0.2 - 0.2 × 0.3 - 0.3 × 0.1}$ is
a) 0.2
b) 0.006
c) 0
d) 0.6
Answer »Answer: (d)
Using (x) of Basic Formulae
Let 0.1 = a, 0.2 = b and 0.3 = c
Then, we have,
${a×a×a+b×b×b+c×c×c-3abc}/{a×a+b×b+c×c - ab - bc - ac}$
= ${a^3+b^3+c^3 - 3abc}/{a^2+b^2+c^2 - ab -bc - ac}$
= a + b + c
= 0.1 + 0.2 + 0.3 = 0.6
Question : 6
Simplify : $8{1/2} - [3{1/4} ÷(1{1/4} - 1/2(1{1/2} - 1/ 3 - {1/6}))]$
a) $2/9$
b) 4$1/2$
c) 9$1/2$
d) 4$1/6$
Answer »Answer: (d)
Using Rule 1,
$8{1/2} - [3{1/4} ÷(1{1/4} - 1/2(1{1/2} - 1/ 3 - {1/6}))]$
$17/2 - [{13/4} ÷({5/4} - 1/2({3/2} - 1/ 3 - {1/6}))]$
$17/2 - [{13/4} ÷({5/4} - 1/2({9 - 2 - 1}/6))]$
$17/2 - [{13/4} ÷({5/4} - 1/2 × {6/6})]$
$17/2 - [{13/4} ÷({5/4} - 1/2)]$
$17/2 - [{13/4} ÷({5 - 2}/4)]$
$17/2 - [{13/4} ÷ {3/4}]$
$17/2 - [{13/4} × 4/3] = 17/2 - 13/3$
= ${51 - 26}/6 = 25/6 = 4{1/6}$
GET simplification PRACTICE TEST 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
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