Practice Problems based on vbodmas - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The value of $5/{1{7/8} of 1{1/3}} × {2{1/10}}/{3{1/2}} of 1{1/4}$
(a)
(b)
(c)
(d)
Using Rule 1,
$5/{15/8 × 4/3} × {21/10}/{7/2} of {5/4}$
=$5 × 2/5 × 21/10 × 2/7 × 5/4$
=$3/2 = 1{1}/2$
Q-2) When simplified, the expression $(100)^{1/2} × (0.001)^{1/3} - (0.0016)^{1/4} × 3^{0} + (5/4)^{- 1}$ is equal to:
(a)
(b)
(c)
(d)
Using Rule 1,
$(100)^{1/2} × (0.001)^{1/3} - (0.0016)^{1/4} × 3^{0} + (5/4)^{- 1}$
=10 × 0.1 - 0.2 × 1 + $4/5$
= 1 - 0.2 + 0.8 = 1.6
Q-3) Find the value of * in the following $1{2/3} ÷ {2/7} × \text"*"/7 = 1{1/4} × 2/3 ÷ {1/6}$
(a)
(b)
(c)
(d)
Using Rule 1,
We have
${5/3} ÷ {2/7} × \text"*"/7 = {5/4} × 2/3 × 6$
${5/3} ÷ {2/7} × \text"*"/7 = {5 × 2 × 6}/{4 × 3}$
* = ${5 × 2 × 6 × 3 × 2 × 7}/{5 × 7 × 4 × 3} = 6$
Q-4) Simplify $[3{1/4} ÷ (1{1/4} - 1/2(2{1/2} - \ov{1/4 - 1/6}))] ÷ (1/2 of 4{1/3})$
(a)
(b)
(c)
(d)
$[3{1/4} ÷ (1{1/4} - 1/2(2{1/2} - \ov{1/4 - 1/6}))] ÷ (1/2 of 4{1/3})$
Using Rule 1,
= $[{13/4} ÷ ({5/4} - 1/2({5/2} - {3 - 2}/12))] ÷ {13/6}$
= $[{13/4} ÷ ({5/4} - 1/2({5/2} - 1/12))] ÷ {13/6}$
= $[{13/4} ÷ ({5/4} - 1/2({30 - 1}/12))] ÷ {13/6}$
= $[{13/4} ÷ ({5/4} - 1/2 × 29/12)] ÷ {13/6}$
= $[{13/4} ÷ ({30 - 29}/24)] ÷ {13/6}$
= $[{13/4} ÷ {1/24}] ÷ {13/6}$
$[{13/4} × 24] ÷ {13/6}$
= $13 × 6 × 6/13 = 36$
Q-5) 1– [5 - {2 + (– 5 + 6 - 2) 2}] is equal to :
(a)
(b)
(c)
(d)
Using Rule 1,
? = 1 - [5 - {2 + (–1)2}]
= 1 - [5 - {2 - 2}]
= 1 - [5 - 0]
= 1 - 5 = - 4
Q-6) For what value of *, statement $[(\text"*")/21 × (\text"*")/189]$ = 1 is correct ?
(a)
(b)
(c)
(d)
Let '*' be H
$[(H)/21 × (H)/189]$ = 1
$(H)^2$ = 21 × 189
H = $√{21 × 189}$ = 63
Q-7) The value of 25 - 5 [2 + 3 (2 - 2 (5 - 3) + 5) - 10] ÷ 4 is :
(a)
(b)
(c)
(d)
Using Rule 1,
Expression
= 25 - 5 [2 + 3 {2 - 2(5 - 3) + 5} - 10] ÷ 4
= 25 - 5 [2 + 3 {2 - 2 × 2 + 5} - 10] ÷ 4
= 25 - 5 [2 + 9 - 10] ÷ 4
= 25 - 5 ÷ 4 = 25 - $5/4$
= ${100 - 5}/4 = 95/4 = 23.75$
Q-8) If $50/\text"*" = \text"*"/{12{1/2}}$, then the value of * is :
(a)
(b)
(c)
(d)
Let the value of * be x.
$50/x = x/{12{1/2}}$
$50/x = {2x}/25$
$2x^2$ = 50 × 25
$x^2$ = 25 × 25 ⇒ x = 25
Q-9) 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)
(b)
(c)
(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
Q-10) 5 - [4 - {3 - (3 - 3 - 6)}] is equal to :
(a)
(b)
(c)
(d)
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