# Model 1 Simplification using VBODMAS Practice Questions Answers Test with Solutions & More Shortcuts

#### simplification PRACTICE TEST [4 - EXERCISES]

Question : 1

When $(1/2 - 1/4 + 1/5 - 1/6)$ is divided by $(2/5 - 5/9 + 3/5 - 7/18)$,the result is :

a) 3$3/10$

b) 5$1/10$

c) 3$1/6$

d) 2$1/18$

Using Rule 1,

? = $(1/2 - 1/4 + 1/5 - 1/6)÷(2/5 - 5/9 + 3/5 - 7/18)$

= $({30 - 15 + 12 -10}/60) ÷ ({36 - 50 +54 - 35}/90)$

= $(17/60) ÷ (5/90) = 17/60 × {18}$

= $51/10 = 5{1/10}$

Question : 2 [SSC CGL Prelim 2002]

For what value of *, statement $[(\text"*")/21 × (\text"*")/189]$ = 1 is correct ?

a) 21

b) 3969

c) 63

d) 147

Let '*' be H

$[(H)/21 × (H)/189]$ = 1

$(H)^2$ = 21 × 189

H = $√{21 × 189}$ = 63

Question : 3 [SSC CGL Prelim 2002]

The value of 25 - 5 [2 + 3 (2 - 2 (5 - 3) + 5) - 10] ÷ 4 is :

a) 25

b) 5

c) 23.75

d) 23.25

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 - $5/4$

= ${100 - 5}/4 = 95/4 = 23.75$

Question : 4 [SSC CGL Prelim 2000]

When simplified, the expression $(100)^{1/2} × (0.001)^{1/3} - (0.0016)^{1/4} × 3^{0} + (5/4)^{- 1}$ is equal to:

a) 0

b) 1.6

c) 1.0

d) 0.8

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

Question : 5 [SSC CGL Prelim 2002]

${0.8\ov{3} ÷ 7.5}/{2.3\ov{21} - 0.0\ov{98}}$ is equal to

a) 0.05

b) 0.6

c) 0.06

d) 0.1

${0.8\ov{3} ÷ 7.5}/{2.3\ov{21} - 0.0\ov{98}} = {{{83 - 8}/90} ÷ {7.5}}/{2{{321 - 3}/990} - {98/990}}$

= ${{75/90} + {7.5}}/{{{2}318/990} - 98/990} = {{75/90} + {7.5}}/{{2}220/990}$

= ${7.5}/{90 × {7.5}} × 990/2200 = 1/20 = 0.05$

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