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 : 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$1/18$
Answer »Answer: (b)
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
Answer »Answer: (c)
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
Answer »Answer: (c)
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
Answer »Answer: (b)
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
Answer »Answer: (a)
${0.8\ov{3} ÷ 7.5}/{2.3\ov{21} - 0.0\ov{98}} = {{{83 - 8}/90} ÷ {7.5}}/{2{{321 - 3}/990} - {98/990}}$
= ${{75/90} + {7.5}}/{{{2}318/990} - 98/990} = {{75/90} + {7.5}}/{{2}220/990}$
= ${7.5}/{90 × {7.5}} × 990/2200 = 1/20 = 0.05$
IMPORTANT quantitative aptitude EXERCISES
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