Simplification online test - Aptitude

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Simplification Questions

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Simplification formula

1. BODMAS rule

The BODMAS rule refers to the rule that is followed to solve mathematical expressions. The BODMAS stands for B - Brackets, O - Order of powers, D - Division, M - Multiplication, A - Addition, and S - Subtraction

2 .Modulus of a Real Number

Thus, |5| = 5 and |-5| = -(-5) = 5

3. Virnaculum (or Bar)

When an expression contains Virnaculum, before applying the 'BODMAS' rule, we simplify the expression under the Virnaculum.

Example - $\ov\text x$

Solved Simplification Questions for upcoming Exams

Simplification using BODMAS

Some Questions With Solutions:-

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

Ans - Let '*' be H

$[(H)/21 × (H)/189]$ = 1

$(H)^2$ = 21 × 189

H = $√{21 × 189}$ = 63

Q-2: ${0.8\ov{3} ÷ 7.5}/{2.3\ov{21} - 0.0\ov{98}}$ is equal to

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

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

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

Q-3: Find the value of * in the following $1{2/3} ÷ {2/7} × \text"*"/7 = 1{1/4} × 2/3 ÷ {1/6}$

Ans - 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: Simplification of $-|-48|$

Ans - Since ∣−x∣=x

−∣−48∣

=−(48)=−48

Simplification based on square & square root

Q-1:The value of ${√{80} - √{112}}/{√{45} - √{63}}$ is

Ans - ${√{80} - √{112}}/{√{45} - √{63}}$

= ${√{16 × 5} - √{16 × 7}}/{√{9 × 5} - √{9 × 7}}$

= ${4√{5} - 4√{7}}/{3√{5} - 3√{7}} = {4(√{5} - √{7})}/{3(√{5} - √{7})$

= $4/3 = 1{1}/3$

Q-2:The value of ${(75.8)^2 - (55.8)^2}/20$ is

Ans - ${(75.8)^2 - (55.8)^2}/20$

= ${(75.8 - 55.8)(75.8 + 55.8)}/20$

= ${20 × 131.6}/20$ = 131.6

Q-3: The value of $√{{(6.1)^2 + (61.1)^2 + (611.1)^2}/{(0.61)^2 + (6.11)^2 + (61.11)^2}}$ is

Ans - $√{{(6.1)^2 + (61.1)^2 + (611.1)^2}/{(0.61)^2 + (6.11)^2 + (61.11)^2}}$

= $√{{(10 × 0.61)^2 + (10 × 6.11)^2 + (10 × 61.11)^2}/{(0.61)^2 + (6.11)^2 + (61.11)^2}}$

=$√{100}$ = 10

Simplification based on cube & cube root

Q-1: $√^3{8}/√{16} ÷ √{100/49} × √^3{125}$ is equal to :

Solution -Expression = $2/4 × 7/10 × 5$

= $7/4 = 1{3}/4$

Q-2: $√^3{{72.9}/{0.4096}}$ is equal to :

Solution-

$√^3{{72.9}/{0.4096}} = √^3{{729000}/{4096}}$

= $√^3{(90)^3/(16)^3} = 90/16 = 45/8 = 5.625$

Simplification with continued fraction

Q-1: $2/{2 + 2/{3 + 2 /{3 + {2/3}}}× {0.39}}$is simplified to

Solution-

$2/{2 + 2/{3 + 2 /{3 + {2/3}}}× {0.39}}$

= $2/{2 + 2/{3 + 2/{11/3}}× {0.39}}$

= $2/{2 + 2/{3 + {6/11}}× {0.39}}$

= $2/{2 + 2/{{33+6}/11}× {0.39}}$

= $2/{2 + {{11×2}/39} × {0.39}}$

= $2/{2 + {{11×2}/39} × {39/100}}$

= $2/{2+{11/50}} = 2/{{100+11}/50} = 100/111$

Q-2: ${4{2/7} - {1/2}}/{3{1/2} + 1{1/7}}$ ÷ $1/{2+1/{2+1/{5-{1/5}}}}$ is equal to

Solution-

First part = ${30/7 - 1/2}/{7/2 + 8/7}$

= ${{60 - 7}/14}/{{49+16}/14} = {53/14} × 14/65 = 53/65$

Second part = $1/{2+1/{2+{1/{{25 - 1}/5}}}}$

= $1/{2+1/{2+{5/24}}} = 1/{2+{1/{{48+5}/24}}}$

= $1/{2+{24/53}} = 1/{{106+24}/53} = 53/130$

Expression = ${53/65} ÷ {53/130} = 53/65 × 130/53 = 2$