# Simplification online test - Aptitude

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

### Simplification Questions and Practice Problems

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Simplification:

Read the following points carefully.

TYPES OF NUMERALS:

1). Natural Numbers: These are counting numbers. For Ex. 1,2,3,4,5……

a) Natural numbers are denoted by ‘N’.

b)  All natural numbers are positive only.

c)  0 is not a natural number. The smallest natural number is ‘1’

2). Whole Numbers: Whole numbers having ‘0’ and natural numbers. Ex.0, 1, 2, 3, 4…….

a) Denoted by ‘W’

3) Integers: Negative numbers and whole numbers forms integers. Ex. -3, -2, -1, 0, 1, 2, 3….

a) Denoted by ‘I’

b) ‘0’ number is neither positive nor negative.

4) Even and Odd Numbers: The counting numbers which are divisible by 2 then those numbers called Even numbers otherwise Odd numbers.

a) Ex. For even numbers: 2, 4, 6, 8,10 ….

b) Ex. For odd numbers: 1, 3, 5, 7,9 ….

5) Prime Numbers: These numbers having only two factors ‘1’ and itself.

Ex. 2, 3, 5, 7, 11, 13 …. 2 is the even number which is prime and prime number is always greater than 1.

6) Rational and Irrational Numbers: Rational numbers are represented in the form of (a/b) where b is not equal to ‘0’, a and b are integers. Ex. (1/2), (2/5), (3/8)…

Irrational numbers cannot be represent in the form of (a/b). Ex. Square roof of 2

7) Composite Numbers: These numbers are not prime numbers and having at least one factor other than ‘1’ and itself.For Ex.  4, 8, 12…

Divisibility Rules:

1. Divisibility by 2: If the number is having either ‘0’ or even number as a last digit then that number is divisible by 2.

Example: 24, 40, 68, 122

2. Divisibility by 3: When sum of the digits of a number is divisible by 3, then the number is divisible by 3.

Example: 159     1 + 5 + 9 = 15     15 is divisible by 3.

3. Divisibility by 4: If the last two digits of a number is divisible by 4, then that number is divisible by 4.

Example: 124     24 is divisible by 4

If the number is having two or more zeroes at the end then also its divisible by 4.

Example: 1200   it’s having two zeroes at the end so its divisible by 4.

4. Divisibility by 5: If the number is having ‘0’ or ‘5’ at the end then it’s divisible by 5.

Example: 150, 205, 300

5. Divisibility by 6: If the number is divisible by 2 and 3then that number should divisible 6.

Example: 36, 54, 60

6. Divisibility by 7: when the difference between twice the digit at last place and the number formed by other digits is either ‘0’ or multiple of ‘7’.

Example:  147    14-(2*7) = 0        144 is divisible by 7

7. Divisibility by 8: If the number made by last three digits is divisible by 8, the number is divisible by 8.

Example: 24032                 032 is divisible 8

8. Divisibility by 9: It’s similar to divisible by 3 rule. When sum of the all the digits of a number is divisible by 9, then the number is divisible by 9.

Example: 4374                   4 + 3 + 7 + 4 = 18               18 is divisible by 9

9. Divisibility by 10: If the number is end with ‘0’ then it’s divisible by 10.

Example: 50, 110, 2000

10. Divisibility by 11: If the sum of digits at odd and even places are equal or differ by a number which is divisible by 11 then the number is divisible by 11.

Example: 216282              sum at even places(2+6+8) = 16;                               sum at odd places (1+2+2) = 5

Difference: (16-5) = 11. So 216282 is divisible by 11.