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Given expression is,

(77.80 ÷ 2.10) + 60.94 = 22.11 × 3.95 + ?

We can write the given VALUES as:

77.80 ≈ 78 and 2.10 ≈ 2

60.94 ≈ 61 and 22.11 ≈ 22 and 3.95 ≈ 4

Then,

⇒ (78 ÷ 2) + 61 = 22 × 4 + ?

⇒ 39 + 61 = 22 × 4 + ?

⇒ 39 + 61 = 88 + ?

⇒ 100 = 88 + ?

⇒ ? = 100 - 88

GIVEN expression

$(\Rightarrow \;? = 12\frac{1}{3} + 10\frac{5}{6} - 7\frac{2}{3})$

$(\Rightarrow \;? = \frac{{37}}{3} + \frac{{65}}{6} - \frac{{23}}{3} )$

$(\Rightarrow \;? = \frac{{74 + 65 - 46}}{6}\;)$

$(\Rightarrow \;? = \frac{{93}}{6})$

$(\Rightarrow \;? = 15\frac{1}{2})$

$(\SQRT[3]{{729}}{\rm{}} + {\rm{}}\sqrt {441}- \sqrt[3]{{343}}{\rm{}} = {\rm{}}\sqrt[3]{{{9^3}}}{\rm{}} + {\rm{}}\sqrt {{{21}^2}}- \sqrt[3]{{{7^3}}}{\rm{}} = {\rm{}}9 + 21 - 7 = 23)$

((√3)^3/2)⁴ + ((√3)^-3/2)⁴ = 3³ + 3^-3 = 3³ + 1/3³ = (3⁶ + 1)/3³ = (729 + 1)/27 = 730/27

$(\BEGIN{array}{l}\frac{{\LEFT( {0.0112 - 0.0012} \RIGHT)\;of\;0.14\; + \;0.25\; \TIMES \;0.2}}{{0.02\; \times \;0.01}}\\ = \;\frac{{\left( {0.01} \right)of\;0.14\; + \;0.05}}{{0.0002}}\\ = \;\frac{{0.01\; \times \;0.14\; + \;0.05}}{{0.0002}}\\\; = \;\frac{{0.0014\; + \;0.05}}{{0.0002}}\\ = \;\frac{{0.0514}}{{0.0002}}\\ = \;257\end{array})$

FOLLOW BODMAS rule to solve this question, as PER the order given below,

Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, the BODMAS rule must be followed,

(60% of 150)% of (80% of 600) = ?

⇒ (60/100 × 150)% of (80/100 × 600) = ?

⇒ 90% of 480 = ?

Step-2- Any MATHEMATICAL 'Of' or 'Exponent' must be solved next,

⇒ 90/100 × 480 = ?

$(\sqrt {50 - 7 \times \frac{{1\frac{2}{7} - 2\frac{3}{7}}}{{\frac{1}{{1 + \frac{3}{7}}} - \frac{9}{{70}}}}} )$

= $(\sqrt {50 - 7 \times \frac{{\frac{9}{7} - \frac{{17}}{7}}}{{\frac{1}{{\frac{{10}}{7}}} - \frac{9}{{70}}}}} )$

= $(\sqrt {50 - 7 \times \frac{{ - \frac{8}{7}}}{{\frac{7}{{10}} - \frac{9}{{70}}}}} )$

= $(\sqrt {50 - 7 \times \frac{{ - \frac{8}{7}}}{{\frac{4}{7}}}} )$

= $(\sqrt {50 - 7 \times - 2} )$

= √64

= 8

FOLLOW BODMAS rule to solve this question, as PER the order given below,

Step-1-Parts of an EQUATION enclosed in 'Brackets' MUST be solved first,

Step-2-Any mathematical 'Of' or 'Exponent' must be solved NEXT,

Step-3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

The given expression:

8500 + (2350 ÷ 50)

= 8500 + 47

Follow BODMAS rule to solve this QUESTION, as per the order given below,

STEP - 1 - Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step - 2 - Any mathematical 'Of' or 'Exponent' must be solved next,

Step - 3 - Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step - 4 - Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is,

(167 × 3 + 7) ÷ 4 = 85 + ?

⇒ ? + 85 = (501 + 7) ÷ 4

⇒ ? + 85 = 508 ÷ 4 = 127

⇒ ? = 127 - 85

∴ ? = 42

CONJUGATE of (√a + √b) is (√a - √b) and vice versa.

$(\begin{array}{l}\left[ {\frac{{3\sqrt 2 }}{{\sqrt 5+ \sqrt 8 }} - \frac{{4\sqrt 3 }}{{2 + \sqrt 8 }} + \frac{{\sqrt 8 }}{{\sqrt 2+ \sqrt 3 }} + \frac{4}{{\sqrt {10}- \sqrt 6 }}} \right] = \left[ {\frac{{3\sqrt 2 \left( {\sqrt 8- \sqrt 5 } \right)}}{{8 - 5}} - \frac{{4\sqrt 3 \left( {\sqrt 8- 2} \right)}}{{8 - 4}} + \frac{{\sqrt 8 \left( {\sqrt 3- \sqrt 2 } \right)}}{{3 - 2}} + \frac{{4\left( {\sqrt {10}+ \sqrt 6 } \right)}}{{10 - 6}}} \right]\\ = \sqrt {16}- \sqrt {10}- \sqrt {24}+ \sqrt {12}+ \sqrt {24}- \sqrt {16}+ \sqrt {10}+ \sqrt 6= \sqrt {6\;}+ \sqrt {12}\END{array})$

$(\begin{array}{l}\FRAC{{0.7\; \TIMES 0.7 \times 0.7\; + \;0.3 \times 0.3 \times 7.3}}{{0.7 \times 0.7\; + \;0.3 \times 0.3\; + \;0.42}}\\= \;\frac{{0.343\; + \;0.09 \times 7.3}}{{{{0.7}^2}\; + \;{{0.3}^2}\; + \;2 \times 0.3 \times 0.7}}\;\\= \;\frac{{0.343\; + \;0.657}}{{{{\left( {0.7\; + \;0.3} \RIGHT)}^2}}}\end{array})$

= 1/1 = 1

The expression can be solved as following:

$(\begin{array}{L} \RIGHTARROW \left( {4.42{\RM{}} \div {\rm{}}1.7} \right) + {\rm{}}\left( {3.36{\rm{}} \div {\rm{}}1.4} \right)-{\rm{}}\left( {4.32{\rm{}} \div {\rm{}}1.2} \right)\\ = \frac{{44.2}}{{17}} + \frac{{33.6}}{{14}} - \frac{{43.2}}{{12}} \end{array})$

= 2.6 + 2.4 – 3.6

Follow BODMAS RULE to solve this QUESTION, as per the order given below,

Step-1- Parts of an equation enclosed in 'Brackets' must be SOLVED first,

Step-2-Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3- Next, the parts of the equation that contain 'Division' and 'Multiplication' are CALCULATED,

Step-4- Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now, the given expression,

14% of 250+ ?% of 300 = 125

⇒35 + ?% of 300 = 125

⇒ 3× ? = 90

Approximating the TERMS as,

⇒ 90.195 ≈ 90

⇒ 2.942 ≈ 3

⇒ 121.012 ≈ 120

⇒ 3.963 ≈ 4

Hence, the EQUATION is approximated to,

⇒ 90 ÷ 3 × 120 ÷ 4

⇒ 30 × 30

⇒ 900

$(\begin{array}{l}5{\left( {\sqrt 5 } \right)^{X + 6}} = {\left( {\sqrt 5 } \right)^{2x + 7}}\\{\left( {\sqrt 5 } \right)^2}{\left( {\sqrt 5 } \right)^{x + 6}} = {\left( {\sqrt 5 } \right)^{2x + 7}}\\{\left( {\sqrt 5 } \right)^{x + 6 + 2}} = {\left( {\sqrt 5 } \right)^{2x + 7}}\end{array})$

From laws of INDICES: x + 8 = 2x + 7

GIVEN that,

$(2\SQRT[3]{{243}}\; + \;3\sqrt[3]{9}\; + \;\sqrt[3]{{1125}})$ 

$(\RIGHTARROW \;2\sqrt[3]{{27\; \times 9\;}}\; + \;3\sqrt[3]{9}\; + \;\sqrt[3]{{125\; \times 9}})$

GIVEN, (1/3.197) = 0.3127

ACCORDING to BODMAS

We will RESOLVE DIVISION and multiplication first

90 × 9 = 810

9000 ÷ 900 = 10

Now the EXPRESSION becomes

810 – 90 + 10

$(\begin{array}{l} \Rightarrow \FRAC{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}} + \frac{1}{{56}}\\ \Rightarrow \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \frac{1}{{4 \times 5}} + \frac{1}{{5 \times 6}} + \frac{1}{{6 \times 7}} + \frac{1}{{7 \times 8}}\\ \Rightarrow \LEFT( {\frac{1}{2}-\frac{1}{3}} \RIGHT) + \left( {\frac{1}{3}-\frac{1}{4}} \right) + \left( {\frac{1}{4}-\frac{1}{5}} \right) + \left( {\frac{1}{5}-\frac{1}{6}} \right) + \left( {\frac{1}{6}-\frac{1}{7}} \right) + \left( {\frac{1}{7}-\frac{1}{8}} \right)\\ \Rightarrow \frac{1}{2}-\frac{1}{3} + \frac{1}{3}-\frac{1}{4} + \frac{1}{4}-\frac{1}{5} + \frac{1}{5}-\frac{1}{6} + \frac{1}{6}-\frac{1}{7} + \frac{1}{7}-\frac{1}{8}\\ \Rightarrow \frac{1}{2}-\frac{1}{8}\\ \Rightarrow \frac{{4-1}}{8}\\\therefore \frac{3}{8}\end{array})$

$( \Rightarrow \frac{1}{2} + \frac{1}{4} + \frac{1}{{10}} + \frac{1}{{20}})$

$( \Rightarrow \frac{{10 + 5 + 2 + 1}}{{20}})$

⇒ 9/10

CONVERTING FRACTIONS to DECIMALS,

⇒ 3/4 = 0.75

⇒ 5/12 = 0.4167

⇒ 13/16 = 0.8125

⇒ 16/29 = 0.5517

⇒ 3/8 = 0.375

? 0.375 < 0.4167 < 0.5517 < 0.75 < 0.8125

$(\frac{1}{{\sqrt 9- \sqrt 8 }} - \frac{1}{{\sqrt 8- \sqrt 7 }} + \frac{1}{{\sqrt 7- \sqrt 6 }} - \frac{1}{{\sqrt 6- \sqrt 5 }} + \frac{1}{{\sqrt 5- \sqrt 4 }})$ ----(1)

In equation 1, we will rationalize the DENOMINATOR of each of the TERMS to remove root from the denominator and MAKE the calculations easier.

$(\frac{1}{{\sqrt 9- \sqrt 8 }} = \frac{1}{{\sqrt 9- \sqrt 8 }} \times \frac{{\sqrt 9+ \sqrt 8 }}{{\sqrt 9+ \sqrt 8 }} = \frac{{\sqrt 9+ \sqrt 8 }}{{{{\left( {\sqrt 9 } \right)}^2} - {{\left( {\sqrt 8 } \right)}^2}}} = \frac{{\sqrt 9+ \sqrt 8 }}{{9 - 8}} = \sqrt 9+ \sqrt 8 )$

Similarly, $(\frac{1}{{\sqrt 8- \sqrt 7 }} = \sqrt 8+ \sqrt 7)$

$(\BEGIN{array}{l}\frac{1}{{\sqrt 7- \sqrt 6 }} = \sqrt 7+ \sqrt 6 \\\frac{1}{{\sqrt 6- \sqrt 5 }} = \sqrt 6+ \sqrt 5 \\\frac{1}{{\sqrt 5- \sqrt 4 }} = \sqrt 5+ \sqrt 4\end{array})$

Equation 1 will reduce to

$(\begin{array}{l}\left( {\sqrt 9+ \sqrt 8 } \right) - \left( {\sqrt 8+ \sqrt 7 } \right) + \left( {\sqrt 7+ \sqrt 6 } \right) - \left( {\sqrt 6+ \sqrt 5 } \right) + \left( {\sqrt 5+ \sqrt 4 } \right)\\ = \sqrt 9+ \sqrt 4\end{array})$ 

= 3 + 2 = 5

Given Equation is,

$(\begin{ARRAY}{L}\frac{{\sqrt {0.441} }}{{\sqrt {0.625} }}\\ = \sqrt {\frac{{441}}{{625}}} \\ = \sqrt {\frac{{{{21}^2}}}{{{{25}^2}}}} \\ = \frac{{21}}{{25}}\END{array})$ 

FOLLOW BODMAS rule to solve this question, as per the ORDER given below,

Step-1: Parts of an equation enclosed in 'Brackets' must be solved FIRST, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated

⇒ (10 × 6 ÷ 6 ÷ 6 ÷ 6 ÷ 6 × 10)

= (10 × 1 ÷ 6 ÷ 6 ÷ 6 × 10)

= (10 × 1/6 ÷ 6 ÷ 6 × 10)

= (10 × 1/36 ÷ 6 × 10)

= (10 × 1/216 × 10)

= 100/216

∴ (10 × 6 ÷ 6 ÷ 6 ÷ 6 ÷ 6 × 10) + (10 × 6 ÷ 6 ÷ 6 ÷ 6 ÷ 6 × 10) = 100/216 + 100/216

Out of 2/3, 7/13, 4/11, 5/9

2/3 is the LARGEST NUMBER followed by 5/9 then 7/13 and the SMALLEST is 4/11.

The option c and d are eliminated SINCE if 12 and 27 are ADDED to 64500, the last digit becomes 2 and 7 respectively. And a perfect never ends in 2, 3, 7, 8.

$(\begin{ARRAY}{l} \FRAC{1}{{\sqrt {729.23} - \sqrt {624.89} }} + \sqrt {729.23} - \sqrt {624.89} \approx \frac{1}{{\sqrt {729} - \sqrt {625} }} + \sqrt {729} - \sqrt {625} \\ \approx \frac{1}{{27 - 25}} + 27 - 25 \approx \frac{1}{2} + 2 \approx \frac{5}{2} \approx 2.5 \end{array})$

858.231 ÷ 39.345 × 74.154 – 1499.98 + 31.798 = (2)^? × 9.879

Taking their APPROX. values

⇒ 858/39 × 74 – 1500 + 32 = (2)^? × 10

⇒ 1628 – 1500 + 32 = (2)^? × 10

⇒ 128 + 32 = (2)^? × 10

⇒ 160/10 = (2)^?

⇒ 2^? = 16

⇒ 2^? = 2⁴

⇒ ? = 4

By APPLYING BODMAS

⇒ (857 × 0.14) – 5.6 × 12.128 = ?

By TAKING APPROXIMATE value

⇒ (857 × 0.14) – (5.6 × 12) = ?

⇒ 119.98 – 67.2 = ?

Taking approximate value

On MULTIPLYING 0.15 and 0.15 , we get

Follow the BODMAS rule to SOLVE the question,

435 ÷ 7.5 – 40% of 130 = ?

Any mathematical 'Of' or 'Exponent' MUST be solved first,

⇒ 435 ÷7.5 – [(40/100) × 130] = ?

The parts of the equation that contain 'Division' and 'Multiplication' are calculated,

⇒ (435 ÷ 7.5) – 52 = ?

The parts of the equation that contain 'Addition' and 'Subtraction' should be calculated,

⇒ 58 – 52 = ?

∴ ? = 6

5/9 = 0.555

√(9/49) = 3/7 = 0.428

0.43 = 0.43

(0.7)² = 0.49

So the LEAST no. is √(9/49)

Let the no. be X. then,CORRECT ANSWER = 8x/7 and wrong answer = 7x/8. Then

8x/7 - 7x/8 = 45 

⇒ (64x - 49x)/56 = 45

⇒ 15x/56 = 45

x = 168

(1501.01 ÷ 98.89)(6.979 × 1.984)

⇒ (1500 ÷ 100)(7 × 2)

⇒ 15 × 14

⇒ 210

The percentage is 75%

= 75/100

= 15/20[when DIVIDED by 5 both]

= 3/4 [when divided by 5 in both]

= 6/8 [MULTIPLYING 2 in both]

Both the expressions are same except for the last part 5 × 2 in first ONE and (5 × 2) in the second one

Finding out the value of the expression except that part:

⇒ [5 + 3² × 1 ÷ 2 + 4² - 7 ÷ (6 + 4^1.5)]

⇒ [5 + 3² × 1 ÷ 2 + 4² - 7 ÷ (6 + 8)] [? BRACKET is operated first, ORDER is next]

⇒ [5 + 3² × 1 ÷ 2 + 4² - 7 ÷ 14]

⇒ [5 + 9 × 1 ÷ 2 + 16 - 7 ÷ 14] [? Order is operated next]

⇒ [5 + 9 × 0.5 + 16 - 0.5] [? Division is operated next]

⇒ [5 + 4.5 + 16 - 0.5] [? Multiplication is operated next]

⇒ [25.5 - 0.5] [? addition is operated next]

⇒ 25 [? subtraction is operated next]

⇒ First expression is (25 ÷ 5) × 2 = 5 × 2 = 10

⇒ Second expression is 25 ÷ (5 × 2) = 25 ÷ 10 = 2.5

$(\frac{{5.6 \TIMES 0.36 + 0.42 \times 3.2}}{{0.8 \times 2.1}})$

$( \Rightarrow \frac{{2 \times 7\left( {8 \times 18 + 6 \times 16} \right)}}{{8 \times 210}})$

$(\Rightarrow \frac{{2\left( {144 + 96} \right)}}{{8 \times 30}})$

$(\Rightarrow \frac{{240}}{{4 \times 30}})$

LAWS of Indices:

1. a^m × a^N = a^{{m + n}}

2. a^m ÷ aⁿ = a^{{m – n}}

3. (a^m)ⁿ = a^mn

4. (a)^-m = 1/a^m

5. (a)^m/n = ⁿ√a^m

6. (a)⁰ = 1

Given, (3⁵)^x ÷ (9)^{2x – 1} = 243

⇒ (3)^5X ÷ (3²)^{2x – 1} = (3)⁵

⇒ (3)^5x ÷ (3)^{4x – 2} = (3)⁵

⇒ (3)^{{5x – 4x + 2}} = (3)⁵

⇒ (3)^{x + 2} = (3)⁵

Equating POWERS,

⇒ x + 2 = 5

⇒ x = 5 – 2 = 3

Also, (5)^{x – 2y} × (5)^{x + y} = 625

⇒ (5)^{(x – 2y + x + y)} = (5)⁴

⇒ (5)^{2x – y} = (5)⁴

Equating powers,

⇒ 2x – y = 4

Substituting for ‘x’,

⇒ 2(3) – y = 4

⇒ y = 6 – 4 = 2

⇒ $(\sqrt {2\sqrt {3\sqrt {3\sqrt {3\sqrt 3 } } \;} } )$

⇒ $(\sqrt {2\sqrt {3\sqrt {3\sqrt {3 \TIMES {3^{\FRAC{1}{2}}}} } \;} } )$

⇒ $(\sqrt {2\sqrt {3\sqrt {3\sqrt {{3^{\frac{3}{2}}}} } \;} } )$ ⇒ $(\sqrt {2\sqrt {3\sqrt {3 \times {3^{\frac{3}{4}}}} \;} } )$ ⇒ $(\sqrt {2\sqrt {3\sqrt {{3^{\frac{7}{4}}}} \;} } )$

⇒ $(\sqrt {2\sqrt {3 \times {3^{\frac{7}{8}}}\;} } )$⇒ $(\sqrt {2 \times {3^{\frac{{15}}{{16}}}}} )$

⇒ √2 × √(3^(15/16))

⇒ 2^(1/2) × 3^(15/32)

∴ POWER of 3 is 15/32.

Solving the FRACTIONS and CONVERTING then into decimal,

⇒ (12/5 + 8/3) = 76/15 = 5.067

⇒ (11/4 + 7/2) = 25/4 = 6.25

⇒ (10/3 + 9/4) = 67/12 = 5.583

⇒ (9/2 + 6/5) = 57/10 = 5.7

∴ The SMALLEST FRACTION is (12/5 + 8/3)

Value of $(\sqrt[3]{{1331}}{\rm{}} + {\rm{}}\sqrt {729}- \sqrt[3]{{512}})$ = 11 + 27 – 8 = 30

Statement I:

2√3 > 3√2

⇒ (2√3)² > (3√2)²

⇒ 12 > 18 which is not TRUE

So, statement I is not true

Statement II:

4√2 > 2√8 = 4√2

⇒ 4√2 > 4√2 which is not true

$(\RIGHTARROW .67777 = .6\bar 7 = \frac{{67 - 6}}{{90}} = \frac{{61}}{{90}})$

Given expression is,

84% of 2500 – 420 = 350% of ?

$(\begin{array}{l} \Rightarrow \frac{{84}}{{100}} \times 2500 - 420 = \frac{{350}}{{100}} \times ?\\ \Rightarrow 84 \times 25 - 420 = \frac{{350}}{{100}} \times \;?\\ \Rightarrow 2100 - 420 = \frac{{35}}{{10}} \times \;?\\ \Rightarrow 1680= \frac{{35}}{{10}} \times \;?\\ \Rightarrow {\RM{\;}}? = \frac{{1680\times 10}}{{35}} = 480 \end{array})$

∴ ? = 480

⇒ 9876 + 34.567 - ? = 9908.221

⇒ 9910.567 - ? = 9908.221

GIVEN EXPRESSION,

⇒ (5981 - √3136) × 0.75 =?

⇒ ? = (5981 - 56) × 0.75

⇒ ? = 5925 × 0.75

⇒ ? = 4443.75

$(\BEGIN{array}{L}{\left[ {{{32}^{\frac{4}{5}}} \times {2^{ - 2}} \div {8^0}} \right]^{\frac{1}{2}}}\\ = \;{\left[ {{2^{5 \times \frac{4}{5}}} \times {2^{ - 2}} \div {8^0}} \right]^{\frac{1}{2}}}\\ = {\left[ {{2^4} \times {2^{ - 2}} \div 1} \right]^{\frac{1}{2}}}\end{array})$

= 2^{2 × 1/2}

= 2

97846 – 43241 + 50253 =? – 3727

⇒ 54605 + 50253 =? – 3727

⇒ 104858 + 3727 =?

The EQUATION can be SIMPLIFIED as

$(\frac{8}{{25}} \times \;\frac{1}{2} \times 9 \times \;\frac{{25}}{16})$