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$(\eqalign{ & 2\sqrt {16 - 6\sqrt {16 - 6\sqrt {16 - \ldots } } } = k \cr & \therefore\ 2\sqrt {16 - 3 \times 2\sqrt {16 - 3 \times 2\sqrt {16 - 3 \times 2} \sqrt {16 - \ldots } } } = k \cr & \therefore \ 2\sqrt {16 - 3K} = k \cr})$

∴ 4 (16 - 3k) = k²

 ∴ k² + 12k – 64 = 0

∴ k² + 16K – 4k – 64 = 0

∴ (k + 16)(k – 4) = 0

k ≠ -16

∴ k = 4

(√64.09 × 23.95 - 31) ÷ 6.89 = ?

APPROXIMATING the values to the nearest INTEGER:

⇒ (8 × 24 - 31) ÷ 7 = ?

⇒ (192 - 31) ÷ 7 = ?

⇒ ? = 161/7

⇒ ? = 23

⇒ 623.697 + 721.36 – 61.006 = ? + 27.271

⇒ 1345.057 – 61.006 = ? + 27.271

⇒ 1284.051 = ? + 27.271

⇒ ? = 1284.051 – 27.271 = 1256.78

∴ ? = 1257 (APPROX.)

(13)² - (14)² + (17)² - 250 = √?

⇒ (13 + 14)(13 - 14) + 289 - 250 = √?

⇒ -27 + 289 - 250 = √?

⇒ -27 + 39 = √?

⇒ √? = 12

⇒ Let {(0.0347)³ + (0.9653)³}/(0.0347)² - (0.0347)(0.9653) + (0.9653)² = X

⇒ We KNOW that a³ + b³ = (a + b) (a² - AB + b²)

⇒ (0.0347 + 0.9653) {(0.0347)² - (0.0347) (0.9653) + (0.9653)}/(0.0347)² - (0.0347) (0.9653) + (0.9653)²

⇒ x = (0.0347 + 0.9653)

The SMALLEST NUMBER is 1/√25 = 1/5

Other NUMBERS are √25 = 5

5/√25 = 5/5 = 1

√25/5 = 5/5 = 1

⇒ 427.01 - 228.99 + 1672.97 = ? + 1211.95

⇒ 427 - 229 + 1673 = ? + 1212

⇒ ? = 427 - 229 + 1673 - 1212

⇒ ? = 659 ≈ 660

⇒ $(\left( {\frac{1}{{2197}}} \right){^{\frac{4}{3}}} \DIV \left( {\frac{1}{{4913}}} \right){^{\frac{2}{3}}})$

⇒ $({2197^{ - \left( {\frac{4}{3}} \right)}} \div {4913^{ - \left( {\frac{2}{3}} \right)\;}})$

⇒ $({4913^{\left( {\frac{2}{3}} \right)}} \div {2197^{\left( {\frac{4}{3}} \right)\;}})$

⇒ $(\left( {{{17}^3}} \right){^{\left( {\frac{2}{3}} \right)}} \div ({13^3}){\;^{\left( {\frac{4}{3}} \right)\;}})$

⇒ $(\frac{{{{17}^2}}}{{{{13}^4}}})$

[(5.168 × 4453 × 3.194)/(67.999 × 4224.017)]

= 73503.85/287228.93

= 0.255

∴ ANSWER is 0.2

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

(6 of 9) + [(3 × 4) - (5 + 1 of 9)] - 25 ÷ 2.5 = ?

⇒ ? = 54 + [12 - (5 + 9)] - 25 ÷ 2.5

⇒ ? = 54 + [12 - 14] - 25 ÷ 2.5

⇒ ? = 54 - 2 - 25 ÷ 2.5

⇒ ? = 54 - 2 - 10

Let x be the NUMBER then

$(\frac{1}{5} \times \frac{1}{6} \times x = 21 \RIGHTARROW x = 21 \times 5 \times 6\therefore \frac{2}{9} \times x = \frac{2}{9} \times 21 \times 5 \times 6 = 140)$

In this type of question, we are EXPECTED to calculate APPROXIMATE value (not EXACT value), so we can replace the GIVEN numbers by their nearest perfect places which makes the calculation easy.

Now, the given expression:

21 + 3.7 × 2.9 =?

⇒ ? ≈ 21 + 4 × 3

⇒ ? ≈ 21 +12 ≈ 33 ≈ 32

ROUNDING off the VALUES in the GIVEN problem

⇒ (√1600 - √169)²/(√196 - √121)²

⇒ (40 - 13)²/(14 - 11)²

⇒ 27²/3²

⇒ 729/9

⇒ 81

∴ Approximate value is 81

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.

[(1326 ÷ 26) × 9 – 64] × 3 = [(? + 13) – 3 × 7]

⇒ [51 × 9 – 64] × 3 = [(? + 13) – 21]

⇒ [459 – 64] × 3 = [(? + 13) – 21]

⇒ 395 × 3 = [(? + 13) – 21]

⇒ 1185 + 21 = (? + 13)

⇒ ? = 1206 – 13

∴ ? = 1193

19750.015 ÷ 979.82 × 201.04 = ?

APPROXIMATING the VALUES to the NEAREST integer:

19750 ÷ 980 × 201 = ?

19750 × (1/980) × 201 = ?

20.15 × 201 = ?

? = 4050.76 ≈ 4050

$(\BEGIN{array}{l} \left( {\frac{2}{{\sqrt 5 + \sqrt 3 }} - \frac{3}{{\sqrt 6 - \sqrt 3 }} + \frac{1}{{\sqrt 6 + \sqrt 5 }}} \right)\\ \RIGHTARROW \left( {\frac{2}{{\sqrt 5 + \sqrt 3 }}} \right) \TIMES \left( {\frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}} \right) - \left( {\frac{3}{{\sqrt 6 - \sqrt 3 }}} \right) \times \left( {\frac{{\sqrt 6 + \sqrt 3 }}{{\sqrt 6 + \sqrt 3 }}} \right) + \left( {\frac{1}{{\sqrt 6 + \sqrt 5 }}} \right) \times \left( {\frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 - \sqrt 5 }}} \right)\\ \Rightarrow \left[ {\frac{{\left\{ {2\left( {\sqrt 5 - \sqrt 3 } \right)} \right\}}}{{\left\{ {5 - 3} \right\}}}-\frac{{\left\{ {3\left( {\sqrt 6 + \sqrt 3 } \right)} \right\}}}{{\left\{ {6 - 3} \right\}}} + \frac{{\left\{ {\sqrt 6 - \sqrt 5 } \right\}}}{{\left\{ {6 - 5} \right\}}}} \right] \end{array})$

⇒ √5 - √3 - √6 - √3 + √6 - √5

⇒ -2√3

⇒ Y - [Y - (X - Y) - {Y - (Y - X - Y)} + 2X]

⇒ Y - [Y - (X - Y) - {Y - (-X)} + 2X]

⇒ Y - [Y - (X - Y) - {Y + X} + 2X]

⇒ Y - [Y - X + Y - Y - X + 2X]

⇒ Y - Y

GIVEN expression,

2106.48 + 3296.52 + 8560.73 + 256.4 × 1.8

= 2106.48 + 3296.52 + 8560.73 + 461.52

= 5403 + 9022.25

= 14425.25

Numerator of expression = √ (√ (21 - 9√5) × √ (21 + 9√5))

⇒ √ (√ (21² – (9√5)²)) = √ (√ (441 – 405)) = √ (√ 36) = √6

Denominator of equation = √ (12 × √ (18 - 8√5) × √ (18 + 8√5))

⇒ √ 12 × √ (18² - (8√5)²) = √12 × √ (324 - 320) = √12 × √4 = 2√12

$( \RIGHTARROW 13\frac{1}{5} + 12\frac{2}{3} + 13\frac{1}{3} + 15\frac{4}{5})$ 

$( \Rightarrow 13\frac{1}{5} + 15\frac{4}{5} + 12\frac{2}{3} + 13\frac{1}{3})$ 

⇒ 29 + 26

⇒ 55

We KNOW that one kilometre = 10³ metre = 10⁶ millimetre

⇒ 10 kilometre = 10⁷ millimetre

$(9.3\bar 1)$ = 9.311111

$(0.\bar 4 )$ = 0.444444

$(0.00\bar 3)$ = 0.0033333

$(9.3\bar 1 + 0.\bar 4 + 0.00\bar 3)$

= 9.311111 + 0.444444 + 0.003333

= 9.758888

Given EXPRESSION is,

⇒ 272.80 - 53.051 = 110.64 + ?

We can write the given values as:

⇒ 272.80 ≈ 273 and 53.051 ≈ 53 and 110.64 ≈ 111

Then, the given expression becomes:

⇒ 273 - 53 = 111 + ?

⇒ 220 = 111 + ?

⇒ ? = 109

15 - (-5) {4 - (7 - 3)} ÷ [3 {5 + (-3) × (-6)}]

= 15 - (-5) {4 - 4} ÷ [3 {5 + 18}]

= 15 - (-5) × 0 ÷ 3 × 23

= 15 - (-5) × 0 ÷ 69

= 15 - (-5) × 0 [PERFORMING division 0 ÷ 69 = 0] 

= 15 

Given,

⇒ (5²)^{X + 2} × (5^{- x})⁴ = (5)^{x – 5}

⇒ (5)^{(2x + 4 – 4x)} = (5)^{x – 5}

⇒ (5)^{(4 – 2x)} = (5)^{x – 5}

⇒ 4 – 2x = x – 5

Given expression,

(11.25 + 17.25) × 1.5 + (96.5 - 65.25) × 0.5 = ?

⇒ 28 × 1.5 + 31 × 0.5

⇒ 42 + 15.5

A = 2⁴ × 3⁵

B = 2⁷ × 3⁵

⇒ A X B = 2⁴ × 3⁵ × 2⁷ × 3⁵

ATQ, x = 2 - √6

$(\frac{1}{x} = \frac{1}{{2 - \sqrt 6 }})$

Now, putting the above values in place of x², we get,

$(\begin{array}{L} \Rightarrow {\LEFT( {2 - \sqrt 6 } \right)^2} + \frac{{12}}{{{{\left( {2 - \sqrt 6 } \right)}^2}}}\\ \Rightarrow 4 + 6 - 2 \TIMES 2 \times \sqrt 6+ \frac{{12}}{{4 + 6 - 2 \times 2 \times \sqrt 6 }}\\ \Rightarrow 10 - 4\sqrt 6+ \frac{{12}}{{10 - 4\sqrt 6 }}\\ \Rightarrow \frac{{{{\left( {10 - 4\sqrt 6 } \right)}^2} + 12}}{{10 - 4\sqrt 6 }}\\ \Rightarrow \frac{{{{\left( {10 - 4\sqrt 6 } \right)}^2} + 12}}{{10 - 4\sqrt 6 }}\\ \Rightarrow \frac{{100 + 16 \times 6 - 2 \times 10 \times 4\sqrt 6+ 12}}{{10 - 4\sqrt 6 }}\\ \Rightarrow \frac{{208 - 80\sqrt 6 }}{{10 - 4\sqrt 6 }}\\\therefore \frac{{8(13 - 5\sqrt {6)} }}{{\left( {5 - 2\sqrt 6 } \right)}}\end{array})$

⇒ To FIND the VALUE of √(5/6) (Multiply the NUMERATOR and DENOMINATOR with 6 we get

√((5 × 6)/(6 × 6)) = (1/√306)

⇒ (1/6) × 5.477 = 0.913

Rounding off the VALUES in the given problem

⇒ (40% of 5500) - (32% of 5600) = ? + 226

⇒ (40 × 55) - (32 × 56) = ? + 226

⇒ (2200) - (1792) = ? + 226

⇒ ? = (2200) - (1792) - 226

⇒ ? = 2200 - (1792 + 226)

⇒ ? = 2200 - (2018)

⇒ ? = 182

∴ The approximate VALUE is 182

The given EXPRESSION:

[(1.3)² × (4.2)²] ÷ 2.7 =?

⇒ ? = [1.69 × 17.64] ÷ 2.7

⇒ ? ≈ 29.81/2.7

⇒ ? ≈ 11.04 ≈ 11

Given expression is,

(180.18 ÷ 3.16) + 44.14 = 15.96 × ? + 7.98

We can WRITE the given values as:

180.18 ≈ 180 and 3.16 ≈ 3

44.14 ≈ 44 and 15.96 ≈ 16 and 7.98 ≈ 8

Then,

⇒ (180 ÷ 3) + 44 = 16 × ? + 8

⇒ 60 + 44 = 16 × ? + 8

⇒ 104 = 16 × ? + 8

⇒ 104 - 8 = 16 × ?

⇒ 96 = 16 × ?

⇒ ? = 96 /16

339.98 ÷ ? = √143.98 + √64.02

By APPROXIMATION;

⇒ 340 ÷ ? = √144 + √64 = 20

Addition takes priority over subtraction, HENCE: 7 + 8 + 9 - 1 + 2 - 21 + 53

= 15 + 9 - 1 + 2 - 21 + 53

= 24 - 1 + 2 - 21 + 53

= 26 - 1 - 21 + 53 = 79 - 1 - 21

Simple subtraction: 79 - 1 - 21 = 57

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,

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 expression is,

⇒ 4 – 1 + 5 × 6 ÷ 2 – 1

⇒ 4 – 1 + 5 × 3 – 1

⇒ 4 – 1 + 15 – 1

⇒ 17

LET the fraction be a/b, From the questions statement

⇒ (a­ + 5)/b = 6/5

⇒ (5a + 25)/6 = b… (i)

When 4 is added to its denominator

⇒ a/(b + 4) = 1/2

⇒ 2A = b + 4 Using (i)

⇒ 2a = (5a + 25)/6 + 4

⇒ 7a = 49

⇒ a = 7 Put this in (i)

⇒ b = (5 × 7 + 25)/6

⇒ b = 10

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,

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.

$(\begin{array}{l} \left( {3\frac{1}{5}of\;\left( {2\frac{2}{3}of3\frac{1}{2}} \right)} \right) \div 1\frac{1}{3}:\left( {2\frac{1}{3}of\left( {4\frac{1}{5}of1\frac{1}{3}} \right)} \right)of\;2\frac{2}{3}\\ = \left( {3\frac{1}{5}of\;\left( {\frac{8}{3}\; \times \;\frac{7}{2}} \right)} \right) \div 1\frac{1}{3}:\left( {2\frac{1}{3}of\left( {\frac{{21}}{5}\; \times \;\frac{4}{3}} \right)} \right)of\;2\frac{2}{3}\\ = \left( {3\frac{1}{5}\;of\;\frac{{28}}{3}} \right) \div 1\frac{1}{3}:\left( {2\frac{1}{3}of\frac{{28}}{5}} \right)of\;2\frac{2}{3}\\ = \left( {\frac{{16}}{5}\; \times \;\frac{{28}}{3}} \right) \div \frac{4}{3}:\left( {\frac{7}{3}\; \times \;\frac{{28}}{5}} \right)of\;\frac{8}{3}\; = \;\left( {\frac{{16}}{5}\; \times \;\frac{{28}}{3}} \right)\; \times \;\frac{3}{4}:\left( {\frac{7}{3}\; \times \;\frac{{28}}{5}} \right)\; \times \;\frac{8}{3}\\ = \frac{{\left( {\frac{{16}}{5}\; \times \;\frac{{28}}{3}} \right)\; \times \;\frac{3}{4}}}{{\left( {\frac{7}{3}\; \times \;\frac{{28}}{5}} \right)\; \times \;\frac{8}{3}}}\; = \;\frac{9}{{14}} \end{array})$

Approximating the TERMS as,

⇒ 69.987 ≅ 70

⇒ 27.872 ≅ 28

Hence, the expression approximately becomes,

⇒ 70 × 2 – 28 × ? = 0

⇒ 140 = 28 × ?

$(\begin{array}{L}\SQRT {1 + \;\frac{X}{{361}}}= \frac{{21}}{{19}}\\ \Rightarrow 1 + \frac{x}{{361}}\; = \frac{{{{21}^2}}}{{{{19}^2}}}\\ \Rightarrow \frac{{361 + x}}{{361}} = \frac{{441}}{{361}}\END{array})$ 

⇒ 361 + x = 441

⇒ x = 441 – 361 = 80

 $(122\frac{{56}}{{10000}})$

For CONVERTING it into decimal

⇒ {(122 × 10000) + 56} ÷ 10000

⇒ (1220000 + 56) ÷ 10000

⇒ 122.0056

$(\sqrt[3]{{\frac{{4913}}{{2197}}}} = \sqrt[3]{{\frac{{17 \TIMES 17 \times 17}}{{13 \times 13 \times 13}}}} = \frac{{17}}{{13}})$

$(\Rightarrow 9{8\ \over 9} \TIMES 9 = 9 \times 9 + \frac{8}{9} \times9 = 89)$

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,

$(17\frac{1}{5} + 21\frac{1}{2} + 4\frac{4}{5} + 2\frac{1}{2} = 177\frac{3}{3} - ?)$

⇒ 86/5 + 43/2 + 24/5 + 5/2 = 534/3 - ?

⇒ 86/5 + 43/2 + 24/5 + 5/2 = 178 - ?

⇒ 110/5 + 48/2 = 178 - ?

⇒ 22 + 24 = 178 - ?

⇒ 46 = 178 - ?

⇒ ? = 178 – 46

5.36 × 3.6 = 536/100 × 36/10

$(\therefore \sqrt {\FRAC{{{{\LEFT( {0.04} \RIGHT)}^2} + {{\left( {0.21} \right)}^2} - 0.0084}}{{{{\left( {0.04} \right)}^3} + {{\left( {0.21} \right)}^3}}}} = \sqrt {\frac{{{{\left( {0.04} \right)}^2} + {{\left( {0.21} \right)}^2} - \left( {0.04 \times 0.21} \right)}}{{\left( {0.04 + 0.21} \right)({{0.04}^2} + {{0.21}^2} - \left( {0.04 \times 0.21} \right)}}} = \sqrt {\frac{1}{{0.25}}} = \frac{1}{{0.5}} = 2)$

$(\begin{ARRAY}{L} \left[ {2\frac{1}{3} - 1\frac{1}{2}} \RIGHT]of\frac{3}{5}\; + \;1\frac{2}{5} \div 2\frac{1}{3}\; = \;\left[ {\frac{7}{3} - \frac{3}{2}} \right]of\frac{3}{5}\; + \;\frac{7}{5} \div \frac{7}{3}\\ = \;\frac{5}{6}of\frac{3}{5}\; + \;\frac{{\frac{7}{5}}}{{\frac{7}{3}}}\\ = \;\frac{5}{6}\; \times \;\frac{3}{5}\; + \;\frac{3}{5}\\ = \;\frac{1}{2}\; + \;\frac{3}{5}\\ = \;\frac{{11}}{{10}}\; = \;1\frac{1}{{10}}\\ \therefore \left[ {2\frac{1}{3} - 1\frac{1}{2}} \right]of\frac{3}{5}\; + \;1\frac{2}{5} \div 2\frac{1}{3}\; = \;1\frac{1}{{10}} \end{array})$

Solving for STATEMENT I,

√64 + √(0.0064) + √(0.81) + √(0.0081) = 9.07

⇒ 8 + 0.08 + 0.9 + 0.09 = 9.07

⇒ 9.07 = 9.07

⇒ L.H.S. = R.H.S.

Solving for statement II,

√(0.010201) + √(98.01) + √(0.25) = 11.51

⇒ 0.101 + 10(APPROX.) + 0.5 = 11.51

⇒ 10.601(approx.) ≠ 11.51

⇒ L.H.S. ≠ R.H.S.

$(\frac{3}{7} = 0.4284,\frac{4}{{11}} = 0.3636,\frac{5}{8} = 0.625)$ 

So smallest is $(\frac{4}{{11}})$

It can also be solved by making the denominators equal

LCM of 7, 11 & 8 = 616

$(\frac{{264}}{{616}} = \frac{3}{7},\frac{{224}}{{616}} = \frac{4}{{11}},\frac{{385}}{{616}} = \frac{5}{8})$ 

When, based are equal, we can CLEARLY see.

$(\frac{{224}}{{616}})$ is the smallest i.e. $(\frac{4}{{11}})$

$(\Rightarrow \sqrt[4]{6} = {6^{\frac{1}{{4}}}} = {6^{\frac{3}{{12}}}} = \left( {216} \right){^{\frac{1}{{12}}}})$

$(\Rightarrow \sqrt[3]{3} = {3^{\frac{1}{3}}} = {3^{\frac{4}{{12}}}} = {\left( {81} \right)^{\frac{1}{{12}}}})$

$(\begin{array}{l} \Rightarrow \sqrt[3]{4} = {4^{\frac{1}{3}}} = {4^{\frac{4}{{12}}}} = {\left( {256} \right)^{\frac{1}{{12}}}}\\ \therefore \sqrt[3]{3} < \sqrt[4]{6} < \sqrt[3]{4} \end{array})$