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The given equation is:

(7/2) + 6 × (1/3) ÷ 4 = ?

(7/2) + 6 × (1/3) × (1/4) = ?

(7/2) + 6 × (1/12) = ?

(7/2) + (1/2) = ?

? = (7/2) + (1/2)

? = 4

∴ The required value of "?" in the given equation is 4

Given: 

(14 + 6√5)

Concept:

(a + b)² = a² + b² + 2ab

Calculation:

First we convert given equation into given concept form

⇒ 14 + 6√5 = 14 + (2 × 3 × √5)

⇒ 14 + 6√5 = 9 + 5 + (2 × 3 × √5)

⇒ 14 + 6√5 = (3)² + (√5)² + (2 × 3 × √5)

Now the given equation become in the form of given concept so,

⇒ a = 3, b = √5 

⇒ (3)2 + (√5)2 + (2 × 3 × √5) = (3 + √5)²

⇒ 14 + 6√5 = (3 + √5)²

Now square root of (14 + 6√5) is 

⇒ √(14 + 6√5) = √(3 + √5)²

⇒ √(14 + 6√5) = 3 + √5

∴ √(14 + 6√5) is 3 + √5 

Let no of passengers in the beginning be x 

After first station no passengers=(x-x/3)+280=2x/3 +280 

After second station no passengers =1/2(2x/3+280)+12 

1/2(2x/3+280)+12=248 

2x/3+280=2*236 

2x/3=192 

x=288

(a+b)² =a² +b² +2ab=117+2*24=225 

a+b=15 

(a-b)² =a² +b² -2ab=117-2*54 

a-b=3 

a+b/a-b=15/3=5

(a) Let the the number be = x
According o the question,
x² – 262 = 549
or, x² – 676 = 549
or, x² = 549 + 676 = 1225
or, x = √1225 =35

Let the no of fifty rupee notes be x 

Then,no of 100 rupee notes =(85-x) 

50x+100(85-x)=5000

x+2(85-x)=100 

x=70 

so,,required amount=Rs.(50*70)= Rs.3500

Given:

x% of 50 = 40

Calculations:

x% of 50 = 40

⇒ (x/100) × 50 = 40

⇒ x/2 = 40

⇒ x = 80

∴ The value of x is 80.

Calculation:

\(3\frac{1}{2} + 4\frac{1}{4} + 6\frac{1}{6}+x = 60 / 2\)

⇒ 7/2 + 17/4 + 37/6 + x = 60/2

⇒ x = 60/2 – 7/2 – 17/4 – 37/6 = (360 – 42 – 51 – 74)/12

⇒ x = 193/12

∴ The value of x is 193/12

Given:

21^√x + 20^√x  = 29^√x

Formula used:

Using phythagorus theorem

a² + b² = c²

Calculation:

21^√x + 20^√x = 29^√x

Compare the whole equation a² + b² = c²

So √x = 2

x = 4

21² + 20² = 29²

441 + 400 = 841

841 = 841

So x =  4 satisfy this equation

 ∴ The value of x is 4.

Given:

The numerator of the fraction is 6 less than its denominator.

If the numerator is decreased by 1 and the denominator is increased by 5, then the denominator becomes 4 times the numerator.

Calculations:

Let the numerator of the fraction be 'x'.

According to question,

Denominator = x + 6

Now,

⇒ 4 × (x - 1) = (x + 6) + 5

⇒ 4x - 4 = x + 11

⇒ x = 5

⇒ Numerator of the fraction = 5

⇒ Denominator of the fraction = x + 6 

⇒ 5 + 6

⇒ 11

∴ The fraction is 5/11.

Topper's approach:

Given:

The numerator of the fraction is 6 less than its denominator.

If the numerator is decreased by 1 and the denominator is increased by 5, then the denominator becomes 4 times the numerator.

Concept used:

Directly satisfy the conditions given in the question through options.

Explanation:

Only option 3 is there which satisfied the first condition that the numerator is 6 less than the denominator.

∴ The fraction is 5/11.

Given-   

\(\frac{1}{{{3^3}}} × \frac{2}{{{3^3}}} × {\left( {2\frac{1}{4}} \right)^2} = \frac{x}{{18}}\)

Calculation-

1/27 × 2/27 × (9/4)² = x/18

⇒ x/18 = 81/16 × 1/27 × 2/27

⇒ x/18 = 1/72

∴  x = 1/4

Calculation:

144 – 9 = a²

⇒ 135 = a²

⇒ √135 = a

Given:

\(\frac 5 {13} + \frac {13} 5\)

Calculation:

\(\frac 5 {13} + \frac {13} 5\)

⇒ (5/13) + (13/5)

⇒ (25 + 169)/65

⇒ 194/65

⇒ \(2\frac {64}{65}\)

∴ The value is \(2\frac {64}{65}\).

Calculation:

[(100 ? 2) ? 2] ? 10 = 59

Checking options

Option 1

[(100 – 2) ÷ 2] + 10 = (98/2) + 10 = 59

Option 2

[(100 – 2) × 2] ÷ 10 = 19.6

Option 3

[(100 + 2) × 2] × 10 = 2040

Option 4

[(100 + 2) ÷ 2] ÷ 10 = 5.1

∴ Option 1 is correct

Concept:

(a + b)² = a² + b² + 2ab

Given:

a = \(\sqrt {{{\left( {2013} \right)}^2} + 2013 + 2014} \)

Calculation:

After rearranging the given question,

a = \(\sqrt {{{\left( {2013} \right)}^2} + 2013 + {(2013~+~1)}} \)

a = \(\sqrt {{{\left( {2013} \right)}^2} + 2 \times 2013 + 1} \)

a = \(\sqrt {{{\left( {2013} \right)}^2} +~1^2~+~ 2 \times 1 \times 2013} \)

a = \(\sqrt {{{\left( {2013~+~1} \right)}^2}} \)

a = 2014

Calculation :

Let the value at question mark be x 

⇒ (23/4) + x + (5/2) = (81/8)

⇒ x = (81/8) - (5/2) - (23/4)

⇒ x = (81 - 20 - 46)/ 8

⇒ x = 15/8

\( \Rightarrow x\; = \;1\frac{7}{8}\)

∴ the required value of x is \(1\frac{7}{8}\)

Given:

 \(\frac{1}{{x + \frac{1}{{y + \frac{2}{{z + \frac{1}{4}}}}}}} = \frac{{29}}{{79}}\) 

Calculation:

RHS can be written as:

\(⇒ \;\frac{1}{{\frac{{79}}{{29}}}} = \frac{{29}}{{79}}\)

\( ⇒ \;\frac{1}{{2\; +\ \frac{{21}}{{29}}\;}} = \frac{{29}}{{79}}\)

\(⇒ \;\frac{1}{{2\; +\ \frac{1}{{\frac{{21}}{{29}}}}\;}} = \frac{{29}}{{79}}\)

\(⇒ \;\frac{1}{{2\; +\ \frac{1}{{1 \ +\ \frac{8}{{21}}}}\;}} = \frac{{29}}{{79}}\)

\(⇒ \;\frac{1}{{2\; +\ \frac{1}{{1\ +\ \frac{2}{{\frac{{21}}{4}}}}}\;}} = \frac{{29}}{{79}}\)

\(⇒ \;\frac{1}{{2\; +\ \frac{1}{{1\ +\ \frac{2}{{5\ +\ \frac{1}{4}}}}}\;}} = \frac{{29}}{{79}}\)

After comparing LHS with the question,

x = 2, y = 1, z = 5

Substituting the value of x, y and z in (2x + 3y - z),

⇒ 2(2) + 3(1) - 5

⇒ 4 + 3 - 5

⇒ 2

∴ The value of (2x + 3y - z) is 2.

Concept:

Use BODMAS rule

Explanation

3 + 6 of 3 – 2 × 5{25 ÷ 5}     

⇒  3 + 6 of 3 – 2 × 5{5}

⇒  3 + 6 of 3 – 50

⇒  3 + 6 × 3 – 50

⇒  3 + 18 – 50

⇒  21 – 50

Given: 

\(\frac{[(1.3)^2+(1.3\times1.5)+(1.5)^2]\times[(1.3)^2-1.95+(1.5)^2]}{(1.3)^4+(1.3)^2(2.25)+(1.5)^4}\)

Concept used:

The numerator of the given expression is of the form = (a² + ab + b²) × (a² – ab + b²)

Expanding and solving we get:

(a² + ab + b²) × (a² – ab + b²) = a⁴ – a³b + a²b² + a³b – a²b² + ab³ + a²b² – ab³ + b⁴

⇒ (a² + ab + b²) × (a² – ab + b²) = a⁴ + a²b² + b⁴

Denominator is of the form = a⁴ + a²b² + b⁴

Calculation:

Numerator:

[(1.3)² + (1.3 × 1.5) + (1.5)²] × [(1.3)² – 1.95 + (1.5)²] = [(1.3)² + (1.3 × 1.5) + (1.5)²] × [(1.3)² – (1.3 × 1.5) + (1.5)²]

⇒ [(1.3)² + (1.3 × 1.5) + (1.5)²] × [(1.3)² – (1.3 × 1.5) + (1.5)²] = (1.3)⁴ + (1.3)² × (1.5)² + (1.5)⁴

Denominator:

(1.3)⁴ + (1.3)² (2.25) + (1.5)⁴ = (1.3)⁴ + (1.3)² × (1.5)² + (1.5)⁴

Now, Numerator/Denominator = [(1.3)⁴ + (1.3)² × (1.5)² + (1.5)⁴]/ [(1.3)⁴ + (1.3)² × (1.5)² + (1.5)⁴]

⇒ Numerator/Denominator = 1

∴ The value of the given expression is 1.

Given:

\(\frac{72}{100}\)

\(\frac{175}{108}\)

Formula Used:

\( \frac{a}{b} \times \frac{c}{d} = \;\frac{{ac}}{{bd}}\)

Calculation:

\(\frac{{72}}{{100}} \times \frac{{175}}{{108}}\)

\( \Rightarrow \frac{2}{{100}} \times \frac{{175}}{3}\)

\( \Rightarrow \frac{2}{4} \times \frac{7}{3}\)

\( \Rightarrow \frac{1}{2} \times \frac{7}{3}\)

\( \Rightarrow \;\frac{7}{6}\)

∴ The Multiplication of \(\frac{72}{100}\) and  \(\frac{175}{108}\) is \(\frac{7}{6}\)

Given

a + b + c = 0

Formula used

(a + b)² = a² + b² + 2ab

Calculation

a + b + c = 0

⇒ a + b = -c

On squaring both sides, we get

(a + b)² = c²

⇒ a² + b² + 2ab = c²

⇒ a² + b² - c² = -2ab      ______(1)

Similarly, b² + c² - a² = -2bc      _____(2)

And c² + a² - b² = -2ca      _____(3)

Then using equation (1), (2) and (3)

\(\Rightarrow \frac{1}{{{b^2} + {c^2} - {a^2}}} + \frac{1}{{{c^2} + {a^2} - {b^2}}} + \frac{1}{{{a^2} + {b^2} - {c^2}}} = {1 \over -2bc} + {1 \over -2ca} + {1 \over -2ab}\)

\(\Rightarrow {-1 \over 2}({1 \over bc} + {1 \over ca} + {1 \over ab})\)

\(\Rightarrow {-1 \over 2}({a + b + c \over abc}) = 0\)

Given:

22x-1 = 8^3x

Calculation:

22x-1 = 8^3x

⇒ 2^{2x - 1} = 2^9x

⇒ 2x - 1 = 9x

⇒ 7x = - 1 

∴ Value of x = - 1/7

Given:

16x - 2 × 22x + 12 = 42x + 6

Concept Used:

a^m × aⁿ = a^{m + n}

If, x^p = x^q, then

p = q

Calculations:

16x - 2 × 22x + 12 = 42x + 6

16 = 2⁴, 4 = 2²

⇒ 24 × (x - 2) × 22x + 12 = 22 × (2x + 6)

⇒ 2^{(4x - 8)} × 22x + 12 = 2^{(4x + 12)}

am × an = am + n

⇒ 2^{(4x - 8 + 2x + 12)} = 2(4x + 12)

⇒ 2^{(6x + 4)} = 2(4x + 12)

Now, xp = xq

⇒ 6x + 4 = 4x + 12

⇒ 2x = 8

⇒ x = 4

∴ The value of x is 4.