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Solving for STATEMENT I,

(0.7)² + (0.07)² + (11.1)² > 123.8

⇒ 0.49 + 0.0049 + 123.21 > 123.8

⇒ 123.7049 > 123.8 (wrong)

⇒ L.H.S. not greater than R.H.S.

Solving for statement II,

(1.12)² + (10.3)² + (1.05)² > 108.3

⇒ 1.2544 + 106.09 + 1.1025 > 108.3

⇒ 108.4469 > 108.3

⇒ L.H.S. > R.H.S.

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,

15 ÷ 12.5 × 35 + 42.8 × 2.5 = ?² + 10²

⇒ (15 ÷ 12.5) × 35 + 42.8 × 2.5 = ?² + 10²

⇒ (1.2 × 35) + (42.8 × 2.5) = ?² + 10²

⇒ 42 + 107 = ?² + 10²

⇒ 149 – 100 = ?²

⇒ ?² = 49

⇒ 3565/1495

⇒ 713/299

Given expression is,

31.87 + 87.80 – 55.11 = 12.84 × ? – 25.88

We can WRITE the given values as:

31.87 ≈ 32 and 87.80 ≈ 88

55.11 ≈ 55 and 12.84 ≈ 13 and 25.88 ≈ 26

Then,

⇒ 32 + 88 – 55 = 13 × ? – 26

⇒ 120 – 55 = 13 × ? – 26

⇒ 65 = 13 × ? – 26

⇒ 65 + 26 = 13 × ?

⇒ 13 × ? = 91

⇒ ? = 91/13

$(\sqrt {676}+ \sqrt {6.76\;}+ \sqrt {0.0676}= 27.76)$

⇒ 26 + 2.6 + 0.26 = 27.7

⇒ 28.86 = 27.7

So, statement 1 is false.

$(\sqrt {339 + 36 + 49 + 81}= 19)$

$(\Rightarrow \sqrt {339 + 6 + 7 + 9}= 19)$

⇒ √361 = 19

⇒ 19 = 19

So, statement 2 is true.

$(\frac{1}{{\sqrt {729.23} - \sqrt {624.89} }} + \sqrt {729.23} - \sqrt {624.89} \approx \frac{1}{{\sqrt {729} - \sqrt {625}}} + \sqrt {729} - \sqrt {625} \approx \frac{1}{{27 - 25}} + 27 - 25 \approx \frac{1}{2} + 2 \approx \frac{5}{2} \approx 2.5)$

<P>Here, p = √5q

⇒ R = √5s

⇒ t = √5u

Now, substituting above values in REQUIRED value

⇒ [(3(√5q) ² + 4(√5s) ² + 5(√5u) ²) /(3q² + 4s² + 5u²) ]

⇒ 5[(3q² + 4s² + 5u²)/(3q² + 4s² + 5u²) ]

∴ answer is 5

USING BODMAS RULE we get,

⇒ 5 + (5 × 5) – 5 + 5

⇒ 5 + 25 – 5 + 5

⇒ 30 

Follow BODMAS 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 contains ‘Division’ and ‘Multiplication’ are calculated,

Step-4: Last but not the least, the parts of the equation that contains ‘Addition’ and Subtraction’ should be calculated.

Given expression,?

⇒ 4.25% of 740 – ?% of 680 = 7.65

$(\BEGIN{array}{L} \Rightarrow \frac{{4.25}}{{100}}\; \times \;740 - \;\frac{?}{{100}}\; \times \;680\; = \;7.65\\ \Rightarrow {\rm{\;}}31.45\; - \;\frac{?}{{100}}\; \times \;680\; = \;7.65\\ \Rightarrow \frac{?}{{100}}\; \times \;680\; = \;31.45\; - \;7.65\\ \Rightarrow ?\; = \;23.8\; \times \;\frac{{100}}{{680}}\; \end{array})$

Solving the GIVEN options:

(6 + 6 + 6)² = 18²  = 324

{(6 + 6)²}² =(12²)² = 20736

(6 × 6 × 6)² = (216)² = 46656

6 + 6² + (6)³ = 6 + 36 + 216 = 258

∴ On COMPARING all the options, (6 × 6 × 6)² is the largest number.

LET the two fractions be ‘x’ and ‘y’.

⇒ xy = 4and x + y = 148/35

⇒ x = 4/y

⇒ (4/y) + y = 148/35

$(\begin{array}{l} \Rightarrow {\rm{\;}}\frac{{4 + {y^2}}}{y}\; = \;\frac{{148}}{{35}}\\ \Rightarrow {\rm{\;}}140 + 35{y^2}\; = \;148y\\ \Rightarrow {\rm{\;}}35{y^2} - 148y + 140\; = \;0\\ \Rightarrow {\rm{\;}}y\; = \;\frac{{148\; \PM \SQRT {{{148}^2} - 4 \times 35 \times 140} }}{{70}}\; = \;\frac{{148\; \pm \sqrt {2304} }}{{70}}\; = \;\frac{{148 \pm 48}}{{70}}\\ \Rightarrow {\rm{\;}}y\; = \;\frac{{100}}{{70}}\;or\frac{{196}}{{70}}\; = \;\frac{{10}}{7}\;or\frac{{14}}{5}\\ \Rightarrow {\rm{\;}}x\; = \;\frac{4}{y}\; = \;\frac{4}{{\frac{{10}}{7}}}\; = \;\frac{{28}}{{10}}\; = \;\frac{{14}}{5}\;or\;x\; = \;\frac{4}{y}\; = \;\frac{4}{{\frac{{14}}{5}}}\; = \;\frac{{20}}{{14}}\; = \;\frac{{10}}{7} \end{array})$

$({\LEFT( {\frac{{4 + 2\sqrt 3 }}{{4 - 2\sqrt 3 }} + \;\frac{{3 - \sqrt 3 }}{{2 + 2\sqrt 3 }} - \frac{{11}}{2} - 5\sqrt 3 } \right)^{9999}})$

$( = {\left( {\left( {\frac{{4 + 2\sqrt 3 }}{{4 - 2\sqrt 3 }} \times \frac{{4 + 2\sqrt 3 }}{{4 + 2\sqrt 3 }}} \right) + \;\left( {\frac{{3 - \sqrt 3 }}{{2 + 2\sqrt 3 }} \times \frac{{2 - 2\sqrt 3 }}{{2 - 2\sqrt 3 }}} \right) - \frac{{11}}{2} - 5\sqrt 3 } \right)^{9999}})$

$( = {\left( {\frac{{16 + 12 + 16\sqrt 3 }}{{16 - 12}} + \frac{{6 - 6\sqrt 3 - 2\sqrt 3 + 6}}{{4 - 12}} - \frac{{11}}{2} - 5\sqrt 3 } \right)^{9999}})$

$( = {\left( {\frac{{28 + 16\sqrt 3 }}{4} + \frac{{12 - 8\sqrt 3 }}{{ - 8}} - \frac{{11}}{2} - 5\sqrt 3 } \right)^{9999}}{\rm{\;}})$

$( = {\left( {\frac{{28 + 16\sqrt 3 }}{4} - \frac{{12 - 8\sqrt 3 }}{8} - \frac{{11}}{2} - 5\sqrt 3 } \right)^{9999}})$

$( = {\left( {\frac{{44 + 40\sqrt 3 }}{8} - \frac{{11}}{2} - 5\sqrt 3 } \right)^{9999}})$

$( = {\left( {\frac{{11}}{2} + 5\sqrt 3 - \frac{{11}}{2} - 5\sqrt 3 } \right)^{9999}})$

0.06 × ? × 0.216 = 1.944

⇒ ? × 0.01296 = 1.944

⇒ ? = 1.944/0.01296 = 150

∴ ? = 150

The GIVEN expression is:

1301 ÷ 14.99 × 19.91 + 500.99 = ?

Here, 14.99 ≈ 15

19.91 ≈ 20

500.99 ≈ 501

Now, the expression will become:

? ≈ 1301 ÷ 15 × 20 + 501

⇒ ? ≈ 433.6 × 4 + 501

⇒ ? ≈ 1735 + 501

⇒ ? ≈ 2236

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,

⇒ (?)³ × 10 = 13230 ÷ (9.261 ÷ 7)

⇒ (?)³ × 10 = 13230 ÷ (1.323)

⇒ (?)³ × 10 = 10000

⇒ (?)³ = 1000

⇒ (?)³ = (10)³

Follow BODMAS rule to solve this question, as per the order is 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 CONTAINS 'Division' and 'Multiplication' are calculated,

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

25% of 100 = 25

45% of 90 = 40.5

60% of 100 = 60

5% of 100 = 5

⇒ (25 – 40.5 + 60) × 5

⇒ 44.5 × 5

⇒ 222.5

⇒ Given: $(a = 124,)$

$(\RIGHTARROW \;\sqrt[3]{{a\LEFT( {{a^2} + 3a + 3} \right) + 1}} = ?)$

⇒ We PUT the value of a in above equation

$(\Rightarrow \sqrt[3]{{\;124\left( {{{124}^2} + \left( {3 \times 124} \right) + 3} \right) + 1}})$

$(\Rightarrow \sqrt[3]{{124\left( {15376 + 372 + 3} \right) + 1}})$

⇒ 125

4082 ÷ 157 – 23

⇒ (4082/157) – 23 = 26 – 23 = 3

0.00092 ÷ 0.008

$(= \frac{{920}}{8} \times \frac{{{{10}^{ - 6}}}}{{{{10}^{ - 3}}}})$

= 115 × 10^-3

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.

Using BODMAS rule

3 – {16 ÷ 5 – (6 ÷ (7 - 2))}

⇒ 3 – {16 ÷ 5 – (6 ÷ (5))}

⇒ 3 – {(16/5) – (6/5)}

⇒ 3 – (10/5) = 3 - 2 = 1

GIVEN expression,

2434.56 + 6543.42 - 1298.42 - 1046.38 = ?

⇒ 2434.56 + 6543.42 - (1298.42 + 1046.38) = ?

⇒ 8977.98 - 2344.8 = ?

⇒ ? = 6633.18

LET initial PRICE of banana be Rs. x

Initial number of bananas in Rs. 18 = 18/x

New price of banana = x – ((100/3)/100) × x = 2x/3

New number of bananas in Rs. 18 = 18/(2x/3) = 27/x

27/x – 18/x = 3

x = 3

8796 × 223 + 8796 × 77 = 8796 × (223 + 77)[by DISTRIBUTIVE LAW]

= (8796 × 300) = 2638800.

$(\begin{array}{l} 6\FRAC{1}{2} - 2\frac{3}{6}\; = \;? - 1\frac{5}{{12}}\\ \Rightarrow \frac{13}{2} - \frac{{15}}{6} + \frac{{17}}{{12}} = ?\\ \Rightarrow ? = \frac{{13 \times 6 - 15 \times 2\; + \;17}}{{12}}\\ \Rightarrow ? = \frac{{65}}{{12}} = 5\frac{5}{{12}} \end{array})$

Rounding off the values in the GIVEN problem

⇒ (340% of 800) + (78% of 1100)

⇒ (340 × 8) + (78 × 11)

⇒ (340 × 8) + (78 × 11)

⇒ 2720 + 858

⇒ 3578

∴ Approximate value is 3578

$(\LEFT( {\sqrt {\sqrt[4]{{\sqrt[3]{{{3^8}}}}}} } \right)\left( {\sqrt {\sqrt[4]{{\sqrt[3]{{{3^8}}}}}} } \right) = \left( {\sqrt {\sqrt[4]{{\sqrt[3]{{{3^8}}}}}} } \right)\left( {\sqrt {\sqrt[4]{{\sqrt[3]{{{3^8}}}}}} } \right) = {\left( {\sqrt {\sqrt[4]{{\sqrt[3]{{{3^8}}}}}} } \right)^2} = {\left( {{{\left( {{3^8}} \right)}^{1/3}}} \right)^{1/4}} ={\left({3^{8/3}} \right)^{1/4}} = {3^{2/3}})$

The Given expression,

⇒ $(? = \FRAC{{{{\left( {0.150 + 0.225} \right)}^2} - {{\left( {0.150 - 0.225} \right)}^2}\;}}{{0.150 \times 0.225}})$

USING,

(A + B)² = A² + 2AB + B²

(A - B)² = A² - 2AB + B²

⇒ $(? = \frac{{{{0.150}^2} + 2 \times 0.150 \times 0.225 + {{0.225}^2} - {{0.150}^2} + 2 \times 0.150 \times 0.225 - {{0.225}^2}}}{{0.150 \times 0.225}})$

⇒ $(? = \frac{{4 \times 0.150 \times 0.225}}{{0.150 \times 0.225}})$

$(\Rightarrow 111\frac{1}{2} + 111\frac{1}{6} + 111\frac{1}{{12}} + 111\frac{1}{{20}} + 111\frac{1}{{30}})$

⇒ 111 + ½ + 111 + 1/6 + 111 + 1/12 + 111 + 1/20 + 111 + 1/30

⇒ (111 + 111 + 111 + 111 + 111) + (1/2 + 1/6 + 1/12 + 1/20 + 1/30)

⇒ (555) + (30 + 10 + 5 + 3 + 2)/60

⇒ (555) + 50/60

⇒ 555 + 5/6

$(\Rightarrow 555\frac{5}{6})$

(1/4 of X) - (4/5 of 6/7) = -9/7

¼ × x – (4/5) × (6/7) = –9/7

x/4 = 24/35 – 9/7

x/4 = –21/35

LET the total marks be ‘X’.

A student who gets 190 marks failed by 35 marks.

⇒ 190 + 35 = (36x/100)

⇒ 225 = (36x/100)

⇒ x = (225 × 100)/36

⇒ x= 625

CONVERTING each FRACTION into decimal.

8/9 = 0.89

1/4= 0.25

Converting the options

1/5 = 0.2

23/24 = 0.96

11/12 = 0.92

17/24 = 0.71

∴ Only option 4 is less than 8/9 and GREATER than 1/4

Let the total number of LED lights in the box be ‘m’

% of LED lights in good CONDITION = (100 – 12)% = 88%

⇒ 88% of m = 66

⇒ m = (66 × 100)/88 = 75

On solving the equation by taking LCM and using BODMAS:

$(\BEGIN{array}{l} = 4\frac{1}{6} + 7\frac{1}{4} - 5\frac{7}{8}\\ = \frac{{25}}{6} + \frac{{29}}{4} - \frac{{47}}{8}\END{array})$

Lcm OF 6, 4 and 8 is 24

$(= \frac{{100 + 174 - 141}}{{24}})$

=133/24

ACCORDING to the GIVEN information,

5.991 ≈ 6

√439 ≈ 21

6² × 21

= 756

We know that

(a – B)³ = a³ – b³ – 3AB(a – b)

So, (11.998)³

⇒ (12 – 0.002)³

⇒ (12)³ – (0.002)³ – 3(12)(0.002)(12 – 0.002)

⇒ 1728 – 0.000000008 – 0.863856

Let the PRESENT AGE of X and Y are x and y respectively.

Sum of X and Y age 7 YEAR ago is 86

⇒ (x - 7) + (y - 7) = 86

⇒ x + y = 100-----(i)

X age 20 year ago is equal to Y age 4 year ago.

x - 20 = y - 4

⇒ x - y = 16-----(ii)

Solving the equation (i) and (ii)

Present age of X = 58 years

Present age of Y = 42 years

GIVEN,

⇒ 2^{3x + 1} × 4^{2y + 1} = 2^{z – X}

⇒ 2^{3x + 1} × 2^{2(2y + 1)} = 2^{z – x}

⇒ 2^{(3x + 1 + 4y + 2)} = 2^{z – x}

Equating the powers of 2,

⇒ 3x + 4y + 3 = z – x

⇒ z = 4x + 4y + 3

⇒ z = 4(x + y) + 3

Putting (x + y) = 1,

25² ÷ 10² × 4² – 3 × 5² = 5^?

⇒ 5⁴ ÷ (5² × 2²) × 2⁴ – 75 = 5^?

⇒ 5⁴/ (5² × 2²) × 2⁴ – 75 = 5^?

⇒ 5²/2² × 2⁴ – 75 = 5^?

⇒ 5² × 2² – 75 = 5^?

⇒ 100 – 75 = 5^?

⇒ 25 = 5^?

∴ ? = 2

((√3)⁶ + (√3)^-6)? = 3³ + 3^-3 = 3³ + 1/3³ = (3⁶ + 1)/3³ = (729 + 1)/27 = 730/27

GIVEN,

⇒ √12.96 × √7.84 ÷ √3.24 × √31.36 = X

⇒ X = 3.6 × 2.8 ÷ 1.8 × 5.6

⇒ X = 31.36

$(\begin{array}{L}\FRAC{5}{{28}} \div \frac{{28}}{{35}} \div \frac{{20}}{{112}}\; = \;\frac{5}{{28}} \div \frac{4}{5} \div \frac{5}{{28}}\; = \;\frac{{\frac{5}{{28}} \div \frac{4}{5}}}{{\frac{5}{{28}}}}\; = \;\frac{{\frac{{25}}{{112}}}}{{\frac{5}{{28}}}}\; = \;\frac{5}{4}\\\therefore \;\frac{5}{{28}} \div \frac{{28}}{{35}} \div \frac{{20}}{{112}}\; = \;\frac{5}{4}\end{array})$

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.

Given expression,

2(11.25 × 6.5) + (5 × 85.87) –$ (33.52 + 46.15)

⇒ 2(73.125) + 429.35 –$ 79.67

⇒ 146.25 + 429.35 –$ 79.67

⇒ 495.93 ≈ 496

∴ $? = 496

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

(5^1/4 – 1)(5^3/4 + 5^1/2 + 5^1/4 + 1)

⇒ (5^1/4 × 5^3/4) + (5^1/4 × 5^1/2) + (5^1/4 × 5^1/4) + 5^1/4 – 5^3/4 – 5^1/2 – 5^1/4– 1

⇒ 5 + 5^3/4 + 5^1/2 + 5^1/4 – 5^3/4 – 5^1/2 – 5^1/4– 1

⇒ 5 – 1

? = 8.97 – 4.05 + 9.02 ÷ 2.99 × 4.04 + (1.56)²

Rewriting equation with approximate values:

⇒ ? ≈ 9 – 4 + 9 ÷ 3 × 4 + (1.6)²

⇒ ? = 9 – 4 + 3 × 4 + 2.56

⇒ ? = 5 + 12 + 2.6

⇒ ? = 19.6 ≈ 20

∴ ? = 20