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Let, the number he added twice = X

Sum of all natural NUMBERS from 1 to 20,

⇒ 20 × (20 + 1)/2

⇒ 10 × 21

⇒ 210

According to problem,

⇒ 210 + x = 215

⇒ x = 5

Let the other number be X

Now

LCM × HCF = Product if TWO numbers

As LCM = 6 × HCF

∴ 6 × HCF × HCF = 6 × X

∴ HCF² = X

Now as HCF is SQUARE root of other number

∴ From options we can say that the answer is 9

We can also check that the LCM of 9 and 6 is 18 and HCF of 9 and 6 is 3

Prime number: A natural number LARGER than unity is a prime number if it does not have other divisors except for itself and unity.

As PROPERTIES of prime number (p):

p > 3, p² - 1 is COMPLETELY DIVISIBLE by 24.

From options:

1) 61

⇒ (61² - 1)/24 = (3721 - 1)/24 = 155

2) 71

⇒ (71² - 1)/24 = (5041 - 1)/24 = 210

3) 69

⇒ (69² - 1)/24 = (4761 - 1)/24 = 198.33

4) 67

⇒ (67² - 1)/24 = (4489 - 1)/24 = 187

4X - 5(2x - 1) > 2x + 3

⇒ 4x – 10x + 5 > 2x + 3

⇒ 5 > 8x + 3

x < 1/4

2x + 3 > 2 - 3X

5x > - 1

x > - 1/5

∴ x = 0

$(\FRAC{1}{{\sqrt 2+ \sqrt 3 }} + \frac{1}{{\sqrt 3+ \sqrt 4 }} + \frac{1}{{\sqrt 4+ \sqrt 5 }} + \frac{1}{{\sqrt 5+ \sqrt 6 }})$

$(= \frac{{(\sqrt 2- \sqrt {3)} }}{{\left( {\sqrt 2+ \sqrt 3 } \right)(\sqrt 2- \sqrt {3)} }} + \frac{{\left( {\sqrt 3- \sqrt 4 } \right)}}{{\left( {\sqrt 3+ \sqrt 4 } \right)\left( {\sqrt 3- \sqrt 4 } \right)}} + \frac{{(\sqrt 4- \sqrt {5)} }}{{(\sqrt 4+ \sqrt {5)} (\sqrt 4- \sqrt {5)} }} + \frac{{(\sqrt 5- \sqrt {6)} }}{{(\sqrt 5+ \sqrt {6)} (\sqrt 5- \sqrt {6)} }})$

[Dividing the NUMERATOR & DENOMINATOR of each fraction by same term]

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

$(= \frac{{(\sqrt 2- \sqrt {3)} }}{{ - 1}} + \frac{{\left( {\sqrt 3- \sqrt 4 } \right)}}{{ - 1}} + \frac{{(\sqrt 4- \sqrt {5)} }}{{ - 1}} + \frac{{(\sqrt 5- \sqrt {6)} }}{{ - 1}})$

= - (√2 - √3) - (√3 - √4) - (√4 - √5) - (√5 - √6)

= √3 - √2 + √4 -√3 + √5 - √4 + √6 - √5

= √6 - √2

= √3 × √2 - √2

$(\frac{1}{X} + \frac{1}{y} = \frac{{x + y}}{{xy}} = \frac{{35}}{{xy}})$

Using the known identity: product of TWO NOS. = LCM × HCF,

we have:

xy = 7 × 42

LCM of 5,6,7,9 is 630

Check STEP 1:630*k +4 

Step 2: 630*2+4 =1264 and divide by 8 ie) 1264 /8 = 158 so the remainder is zero ,hence the answer is 1264. Or move from option which ONE is divisible by the GIVEN divisor.

Factors of 480 = 2⁵ × 3 × 5

Sum of Factors = (2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵) × (3⁰ + 3¹) × (5⁰ + 5¹)

$(\therefore {\rm{}}\left( {\sqrt 3- \FRAC{{10}}{{\sqrt 3 }} + \sqrt {27} } \right) = \frac{{3 - 10}}{{\sqrt 3 }} + 3\sqrt 3 = \frac{{ - 7 + 9}}{{\sqrt 3 }} = \frac{2}{{\sqrt 3 }} = \frac{{2\sqrt 3 }}{3} = 2 \times \frac{{1.732}}{3} = 1.154)$

N = 9⁹ = 3¹⁸

Positive perfect cubes that can DIVIDE N are 1, 3³, 3⁶, 3⁹, 3¹², 3¹⁵, 3¹⁸

LET The number of toffees with P be ‘x’

∴ S has x/2 toffees

And, Q has 4/5 × (P + S) =(4/5)(x + x/2) = 6x/5 toffees

∴ R has 2/3 ×(P + Q + S) = 2/3 × (x + x/2 + 6x/5) = 2/3 × (3x/2 + 6x/5) = 9x/5 toffees

∴ x + x/2 + 6x/5 + 9x/5 = 135

∴ (10x + 5x + 12x + 18x) = 1350

∴ 45X = 1350

∴ x = 30

LCM = (LCM of 2, 8, 16, 32)/(HCF of 3, 9, 27, 81) = 32/3

LET suppose first no as x,

So the series is LIKE,

⇒ x, x + 1, x + 2, ....., x + 5

⇒ Sum of six NUMBER = 483

⇒ x + x + 1 + x + 2 + x + 3 + x + 4+ x + 5 = 483

⇒ 6x + 15 = 483

$(\Rightarrow x = 78)$

⇒ Fourth no. = x + 3 = 81

⇒ Sixth no. = x + 5 = 83

⇒ Sum = 81 + 83 = 164

∴ Sum of fourth and sixth number is 164

The greatest number that will DIVIDE 149, 247 and 622 LEAVING remainder 5, 7 and 10

Factor of (149 – 5) = 144 = 12 × 12

Factor of (247 – 7) = 240 = 12 × 20

Factor of (622 – 10) = 612 = 12 × 51

The largest 4 digit NUMBER is 9999

When 9999 is divided by 81, a remainder of 36 is obtained

∴ The largest 4 digit number EXACTLY divisible by 81 = 9999 - 36 = 9963

24b⁶c⁸d² = 2 × 2 × 2 × 3 × b² × b⁴ × c² × c⁶ × d²

18a⁶c²d⁴ = 2 × 3 × 3 × a² × a⁴ × c² × d² × d²

12a⁴b⁴ = 2 × 2 × 3 × a⁴ × b⁴

2/3, 3/4, 4/5, 1/2

⇒ 40/60, 45/60, 48/60, 30/60

Arranging in ASCENDING ORDER?

⇒ 30/60, 40/60, 45/60, 48/60

As, 5 × 2 will give a zero and nearest multiple of of 5 below 129 is 125

Number of multiples of 5 in 129! = 125/5 = 25

Number of multiples of 5² ie 25 in 129! = 125/25 = 5

Number of multiple of 5³ ie 125 in 129! = 125/125 = 1

LCM of 15, 18, 27, 30 = 2 × 3 × 3 × 3 × 5 × 1 × 1 × 1 = 270

After dividing 999 by 270, we get

Remainder = 189

∴ Answer = 999 - 189 = 810

Here is losing total of 67 MARKS in test (from the all CORRECT situation)

If he answers a question wrongly, he should drop by 1.25 marks

If he leaves a question unanswered, he is losing 1.125 marks

Going through OPTIONS, we find that

For option 1) 

⇒ Marks lost by questions = 2.5

⇒ Marks required to be lost by unanswered ques. = 64.5

But 64.5 / 1.125 is not an integer.

Hence, it is not possible

For option 2)

⇒ Marks lost by question = 6.25

⇒ Marks required to be lost by unanswered ques. = 60.75

Here, 60.75 / 1.125 is an integer.

Hence, it is possible

Other two options ALSO doesn’t give an integer.

∴ The minimum number of WRONG answers is 5

OPTION A? 0.5

Option B? 110.5

Options C? 0.5 × 0.5 = 0.25

Option D? 0.5 × 2 = 1

Zero is NEITHER positive nor NEGATIVE. HENCE, it is included in both the sets of non-negative and non-positive INTEGERS.

LET TWO consecutive number be x and x + 1

According to QUESTION,

⇒ 3x - 2(x + 1) = 5

⇒ 3x - 2x - 2 = 5

⇒ x = 7

$(\BEGIN{ARRAY}{l} \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + B\sqrt 6 \\ \RIGHTARROW \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {16 \times 3} + \sqrt {9 \times 2} }} = a + b\sqrt 6 \\ \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \\ \Rightarrow \frac{{4\sqrt 3 + 3\sqrt 2 + 2\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \\ \Rightarrow 1 + \frac{{2\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \end{array})$

Multiplying and dividing $(\frac{{2\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }})$ by $(4\sqrt 3 - 3\sqrt 2)$

$(\Rightarrow 1 + \frac{{2\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} \times \frac{{4\sqrt 3 - 3\sqrt 2 }}{{4\sqrt 3 - 3\sqrt 2 }} = a + b\sqrt 6)$

$(\begin{array}{l} \Rightarrow 1 + \frac{{8\sqrt 6 - 6 \times 2}}{{48 - 18}} = a + b\sqrt 6 \\ \Rightarrow 1 + \frac{{8\sqrt 6 - 12}}{{30}} = a + b\sqrt 6 \\ \Rightarrow 1 + \frac{{4\sqrt 6 }}{{15}} - \frac{2}{5} = a + b\sqrt 6 \\ \Rightarrow \frac{3}{5} + \frac{{4\sqrt 6 }}{{15}} = a + b\sqrt 6 \end{array})$

Comparing both sides we have,

Let us TRY to make sense of the DATA given,

A₁ = 2 = 2 × 1

A₂ = 22 = 2 × 11

A₃ = 222 = 2 × 111

.

.

.

.

A_n = 2222…..(n times) = 2 × 1111…..(n times)

Looking at the options given, the first and the second are straightaway ruled out, because they do not GIVE us the required answer.

Looking at the fourth option, let us try n = 2,

$({\left( {\frac{2}{9}} \right)^n}\left( {{{10}^n} - 1} \right) = \frac{4}{{81}}\left( {100 - 1} \right) = \frac{{44}}{9})$

This is also not the right answer.

Now, look at option 3, and trying for n = 3,

$(\frac{2}{9}\left( {{{10}^n} - 1} \right) = \frac{2}{9}\left( {1000 - 1} \right) = 222)$

Time taken by three PHONES to RING TOGETHER = LCM of 20, 24 and 30

⇒ Time taken by three phones to ring together = 120 seconds

(? 1 min = 60 seconds)

⇒ Time taken by three phones to ring together = 120/60 = 2 min

If the three phones ring together at 11:25 am then they will next ring together after 2 minutes i.e. at 11:27 am.

The largest NUMBER which when divides 364, 453 and 548 leaves remainders as 8, 8 and 14 respectively.

= HCF (364 - 8, 453 - 8, 548 - 14)

= HCF (356, 445, 534)

= 89

Let the QUOTIENT and the dividend be ‘q’ and ‘X’ respectively

⇒ x = 256q + 141

⇒ x = (16 × 16)q + (16 × 8) + 13

⇒ x = 16(16q + 8) + 13

∴ When the NUMBER is DIVIDED by 16, the remainder is 13

⇒ DIVISOR = (DIVIDEND – REMAINDER)/Quotient

⇒ Divisor = (13501 – 7)/78 = 13494/78

∴ Divisor = 173

Dividing 80 by 7 and 11

⇒ Remainder when DIVIDED by 7 = 3

⇒ Remainder when divided by 11 = 3

Dividing 73 by 7 and 11

⇒ Remainder when divided by 7 = 3

⇒ Remainder when divided by 11 = 7

Dividing 93 by 7 and 11

⇒ Remainder when divided by 7 = 2

⇒ Remainder when divided by 11 = 5

Dividing 150 by 7 and 11

⇒ Remainder when divided by 7 = 3

⇒ Remainder when divided by 11 = 7

LETS factorize each one them

420 = 2 × 2 × 3 × 5 × 7

525 = 3 × 5 × 5 × 7

500 = 2 × 2 × 5 × 5 × 5

Hence we can see only number common two all three is 5.

SUM of first 9 terms can be given as

⇒ sum = 9(7 + 55)/2 = 9 × 31 = 279

∴ the sum of first 9 terms is 279

Let a and B be the TWO numbers

We know that,

a × b = L.C.M × H.C.F

⇒ 504 = 168 × H.C.F

⇒ H.C.F = 504/168 = 3

Greatest three DIGIT number = 999

To find the divisible number, LCM of three number is require,

Factors of 12 = 3 × 4 = 2² × 3

Factors of 30 = 6 × 5 = 2 × 3 × 5

Factors of 50 = 25 × 2 = 2 × 5²

LCM of (12, 30, 50) = 2² × 3 × 5² = 300

On dividing 999 by 300, the REMAINDER 99,

FACTORS of 256 = 1, 2, 4, 8, 16, 32, 64, 128, 256

And the factors of 256 which are perfect squares = 1, 4, 16, 64 and 256

Now, in the above EXPRESSION, 651 is COMPLETELY divisible by 7

⇒ 651/7 = 93

LET the no of votes casted be X

Required votes = 2x/3

Counted votes = x/5

Uncounted votes = x - (x/5) = 4x/5

Votes won by candidate = (5/18) of (2x/3) = 5x/27

Remaining votes required = (2x/3) – (5x/27) = 13x/27

We all know that largest 5 digit number = 99999

LCM of 8, 9, 10 and 12 = 360

99999 ÷ 360 = Remainder = 279

5 digit largest number DIVISIBLE by 8, 9 , 10 and 12 = 99999 – Remainder

5 digit largest number divisible by 8, 9, 10 and 12 = 99999 – 279 = 99720

∴ Required number = 99720 + 3 = 99723

LET the DIVISOR be X

QUOTIENT = x/50

Remainder = x/10

x/50 = 32 ⇒ x = 1600

The number/dividend is (Divisor × Quotient) + Remainder

Number = x²/50 + x/10

The MINIMUM no. of GUAVAS required is the LCM of 10, 13 and 52.

10 = 2 × 5

13 = 1 × 13

52 = 2 × 2 × 13

CHECK through OPTIONS:

⇒ 9999/19 = 526.26

⇒ 9989/19 = 525.73

⇒ 9994/19 = 526

⇒ 9981/19 = 525.31

Let the NUMBER be ‘x’.

Given, 3X + 2/x = 97/4

⇒ 3x² + 2 = 97x/4

⇒ 12X² + 8 = 97x

⇒ 12x² – 97x + 8 = 0

⇒ 12x² – 96x – x + 8 = 0

⇒ 12x(x – 8) – 1(x – 8) = 0

⇒ (x – 8)(12x – 1) = 0

⇒ x = 8 or x = 1/12

? All the options are whole numbers

Considering each statement one by one:

Statement I:

Multiples of 9 between 7 and 109 are 9, 18, 27, …., 108

⇒ The above series is an AP with FIRST term = 9 and common difference = 9 and last term = 108

In an AP, A_n = a + (n - 1) × d

$(\Rightarrow 108 = 9 + \left( {n - 1} \right)9 = 9 + 9n - 9)$

⇒ 108 = 9n

⇒ n = 108/9 = 12

∴ Statement I is correct.

Statement II:

Multiples of 13 from 19 to 119 are 26, 39, …., 117

The above series is an AP with first term = 26 and common difference = 13 and last term = 117

In an AP, A_n = a + (n - 1) × d

⇒ $(117 = 26 + \left( {n - 1} \right)13 = 26 + 13n - 13)$

⇒ 117 - 13 = 13n

⇒ 13n = 104

⇒ n = 104/13 = 8

GIVEN, 1 + 2 + 3 + … + N $(= \frac{{N\left( {N + 1} \right)}}{2})$

Now, 1 + 3 + 5 + ….. + 99

= (1 + 2 + 3 + 4 + …….. + 100) – (2 + 4 + 6 + 8 + 10 ….. + 100)

$(\begin{array}{l} = \frac{{100\left( {100 + 1} \right)}}{2} - 2\left( {1 + 2 + 3 + 4 \ldots+ 50} \right)\\ = 5050 - 2 \times \frac{{51 \times 50}}{2}\END{array})$

= 5050 – 2550

LET 2³⁷ be X

⇒ 2³⁷ + 1 = x +1

Let x + 1 be completely DIVISIBLE by the number, N

⇒ (2¹¹¹ + 1) = (2³⁷)³ + 1

⇒x³ + 1³ = (x+1)(x² – x + 1)

If x + 1 is completely divisible by N then (x+1)(x² – x + 1) is also divisible by N

Let the NUMBER be 45x + 2

Required remainder = Remainder obtained by dividing [15(3x) + 2] by 15 is 2