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As known,

Product of two numbers = HCF × LCM

Given,

⇒ (X - 7) × (x² + 8x + 12) = (x + 6) × LCM

⇒ (x - 7) × (x + 6)(x + 2) = (x + 6) × LCM

⇒ (x - 7)(x + 2) = LCM

⇒ (x² + 2x - 7x - 14) = LCM

Let the two numbers be 4x and 5x respectively.

The HCF of 4x and 5x is x.

⇒ x = 13

We know that,

Product of two numbers x and y = LCM(x, y) × HCF(x, y)

⇒ 4 × 13 × 5 × 13 = LCM × 13

⇒ LCM = 260

$(3 + \FRAC{1}{3} + 33 + \frac{1}{3}...\:333333 + \frac{1}{3})$

⇒ 3 + 33 + 333 + 3333 + 33333 + 333333 + $(\frac{1}{3} \times 6)$ 

As per the given data,

HCF of 126 cm, 198 cm and 162 cm is the greatest length which can be USED to exactly measure three cloth pieces.

Factors of 126 = 1, 2, 3, 6, 7, 9, 12, 14, 18, 21, 42, 63, 126

Factors of 198 = 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198

Factors of 162 = 1, 2, 3, 6, 9, 18, 27, 54, 81, 162

⇒ HCF of 126, 198 and 162 = 18 cm

∴ Greatest length which can be used to measure exactly three cloth pieces of lengths 1.26 m, 1.98 m and 1.62 m is 18 cm

⇒ (√16) × (4)² = 4 × 16 = 64

Smallest and largest 3 digit number divisible by 13 is 104 and 988.

SUM = 104 + 117 + 130 + 143 + …………+ 988

Using sum of arithmetic series,

$(\frac{n}{2}\left( {FIRST\;TERM + last\;term} \right))$ 

We can write,

988 = 104 + (n -1)13 [nth term = a + (n - 1)d]

$(\Rightarrow {\rm{\;n}} = \frac{{884}}{{13}} + \;1 = 68 + 1 = 69)$

From given data,

Factors of 16 are 1, 2, 4, 8, 16

Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24

∴ Highest common FACTOR (HCF) is 8

LCM of 16, 24 = 2 × 2 × 2 × 2 × 3 = 16 × 3 = 48

⇒ 1^st number = 1125

⇒ LAST number = 4950

⇒ Number of TERMS = (4950 – 1125)/225 + 1

⇒ 3825/225 + 1

⇒ 17 + 1 = 18

Let us ASSUME he needs X chocolate and y fruits more to distribute equal number of chocolate and fruits

Then total chocolate = 4630 + x

And total fruits = 2364 + y

Number of chocolate left after saving two chocolates for himself = 4630 + x – 2 = 4628 + x

Number of chocolate left after saving one fruit for himself = 2364 + y – 1 = 2363 + y

As we have already took away 2 chocolates and 1 fruit for Atif now after distributing remaining things among 23 children no chocolate or fruit should left, i.e.

⇒ REMAINDER of (4628 + x)/23 = 0

And remainder of (2363 + y)/23 = 0

We KNOW that,

If N = A+ B + C…, then the remainder when N is divided by X is equal to the sum of the remainders when A, B, C ... are divided by X.

⇒ Remainder of (N/X)= Remainder of (A/X) + Remainder of (B/X) + Remainder of (C/X)……

∴Remainder of (4628 + x)/23 = remainder of 4628/23 + remainder of x/23 = 0

⇒ 5 + remainder of x/23 = 0

⇒ Remainder of x/23 = –5

Here negative remainder indicates that x is 5 shorter than 23 and minimum value of such x is 23 – 5 = 18

Similarly, ∴Remainder of (2363 + y)/23 = remainder of 2363/23 + remainder of y/23 = 0

⇒ 17 + remainder of y/23 = 0

⇒ Remainder of y/23 = – 17

Here negative remainder indicates that y is 17 shorter than 23 and minimum value of such y is 23 – 17 = 6

Using the standard identity that the sum of first N CUBES is $({\left[ {\FRAC{{n\left( {n + 1} \right)}}{2}} \right]^2})$, we have

On DIVIDING each option by 7, only 999964 is divisible by 7.

HCF of (115 – 3, 149 – 5, 183 – 7)

⇒ HCF of (112, 144, 176) = 16

∴ 16 is greatest number that will DIVIDE 115, 149 and 183 leaving reminder 3, 5, 7 respectively

Given data-

4 bells ring at INTERVALS of 30 MINUTES, 1 hour, $(1\frac{1}{2})$ hour and 1 hour 45 minutes respectively

Converting them into fractions

30min = 1/2 hour

1 hour = 1/1 hour

$(1\frac{1}{2})$ Hour = 3/2 hour

1 hour 45 minutes = 7/4 hour

To find NEXT simultaneous ring of 4 bells

Let’s find L.C.M. of these fractions:

L.C.M of fractions = (L.C.M of numerators)/(H.C.F of denominators)

⇒ LCM of fractions $(= {\rm{\;}}\frac{{{\rm{L}}.{\rm{C}}.{\rm{M\;of\;}}\left( {1,1,3,7} \right)}}{{{\rm{H}}.{\rm{C}}.{\rm{F\;of}}\left( {2,1,2,4} \right)}})$

⇒ LCM of fractions = 21/1 = 21

∴ Bells will ring simultaneously after 21 HOURS.

Counting 21 hours after 12 noon.

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

270 sec = 270/60 = 4 min 30 sec = 0 : 04 : 30

∴ They will again change simultaneously at 6 : 10 : 00 + 0 : 04 : 30 = 6 : 14 : 30 hours

Formula to be used:-

(a + b)² = a² + b² +2ab

We have √6 + √2 and √5 + √3

Squaring both of them, we get

(√6 + √2)² = 6 + 2 + 2√12

= 8 + 2√12

And (√5 + √3)² = 5 + 3 + 2√15

= 8 + 2√15

Thus (√6 + √2)² – (√5 + √3)² = 2(√12 - √15) < 0

∴ (√6 + √2)² – (√5 + √3)² < 0

⇒ (√6 + √2)² < (√5 + √3)²

For a² < b² we get

a < b if both a and b are positive

Thus (√6 + √2) < (√5 + √3).

20 = 2 × 2 × 5

28 = 2 × 2 × 7

108 = 2 × 2 × 3 × 3 × 3

105 = 3 × 5 × 7

We can WRITE, 2⁵⁵ = (2⁴)¹³ × 2³

⇒ Unit digit (2⁵⁵) = Unit digit (6¹³ × 8)

⇒ Unit digit (2⁵⁵) = Unit digit (6 × 8) = 8

Similarly,

We can write, 3²³ = (3⁴)⁵ × 3³

⇒ Unit digit (3²³) = Unit digit (1⁵ × 7)

⇒ Unit digit (3²³) = 7

Now,

Unit digit (2⁵⁵ + 3²³) = Unit digit (8 + 7) = 5

? When 5 is DIVIDED by 5, the remainder is 0

LCM of 42, 36 and 45 = 1260

This is the LEAST NUMBER which will be divisible by 42, 36 and 45.

If p & q = p² + 4PQ - q²

Put the value p = 3 and q = 6

⇒ 3 & 6 = 3² + 4 × 3 × 6 – 6² = 9 + 72 – 36 = 45

Similarly, we put p = 4 and q = 5

⇒ 4 & 5 = 4² + 4 × 4 × 5 – 5² = 16 + 80 – 25 = 71

let the HCF be K

∴ LCM = 8112k

Then, k + 8112k = 40565

⇒ 8113k = 40565

⇒ k = 5

∴ LCM =8112k = 8112 × 5 = 40560

⇒ First number × SECOND number = LCM × HCF

⇒ 240 × second number = 40560 × 5

The HCF of fractional NUMBER = HCF of numerator/LCM of denominator

The HCF of (6/19) and (16/57) $( = \frac{{HCF\;of\;6\;and\;16}}{{LCM\;of\;19\;and\;57}} = \frac{2}{{57}})$

We can WRITE,

⇒ 420 = 2² × 3 × 5 × 7

⇒ 490 = 2 × 5 × 7²

⇒ 630 = 2 × 3² × 5 × 7

? HCF is the product of the common factors and LCM is the product of the highest powers of all the factors

⇒ HCF of 420, 490 & 630 = 2 × 5 × 7 = 70

⇒ LCM of 420, 490 & 630 = 2² × 3² × 5 × 7² = 8820

⇒ LCM/HCF = 8820/70 = 126

SINCE the number is of 100 digits and 3446, 5887 and 2445 are put side by side in the same ORDER;

∴ 3446 must be repeated 9 times, 5887 & 2445 must be repeated 8 times;

∴ SUM of digits of 100 digits number = [3 + 4 + 4 + 6] × 9 + [5 + 8 + 8 + 7 + 2 + 4 + 4 + 5] × 8

⇒ 153 + 344

⇒ 497

Reminder when 497 is divided by 9 = 2

PRODUCT of TWO NUMBERS = HCF × LCM

⇒ 35828 = 26 × LCM

WEIGHT of GEETA = 11.235 KG

Her SISTER’s weight = 1.4 × 11.235 = 15.729 kg

⇒ Their COMBINED weight = 11.235 + 15.729 = 26.964 kg

Using the CONCEPT a – b = (√a - √b) (√a + √b)

MULTIPLY and dividing each fraction by its conjugate we get,

$(\frac{1}{{\sqrt 7- \sqrt 6 }} \TIMES \frac{{\sqrt 7+ {\rm{\;}}\sqrt 6 }}{{\sqrt 7+ {\rm{\;}}\sqrt 6 }} - \frac{1}{{\sqrt 6- \sqrt 5 }} \times \frac{{\sqrt 6+ {\rm{\;}}\sqrt 5 }}{{\sqrt 6+ {\rm{\;}}\sqrt 5 }} + \frac{1}{{\sqrt 5- 2}} \times \frac{{\sqrt 5+ {\rm{\;}}2}}{{\sqrt 5+ {\rm{\;}}2}} - \frac{1}{{\sqrt 8- \sqrt 7 }} \times \frac{{\sqrt 8 {\rm{\;}} + {\rm{\;}}\sqrt 7 }}{{\sqrt 8+ \sqrt 7 }} + \frac{1}{{3 - \sqrt 8 }} \times \frac{{3{\rm{\;}} + {\rm{\;}}\sqrt 8 }}{{3{\rm{\;}} + \sqrt 8 }})$

= √7 + √6 - √6 - √5 + √5 + 2 - √8 -√7 + 3 + √8

LCM of 3, 4, 5 and 6 = 2 × 3 × 2 × 5 × 1 × 1 = 60

On dividing 60 with 3105, we get remainder 45

∴ Number to be added = 60 - 45 = 15

Quantity 1:

LET the number be 10x + y and when you reversed the DIGIT, number becomes 10y + x.

Given: When you reverse the digits of a number, the number increases by 27

⇒ (10y + x) - (10x + y) = 27

⇒ (10y + x) - 10x - y = 27

⇒ 9y - 9x = 27

⇒ 9(y - x) = 27

⇒ y - x = 3

⇒ All the possible combination of y and x are = (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)

⇒ Possible numbers other than 14 are = 25, 36, 47, 58 and 69

∴ Quantity 1 = 5

Quantity 2:

Let the number is 10x + y and when you reversed the digit number becomes 10y + x

Given: When you reverse the digits of the number the number increases by 18

⇒ (10y + x) - (10x + y) = 18

⇒ (10y + x) - 10x - y = 18

⇒ 9y - 9x = 18

⇒ 9(y - x) = 18

⇒ y - x = 2

All the possible combination of y and x are = (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9).

Possible numbers other than 13 are = 24, 35, 46, 57, 68 and 79.

∴ Quantity 2 = 6

∴ We can see here that Quantity 2 > Quantity 1

Since the HCF TWO numbers is 45 and the RATIO of the two numbers is 2 ? 3,

⇒ Smaller number X = 2 × 45 = 90

Let the two numbers be 5x and 7x respectively.

LCM of 5x and 7x is 35x.

⇒ 35x = 105

⇒ x = 105/35 = 3

⇒ The SUM of the two numbers = 5 × 3 + 7 × 3 = 15 + 21 = 36

Difference of the two numbers = 21 – 15 = 6

X = 36/6 = 6

LET the two numbers be 7X and 9x respectively.

7x × 9x = 4032(GIVEN)

⇒ 63x² = 4032

⇒ x² = 64 = 8²

⇒ x = 8

The difference between the squares of these two numbers = (9x)² – (7x)² = 81x² – 49x² = 32x² = 32 × 64 = 2048

∴ The difference between the squares of these two numbers = 2048

SINCE we NEED to find how MANY chocolates left, we have to find the reminder when 4¹¹ is divided by 21;

⇒ 4¹¹/21 = (64 × 64 × 64 × 16)/21 = {(63 + 1)³ × 16}/21

⇒ Reminder = 16

We know that,

PRODUCT of two numbers X and y = LCM(x, y) × HCF(x, y)

⇒ 2522 = 97 × HCF

⇒ HCF = 2522/97 = 26

LCM of 34, 27 and 18 = 918

? 1 feet = 12 inches

So, Length of string = 8 feet 9 inches = (8 × 12) + 9 = 105 inches

 Now, as it is DIVIDED into 3 parts,

If we write all the given numbers in a form (10A + B), b will turn out to be the unit’s digit here.

So when we multiply all the given numbers, the unit’s digit in the PRODUCT will be nothing but the product of all unit’s digits, irrespective of what the other digits in the NUMBER are.

3 × 9 = 27 → 7

7 × 1 = 7 → 7

7 × 2 = 14 → 4

HCF of 2³ × 7² and 2⁵ × 3¹

Highest COMMON FACTOR = 2³

Let the number be x

From GIVEN data-

Number when divided by 5 leaves a REMAINDER 3

∴ x = 5q + 3→ dividend= divisor× quotient+ remainder

When SQUARE of number is divided by 5

∴ (5q+3)² ÷ 5

⇒ (25q² + 30q + 9) ÷ 5→ (a + b)² = a² + 2AB + b²

⇒ [5(5q² + 6q + 1) + 4] ÷ 5

We can see that 7 × 47 = 329,

So 0 must be added to 329 to MAKE it DIVISIBLE by 7

We know that,

Dividend = DIVISOR × QUOTIENT + Remainder

⇒ 14528 = Divisor × 83 + 3

⇒ 14528 - 3 = Divisor × 83

⇒ 14525/83 = Divisor

⇒ Divisor = 175

As per the given data,

LCM of 12, 15, 18 and 27 = 2 × 2 × 3 × 3 × 3 × 5 × 1 = 540

We NEED the remainders 10, 13, 16 and 25

⇒ 12 - 10 = 2, 15 - 13 = 2, 18 - 16 = 2, 27 - 25 = 2

∴ Difference is 2 in each case

⇒ We need to subtract 2 to from the LCM of 12, 15, 18, 27 = 540 - 2 = 538

We KNOW that (50)² = 2500 and (55)² = 3025 and the no. 291N is < (55)²

Also,

⇒ (54)² = 2916

∴ N = 6

It can be written as

⇒ 1 × 2⁶ + 1 × 2⁵ + 0 × 2⁴ + 1 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰ = 2⁶ + 2⁵ + 0 + 2³ + 2² + 2¹ + 1

⇒ 64 + 32 + 8 + 4 + 2 + 1 = 111

∴ The decimal value of 1101111 is 111

1/ 5 = 0.2

20% = 0.2

Concept-$(LCM\ of\ \left( {\frac{a}{b},\frac{c}{d},\frac{e}{F}, \ldots } \right) = \frac{{LCM\ of\ \left( {a,c,e, \ldots } \right)}}{{HCF\ of\ \left( {b,d,f, \ldots } \right)}})$

The smallest sum of money will be the LCM of 1.80, 9, 3.6 and 1.2 as LCM would be the NUMBER which will be multiple of all given four numbers.

i.e we have to FIND the LCM of 1.80, 9, 3.6 and 1.2

Or, we have to find the LCM of $(\left( {\frac{9}{5},\frac{9}{1},\frac{{18}}{5}\ and\ \frac{6}{5}} \right))$

$(\BEGIN{array}{l} \Rightarrow LCM\ of\ \left( {\frac{9}{5},\frac{9}{1},\frac{{18}}{5}and\frac{6}{5}} \right) = \frac{{LCM\ of\ \left( {9,9,18,6} \right)}}{{HCF\ of\ \left( {5,1,5,5} \right)}}\\ \Rightarrow LCM\ of\ \left( {\frac{9}{5},\frac{9}{1},\frac{{18}}{5}and\frac{6}{5}} \right) = 18\end{array})$

FACTORIZING each number,

⇒ 16 = 2 × 2 × 2 × 2

⇒ 40 = 2 × 2 × 2 × 5

⇒ 64 = 2 × 2 × 2 × 2 × 2 × 2

LET two NUMBERS be A and B respectively.

Given,

⇒ A × B = 144 × 2

And, A : B = 8 : 9

⇒ A = 9B/8

Solve above equation,

⇒ 9B/8 × B = 288

⇒ 9B² = 288 × 8

⇒ B = 16

⇒ A = 16 × 9/8 = 18

X⁴ – 1 = (X²)² – (1)²

⇒ (X² – 1) (X² + 1)

⇒ (X² – 1²) (X² + 1)

⇒ (X – 1) (X + 1) (X² + 1)...(1)

 X⁴ + 5X² + 4

⇒ X⁴ + X² + 4X² + 4

⇒ X² (X² + 1) + 4(X² + 1)

⇒ (X² + 1) (X² + 4)...(2)

Common factor from both the EQ. (1) and (2)

As,

100 × 100 = 10000

So, the square of any three-digit number is of 5 digit or more

⇒ 99 × 99 = 9801

The greatest number,which when divided by 494, 1256, 1568 LEAVES REMAINDERS 14, 16and 8 is the greatest number,which when divided by 494 - 14, 1256 - 16, 1568 - 8 leaves remainders 0 in each case.

∴ The greatest number is the HCF(494 - 14, 1256 - 16, 1568 - 8) = HCF(480, 1240, 1560) = 40