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(2) a rational number

Irrational number have non-terminating and non recurring decimal expansion.

(i) \(0.\overline{24}\)

Let x = \(0.\overline{24}\) = 0.24242424 … (1)

(Here period of decimal is 2, multiply equation (1) by 100)

100x = 24.242424 … (2)

(2) – (1)

100x – x = 24.242424… – 0.242424…

99x = 24

x = \(\frac{24}{99}\)

(ii) \(2.\overline{327}\)

Let x = 2.327327327… (1) 

(Here period of decimal is 3, multiply equation (1) by 1000) 

1000x = 2327.327… (2) 

(2) – (1) 

1000x – x = 2327.327327… – 2.327327… 

999x = 2325 

x = \(\frac{2325}{999}\)

(iii) -5.132

x = -5.132 = \(\frac{-5132}{1000}=\frac{-1283}{250}\)

(iv) \(3.1\bar{7}\)

Let x = 3.1777 … (1) 

(Here the repeating decimal digit is 7, which is the second digit after the decimal point, multiply equation (1) by 10) 

10x = 31.7777 … (2) 

(Now period of decimal is 1, multiply equation (2) by 10) 

100x = 317.7777… (3) 

(3) – (2) 

100x – 10x = 317.777… – 31.777…

90x = 286

x = \(\frac{286}{90}=\frac{143}{45}\)

(v) \(17.\overline{215}\)

Let x = 17.215215 … (1) 

1000x = 17215.215215 … (2)

(2) – (1) 

1000x – x = 17215.215215… – 17.215… 

999x = 17198 

x = \(\frac{17198}{999}\)

(vi) \(-21.213\bar{7}\)

Let x = -21.2137777… (1) 

10x = -212.137777… (2) 

100x = -2121.37777… (3) 

1000x = -21213.77777… (4) 

10000x = 212137.77777… (5) 

(Now period of decimal is 1, multiply equation (4) it by 10) 

(5) – (4) 

10000x – 1000x = (-212137.7777…) – (-21213.7777…) 

9000x = -190924

x = \(-\frac{190924}{9000}\)

Let my friend choosen 73; then my questions may be like this.

Q : Does it lie in the first 50 numbers ? 

A: No . 

Q : Does it lie between 50 and 60 ? 

A: No 

Q : Does it lie between 60 and 70 ? 

A: No 

Q : Does it lie between 70 and 80 ? 

A: Yes 

[then my guess would be “it is a number from 70 to 80] 

Q : Is it an even number ? 

A: No 

[my guess : it should be one of the numbers 71, 73, 75, 77 and 79]

Q : Is it a prime number ? 

A: Yes 

[my guess : it may be 71, 73 or 79] 

Q : Is it less than 75 ? 

A: Yes 

[my guess : it may be either 71 or 73] 

Q : Is it less than 72 ? 

A: No 

Then the number is 73. 

Therefore the strategy is → Use number properties such as even, odd, composite or prime to determine the number.

Given that log₅ \(\sqrt{625}\)

= log₅25 = logs₅5² 

= 2 log₅5 { ∵ log_axⁿ = n log_ax} 

= 2 .1 = 2

The rational numbers between 0.2 and 0.3 is 0.25 and 0.28.

It means 25/100, 28/100 = 1/4, 7/25

Two irrational numbers between 0.5 and 

0.55 are (0.52515345 ……………….. and 

0.541656475 …………….. ).

\(15\sqrt[5]{32} + \sqrt{225} - 8\sqrt[3]{343} + \sqrt[4]{81}\)

\(15\sqrt[5]{2^5} + \sqrt{15^2} - 8\sqrt[3]{7^3} + \sqrt[4]{3^4}\)

= (15 × 2) + 15 – (8 × 7) + 3 

= 30 + 15 – 56 + 3 = – 8.

For the correct process of division. 

For expressing = 0.5222222 …………

= \(0.5\overline2\)

Given:

The size of the bathroom is 10 ft. by 8 ft.

To find:

the size in inches of the tile required that has to be cut and how many such tiles are required?

Solution:

To find the largest size of tile, we should find HCF of 10 and 8

Prime factors of 10 = 2 × 5

Prime factors of 8 = 2 × 2 × 2

Therefore HCF of 10 and 8 is: 2 ft

Since i ft = 12 inches

Therefore the largest size of tile is:

2 × 12 inches = 24 inches

Area of bathroom = length of bathroom × breadth of bathroom

= 10 × 8

= 80 sq. ft

Area of 1 tile = length of tile × breadth of tile

= 2 × 2

= 4 sq. ft

Number of tiles = \(\frac{area\,of\,bathroom}{area\,of\,one\,tile}\)

= \(\frac{80}{4}\)

= 20 tiles

NOTE:

Always find the HCF of the given values to find their maximum.

Solution :

i) 140

Take  LCM of 140 i.e  2×2×5×7×1

Hence,  140=2×2×5×7×1

ii) 156

Take LCM of 156 i.e 2×2×13×3×1

Hence, 156=2×2×13×3×1

iii)3825

Take  LCM of 3825 i.e 3×3×5×5×17×1

Hence, 3825=3×3×5×5×17×1

iv)5005

Take LCM of 5005 i.e 5×7×11×13×1

Hence, 5005=5×7×11×13×1

v)7429

Take LCM of 7429 i.e 17×19×23×1

Hence, 7429=17×19×23×1

Solution:

Given : mⁿ =  32  , So

mⁿ = 2⁵                              ( As we know 2⁵ =  32 )

By comparing both side we get

m  = 2 and  n = 5  , So
Value of n^m , So
⇒5² 
⇒25

Solution :

i) 26 and 91

26=2×13×1(expressing as product of it’s prime factors)

91=7×13×1(expressing as product of it’s prime factors)

So, LCM(26,91)=2×7×13×1=182

HCF(26,91)=13×1=13

Verification:

LCM×HCF=13×182=2366

Product of 26 and 91 =2366

Therefore,LCM×HCF=Product of the two numbers .

i) 510 and 92

510=2×3×17×5×1(expressing as product of it’s prime factors)

92=2×2×23×1(expressing as product of it’s prime factors)

So,

LCM(510,92)=2×2×3×5×17×23=23,460

HCF(510,92)=2

Verification:

LCM×HCF=23,460×2=46,920

Product of 510 and 92 =46,920

Therefore,LCM×HCF=Product of the two numbers .

iii) 336 and 54

336=2×2×2×2×7×3×1(expressing as product of it’s prime factors)

54=2×3×3×3×1(expressing as product of it’s prime factors)

So,

LCM(336,54)=2⁴×3³×7=3024

HCF(336,54)=2×3=6

Verification:

LCM×HCF=3024×6=18,144

Product of 336 and 54=18,144

Therefore,LCM×HCF=Product of the two numbers .

Solution:

The product of two numbers = LCM x HCF

= 4*9696= 38784

The prime factorization of 144 is as follows:

144 = 2 × 2 × 2 × 2 × 3 × 3

⇒ 144 = 2⁴ × 3²

We know that the exponent of a number a^m is m.

∴ The exponent of 2 in the prime factorization of 144 is 4.

If any number ends with the digit 0, it should be divisible by 10,

i.e. it will be divisible by 2 and 5.

Prime factorization of n is given as 2³ × 3⁴ × 5⁴ × 7.

It can be observed that there is (2 × 5) × (2 × 5) × (2 × 5)

⇒ 10 × 10 × 10 = 1000

Thus, there are 3 zeros in n.

Let x be \(0.\bar4\) 

x = \(0.\bar4\)......(1)

Multiplying both sides by 10, we get 

10x = \(4.\bar4\).......(2)

Subtracting (1) from (2), we get

10x – x = \(4.\bar4\) = \(0.\bar4\)

⇒ 9x = 4

⇒ x = \(\frac{4}9\)

Thus, simplest form of \(0.\bar4\) as a rational number is \(\frac{4}9\).

Given: LCM of two numbers = 1200

HCF should divide LCM exactly.

Using Euclid’s division lemma – a = bq + r. where q is the quotient, r is the remainder and b is the divisor.

Let us say, a = 1200 and b = 500.

If HCF divides LCM completely, then remainder is zero.

Here 1200 = 500(2) + 200

r = 200 ≠ 0

Correct option is (A) 600

HCF of two numbers is always a divisor of their LCM. Among all given numbers only 600 is a divisor of 1200.

\(\therefore\) HCF of both numbers can be 600.

Correct option is A) 600

We know that LCM of two or more numbers is always divisible by their HCF.

1200 is divisible by 600, 200 and 400 but not by 500.

If the LCM of two numbers is 1200 then, it is not possible to have their HCF equals to 500. Since, HCF must be a factor of LCM, but 500 is not a factor of 1200. 

If the LCM of two numbers is 1200 then, it is not possible to have their HCF equals to 500. 

Since, HCF must be a factor of LCM, but 500 is not a factor of 1200.

If two numbers are relatively prime then their greatest common factor will be 1. 

∴ HCF (a,b) = 1 

Using the formula, Product of two numbers = HCF × LCM 

we conclude that, 

a × b = 1 × LCM 

∴ LCM = ab 

Thus, LCM (a,b) is ab.

We know that LCM = Product of the greatest power of each prime factor, involved in the numbers.

If a and b are two prime numbers, then a × b is their LCM.

If two numbers are relatively prime then their greatest common factor will be 1. 

Thus, HCF (a, b) = 1

Given, we need to express the H.C.F. of 468 and 222 as 468 x + 222 y where x, y are integers in two different ways.

So, here the integers are: 468 and 222, and 468 > 222 

Then, by applying Euclid’s division lemma, we get 

468 = 222 x 2 + 24……… (1) 

Since the remainder ≠ 0, so apply division lemma on divisor 222 and remainder 24 

222 = 24 x 9 + 6………… (2) 

Since the remainder ≠ 0, so apply division lemma on divisor 24 and remainder 6 

24 = 6 x 4 + 0……………. (3) 

We observe that remainder is 0. 

So, the last divisor 6 is the H.C.F. of 468 and 222 

Now, in order to express the HCF as a linear combination of 468 and 222, we perform 

6 = 222 – 24 x 9 [from (2)] 

= 222 – (468 – 222 x 2) x 9 [from (1)] 

= 222 – 468 x 9 + 222 x 18 

6 = 222 x 19 – 468 x 9 

= 468(-9) + 222(19) 

∴ 6 = 468 x + 222 y, where x = -9 and y = 19.

44 = 2 x 2 x 11= 2² x 11

and 99 = 3 x 3 x 11 = 3² x 11

Now product of prime factors of both numbers = 11

(H.C.F.) = 11.

Answer is (C) 12 and 30

(i) HCF of 10 and 36 = 2

(ii) HCF of 24 and 15 = 3

(iii) HCF of 12 and 30 = 6

(iv) HCF of 18 and 20 = 2.

So, L.C.M. x H.C.F. = 60 x 6 = 360

First number x second number 

= 12 x 30 = 360.

Answer is (d) 4

Since two numbers  are and 18

H.C.F. = 2

L.C.M. = 36

Product of both = H.C.F. x L.C.M.

a x 18 = 2 x 36

a = 4

Given numbers are 250, 175 and 425

∴ 425 > 250 > 175

On applying Euclid’s division lemma for 425 and 250, we get

425 = 250 × 1 + 175

Here, r = 175 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 250 and new divisor 175, we get

250 = 175 × 1 + 75

Here, r = 75 ≠ 0

So, on taking 175 as dividend and 75 as the divisor and again we apply Euclid’s division lemma, we get

175 = 75 × 2 + 25

Here, r = 25 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 75 and new divisor 25, we get

75 = 25 × 3 + 0

Here, r = 0 and divisor is 25.

So, HCF of 425 and 225 is 25.

Now, applying Euclid’s division lemma for 175 and 25, we get

175 = 25 × 7 + 0

Here, remainder = 0

So, HCF of 250, 175 and 425 is 25.

We know that LCM divides by HCF.

But LCM 175 does not divide by HCF 15.

Hence, HCF and LCM of two numbers can not 15 and 175 respectively

Since, 15 does not divide 175 and

LCM of two numbers should be exactly divisible by their HCF

.'. Two numbers cannot have their HCF as 15 and

LCM as 175

Correct answer is (C) 2m

Solution:

We know that, even integers are 2, 4, 6 …

Where,

m = integer 

[Since, here integer is represented by m]

=⋯,−1,0,1,2,3,…

∴ 2=⋯,−2,0,2,4,6,…

7ⁿ = (1 x 7)ⁿ = 1ⁿ x 7ⁿ 

So  the only prime in the factorization of 7ⁿ is 7, not 2 or 5.

:.  7ⁿ can not end with the digit zero.

The largest number divides 438 and 606, leaving remainder 6 is the largest number divides 432 (i.e. 438 – 6 = 432) and 600 (i.e. 606 – 6 = 600)

HCF of 432 and 600 gives the required number.

To find: HCF of 432 and 600

Find HCF of 432 and 600 using prime factorization

432 = 2⁴ × 3³

600 = 2³ × 3 × 5²

HCF = 2³ × 3 = 24

Therefore, the required number is 24.

By applying Euclid’s division lemma on 1288 and 575, we get 

1288 = 575 x 2+ 138………… (1) 

As the remainder ≠ 0, apply division lemma on divisor 506 and remainder 143 

575 = 138 x 4 + 23……………. (2) 

As the remainder ≠ 0, apply division lemma on divisor 143 and remainder 77 

138 = 23 x 6 + 0……………….. (3) 

Thus, we can conclude the H.C.F. = 23. 

Now, in order to express the found HCF as a linear combination of 1288 and 575, we perform 

23 = 575 – 138 x 4 [from (2)] 

= 575 – [1288 – 575 x 2] x 4 [from (1)]

= 575 – 1288 x 4 + 575 x 8 

= 575 x 9 – 1288 x 4 

Firstly, the required numbers which on dividing doesn’t leave any remainder are to be found. 

This is done by subtracting 6 from both the given numbers. 

So, the numbers are 615 – 6 = 609 and 963 – 6 = 957. 

Now, if the HCF of 609 and 957 is found, that will be the required number. 

By applying Euclid’s division lemma 

957 = 609 x 1+ 348 

609 = 348 x 1 + 261 

348 = 216 x 1 + 87 

261 = 87 x 3 + 0. 

⇒ H.C.F. = 87. 

Therefore, the required number is 87.

^{LCM (p, q) =} a³b³

and HCF(P, q) = a²b

LCM(p, q) x HCF(p, q) =  a³b³ x a³b

= a⁵b⁴ 

=  a²b³ x a³b

= pq

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

So,

HCF(a, b) = pq

A composite number has more than two factors.

Another definition: A composite number is a number which is not a prime.

Product of two numbers = HCF x LCM …(1)

Given: Product of two numbers = 1050

and HCF = 25

From equation (1), we have

1050 = 25 x LCM

or

LCM = 1050 / 25 = 42

π  is an irrational number.

LCM of two prime numbers is always the product of these two numbers.

If a and b are two prime numbers, then

LCM (a, b) = ab

HCF of two numbers = 25 

Product of two numbers = 1050 

Let their LCM be x. 

Using the formula, Product of two numbers = HCF × LCM 

We conclude that, 1050 = 25 × x 

x = 1050/25 

 = 42 

Hence, their LCM is 42.

HCF of two primes is always 1.

HCF (a, b) = 1

Prime factorization: 

a = a 

b = b 

LCM = product of greatest power of each prime factor involved in the numbers = a × b 

Thus, LCM (a, b) = ab 

(2) \(\frac{3}{10}\)

\(\frac{3}{10}\) is the small number.

The fundamental theorem of arithmetic, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique.

Prime factorization: 

a = a 

b = b 

HCF = product of smallest power of each common factor in the numbers = 1 

Thus, HCF(a,b) = 1 

Since      1 / 7 x 7 / 100 = 1 / 100 = 0.01.

Thus smallest rational numbers is  7 / 00

Prime factorization: 

a = a 

b = b 

HCF = product of smallest power of each common factor in the numbers = 1 

Thus, HCF(a,b) = 1