Rational & Irrational Numbers Notes for RRB ALP Maths Trade Exam

By Rahul Chadha|Updated : November 28th, 2018

Rational & Irrational Numbers Notes for RRB ALP Maths Trade Exam. These Notes will help you score good marks in the upcoming RRB ALP Maths & Physics Trade Exam.

Introduction: Rational & Irrational Numbers

1. Euclid’s Division lemma:- Given Positive integers a and b there exist unique integers q and r satisfying a=bq +r, where 0≤r<b, where a, b, q and r are respectively called as dividend, divisor, quotient and remainder.

2. Euclid’s division Algorithm:- To obtain the HCF of two positive integers say c and d, with c>0, follow the steps below:

  • Step I: Apply Euclid’s division lemma, to c and d, so we find whole numbers, q and r such that c =dq +r,  0≤r<d
  • Step II: If r=0, d is the HCF of c and d. If r is not equal to 0, apply the division lemma to d and r.
  • Step III: Continue the process until the remainder is zero. The divisor at this stage will be the required HCF

3. The Fundamental theorem of Arithmetic:- Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur. Ex.: 24= 2x2x2x3 = 3x2x2x2

Theorem: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form of p/q where p & q are co-prime and the prime factorization of q in the form of 2n.5m , where n, m are non-negative integers. Ex. 7/10 = 7/2x5 = 0.7

Theorem: Let x = p/q be a rational number such that the prime factorization of q is not of the form of 2n.5m, where n, m are non-negative integers. Then x has a decimal expansion which is non terminating repeating (recurring). Ex. 7/6 = 7/2x3 = 1.1666...

Theorem: For any two positive integers a and b, HCF (a,b) X LCM (a,b)=a X b

Ex.: 4 & 6; HCF (4,6) = 2, LCM (4,6) = 12; HCF X LCM = 2 X 12 =24 Ans. : a X b = 24

Questions for Practice

Q1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225

Solution:

We know that,

 = 225 > 135 

 Applying Euclid’s division algorithm

225 = 135 × 1+90                 byjusexamprep

Here remainder = 90,

So, Again Applying Euclid’s division algorithm

135 = 90 × 1+45 

Here  remainder = 45,

So, Again Applying Euclid’s division algorithm

90 = 45×2+0

Remainder = 0,

Hence ,

HCF of (135, 225)  = 45

(ii) 196 and 38220

Solution:

We know that,

38220>196

So, Applying Euclid’s division algorithm

38220 = 196×195+0  byjusexamprep

Remainder = 0

Hence, HCF of  (196, 38220) = 196

(iii) 867 and 255

Solution:

We know that,

867>255

So,  Applying Euclid’s division algorithm

867 = 255×3+102  byjusexamprep

Remainder = 102

So, Again Applying Euclid’s division algorithm

255 = 102×2+51

Remainder = 51

So, Again Applying Euclid’s division algorithm

102 = 51×2+0

Remainder = 0

Hence, (HCF 0f 867 and 255) = 51

Q2. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

By using,  Euclid’s division algorithm

616 = 32×19+8    

Remainder ≠ 0

So, again Applying Euclid’s division algorithm

32 = 8×4+0

HCF of (616, 32) is 8.

So ,

They can march in 8 columns each.

Q3. Express each number as a product of its prime factors:

(i) 140

Solution:

140 = 2×2×5×7 = 22×5×7

(ii) 156

Solution:

156 = 2×2×3×13 = 22×3×13

(iii) 3825

Solution:

3825 = 3×3×5×5×17 = 32×52×17

(iv) 5005

Solution:

5005 = 5×7×11×13

(v) 7429

Solution:

7429 = 17×19×23

Q4. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i)   26 and 91

Solution:

26 = 2×13

91 = 7×13

HCF =13

LCM = 2×17×13 =182

Product of the two numbers = 26×91 = 2366

HCF×LCM = 13×182 = 2366

Hence, product of two numbers = HCF×LCM

(ii)  510 and 92

Solution:

510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

Product of the two numbers = 510×92 = 46920

HCF×LCM = 2×23460

HCF×LCM = 46920

Hence, product of two numbers = HCF×LCM

(iii)  336 and 54

Solution:

336 = 2×2×2×2×3×7

336 = 24×3×7

54 = 2×3×3×3

54 =2×33

HCF = 2×3 = 6

LCM = 24×33×7 = 3024

Product of the two numbers = 336×54 = 18144

HCF×LCM = 6×3024 = 18144

Hence, product of two numbers = HCF×LCM

Q5. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i)   12, 15 and 21

Solution:

12 = 2×2×3 = 22×3

15 = 3×5

21 = 3×7

HCF = 3

LCM = 22×3×5×7 = 420

(ii)  17, 23 and 29

Solution:

17 = 1×17

23 = 1×23

29 = 1×29

HCF = 1

LCM = 17×23×29 = 11339

(iii)  8, 9 and 25

Solution:

8=2×2×2

9 = 3×3

25 = 5×5

HCF = 1

LCM = 2×2×2×3×3×5×5 = 1800

Q6. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

HCF of (306, 657) = 9

We know that,

LCM×HCF = product of two numbers

LCM×HCF = 306×657

LCM = byjusexamprep

LCM = 22338

Q7. Check whether 6n can end with the digit 0 for any natural number n.

Solution:

If any number has last digit 0,

Then,  it should be divisible by 10 

Factors of  10 = 2×5

So, Value 6n should be divisible by 2 or 5

Prime factorisation of 6n = (2×3)n

Hence, 6n is divisible by 2 but not by 5.

It can not end with 0.

Q8. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

They will meet again after LCM of both values at starting point.

18 = 2×3×3

And

12 = 2×2×3

LCM of 12 and 18 = 2×2×3×3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Q9. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i)                  byjusexamprep

Solution:

byjusexamprep

Factorize the denominator we get,

3125 = 5×5×5×5×5 = 55

The denominator is of the form 5m

Hence, the decimal expansion of byjusexamprep is terminating.

(ii)                byjusexamprep

Solution:

byjusexamprep

Factorize the denominator we get,

8 = 2 × 2 ×2 = 23

The denominator is of the form 2m

Hence, the decimal expansion of byjusexamprep is terminating.

(iii)              byjusexamprep

Solution:

byjusexamprep

Factorize the denominator we get,

455 = 5×7×13

Since, the denominator is not in the form of 2m × 5n, and it also contains 7 and 13 as its factors,

Its decimal expansion will be non-terminating repeating.

(iv)              byjusexamprep

Solution:

byjusexamprep

Factorize the denominator we get,

1600 = 26×52

The denominator is in the form 2m × 5n

Hence, the decimal expansion of byjusexamprep is terminating.

(v)                byjusexamprep

Solution:

byjusexamprep

Factorize the denominator we get,

343 = 73

Since the denominator is not in the form of 2m × 5n,  it has 7 as its factors.

So, the decimal expansion of byjusexamprep non-terminating repeating.

(vi)              byjusexamprep

Solution:

byjusexamprep

The denominator is in the form 2m×5n

Hence, the decimal expansion of byjusexamprep is terminating.

(vii)            byjusexamprep

Solution:

byjusexamprep

Since, the denominator is not in the form of 2m × 5n, as it has 7 in denominator.

So, the decimal expansion of byjusexamprep  is non-terminating repeating.

(viii)          byjusexamprep

Solution:

byjusexamprep

The denominator is in the form 5n

Hence, the decimal expansion of byjusexamprep is terminating.

(ix)              byjusexamprep

Solution:

byjusexamprep

Factorize the denominator we get,

10 = 2×5

The denominator is in the form 2m×5n

Hence, the decimal expansion of byjusexamprep is terminating.

(x)                byjusexamprep

Solution:

byjusexamprep

Factorize the denominator we get,

30 = 2×3×5

Since the denominator is not in the form of 2m × 5n, as it has 3 in denominator.

So, the decimal expansion of byjusexamprep  is a non-terminating repeating.

Q 10.  Write down the decimal expansions of those rational numbers in which have terminating decimal expansions.

Solution:

(i)byjusexamprep

 

C360_2017-05-23-07-49-23-781.jpg

 

 

(ii) byjusexamprep

 

C360_2017-05-23-07-51-34-339.jpg

 

(i)       byjusexamprep

 

C360_2017-05-23-07-51-02-802.jpg

 

 

(viii)  byjusexamprep

 

C360_2017-05-23-07-51-51-667.jpg

 

 

 

(ix)      byjusexamprep

 

C360_2017-05-23-07-52-02-406.jpg

 

Q 11. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form byjusexamprepwhat can you say about the prime factors of q?

 

Solution:

(i)  43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form byjusexamprep and q is of the form 2m×5n . That is, the prime factor of q will be 2 or 5 or both.

(ii)  0.120120012000120000...

The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii)  43.123456789

Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form byjusexamprep and q is not of the form 2m×5n that is, the prime factors of q will also have a factor other than 2 or 5.

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