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 
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 
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
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 = 
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) 
Solution:
Factorize the denominator we get,
3125 = 5×5×5×5×5 = 55
The denominator is of the form 5m
Hence, the decimal expansion of 
is terminating.
(ii) 
Solution:
Factorize the denominator we get,
8 = 2 × 2 ×2 = 23
The denominator is of the form 2m
Hence, the decimal expansion of 
is terminating.
(iii) 
Solution:
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) 
Solution:
Factorize the denominator we get,
1600 = 26×52
The denominator is in the form 2m × 5n
Hence, the decimal expansion of 
is terminating.
(v) 
Solution:
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 
non-terminating repeating.
(vi) 
Solution:
The denominator is in the form 2m×5n
Hence, the decimal expansion of 
is terminating.
(vii) 
Solution:
Since, the denominator is not in the form of 2m × 5n, as it has 7 in denominator.
So, the decimal expansion of 
is non-terminating repeating.
(viii) 
Solution:
The denominator is in the form 5n
Hence, the decimal expansion of 
is terminating.
(ix) 
Solution:
Factorize the denominator we get,
10 = 2×5
The denominator is in the form 2m×5n
Hence, the decimal expansion of 
is terminating.
(x) 
Solution:
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 
is a non-terminating repeating.
Q 10. Write down the decimal expansions of those rational numbers in which have terminating decimal expansions.
Solution:
(i)
(ii) 
(i) 
(viii) 
(ix) 
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 
what 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 
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
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.Preparing for RRB ALP 2nd Stage? BYJU'S Exam Prep has come up with the complete test series package for upcoming Assistant Loco Pilot & Technicians 2nd Stage Exam. The first test is free. So, what are you waiting for? Check your preparation level by clicking on the below given link:
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