# Real Numbers

Real Numbers Class 10

REAL NUMBERS#1- INTRODUCTION – CLASS XTH MATHS CBSE -(IN HINDI)

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REAL NUMBERS#2 EX 1.1 – Finding HCF using Euclid’s Division Lemma, CBSE CLASS 10 MATHS in HINDI

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Real Numbers#3 EX1.1 – Q.No 3, EXAMPLE 4 – CLASS 10 MATHS CBSE in Hindi

Real Number EX1.1 – Q.No 3, EXAMPLE 4 (Application Of Euclid’s Division Lemma) – CLASS 10 MATHS CBSE in Hindi.
In this video we have discussed two questions based on application of Euclid’s Division Lemma from the chapter Real Numbers CBSE class 10 mathematics. The two questions which have been discussed here are:

Example 4 : A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?
EXERCISE 1.1
3. 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?
Visit our official website http://www.onlinepsa.in for more videos on class 10 mathematics.

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Real Numbers#4 EX 1.1 – Q.No 2,4,5 – CLASS 10 MATHS CBSE (in Hindi)

This is the 4th video of the series Real Numbers in which we have discussed Question number 2,4 and 5 from exercise 1.1. These questions are very important in terms of examination point of view and out of these three similar type of questions you can expect one guaranteed question in board examination. So dont forget to watch the video completely.

EXERCISE 1.1
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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Real numbers#5 Ex1.2 (Q.No-1,2,3,4) CBSE CLASS 10 MATHS (in HINDI)

Real numbers#5 Ex1.2 (Q.No-1,2,3,4) CBSE CLASS 10 MATHS NCERT SOLVED AND EXPLAINED in Hindi.
In this video we have discussed Q.No- 1,2,3,4 from EX1.2 of chapter 1 Real numbers. Following Questions have been discussed in this video which are basically finding HCF and LCM using the concept of Euclid’s Division Lemma/algorithm.

EXERCISE 1.2
1. Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
2. 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 (ii) 510 and 92 (iii) 336 and 54
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
4. Given that HCF (306, 657) = 9, find LCM (306, 657).

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Real numbers#6 Ex1.2 (Q.No-5,6,7) CBSE CLASS 10 MATHS in Hindi

Real numbers#6 Ex 1.2 (Q.No-5,6,7) CBSE CLASS 10 MATH (in Hindi-Eng)
In this video we have discussed Q.No- 5,6,7 from EX 1.2 of chapter 1 Real numbers. Following Questions have been discussed in this video which are basically application based question from the concept of Euclid’s Division Lemma/algorithm and Fundamental Theorem of Arithmetic.

EXERCISE 1.2
5. Check whether 6^n can end with the digit 0 for any natural number n.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
7. 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?

visit our official website for more videos on CBSE class 10 maths http://www.onlinepsa.in

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Real numbers#7- Ex1.3 – Root 2 is Irrational, CBSE class 10 MATHS

In this video we have explained some concepts from the chapter “Real Numbers – CBSE CLASS 10 ” which are particularly required to prove the irrationality of root 2.

The concepts discussed in this video are –
1. what are irrational numbers?
2. what are co-prime numbers?
3. Theorem 1.3: which states that “Let p be a prime number. If p divides a-square, then p divides a, where a is a positive integer.”
-Prove that root 2 is irrational.
– Prove that root 5 is irrational.

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# REAL NUMBERS#8 EX 1.3 (Q.No- 2,3 EXPLAINED) , CBSE CLASS 10 MATH (in Hindi)

In this video we have discussed Q.No 2 & 3 from the Exercise 1.3 of chapter “REAL NUMBERS” class 10 cbse. This video is the continuation part of our previous video on Real numbers#7- Ex1.3 – Root 2 is Irrational, CBSE class 10 MATHS where we have discussed about “PROVING IRRATIONALITY” .

Exercise 1.3

2. Prove that 3 + (2root5) is irrational.
3. Prove that the following are irrationals :
(i) 1/(root2)
(ii) 7(root5) (iii) 6 +(root2)

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# Real Numbers#9 Ex 1.4 – Class 10 Math CBSE (in Hindi)

This is the last video for the chapter Real Numbers cbse class 10. In this video, we have discussed the trick to recognize whether the rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion without actually doing long division .

Following concepts have been discussed in this video:
Theorem1.7 : Let x =p/q be a rational number, such that the prime factorisation of q is not of the form 2^n5^m, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).

Exercise 1.4

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

2. Write down the decimal expansions of those rational numbers in Question 1 which have terminating decimal expansions.

3. 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?
(i) 43.123456789

(ii) 0.1201120012000120000…

(iii)43.123456789 (complete bar after decimal)

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Chapter-1

Real Numbers

Basic concepts and formulas

1. Euclid’s division lemma :

Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r,

0 ≤ r < b.

2. Euclid’s division algorithm : This is based on Euclid’s division lemma. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows:

Step 1 : Apply the division lemma to find q and r where a = bq + r, 0 ≤ r < b.

Step 2 : If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be

HCF (a, b). Also, HCF(a, b) = HCF(b, r).

3. The Fundamental Theorem of Arithmetic :

Every composite number can be expressed (factorised) as a product of primes, and this

factorisation is unique, apart from the order in which the prime factors occur.

4. If p is a prime and p divides a2, then p divides q, where a is a positive integer.

5. Let x be a rational number whose decimal expansion terminates. Then we can express x in the form p/q , where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.

6. Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2n5m,where n, m are non-negative integers. Then x has a decimal expansion which terminates.

7. Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2n5m,where n, m are non-negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring).

QUESTIONS

REAL NUMBER

(ONE Mark Question)

1.Find the LCM and HCF of the integers  17, 23 and 29  by applying the prime factorization

method.

Ans:

17,23 and 29

17=1×17,

23=1×23,

29=1×29.

HCF=1

LCM=17×23×29=11339

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

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6n = (2 ×3)n

It can be observed that 5 is not in the prime factorisation of 6n.

Hence, for any value of n, 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

3: The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, find the other.

Ans:  Other number=×2175=435.

4.Find the [𝐻𝐶𝐹×𝐿𝐶𝑀] for the numbers 100 and 190.

Ans: [𝐻𝐶𝐹×𝐿𝐶𝑀]= product of two numbers=100×190=19000.

5.The decimal expansion of the rational number            will terminate after how many placed of decimals?

Ans:     =0.375. So it’s terminating after 3-dacimal places.

6.. Has the rational number             a terminating or a non-terminating decimal representation?

Ans: = , here q=175=5×5×7≠2m5n . Therefore it has non-terminating decimal expansion.

7. Use Euclid’s division algorithm to find the HCF of  135 and 225 .

Ans:   225=135×1+95,

135=95×1+40,

95=40×2+15,

40=15×2+10,

15=10×1+5,

10=5×2+0. So HCF=5.

8. Express 3825 number as a product of its prime factors.

Ans:3825=3×3×5×5×17=32×52×17.

9.Determine  the  value of p and  q  so  that  the prime factorisation of 2520 is expressed as                                      .

Ans:  2520=2×2×2×3×3×5×7= 23×32×5×7× =.

Therefore p=2,q=5.

10.If A =2n+3 and B=n+7,find HCF of A and B.

Ans: If n=1,then A=5,B=8.

If n=2,then A=7,B=9.

If n=3,then A=9,B=10.

Therefore HCF of A and B is 1.

Two marks Question

1.During a sale, color pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many of each would you need to buy?

Sol: We are given that during a sale, color pencils were being sold in packs of 24 each and crayons in packs of 32 each. If we want full packs of both and the same number of pencils and crayons, we need to find the number of each we need to buy. Given that, Number of color pencils in one pack = 24 Number of crayons in pack = 32. Therefore, the least number of both colors to be purchased L.C.M of 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 = 96.

Hence, the number of packs of pencils to be bought 9624=4, and number of packs of crayon to be bought  9632=3

2: Can two numbers have 16 as their HCF and 380 as their as their LCM? Give reason.

Sol:  On dividing 380 by 16 we get 23 as the quotient and 12 as the remainder Since LCM is not exactly divisible by the HCF, two number cannot have 16 as their HCF and 380 as their LCM

3. 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?

Sol. We are given that 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. We need to fund the maximum number of columns in which they can march. Members in army = 616 Members in band = 32. Therefore, Maximum number of columns = H.C.F of 616 and 32. By applying Euclid’s division lemma 616 = 32 x 19 + 8 32 = 8 x 4 + 0. Therefore, H.C.F. = 8 Hence, the maximum number of columns in which they can march is 8

4. Find the LCM and HCF of the pairs of integers 336 and 54 and verify that LCM × HCF = product of the two numbers.

Ans: 336=2×2×2×2×3×7,    54=2×3×3×3.

LCM=2×2×2×2×3×3×3×7=3024,

HCF=2×3=6. LCM×HCF=18144

Product of two numbers=18144.

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

Ans: HCF (306, 657) = 9.

We know that LCM×HCF= Product of two numbers.

LCM×HCF=306×657

LCM=.

Therefore LCM is 22338.

6.Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer : Numbers are of two types – prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)

= 13 × 78

= 13 ×13 × 6

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 ×1009

1009 cannot be factorized further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

7.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?

Ans:It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

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.

Three marks Question

1:  Prove that the product of two consecutive positive integers is divisible by 2.

Proof: Let n – 1 and n be two consecutive positive integers. Then their product is n (n – 1) = n2 n .We know that every positive integer is of the form 2q or 2q + 1 for some integer q.

So let n = 2q So, n2 – n = (2q) 2 – (2q)

or   n2 – n = (2q) 2 – (2q)

or  n2 – n = 4q2 – 2q

or  n2 – n = 2q (2q – 1)

or  n2 – n = 2r [where r = q (2q – 1)]

So n2 – n is even and divisible by 2

Let n = 2q + 1 So, n2 – n = (2q + 1) 2 – (2q + 1)

or  n2 – n = (2q + 1)   (2q + 1) – 1) n2 – n = (2q + 1)   (2q)

or  n2 – n = 2r

[r = q (2q + 1)]

n2  n is even and divisible by 2 Hence it is proved that that the product of two consecutive integers is divisible by 2.

2. A merchant has 120 liters of oil of one kind, 180 liters of another and 240 liters of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

Sol: The merchant has 3 different oils of 120 liters, 180 liters and 240 liters respectively. So the greatest capacity of the tin for filling three different types of oil is given by the H.C.F. of 120,180 and 240. So first we will calculate H.C.F of 120 and 180 by Euclid’s division lemma. 180 = (120) (1) + 60 120 = (60) (2) + 0 The divisor at the last step is 60. So the H.C.F of 120 and 180 is 60. Now we will fund the H.C.F. of 60 and 240, 240 = (60) (4) + 0 The divisor at the last step is 60. So the H.C.F of 240 and 60 is 60. Therefore, the tin should be of 160 liters.

3. A rectangular courtyard is 18m 72cm long and 13m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Sol: Given: A rectangular yard is 18 m 72 cm long and 13 m 20 cm broad .It is to be paved with square tiles of the same size. To Find: Least possible number of such tiles. Length of the yard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (therefore, 1 m = 100 cm) Breadth of the yard = 13 in 20 cm = 1300 cm + 20 cm = 1320 cm The size of the square tile of same size needed to the pave the rectangular yard is equals the HCF of the length and breadth of the rectangular yard. Prime factorization of 1872 = 24 x 32 x 13 Prime factorization of 1320 = 23 x 3 x 5 x 11 HCF of 1872 and 1320 = 23 x 3 = 24 Therefore, Length of side of the square tile = 24 cm Number of tiles required = Area of the courtyard, Area of each tile = Length x Breadth Side2 = 1872 cm x 1320 cm 24 cm = 4290. Thus, the least possible number of tiles required is 4290.

4. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

Sol:  The greatest 6 digit number be 999999 24, 15 and 36 24 = 2 x 2 x2 x3 15 = 3 x 5 36 = 2 x 2 x 3 x 3 L.C.M of 24, 15 and 36 = 360Since,999999360=2777×360+279 Therefore, the remainder is 279. Hence the desired number is = 999999 – 279 = 9997201 Hence 9997201 is the greatest number of 6 digits exactly divisible by 24, 15 and 36.

5. Prove that 3+2√5 is an irrational number.

Let 3+2√5 is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

that   3+2√5  =

=>   √5 =

Since a and b are integers, will also be rational and therefore,5  is rational.

This contradicts the fact that 5  is irrational. Hence, our assumption that 3+2√5 is rational is false. Therefore, 3+2√5 is irrational.

6. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Case-1

a  = 3q

a2=(3q)2=9q2=3(3q2)=3k1, where k1 is some positive integers

Case-2

a  = 3q+1

a2= (3q+1)2=9q2+6q+1=3 (3q2+2q)+1=3k2+1, where k2 is some positive integers

Case-3

a  = 3q+2

a2= (3q+2)2=9q2+12q+4=3 (3q2+4q+1)+1=3k3+1, where k3 is some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m +1

7. Prove that is an  irrational.

Ans: Let be rational.

=

=, where 5a and 2b are integers and is irrational.

.So our supposition is wrong.

Therefore is  irrational.

8.If ,find the value of m and n.

Ans:m=3,n=5.

Four marks Question

1. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer: Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

a = 6q + rfor some integer q0, and r = 0, 1, 2, 3, 4, 5 because 0 r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,

or 6q + 5

2. Prove that √5 is an irrational number.

Let √5  is a rational number.

Therefore, we can find two integers a, b (b ≠ 0) such that   5  =

Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.

a=√5b

=>a2=5b2

Therefore, a2 is divisible by 5 and it can be said that a is divisible by 5.

Let a = 5k, where k is an integer

his means that b2 is divisible by 5 and hence, b is divisible by 5.

a2=5k2     This implies that a and b have 5 as a common factor.

And this is a contradiction to the fact that a and b are co-prime.

Hence, √5 cannot be expressed as or it can be said that √5 is irrational.

3. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3.

a = 3q or 3q + 1 or 3q + 2

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

a3 = (3q )3== 27q3=9(3q3)=9m,

where m is an integer such that m = 3q3

Case 2: When a = 3q + 1,

a3 = (3q +1)3

a3 = 27q3 + 27q2 + 9q + 1

a3 = 9(3q3 + 3q2 + q) + 1

a3 = 9m + 1

Where m is an integer such that m = (3q3 + 3q2 + q)

Case 3: When a = 3q + 2,

a3 = (3q +2)3

a3 = 27q3 + 54q2 + 36q + 8

a3 = 9(3q3 + 6q2 + 4q) + 8

a3 = 9m + 8

Where m is an integer such that m = (3q3 + 6q2 + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.

4. Prove that the product of three consecutive positive integer is divisible by 6.

Sol:  Let n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5.

If n = 6q, n (n + 1)   (n + 2)  = 6q (6q + 1)   (6q + 2), which is divisible by 6

If n = 6q + 1, n (n + 1)   (n + 2)  = (6q + 1)   (6q + 2)   (6q + 3)

n (n + 1)   (n + 2)  = 6 (6q + 1)   (3q + 1)   (2q + 1)  Which is divisible by 6

If n = 6q + 2, n (n + 1)   (n + 2)  = (6q + 2)   (6q + 3)   (6q + 4) n (n + 1)   (n + 2)

= 12 (3q + 1)   (2q + 1)   (2q + 3), Which is divisible by 6. Similarly we can prove others. Hence it is proved that the product of three consecutive positive integers is divisible by 6.

5. Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is

some integer.

Ans:Let us start with taking a, where a is a positive odd integer. We apply the

division algorithm with a and b = 4.

Since 0 ≤r < 4, the possible remainders are 0, 1, 2 and 3.

That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.

However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).

Therefore, any odd integer is of the form 4q + 1 or 4q + 3.