# NCERT Solutions for class 10th Maths Chapter 1 Real Numbers

#### Exercise:1.1

Question 1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Note: Use Euclid’s division algorithm, a = bq + r; o r < b where a, b, q and rare integer’s. Here a = dividend and b = divisor

Sol. (i) Here, dividend= 225 and divisor= 135 225= 135 X 1 +90 135 =90 X 1 +45 90=45 X 2+0 :. HCF (225, 135) = HCF (135, 90) = HCF (90, 45) = 45 Hence, the HCF (225, 135) = 45

(ii) We have: Dividend= 38220 and Divisor= 196 38220 = 196 X 195 + 0 Hence, HCF (196, 38220) = 196

(iii) We have: Dividend = 867 and Divisor = 255 .’. 867 = 255 X 3 + 102 255 = 102 X 2 + 51 102 = 51 X 2 + 0 Hence, HCF (255, 867) = 51

Question 2.Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5,where q is some integer.

Sol. Let a be positive odd integer.

By division algorithm,

a = 6 x q + r; 0 :s; r < 6

Here the possible integers can be:

a = 6q + 0 or 6q +1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.

We reject 6q + 0, 6q + 2, and 6q + 4 because they are even integers. Hence, the possible form of a is 6q + 1, or 6q + 3, or 6q + 5.

Question 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?

Note: Use Euclid’s division algorithm to find the maximum number which can divide both 616 & 32.

Sol. Let n = number of columns required. Now, we need to calculate maximum number of column in which both army contingent and army band can march, i.e; we need to calculate highest number which can divide both 616 & 32. n=HCF(616,32). Using Euclid’s Division Lemma, 616=32x 19+8 32 =8 X 4+0 Hence, HCF (616, 32) = 8 n=8 Hence, the maximum number of the columns in which they can march= 8.

Question 4. Using E.D.L show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Sol. Let n be a positive integer. Using E.D.L, we can write ‘n’ as n = 3q + r,0::::; r< 3 :. n can be of the form: 3q or 3q + 1 or 3q + 2. If n=3q —> n2=(3q)2=3 x 3q2=3m, wherem=3q2
If n=3q+ 1n2=3 (3q2+2q)+ 1=3m+ 1, wherem=3q2+2q
If n=3q+ 2n2 = 3 (3q2 +4q+ 1) + l= 3m+ 1, wherem= 3q2+4q+ 1 Thus, n2 can be either of the form 3m or 3m + 1.

Question 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.
Sol. Let n be a positive integer.
Using E.D.L, we can write ‘n’ as: n = 3q + r, 0::::; r< 3

:.n can be of the form: 3q or 3q + 1 or 3q + 2

If n=3q,n3 =27q3 =9x3q3 =9m where m=3q3

If n=3q + 1—> n3=(3q +1)3

n3 =9{3q3 +3q2 +q}+l

·:((a+ b)3 =a3 + b3 + 3ab2 + 3a2b)

= 9m +1, where m = 3q3 + 3q2 + q

If n = 3q + 2 n3=(3q+ 2)3 = 27q3 +45q2+ 36q

n3 = 9(3q3 + 6q2 +4q) + 8 = 9m + 8, where m= 3q3 + 6q2+49

Similarly, n3 = 9m + 8

n3 is either of the form 9m or 9m + 1 or 9m + 8.

#### Exercise 1.2

Question 1. Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

(ii) We use the division method as shown below :

156 = 2 X 2 X 3 X l3 = 22 X 3 X 13

(iii) Using division method, we have:

.’. 3825 = 3 X 3 X 5 X 5 X 17 = 32 X 52 X 17

(iv) We use the division method as shown below:

.’. 5005 = 5 X 7 X 11 X l3

(v) We use the division method as shown below:

7429=17x 19×23

Note: Prime factors can be calculated either by Prime Factorisation Method or by Division Method

Question 2. Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = product of the two numbers.

(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Hint: HCF is the product of the smallest power of each common prime factor in the number. LCM is the product of the greatest power of each prime factor involved in the numbers.

Sol. (i) 26 = 2 x 13 and 91 = 7 x 13

HCF(26, 91) = 13,

LCM = (26, 91) = 2 x 7 x 13 = 182

LCM x HCF = 182 x 13 = 2366, and 26 x 91 = 2366.

Therefore, LCM x HCF = 26 x 91 (Hence verified).

(ii) 510 = 2 x 3 x 5 x 17 and 92 = 2 x 2 x 23

:.LCM of 510 and 92 = 2 x 2 x 3 x 5 x 17 x 23 = 23460

and HCF of 510 and 92 = 2

.•. LCM x HCF = 23460 x 2 = 46920 and 510 x 92 = 46920

:. LCM x HCF = 510 x 92 (Hence verified).

(iii) 336 = 2 x 2 x 2 x 2 x 3 x 7 and 54 = 2 x 3 x 3 x 3

:. LCM of 336 and 54 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7 = 3024

and HCF of 336 and 54 = 2 x 3 = 6

Now, HCF x LCM= 3024 x 6 = 18144

and 336 x 54 = 18144

:.HCF x LCM= 336 x 54 (Hence verified).

Question 3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12,15and21 (ii) 17,23and29 (iii) 8,9and25

Sol. (i) First we write the prime factorisation of each of the given numbers.

12 = 22 x 3; 15 =3 x 5 and 21 = 3 x 7

HCF (12, 15, 21) =3

LCM (12, 15, 21) =22 x 3 x 5 x 7 = 420

(ii) First we write the prime factorisation of each of the given numbers.

17 = 17; 23 = 23 and 29 = 29

:. LCM= 17 x 23 x 29 = 11339 and HCF = 1

(iii) First we write the prime factorisation of each of the given numbers.

8 = 2 x 2 x 2 = 23·9 = 3 x 3 = 32 and 25 = 5 x 5 = 52

LCM= 23 x 32 x 2 = 8 x 9 x 25 = 1800 and HCF = 1

Notes: (a) Every integer, n > I has a prime factor. If in is a prime number, then its prime factor is n itself If n is composite number, then it has prime factors less than itself
(b) LCM of two or more numbers having no common prime factor is equal to their product. and HCF of two or more numbers having no common prime factor is I. As we have seen in Q 3 [(ii) and (iii)}

Question 4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Sol. We know that; HCF (a, b) x LCM (a, b) =ax b Here; a= 306, b = 657 & HCF = 9.
Since, HCF (a, b) x LCM (a, b) =ax b
:.9 x LCM=306 x 657LCM= 306 X 657 = 34 x 657 = 22338
:.LCM (306, 657) = 22338

Question 5. Check whether 6n can end with the digit 0 for any natural number n.

Sol. Let if possible 6n ends with digit 0. And let 6n = 10 x q
2n x3n =2x5xq 5 is a primefactor of 2n x3n
which is not possible because 2n x 3n can have only 2 and 3 as prime factors. Hence, 6n cannot end with the digit O (zero) for any natural number n.

Question 6. Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Hint: A composite number is a positive integer which is not prime (i.e; which has factors other then I and itself).
Sol. (i) 7 x 11 x 13 + 13 = 13 x {7 x 11 + 1} = 13 x 78 = 2 x 3 x 132
Here, we get product of primes and this factorisation is unique. Hence, this is a composite number.

(ii) 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5 = 5 X {7 X 6 X 4 X 3 X 2 X 1 + l}
= 5 X 1009
Here, we get product of primes and this factorisation is unique. Hence, this is a composite number.

Question 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?

Sol. Let ‘n’ minutes be the required time.
So, first time, they will meet again in ‘n’ minutes. Second time, they will meet again in ‘2n’ minutes & so on.
n is the minimum time required i.e; for 18 minutes & 12, minutes, ‘n’ will be the LCM of 18 & 12, i.e; lowest number which is the multiple ofl8 & 12 both. Here, 18 = 2 x 3 x 3 and 12 = 2 x 2 x 3
:. LCM(l8, 12)=2 x 2 x 3 x 3 =36.
:. Ravi and Sonia will meet again at the starting point after 36 minutes.

#### Exercise 1.3

Question 1. Prove that root 5 is irrational.
Hint: Use the method of contradiction, i.e., we will assume that 5 is not irrational or 5 is rational

Sol. Suppose root 5 is not irrational,
i.e; 5 is rational & can be written as:5 =p/q, where p & q are co-primes.
,root 5 q=p
Squaring both sides we get :
5xq2=p2 …(i)
5 divides p (·: p & q are co-primes)
p =5 xp1 ; p1 is an integer. …(ii)
Put the value of p from (ii) in (i), we get,
5xq2=(5xp1)2=52xpf q2=5xpf
5 divides q (·: p1 & q are co-primes)
q = 5 x p2 ; p2 is an integer. ….(iii)
From (ii) and (iii), we find 5 as a common factor of p and q. It contradicts that
p and q are co-primes and 5 is rational number. Hence, root 5 is an irrational number.

#### Exercise 1.4

Question 1. 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:

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

Question 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 p/q, what can you say about the prime factors of q?

Sol. (i) Rational:q =109=29 x 59.
(ii) Non-rational.
(iii) Rational, prime factors of q will also have factors other than 2 or 5 because the decimal expansion is non-terminating repeating.

Related Articles: