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Class 10 Maths NCERT Solution Chapter 1 Real Numbers. - Best Maths Practice Material

# Class 10 Maths NCERT Solution 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

Solution 1 (i) —  Using  Euclid’s division algorithm we get,

$225 = 135 \times 1\,+\,90$

$135 = 90 \times 1\,+\,45$

$90 = 45 \times 2\,+\,0$

Hence HCF of  135 and 225  = 45

Solution 1 (ii) —  Using  Euclid’s division algorithm we get,

$38220 = 196 \times 195\,+\,0$

Hence HCF of  196 and 38220  = 196

Solution 1 (iii) —  Using  Euclid’s division algorithm we get,

$867 = 255 \times 3\,+\,102$

$255 = 102 \times 2\,+\,51$

$102 = 51 \times 2\,+\,0$

Hence HCF of  867 and 255  = 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.

Solution 2:-  We know that any integer can be a can be written as $a=6q+r$ for $q\in I \;,\;q\ge 0$ and $r=0,1,2,3,4,5$ by Euclid’s division algorithm

Hence  $a = 6q+r$  when r = 0 , 2 , 4  then $a=6q$, or$a=6q+2=2(3q+1)$ and $a=6q+4=2(3q+2)$ which is divisible by 2 hence are even integer.

Where as when r=1,3,5 then  $a=6q+1$, or $a=6q+3=2(3q+1)+1$, and $a=6q+5=2(3q+2)+1$ which when divided by 2 gives remainder as 1.

Hence all numbers of the form $6q+1$, or $6q+3$, or $6q+5$ are odd positive integer.

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 numbers of columns. What is the maximum number of columns in which they can march?

Question 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.

Solution 4:- Using Euclid’s division lemma we can write any number $a=3q+r$ where  $q\in I$ and $r=0,1,2$

Case I —  when $r=0$ hence $a=3q$

$\therefore a^2=9q=3(3q)=3m$ for some integer $m$

Case II —  when $r=1$ hence $a=3q+1$

$\therefore a^2=(3q+1)^2$

$=9q^2+6q+1=3q(3q+2)+1$ $=3m+1$ for some integer $m=q(3q+2)$

Case III —  when $r=2$ hence $a=3q+2$

$\therefore a^2=(3q+2)^2$ $=9q^2+12q+4$ $=3(3q^2+4q+1)+1$ $=3m+1$ for some integer $m=3q^2+4q+1$

Question 5 :-Use Euclid’s division lemma to show that the cube of any positive integer is of the form $9m$ , $9m + 1$ or $9m+8$ .

Solution 5:- Using Euclid’s division lemma we can write any number $a=3q+r$ where  $q\in I$ and $r=0,1,2$

Case I —  when $r=0$ hence $a=3q$

$\therefore a^3=27q=9(3q)=9m$ for some integer $m$

Case II —  when $r=1$ hence $a=3q+1$

$\therefore a^3=(3q+1)^3$ $=27q^3+3.3q.1(3q+1)+1$ $=9q(3q^2+(3q+1))+1=3m+1$ for some integer $m=q(3q^2+3q+1)$

Case III —  when $r=2$ hence $a=3q+2$

$\therefore a^3=(3q+2)^3$ $=27q^2+3.3q.2(3q+2)+8$  $=9q(3q^2+6q+4)+8$ $=3m+1$ for some integer $m=q(3q^2+6q+4)$

## Exercise 1.2

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

(i)    140

(ii)    156

(iii)    3825

(iv)    5005

(v)    7429

Solution 1 —(i)  140 = $2^2\times 5 \times 7$

(ii)    156 = $2^2\times 3\times 13$

(iii)    3825 = $3^2\times 5^2\times 17$

(iv)    5005 = $5\times 11\times 13\times 17$

(v)    7429 = $3\times 11\times 17\times 19$

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

(i)  26 and 91

(ii)  510 and 92

(iii)  336 and 54

Solution 2 — (i)      26 = $2\times 13$ where as  91 = $7\times 13$

Hence  LCM = $2\times 7\times 13=\,182$  and HCF = 13 .

Therefore LCM X HCF = $182 \times 13=\,2366$

Product of numbers = $26\times 91=\,2366$

Therefore  LCM X HCF = product of numbers , Hence Verified.

Solution 2 — (ii)      510 = $2\times 3\times 5\times 17$ where as  92 = $2^2 \times 23$

Hence  LCM = $2^2\times 3\times 5\times 17\times 23=\,23460$  and HCF = 2 .

Therefore LCM X HCF = $23460 \times 2=\,46920$

Product of numbers = $510\times 92=\,46920$

Therefore  LCM X HCF = product of numbers , Hence Verified.

Solution 2 — (iii)      336 = $2^4\times 3\times 7$ where as  54 = $2 \times 3^3$

Hence  LCM = $2^4\times 3^3\times 7=\,3024$  and HCF = $2\times 3=6$ .

Therefore LCM X HCF = $3024 \times 6=\,18144$

Product of numbers = $336\times 54=\,18144$

Therefore  LCM X HCF = product of numbers , Hence Verified.

Question 3:-  Find LCM and HCF of the following integers by applying the prime factorization method.

(i)  12, 15 and 21

(ii)  17, 23 and 29

(iii)  8, 9 and 25

Solution 3 — (i)  Prime factors of numbers

12 = $2^2\times 3$,  15=$3\times 5$ and  21=$3\times 7$

Hence LCM ( 12, 15, 21) = $2^2\times 3\times 5\times 7= 420$

HCF ( 12, 15, 21) = 3

Solution 3 — (ii)  Prime factors of numbers

17 = 17,  23=23 and  29=29

Hence LCM ( 17, 23, 29) = $17\times 23\times 29 = 11339$

HCF ( 17, 23, 29) = 1

Solution 3 — (iii)  Prime factors of numbers

8 = $2^3$,  9=$3^2$ and  25=$5^2$

Hence LCM ( 8, 9, 25) = $8\times 9\times 25 = 3600$

HCF ( 8, 9, 25) = 1

Question 4:-  Given that HCF( 306, 657)=9. Find LCM (306, 657).

Solution 4:-  Given numbers are 306, and 657 and HCF = 9

We know that LCM X HCF = Product of Numbers.

$\therefore LCM\times 9 = 306\times 657$

Hence  LCM =$\cfrac{306 \times 657}{9}=22338$

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

Solution 5:-  We know that numbers ending with 0 must have two of its prime factors as $2^a.5^b$ for some natural numbers a, b.

Now prime factor of $6^n=2^n.3^n$.

Since $5^b$ for some natural number b does not exist in $6^n$ Hence $6^n$ cannot end with 0.

Question 6:-  Explain why $7\times 11\times 13 +13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1 + 5$ are composite numbers.

Solution 6:-  Number $7\times 11\times 13 + 13$

= $13\times (7\times 11 +1)$

Hence is product of two numbers, Hence number is composite number.

For Number $7\times 6\times 5\times 4\times 3\times 2\times 1 + 5$

= $5\times (7\times 6\times 4\times 3\times 2\times 1 + 1)$

Thus above number is also product of two numbers hence is composite.

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 same direction. After how many minutes will they meet again at the starting point?

Solution 7:-

## Exercise 1.3

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Question 1:-  Prove that $\sqrt{5}$ is irrational.

Solution 1 :-  Let $p,\,q\in N$ such that $p,\,q$ are co-prime numbers such that  $\sqrt{5}=\cfrac{p}{q}$

Now squaring both sides we get  $p^2=5\times q^2\;—(1)$ Hence p is multiple of 5.

Let $p=5m$ in equation (1) we get.

$25\times m^2=5q^2$ therefore  $5\times m^2=q^2$

Hence q is multiple of 5.

Hence p, q are non co-prime number, Hence $\sqrt{5} \notin Q$

Therefore $\sqrt{5}$ is an irrational number.

Question 2:-  Prove that $3+2\sqrt{5}$ is irrational.

Solution 2 :-  Let $p,\,q\in N$ such that $p,\,q$ are co-prime numbers such that  $3+2\sqrt{5}=\cfrac{p}{q}$

$\therefore 2\sqrt{5}=\cfrac{p}{q} – 3$

$\therefore \sqrt{5}=\cfrac{q-3p}{2p}$

Now L.H.S is irrational where as R.H.S is rational.

Where as a rational number cannot be equal to irrational number.

Hence $3+2\sqrt{5}$ is an irrational number.

Question 3:-  Prove that the following are irrational.

(i)  $\cfrac{1}{\sqrt{2}}$

(ii)  $7\sqrt{2}$

(iii)  $6+\sqrt{2}$

Solution 3 (i) :-  Let $p,\,q\in N$ such that $p,\,q$ are co-prime numbers such that  $\cfrac{1}{\sqrt{2}}=\cfrac{p}{q}$

Squaring both side we get

$2 p^2=q^2$–(1)

Hence  $q^2$ is multiple of 2.

Let $q=2m$ put in (1) we get

$p^2=2m^2$ hence p is multiple of 2.

therefore p and q are non co-prime numbers.

Hence $\cfrac{1}{\sqrt{2}}$ is irrational.

Solution 3 (ii) :-  Let $p,\,q\in N$ such that $p,\,q$ are co-prime numbers such that  $7{\sqrt{5}}=\cfrac{p}{q}$

Squaring both side we get

$p^2=49\times 5\times q^2$–(1)

Hence  $p^2$ is multiple of 7 and 5.

Let $p=35m$ put in (1) we get

$49\times 25\times m^2 =49\times 5\times q^2$

hence q is multiple of 5.

therefore p and q are non co-prime numbers.

Hence $7{\sqrt{5}}$ is irrational.

Solution 3 (iii) :-  Let $p,\,q\in N$ such that $p,\,q$ are co-prime numbers such that  $6+\sqrt{2}=\cfrac{p}{q}$

$therefore \sqrt{2}=\cfrac{p-6q}{q}$

Since Right Hand Side is rational and we know that $\sqrt{2}$ is irrational hence, Not possible.

Hence $6+\sqrt{2}$ is irrational.

## Exercise 1.4

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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:

(i) $\cfrac{13}{3125}$

(ii) $\cfrac{17}{8}$

(iii) $\cfrac{64}{455}$

(iv) $\cfrac{15}{1600}$

(v) $\cfrac{29}{343}$

(vi) $\cfrac{23}{2^3 5^2}$

(vii) $\cfrac{129}{2^2 5^7 7^5}$

(viii) $\cfrac{6}{15}$

(ix) $\cfrac{35}{50}$

(x) $\cfrac{77}{210}$

Solution 1.  We know that in its simplest form for rational number $\cfrac{p}{q}$ form if Denominator $q=2^n 5^m$ form for $n,m$ is a whole number. Than the number has a terminating decimal.

Now (i) 3125 = $5^5$ Hence number has terminating decimal.

(ii) 8 = $2^3$ Hence number has terminating decimal.

(iii) 455 = $5\times 7\times 13$ Hence non-terminating repeating.

(iv) $\cfrac{15}{1600}=\cfrac{3}{320}$  where 320 = $2^6 5$  Hence number is terminating decimal.

(v) Denominator is $343\neq 2^n 5^m$ for any whole number n and m. Hence is non-terminating decimal

(vi)  Denominator is $2^3 5^2$ Hence terminating decimal.

(vii) Denominator is $2^2 5^7 7^5$ Hence number non-terminating decimal.

(viii) $\cfrac{6}{15}=\cfrac{2}{5}$. Since Denominator is 5 . Hence number is having terminating decimal.

(ix) Denominator is $2\times 5^2$ hence is terminating decimal.

(x) $\cfrac{77}{210}=\cfrac{11}{30}$ Denominator is of form $2\times 3\times 5$ Hence non-terminating decimal

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

Solution 2.  We know that in its simplest form for rational number $\cfrac{p}{q}$ form if Denominator $q=2^n 5^m$ form for $n,m$ is a whole number. Than the number has a terminating decimal.

Now (i) 3125 = $5^5$ Hence number has terminating decimal.

Number in decimal form will be $=\cfrac{13}{3125}$

$=\cfrac{13}{5^5}(\cfrac{2^5}{2^5})$

=$\cfrac{13\times 32}{10^5}=0.00416$

(ii) Number is $\cfrac{17}{8} =\cfrac{17}{2^3}(\cfrac{5^3}{5^3})$

=$\cfrac{2125}{1000}=2.125$

(iv) $\cfrac{15}{1600}=\cfrac{3}{320}$  where 320 = $2^6 5$  Hence number is terminating decimal.

Hence $\cfrac{15}{1600}=\cfrac{3}{2^6\times 5}(\cfrac{5^5}{5^5})$

=$\cfrac{9375}{10^6}=0.009375$

(vi)  Denominator is $2^3 5^2$ Hence terminating decimal.

Number is $\cfrac{23}{2^3 5^2}=\cfrac{23}{2^3 5^2}(\cfrac{5}{5})$

$=\cfrac{115}{10^3}=0.115$

(viii) $\cfrac{6}{15}=\cfrac{2}{5}$. Since Denominator is 5 . Hence number is having terminating decimal.

The number is $\cfrac{2}{5}(\cfrac{2}{2})$

$=\cfrac{4}{10}=0.4$

(ix) Denominator is $2\times 5^2$ hence is terminating decimal.

The Number is $\cfrac{35}{50}=0.7$

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 $\cfrac{p}{q}$, what can you say about the prime factor of q?

(i) 43.123456789

(ii) 0.120120012000120000….

(iii) $43.\overline{123456789}$

Solution 3. (i)  Number is 43.123456789 a terminating decimal is a rational number.

$\cfrac{p}{q}$ form of number is $=\cfrac{43123456789}{10^9}$

Hence prime factors of q are 2 and 5.

(ii)  Number 0.120120012000120000…. is non-terminating decimal. Hence an irrational number.

(iii)  Number $43.\overline{123456789}$ if expressed in the form $\cfrac{p}{q}$ then $q=3^2\times 11\times 1010101$ Hence prime factors of q are 3, 11, 1010101.