NCERT Solutions for Maths Class 10 Chapter 1

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 \(225\)

Answer:

(i) \(135\) and \(225\)

Since \(225 > 135\), we apply the division lemma to \(225\) and \(135\)  to obtain

\(225 = 135\times 1 + 90\)

Since remainder \(90 ≠ 0\), we apply the division lemma to \(135\) and \(90\) to obtain

\(135 = 90\times 1 + 45\)

We consider the new divisor \(90\) and new remainder \(45\), and apply the division lemma to obtain

\(90 = 2 × 45 + 0 \)

Since the remainder is zero, the process stops.
Since the divisor at this stage is \(45\), Therefore,
the HCF of \(135 \) and \(225\) is \(45\).

Therefore, the \(HCF\) of \(135\) and \(125\) is \(45\).

\(135\begin{vmatrix}225\cr \underline{-135}\end{vmatrix}1\)
                  \(\underline{90}\)

    \(90\begin{vmatrix}135\cr \underline{-90}\end{vmatrix}1\)
                  \(\underline{45}\)

     \(45\begin{vmatrix}90\cr \underline{-90}\end{vmatrix}2\)
                  \(\underline{0}\)

(ii) \(196\) and \(38220\)

Since \(38220 \gt 196\), we apply the division lemma to \(38220\) and \(196\) to obtain

\(38220 = 196 \times 195 + 0\)

Since the remainder is zero, the process stops.

Since the divisor at this stage is \( 196\),

Therefore, \(HCF\) of  \(196\) and \(38220 \) is \(196\).

\(196\begin{vmatrix}38220\cr \underline{-196}\end{vmatrix}195\)
                  \({1862}\)
              \(\underline{-1764}\)
                     \({980}\)
                 \(\underline{-980}\)
                         \(\underline{0}\)

\(255\begin{vmatrix}867\cr \underline{-765}\end{vmatrix}3\)
                \({\underline {102}}\)

  \(102\begin{vmatrix}255\cr \underline{-204}\end{vmatrix}2\)
                    \({\underline {51}}\)

    \(51\begin{vmatrix}102\cr \underline{-102}\end{vmatrix}2\)
                      \({\underline {0}}\)

(iii) 867 and 255

Since \(867 \gt  255\), we apply the division lemma to \(867\) and \(255\)  to obtain

\(867 = 255 \times 3 + 102\)

Since remainder \(102\ne 0\), we apply the division lemma to \(255\) and \(102\) to obtain

\(255 = 102\times2 + 51\)

We consider the new divisor  \(102\)  and new remainder \(51\), and apply the division lemma to obtain

\(102 = 51 \times2 + 0\)

Since the remainder is zero, the process stops. Since the divisor at this stage is \(51\),
Therefore, \(HCF\) of  \(867\) and \(255\) is \(51\).

We consider the new divisor \(102\) and new remainder \(51\), and apply the division lemma to obtain \(102=51×2+ 0\)

Since the remainder is zero, the process stops. Since the divisor at this stage is \(51\),
Therefore, HCF of \(867\) and \(255\) is \(51\).

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.
Answer:

Let a be any positive integer and \(b =\) 6. Then, by Euclid’s algorithm, \(a = 6q + r\) for some integer \(q ≥ 0\), 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, \(6q + 1 = 2 \times 3q + 1 = 2k1 + 1\), 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 \(6q + 5 \)

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?

Answer:

\(HCF\space (616,\space 32)\) will give the maximum number of columns in which they can march.

We can use Euclid’s algorithm to find the \(HCF\).

\(616 = 32 \times 19 + 8\)

\(32 = 8 × 4 + 0 \)

The \(HCF\space (616\space, 32)\) is \(8\)..

Therefore, they can march in \(8\) columns each.

Question 4:

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

[Hint: Let x be any positive integer then it is of the form \(3q\space, 3q + 1\)  or\( 3q + 2\). Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Answer:

Let a be any positive integer and \(b=3\).
Then \(a = 3q + r\) for some integer \(q ≥ 0\)
And \(r = 0,\space 1,\space 2\) because \(0 ≤ r < 3\)
Therefore, \(a = 3q\) or \(3q + 1\) or \(3q + 2\)
Or,

\(a^2=(3q)^2\)  or \((3q+1)^2\)  or \((3q+2)^2\)

\(a^2=(9q)^2\)  or \(9q^2+6q+1\) or \(9q^2+12q+4\)

\(=3×(3q)^2\)  or \(3(3q^2+2q)+1\) or \(3(3q^2+4q+1)+1\)

\(=3k_1\)  or \(3k_2+1\) or \(3k_3+1\)

Where \(k_1,\space k_2\) and \(k_3\) are some positive integers, Hence, it can be said that the square of any positive integer is 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 \(9m,\space 9m + 1\) or \(9m + 8\).

Ans.:   Let a be any positive integer and \(b = 3\)

\(a = 3q + r\), where \(q \ge 0\) and \(0 \le r \gt 3\)

\(∴a=3q\) or \(3q+1\) or \(3q+2\)

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

Case 1:

When \(a = 3q\),

\(a^3=(3q)^3=27q^3=9(3q^3 )=9m\)

Where m is an integer such that \(m=3q^3\)

Case 2:

When \(a = 3q + 1\)

\(a^3 = (3q +1)^3\)

\(a^3 = 27q^3 + 27q^2 + 9q + 1\)
\(a^3 = 9(3q^3 + 3q^2 + q) + 1\)
\(a^3 = 9m + 1\)

Where m is an integer such that \(m = (3q^3 + 3q^2 + q)\)

Case 3:

When \(a = 3q + 2\)

\(a^3 = (3q +2)^3\)

\(a^3 = 27q^3 + 54q^2 + 36q + 8\)
\(a^3 = 9(3q^3 + 6q^2 + 4q) + 8\)
\(a^3 = 9m + 8\)

Where m is an integer such that \(m = (3q^3 + 6q^2 + 4q)\)

Therefore, the cube of any positive integer is of the form \(9m,\space 9m + 1\) or \(9m + 8\)

Exercise 1.2

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

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

Answer: 

i. \(140=2\times2\times5\times7=2^2\times5\times7\)

ii. \(156=2\times2\times3\times13=2^2\times3\times13\)

iii. \(3825=3\times3\times5\times5\times17=3^2\times5^2\times17\)

iv. \(5005=5\times7\times11\times13\)

v. \(7429=17\times19\times23\)

Q.2: Find the \(LCM\) and \(HCF\) of the following pairs of integers and verify that \(LCM \times HCF =\) product of the two numbers.

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

Answer: 

i. 26 and 91

\(29=2\times13\)

\(91=7\times13\)

\(HCF=13\)

\(LCM=2\times7\times13=182\)

Product of two numbers \(=26\times91=2366\)

\(HCF\times LCM=13\times182=2366\)

Hence, product of two numbers \(=HCF\times LCM\)

ii. \(510\) and \(92\)

\(5190=2\times3\times5\times17\)

\(92=2\times2\times23\)

\(HCF=2\)

\(LCM=2\times2\times3\times5\times17\times23=23460\)

Product of the two numbers \(=510\times92=496920\)

\(HCM×LCM=2\times23460\)

\(HCF×LCM=2×23460\)

                            \(=46920\)

Hence, product of two numbers \(=HCF×LCM\)

iii. \(336\) and \(54\)

\(336=2×2×2×2×3×7\)

\(336=2^4×3×7\)

\(54=2×3×3×3\)

\(54=2×3^3\)

\(HCF=2×3=6\)

\(LCM=2^4×3^3×7=3024\)

Product of the two numbers \(=336×54=18144\)

\(HCF×LCM=6×3024=18144\)

Hence, product of two numbers \(=HCF×LCM\)

Q.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

Answer:

(i) \(2, 15\) and \(21\)

\(12=2^2×3\)

\(15=3×5\)

\(21=3×7\)

\(HCF=3\)

\(LCM=2^2×3×5×7=420\)

Correct: \(HCF=3\),     \(LCM=420\)

(ii) \(17,\space 23\) and \(29\)

\(17=1×7\)

\(23=1×23\)

\(29=1×29\)

\(HCF=1\)

\(LCM=17×23×29=11339\)

Correct:     \(HCF=1\),     \(LCM = 11339\)

(iii) \(8, 9\) and \(25\)

\(8=2×2×2\)

\(9=3×3\)

\(25=5×5\)

\(HCF=1\)

\(LCM=2×2×2×3×3×5×5=1800\)

Correct:   \(HCF=1\)         \(LCM=1800\)

Q.4:  Given that \(HCF\space (306,\space 657) = 9\), find \(LCM\space (306,\space 657)\).

Answer:  \(HCF\space(306,\space 657)=9\)

We know that, LCM×HCF=\)Product of two numbers

\(∴LCM×HCF=306×657 \)

\(LCM=\frac{306\times657}{HCF}=\frac{306\times657}{9}\)

\(LCM=22338\)

Q.5: Check whether \(6^n\) can end with the digit \(0\) for any natural number \(n\). 

Answer: 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 \(6^n = (2 ×3)^n\)

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

Hence, for any value of n, \(6^n\) will not be divisible by \(5\).

Therefore, \(6^n\) cannot end with the digit 0 for any natural number \(n\).

Q.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 factorised further. Therefore, the given expression has \(5\) and \(1009\) as its factors. Hence, it is a composite number.

Q.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? 

Answer:

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.

Correct Answer: 36 minute

Exercise 1.3

Q.1:     Prove that \(\sqrt5\) is irrational.

Answer:

Let \(\sqrt5\)  is a rational number.

Therefore, we can find two integers \(a,\space b\space (b ≠ 0)\) such that \(\sqrt5=\frac{a}{b}\)

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.

\(a=\sqrt5b\)

\(a^2=5b^2\)

Therefore, \(a^2\) is divisible by \(5\) and it can be said that a is divisible by \(5\).

Let \(a = 5k\), where k is an integer

\((5k)^2=5b^2\)

\(b^2=5k^2\)

This means that \(b^2\) is divisible by \(5\) and hence, \(b\) is divisible by \(5\).

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.

Here, \(\sqrt5\) cannot be expressed as \(\frac{p}{q}\) or it can be said that \(\sqrt5\) is irrational.

Q.2: Prove that \(3+2\sqrt5\) is irrational.

Answer: 

Let \(3+2\sqrt5\) is rational

Therefore, we can find two integers a, \(b\space (b \ne 0)\) such that

\(3+2\sqrt5=\frac{a}{b}\)

\(2\sqrt5=\frac{a}{b}-3\)

\(\sqrt5=\frac{1}{2}(\frac{a}{b}-3)\)

Since \(a\) and \(b\) are integers, \(\frac{1}{2}(\frac{a}{b}-3)\) will also be rational and therefore, \(\sqrt5\) is rational.

This contradicts the fact that \(\sqrt5\) is irrational.

Hence, our assumption that \(3+2\sqrt5\) is rational is false.

Therefore, \(3+2\sqrt5\) is irrational.

(ii) \(510\) and \(92\)

\(5190=2×3×5×17\)

\(92=2×2×23\)

\(HCF=2\)

\(LCM=2×2×3×5×17×23=23460\)

Product of the two numbers \(=510×92=496920\)

\(HCM×LCM=2×23460\)

\(HCF×LCM=2×23460\)

                  \(=46920\)

Hence, product of two numbers \(=HCF×LCM\)

(iii) \(336\) and \(54\)

\(336=2×2×2×2×3×7\)

\(336=2^4×3×7\)

\(54=2×3×3×3\)

\(54=2×3^3\)

\(HCF=2×3=6\)

\(LCM=2^4×3^3×7=3024\)

Product of the two numbers \(=336×54=18144\)

\(HCF×LCM=6×3024=18144\)

Hence, product of two numbers \(=HCF×LCM\)

Q.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

Amswer: 

(i) \(12,\space 15\) and \(21\)

\(12=2^2×3\)

\(15=3×5\)

\(21=3×7\)

\(HCF=3\)

\(LCM=2^2×3×5×7=420\)

Correct: \(HCF=3\)  and  \(LCM=420\)

(ii) \(17,\space 23\) and \(29\)

\(17=1×7\)

\(23=1×23\)

\(29=1×29\)

\(HCF=1\)

\(LCM=17×23×29=11339\)

Correct:  \(HCF=1\)  and  \(LCM = 11339\)

(iii) \(8,\space 9\) and \(25\)

\(8=2×2×2\)

\(9=3×3\)

\(25=5×5\)

\(HCF=1\)

\(LCM=2×2×2×3×3×5×5=1800\)

Correct:  \(HCF=1\) and \(LCM=1800\)

Q.4:  Given that \(HCF\space (306, 657) = 9\), find \(LCM \space (306, 657)\)

Answer.:  \(HCF\space(306,\space 657)=9\)

We know that, \(LCM×HCF=\)Product of two numbers

\(∴LCM×HCF=306×657\)

\(LCM=\frac{306×657}{HCF}=\frac{306×657}{9}\)

\(LCM=22338\)

Correct: 22338

Q.5: Check whether \(6^n\) can end with the digit \(0\) for any natural number \(n\).

Answer: 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 \(6^n= (2 ×3)^n\)

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

Hence, for any value of n, \(6^n\) will not be divisible by \(5\).

Therefore, \(6^n\) cannot end with the digit \(0\) for any natural number \(n\).

Q.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 factorised further. Therefore, the given expression has \(5\)and \(1009\) as its factors. Hence, it is a composite number.

Q.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?

Answer.:   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.

Correct: \(36\) minute

Exercise 1.3

Q.1: Prove that \(\sqrt5\) is irrational.

Answer: Let \(\sqrt5\) is a rational number.

Therefore, we can find two integers \(a,\space b (b\ne0)\) such that \(\sqrt5=\frac{a}{b}\)

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. 

\(a=\sqrt5b\)

\(a^2=5b^2\)

Therefore, \(a^2\) is divisible by \(5\) and it can be said that a is divisible by \(5\).

Let \(a = 5k\), where k is an integer

\((5k)^2=5b^2\)

\(b^2=5k^2\)

This means that \(b^2\) is divisible by \(5\) and hence, \(b\) is divisible by \(5\).

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, \(\sqrt5\)  cannot be expressed as  \(\frac{p}{q}\) or it can be said that \(\sqrt5\)  is irrational.

Q.2: Prove that \(3+2\sqrt5\) is irrational.

Answer: Let \(3+2\sqrt5\) is irrational.

Therefore, we can find two integers \(a,\space\) (b\ne0)\) such that

\(3+2\sqrt5=\frac{a}{b}\)

\(2\sqrt5=\frac{a}{b}-3\)

\(\sqrt5=\frac{1}{2}(\frac{a}{b}-3)\)

Since \(a\) and \(b\) are integerst, \(\frac{1}{2}(\frac{a}{b}-3)\) will also be rationa and therefore, \(\sqrt5\) is rational.

This contradicts the fact that \(\sqrt5\) is irrationa. Hence, our assumption that \(3+2\sqrt5\) is rational is false. Therefore, \(3+2\sqrt5\) is irrational.

Q.3: Prove that the following are irrationals:

        (i) \(\frac{1}{\sqrt2}\)      (ii) \(7\sqrt5\)       (iii) \(6+\sqrt2\)

Answer: 

(i) \(\frac{1}{\sqrt2}\)

Let \(\frac{1}{\sqrt2}\) is rational.

Therfore, we can find two integers \(a,\space b\space(b\ne0)\) such that 

\(\frac{1}{\sqrt2}=\frac{a}{b}\)

\(\sqrt=\frac{b}{a}\)

\(\frac{b}{a}\) is rational as \(a\) and \(b\) are integers.

Therefore, \(\sqrt2\) is rational which contradicts to the fact that is \(\sqrt2\) irrational.

Hence, our assumption is false and is \(\frac{a}{\sqrt2}\) irrational.

(ii) \(7\sqrt5\)

Let \(7\sqrt5\) is rational.

\(∴ \) we can find two integers \(a,\space b \space(b ≠ 0)\) such that \(7\sqrt5=\frac{a}{b}\) for some integers \(a\) and \(b\).

(iii) \(6+\sqrt2\)

Let |(6+\sqrt2\) be rational.

Therefore, we cam find two integers \(a,\space b\space(b\ne0)\) such that

\(6+\sqrt2=\frac{a}{b}\)

\(\sqrt2=\frac{a}{b}-6\)

Since \(a\) and \(b\) are intergers, \(\frac{a}{b}-6\) is also rational and hence, \(\sqrt2\) should be rational. This contradicts the fact that \(\sqrt2\) is irrational.

Therfore, our assumption is false and hence, \(6+\sqrt2\) is irrational.

\(∴\sqrt5=\frac{a}{7b}\)

\(\frac{a}{7b}\) is rational as \(a\) and \(b\) are integers. Therefore, should be rational.

This contradicts the fact that \(\sqrt5\) is irrational. Therfore, our assumption that \(7\sqrt5\) is rational is false.

Hence, \(7\sqrt5\) is irrational.

Excercise 1.4

Q.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.

(i) \(\frac{13}{3125}\)         (ii) \(\frac{17}{8}\)         (iii) \(\frac{64}{455}\)         (iv) \(\frac{15}{1600}\)(

v) \(\frac{29}{343}\)          (vi) \(\frac{23}{2^35^2}\)         (vii) \(\frac{120}{2^25^77^5}\)

(viii) \(\frac{6}{15}\)          (ix) \(\frac{35}{50}\)         (x) \(\frac{77}{210}\)

Answer: 

(i) \(\frac{13}{3125}\)

\(3125=5^5\)

The denomonator is of the form \(5^m\)

Hence, the decimal expansion of \(\frac{13}{3125}\) is terminating.

(ii) \(\frac{17}{8}\)

\(8=2^3\)

Teh denominator is of the form \(2^m\).

Hence, the decimal expansion of \(\frac{17}{8}\)  is terminating.

(iii) \(\frac{64}{455}\)

\(455=5\times7\times13\)

Since the denominator is not in the form \(2^m\times5^n\), and it also contains \(7\) and \(13\) as its factors, its decimal expansion will be non-terminating repeating.

(iv) \(\frac{15}{1600}\)

\(1600=2^6\times5^2\)

The denominator is of the form \(2^m\times5^n\).

Hence, the decimal expansion of \(\frac{15}{1600}\) is terminating.

(v) \(\frac{29}{243}\)

\(343=7^3\)

Since the denominator is not in the form \(2^m\times5^n\) and it has \(7\) as its factor, the decimal expansion of \(\frac{29}{343}\) is non-terminating repeating.

(vi) \(\frac{23}{2^3\times5^2}\)

The denominator is of the form \(2^m\times5^n\).

Hence, the decimal expansion of \(\frac{23}{2^3\times5^2}\) is terminsting.

(vii) \(\frac{129}{2^2\times5^7\times7^5}\)

Since the denominator is not in the form \(2^m\times5^n\) and it has \(7\) as its factor, the decimal expansion of \(\frac{129}{2^2\times5^7\times7^5}\) is non-terminating repeating.

(viii) \(\frac{6}{15}=\frac{2\times3}{3\times5}=\frac{2}{5}\)

The denominator is of the form \(5^n\).

Hence, the decimal expansion of  \(\frac{6}{15}\) is terminating

(ix) \(\frac{35}{50}=\frac{7\times5}{10\times5}=\frac{7}{10}\)

\(10=2\times5\)

The denominator is of the form \(2^m\times5^n\).

Hence, the decimal expansion of  \(\frac{35}{50}\)  is terminating.

(x) \(\frac{77}{210}=\frac{11\times7}{30\times7}=\frac{11}{30}\)

\(30=2\times3\times5\)

Since the denominator is not of the form \(2^m\times5^n\)and it also has \(3\) as its factors, the decimal expansion of   \(\frac{77}{210}\) is non-terminating repeating.

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

Answer:

Q.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  \(\frac{p}{q}\) what can you say about the prime factor of \(q\)?

(i) \(43.123456789\)   

(ii) \(0.120120012000120000…\)

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

Answer: 

(i) \(43.123456789\)

Since this number has a terminating decimal expansion, it is a rational number of the form  \(\frac{p}{q}\)  and \(q\) is of the form \(2^m×5^m\). i.e., the prime factors of \(q\) will be either \(2\) or \(5\) or both.

(ii) \(120120012000120000 …\)

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

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

Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form  \(\frac{p}{q}\)  and \(q\) is not of the form \(2^m×5^m\)  i.e., the prime factors of \(q\) will also have a factor other than \(2\) or \(5\).