The DAV Class 7 Maths Book Solutions Pdf and **DAV Class 7 Maths Chapter 3 Worksheet 2** Solutions of Rational Numbers as Decimals offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 3 WS 2 Solutions

Question 1.

Express the following rational numbers as decimals by using long division method.

(i) \(\frac{21}{16}\)

Answer:

Hence \(\frac{21}{16}\) = 1.3125 (Terminating decimal)

(ii) \(\frac{129}{25}\)

Answer:

Hence \(\frac{129}{25}\) = 5.16 (Terminating decimal)

(iii) \(\frac{17}{200}\)

Answer:

Hence \(\frac{17}{200}\) = 0.085 (Terminating decimal)

(iv) \(\frac{5}{11}\)

Answer:

Hence \(\frac{5}{11}\) = 0.454545…. = 0.\(\overline{45}\)

(Non – terminating repeating decimal)

(v) \(\frac{22}{7}\)

Answer:

Hence \(\frac{22}{7}\) = 3.1428 (Non – terminating decimal)

(vi) \(\frac{31}{27}\)

Answer:

Hence \(\frac{31}{27}\) = 1.148148……..

= 1.\(\overline{148}\) (Non – terminating and repeating decimal)

(vii) \(\frac{2}{15}\)

Answer:

Hence \(\frac{2}{15}\) = 0.1333……….

= 0.13̄ (Non – terminating repeating decimal)

(viii) \(\frac{63}{55}\)

Answer:

Hence \(\frac{63}{55}\) = 1.14545……….

= 1.1\(\overline{45}\) (Non – terminating repeating decimal)

Question 2.

Without actual division, determine which of the following rational numbers have a terminating decimal representation and which have a non-terminating decimal representation.

(i) \(\frac{11}{4}\)

Answer:

Prime factors of 4 = 2 × 2

The denominator 7 has no prime factors 2^{2}

∴ \(\frac{11}{4}\) represents a terminating decimal.

(ii) \(\frac{13}{80}\)

Answer:

Prime factors of 80 = 2 × 2 × 2 × 2 × 5

= 2^{4} × 5^{1}

Hence \(\frac{13}{80}\) represents a terminating decimal.

(iii) \(\frac{15}{11}\)

Answer:

The denominator 11 has no prime factors as 2^{m} × 5^{n}

Hence \(\frac{15}{11}\) represents a non-terminating decimal.

(iv) \(\frac{22}{7}\)

Answer:

The denominator 11 has no prime factors as 2^{m} × 5^{n}

Hence \(\frac{22}{7}\) represents a non-terminating decimal.

(v) \(\frac{29}{250}\)

Answer:

Prime factors of 250 = 2 × 5 × 5 × 5 = 2^{1} × 5^{3}

Here, the prime factors of 250 has 2 and 5 only

Hence \(\frac{29}{250}\) represents a terminating decimal.

(vi) \(\frac{37}{21}\)

Answer:

Prime factors of 21 = 3 × 7

The denominator has no prime factors as 2^{m} × 5^{n}

Hence \(\frac{37}{21}\) represents a non-terminating decimal.

(vii) \(\frac{49}{14}\)

Answer:

Prime factors of denominator are in the form 2^{1} × 5^{0}

Hence \(\frac{49}{14}\) represents a terminating decimal

(viii) \(\frac{126}{45}\)

Answer:

Prime factors of denominator are in the form 2^{0} × 5^{1}

Hence \(\frac{126}{45}\) represents a terminating decimal.

Question 3.

Find the decimal representation of the following rational numbers.

(i) \(\frac{-27}{4}\)

Answer:

\(\frac{-27}{4}=\frac{-27 \times 25}{4 \times 25}\)

= \(\frac{-675}{100}\)

= -6.75

(ii) \(\frac{-37}{60}\)

Answer:

60 = 2 × 2 × 3 × 5 (not in the form of 2^{m} × 5^{n})

So it is a non-terminating decimal.

∴ We will use long division method.

Hence \(\frac{-37}{60}\) = -0.61666…..

= -0.616̄ (Non – terminating repeating decimal)

(iii) \(\frac{-18}{125}\)

Answer:

125 = 5 × 5 × 5 = 5^{3} × 2^{0} (Terminating decimal)

∴ \(\frac{-18 \times 8}{125 \times 8}=\frac{-144}{1000}\)

= -0.1444

(iv) \(\frac{-15}{8}\)

Answer:

8 = 2 × 2 × 2 = 2^{3} × 5^{0} (Terminating decimal)

= \(\frac{-15}{8}=\frac{-15 \times 125}{8 \times 125}\)

= \(\frac{-1875}{1000}\) = -1.875

Question 4.

If the number y + y non-terminating? Justify your answer.

Answer:

The denominators are 2, 3, 4, 5. They are not all in the form of 2^{m} × 5^{n}.

Therefore, the given numbers is non-terminating decimal representation.

Question 5.

Justify the following statements as true or false.

(i) \(\frac{22}{7}\) can be represented as terminating decimal.

Answer:

(F) \(\frac{22}{7}\), Here prime factors of 7 are not in the form 2^{m} × 5^{n}.

Hence it can not be represented as terminating decimal.

(ii) \(\frac{51}{512}\) can be represented as a terminating decimal.

Answer:

(T) \(\frac{51}{512}\), Here the prime factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29

It is in the form of 2^{m} × 5^{n} i.e. 2^{9} × 5°

Hence it is a terminating decimal representation.

(iii) \(\frac{19}{45}\) can be represented as a non-terminating repeating decimal.

Answer:

(T), Here prime factors of 45 = 3 × 3 × 5 are not in the form of 2^{m} × 5^{n}. So it can not be represented as terminating decimal.

(iv) \(\frac{3}{17}\) cannot be represented as a non-terminating repeating decimal.

Answer:

(T) Here, the denominator 4 × 5 is in the form of 2^{m} × 5^{n}. So, it is a terminating decimal representation.

(v) If \(\frac{3}{2}\) and \(\frac{7}{5}\) are terminating decimals, then \(\frac{3}{2}+\frac{7}{5}\) is also a terminating decimal.

Answer:

(T) \(\frac{3}{17}\), Here, prime factors of 17 are not in the form of 2^{m} × 5^{n}. So it is a non-terminating decimal representation.

(vi) If \(\frac{1}{4}\) and \(\frac{1}{5}\) both have terminating decimal representation, then \(\frac{1}{4} \times \frac{1}{5}\) also has a terminating decimal representation.

Answer:

(T) Here, denominators are 2 and 5, \(\frac{3}{2}+\frac{7}{5}\) is also a terminating decimal.