Class 6 Math Solutions WBBSE Chapter 10 Recurring Decimal Numbers Exercise 10
Question 1. Identify which is a Terminating and Non-terminating decimal number
⇒ \(\frac{1}{3}, \frac{7}{9}, \frac{1}{6}, \frac{7}{11}, \frac{11}{12}, \frac{15}{37}, \frac{2}{15}, \frac{49}{63}, \frac{11}{37}, \frac{12}{70}, \frac{1}{2}, \frac{9}{45}, 11 \frac{10}{12}, \frac{6}{13}\)
Solution:
Given
\(\frac{1}{3}, \frac{7}{9}, \frac{1}{6}, \frac{7}{11}, \frac{11}{12}, \frac{15}{37}, \frac{2}{15}, \frac{49}{63}, \frac{11}{37}, \frac{12}{70}, \frac{1}{2}, \frac{9}{45}, 11 \frac{10}{12}, \frac{6}{13}\)
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⇒ These are non-terminating decimals
⇒ These are non-terminating decimals
⇒ These are terminating decimals
Question 2. Identify which is Mixed recurring and pure recurring decimal number.
⇒ \(\frac{5}{6}, \frac{34}{510}, \frac{1}{15} \frac{52}{41}, \frac{15}{13}, \frac{4}{7}, \frac{6}{7}, \frac{8}{9}, \frac{5}{11}, \frac{7}{11}, \frac{3}{13}, \frac{4}{15}, \frac{13}{15}\)
Class 6 Math Solutions WBBSE
Solution:
Given
\(\frac{5}{6}, \frac{34}{510}, \frac{1}{15} \frac{52}{41}, \frac{15}{13}, \frac{4}{7}, \frac{6}{7}, \frac{8}{9}, \frac{5}{11}, \frac{7}{11}, \frac{3}{13}, \frac{4}{15}, \frac{13}{15}\)
⇒ These are mixed recurring decimals
⇒ These are mixed recurring decimals
Class 6 Math Solutions WBBSE
⇒ These are mixed recurring decimals
Question 3. Let’s directly convert the following pure recurring decimal numbers to corresponding fractions:
1. \(0 . \dot{3}\)
Solution:
\(0 . \dot{3}\) = \(\frac{3}{9}\)
2. \(0 . \dot{8}\) = ______
Solution:
= \(\frac{8}{9}\)
3. \(0 . \ddot{1} \dot{8}\) = _______
Solution:
= \(\frac{18}{99} = \frac{2}{11}\)
4. \(0 . \dot{2} \dot{7}\) = _______
Solution:
= \(\frac{27}{99}=\frac{3}{11}\)
5. \(0 . \dot{1} 6 \dot{2}\) _____
Solution:
= \(\frac{162}{999}=\frac{18}{111}\)
6. \(0 . \dot{2} 9 \dot{7}\) = ______
Solution:
= \(\frac{297}{999}=\frac{33}{111}\)
7. \(0 . \dot{5} 64 \dot{3}\) = ________
Solution:
= \(\frac{5643}{9999}=\frac{627}{1111}\)
Question 4. Let’s express the following pure recurring decimal numbers in fractions
⇒ \(0 . \dot{5}, 0 . \ddot{4} \dot{5}, 0 . \ddot{5} \dot{3}, 0 . \ddot{1}, \quad 0 . \dot{5} \dot{1}\)
Solution:
⇒ \(0 . \dot{5}=\frac{5}{9} ; \quad 0 . \dot{4} \dot{5}=\frac{45}{99}=\frac{5}{11} ; \quad 0 . \dot{5} \dot{3}=\frac{53}{99}\)
⇒ \(0 . \ddot{12}=\frac{12}{99}=\frac{4}{33} ; \quad 0 . \dot{5} 1 \dot{2}=\frac{512}{999}\)
Class 6 Maths Solutions WBBSE
Question 5. Convert the following mixed recurring decimal numbers into fractions:
1. \(0.2 \dot{7}\)
Solution:
= \(\frac{27-2}{90}\)=\(\frac{25}{90}\) = \(\frac{5}{18}\)
2. \(0.08 \ddot{1}\)
Solution:
= \(\frac{81}{990}=\frac{9}{110}\)
3. \(2.8 \dot{2}\)
Solution:
= \(\frac{282-28}{90}=\frac{254}{90}\)
= \(\frac{127}{45}\)
4. \(0.2 \ddot{7} \dot{2}\)
Solution:
= \(\frac{272-2}{990}=\frac{270}{990} =\frac{3}{11}\)
5. \(3.4 \ddot{3} \dot{2}\)
Solution:
= \(\frac{3432-34}{990}=\frac{3398}{990} =\frac{1699}{495}\) .
Class 6 Maths Solutions WBBSE Chapter 10 Recurring Decimal Numbers Exercise 10.1
Question 1. By actual division let us find if the quotients are terminating decimal numbers or recurring decimal numbers
1. A 7 m long ribbon is divided into 8 equal pieces, let’s find the length of each piece.
Solution:
Given
7m long ribbon is divided into 8 equal pieces.
∴ Length of each piece = \(\frac{7}{8}\) m.
= 0.875 m (T erminating)
2. Let’s find the weight of each packet of sugar when 11 kg sugar is divided equally in 12 packets.
Solution:
Given
11 kg sugar is divided in 12 packets.
∴ Each packet contains \(\frac{11}{12}\)kg.
= 0.9166 kg
= 0.916 kg (Recurring)
3. Let’s calculate the amount of water each bottle can hold when 12 litres water is poured into 7 equal-sized bottles.
Solution:
The amount of water in each bottle = \(\frac{12}{7}\) litre.
= 1.7142857
= 1.71428 5 litre (Recurring)
WBBSE Math Solutions Class 6
4. If 15 trees are planted along the side of a 24 m long road including its two ends, let’s find the distance between two consecutive trees.
Solution:
Given
If 15 trees are planted along the side of a 24 m long road including its two ends
The distance between two consecutive trees
= \(\frac{24}{15}\) m= \(\frac{8}{5}\) m = 1.6 m. (Terminating)
Question 2. Express the following fractions in decimal fractions and identify the terminating and recurring decimal numbers:
Solution:
Question 3. Let’s identify the pure and mixed recurring decimal numbers and convert them to fractions:
WBBSE Math Solutions Class 6
1. \(0 . \dot{5} \dot{4}\)
Solution:
= \(\frac{54}{99}=\frac{6}{11}\)
2. \(0 . \ddot{3} \dot{9}\)
Solution:
= \(\frac{39}{99}=\frac{13}{33}\)
3. \(0.0 \dot{2} \dot{4}\)
Solution:
= \(\frac{24}{990}=\frac{4}{165}\)
4. \(0 . \dot{6} \dot{9}\)
Solution:
= \(\frac{69}{99}=\frac{23}{33}\)
5. \(0 . \dot{9} \dot{3}\)
Solution:
= \(\frac{93}{99}=\frac{31}{33}\)
6. \(0.0 \ddot{8} \ddot{1}\)
Solution:
= \(\frac{81}{990}=\frac{9}{110}\)
7. \(0.27 \dot{2}\)
Solution:
= \(\frac{272-27}{900}=\frac{245}{900}\)
= \(\frac{49}{180}\)
WBBSE Math Solutions Class 6
8. \(0 . \dot{5} 1 \dot{3}\)
Solution:
= \(\frac{513}{999}=\frac{19}{37}\)
9. \(0 . \dot{14} \dot{4}\)
Solution:
= \(\frac{144}{999}\)
10. \(3: 4 \ddot{3} \dot{2}\)
Solution:
= \(\frac{3432-34}{990}=\frac{3398}{990}\)
= \(\frac{1699}{495}=3 \frac{214}{495}\)
11. \(7.0 \ddot{2} \dot{8}\)
Solution:
= \(\frac{7028-70}{990}=\frac{6958}{990}\)
= \(\frac{3479}{495}=7 \frac{14}{495}\)
12. \(0 . \dot{3} 7 \dot{5}\)
Solution:
= \(\frac{375}{999}\)=\(\frac{125}{333}\)
13. \(0 . \dot{29} \dot{1}\)
Solution:
= \(\frac{291}{999}=\frac{97}{333}\)
14. \(3 . \dot{200}\)
Solution:
= \(\frac{3205-3}{999}\)=\(\frac{3202}{999}\)
15. 0.375
Solution:
= \(\frac{375}{999}=\frac{125}{333}\)
16. \(0 . \dot{29 \dot{1}}\)
Solution:
= \(\frac{291}{999}=\frac{97}{333}\)
WBBSE Math Solutions Class 6
17. \(3 . \dot{20} \dot{5}\)
Solution:
= \(\frac{3205-3}{999}=\frac{3202}{999}\)
= \(3 \frac{205}{999}\)
18. \(0 . \dot{0} 12 \dot{1}\)
Solution:
= \(\frac{121}{9999}=\frac{11}{909}\)
Pure: 1, 2, 4, 5, 8, 9, 12, 13, 14, 15
Mixed: 3, 6, 7, 10, 11
Question 4. Let’s arrange the following in ascending order:
1. \(0 . \dot{3}, 0 . \dot{1}, 0 . \dot{1}\)
Solution:
⇒ \(0 . \dot{3}=0.333\)
⇒ \(0 . \dot{16}=0.166\)
⇒ \(0 . \dot{1}=0.111\)
∴ In ascending order: 0.111 ; 0.166; 0.333
i.e., \(0 . \dot{1}, 0 . \dot{1}, 0 . \dot{3}\).
2. \(0 . \dot{6} \dot{3} ; \frac{3}{4}, \frac{5}{6}\)
Solution:
⇒ \(0 . \ddot{6} \dot{3}=0.6363\)
⇒ \(\frac{5}{6}=0.8333\)
⇒ \(\frac{3}{4}=0.7500\)
∴ In ascending order: 0.6363; 0.7500; 0.8333
i.e., \(0 . \dot{6} 3 ; \frac{5}{6}, \frac{3}{4}\)
3. \(0.5 \dot{3} ; \frac{2}{25} ; \frac{16}{75}\)
⇒ \(0.5 \dot{3}=0.533\)
⇒ \(\frac{2}{25}=0.80\)
⇒ \(\frac{16}{75}=0.213\)
∴ In ascending order: 0.080; 0.213,0.533
i.e., \(\frac{2}{25}, \frac{16}{75}, 0.5 \dot{3}\)
4. \(0.91 \dot{6}, \frac{1}{121} ; \frac{3}{44}\)
⇒ \(0.916=0.916000\)
⇒ \(\frac{1}{121}\)=0.008264
⇒ \(\frac{3}{44}=0.068181\)
∴ In ascending order: 0.008264; 0.068181; 0.916000
i.e., \(\frac{1}{121} ; \frac{3}{44} ; 0.916\)