RD Sharma Solutions Class 7 Fractions Exercise 2.1

RD Sharma Class 7 Solutions Chapter 2 Ex 2.1 PDF Free Download

RD Sharma Solutions Class 7 Chapter 2 Exercise 2.1

Exercise 2.1

Q1. Compare the following fractions by using the symbol > or < or =;

(i) \(\frac{7}{9} \; and \; \frac{8}{13}\)

(ii) \(\frac{11}{9} \; and \; \frac{5}{9}\)

(iii) \(\frac{37}{41} \; and \; \frac{19}{30}\)

(iv) \(\frac{17}{15} \; and \; \frac{119}{105}\)

Solution:

(i) We have,

\(\frac{7}{9} \; and \; \frac{8}{13}\)

Taking the LCM of 9 and 13, we get,

9 x 13 = 117

Now, we convert the given fractions to equivalent fractions by making the denominators 117,

\(\frac{7\times 13}{9\times 13} \; and \; \frac{8\times 9}{13 \times 9}\)

\(\frac{91}{117} \; and \; \frac{72}{117}\)

As we know, 91 > 72

Therefore, \(\frac{91}{117} \; > \; \frac{72}{117}\)

Hence, \(\frac{7}{9} \; > \; \frac{8}{13}\)

(ii) We have, \(\frac{11}{9} \; and \; \frac{5}{9}\)

The given fractions are equivalent fractions as the denominators are equal,

And we know that, 11 > 5

Therefore, \(\frac{11}{9} > \frac{5}{9}\)

(iii) We have, \(\frac{37}{41} \; and \; \frac{19}{30}\)

Taking the LCM of 41 and 30, we get,

41 x 30 = 1230

Now, we convert the given fractions to equivalent fractions by making the denominators 1230,

\(\frac{37 \times 30}{41 \times 30} \; and \; \frac{19 \times 41}{30 \times 41}\)

\(\frac{1110}{1230} \; and \; \frac{779}{1230}\)

Now, we clearly know 1110 > 779

Therefore, \(\frac{1110}{1230} \; > \; \frac{779}{1230}\)

Hence, \(\frac{37}{41} > \frac{19}{30}\)

(iv) \(\frac{17}{15} \; and \; \frac{119}{105}\)

Taking the LCM of 15 and 105, we get,

5 x 3 x 7 = 105

Now, we convert the given fractions to equivalent fractions by making the denominators 105,

\(\frac{17 \times 7}{15 \times 7}\) and \(\frac{119}{105}\)

= \(\frac{119}{105}\) and \(\frac{119}{105}\)

Now, we clearly know 119 = 119

Therefore, \(\frac{119}{105}\) = \(\frac{119}{105}\)

Hence, \(\frac{17}{15} = \frac{119}{105}\)

Q2. Arrange the following fractions in ascending order:

(i) \(\frac{3}{8}, \frac{5}{6}, \frac{6}{8}, \frac{2}{4}, \frac{1}{3}\)

(ii) \(\frac{4}{6}, \frac{3}{8}, \frac{6}{12}, \frac{5}{16}\)

Solution:

(i) We have, \(\frac{3}{8}, \frac{5}{6}, \frac{6}{8}, \frac{2}{4}, \frac{1}{3}\)

Taking the LCM of 8, 6, 8, 4 and 3, we get,

2 x 4 x 3 = 24

Now, we convert the given fractions to equivalent fractions by making the denominators 24,

\(\frac{3 \times 3}{8\times 3}, \frac{5 \times 4}{6 \times 4}, \frac{6\times 3}{8 \times 3}, \frac{2 \times 6}{4 \times 6}, \frac{1 \times 8}{3 \times 8}\)

\(\frac{9}{24}, \frac{20}{24}, \frac{18}{24}, \frac{12}{24}, \frac{8}{24}\)

We know that, 8 < 9 < 12 < 18 < 20

Therefore, \(\frac{8}{24} < \frac{9}{24} < \frac{12}{24} < \frac{18}{24} < \frac{20}{24}\)

Hence, \(\frac{1}{3} < \frac{3}{8} < \frac{2}{4} < \frac{6}{8} < \frac{5}{6}\)

(ii) We have, \(\frac{4}{6}, \frac{3}{8}, \frac{6}{12}, \frac{5}{16}\)

Taking the LCM of 6, 8, 12 and 16, we get,

2 x 2 x 2 x 2 x 3 = 48

Now, we convert the given fractions to equivalent fractions by making the denominators 48,

\(\frac{4 \times 8}{6 \times 8}, \frac{3 \times 6}{8 \times 6}, \frac{6 \times 2}{12 \times 2}, \frac{5\times 3}{16\times 3} \\ = \frac{32}{48}, \frac{18}{48}, \frac{12}{48}, \frac{15}{48}\\\)

We know that, 12 < 15 < 18 < 32

Therefore, \(\frac{12}{48} < \frac{15}{48} < \frac{18}{48} < \frac{32}{48},\\ Hence, \frac{6}{12} < \frac{5}{16} < \frac{3}{8} < \frac{4}{6}\)

Q3. Arrange the following fractions in descending order:

(i) \(\frac{4}{5}, \frac{7}{10}, \frac{11}{15}, \frac{17}{20}\)

(ii) \(\frac{2}{7}, \frac{11}{35}, \frac{9}{14}, \frac{13}{28}\)

Solution:

(i) We have, \(\frac{4}{5}, \frac{7}{10}, \frac{11}{15}, \frac{17}{20}\)

Taking the LCM of 5, 10, 15 and 20, we get,

5 x 2 x 2 x 3 = 60

Now, we convert the given fractions to equivalent fractions by making the denominators 48

\(\frac{4 \times 12}{5 \times 12}, \frac{7 \times 6}{10 \times 6}, \frac{11 \times 4}{15 \times 4}, \frac{17 \times 3}{20 \times 3} \\ \frac{48}{60},\frac{42}{60}, \frac{44}{60}, \frac{51}{60} \\ As\; we\; know\; 51 > 48 > 44 > 42\\ Therefore,\; \frac{51}{60} < \frac{48}{60} < \frac{44}{60} < \frac{42}{60}\\ Hence, \frac{17}{20} < \frac{4}{5} < \frac{11}{15} < \frac{7}{10}\)

(ii) \(\frac{2}{7}, \frac{11}{35}, \frac{9}{14}, \frac{13}{28}\)

Taking the LCM of 7, 35, 14 and 28, we get,

7 x 5 x 2 x 2 = 140

Now, we convert the given fractions to equivalent fractions by making the denominators 140

\(\frac{2 \times 20}{7 \times 20}, \frac{11 \times 4}{35 \times 4}, \frac{9 \times 10}{14 \times 10}, \frac{13 \times 5}{28 \times 5}\\ \frac{40}{140},\frac{44}{140}, \frac{90}{140}, \frac{65}{140} \\ As\; we\; know\; 40 > 44 > 65 > 90\\ Therefore,\; \frac{90}{140} < \frac{65}{140} < \frac{44}{140} < \frac{40}{140}\\ Hence, \frac{9}{14} < \frac{13}{28} < \frac{11}{35} < \frac{2}{7}\)

Q4. Write the equivalent fractions of \(\frac{3}{5}\)

Solution:

Multiplying or dividing both the numerator and denominator by the same number, so that the fraction keeps its value.

So the equivalent fractions of \(\frac{3}{5}\) are

\(\frac{3\times 2}{5\times 2}, \frac{3\times 3}{5\times 3}, \frac{3\times 4}{5\times 4}, \frac{3\times 5}{5\times 5}, \frac{3\times 6}{5\times 6}\\ \frac{6}{10}, \frac{9}{15}, \frac{12}{20}, \frac{15}{25}, \frac{18}{30}\\\)  are the five equivalent fractions of \(\frac{3}{5}\)

Q5. Find the sum:

(i) \(\frac{5}{8} + \frac{3}{10}\)

(ii) \(4\frac{3}{4} + 9\frac{2}{5}\)

(iii) \(\frac{5}{6} + 3 + \frac{3}{4}\)

(iv) \(2\frac{3}{5} + 4\frac{7}{10} + 2\frac{4}{15}\)

Solution:

(i) We have, \(\frac{5}{8} + \frac{3}{10}\)

Taking the LCM of 8 and 10, we get,

2 x 4 x 5 = 40

Now, we convert the given fractions to equivalent fractions by making the denominators 40

\(\frac{5 \times 5}{8 \times 5} + \frac{3 \times 4}{10 \times 4}\\ \frac{25}{40} + \frac{12}{40}\\ \frac{25 + 12}{40}\\ \frac{37}{40}\)

(ii) We have, \(4\frac{3}{4} + 9\frac{2}{5}\)

\(= \frac{19}{4} + \frac{47}{5}\\\)

Taking out the LCM of 4 and 5, we get,

4 x 5 = 20

Now, we convert the given fractions to equivalent fractions by making the denominators 20

\(= \frac{19 \times 5}{4 \times 5} + \frac{47 \times 4}{5 \times 4}\\ = \frac{95}{20} + \frac{188}{20}\\ = \frac{95 + 188}{20}\\ = \frac{283}{20}\)

(iii) We have, \(\frac{5}{6} + 3 + \frac{3}{4}\)

Taking out the LCM of 6 and 4, we get,

2 x 2 x 3 = 12

Now, we convert the given fractions to equivalent fractions by making the denominators 12

\(= \frac{5 \times 2}{6 \times 2} + \frac{3 \times 12}{12} + \frac{3 \times 3}{4 \times 3}\\ = \frac{10}{12} + \frac{36}{12} + \frac{9}{12}\\\)

\(= \frac{10 + 36 + 9}{12} = \frac{55}{12}\)

(iv) We have, \(2\frac{3}{5} + 4\frac{7}{10} + 2\frac{4}{15}\)

= \(\frac{13}{5} + \frac{47}{10} + \frac{34}{15}\)

Taking out the LCM of 5, 10  and 15 , we get,

5 x 2 x 3 = 30

Now, we convert the given fractions to equivalent fractions by making the denominators 30

\(\frac{78}{30} + \frac{141}{30} + \frac{68}{30}\\ = \frac{287}{30}\)

Q6. Find the difference of

(i) \(\frac{13}{24} \; and \;\frac{7}{16}\)

(ii) 6 and \(\frac{23}{3}\)

(iii) \(\frac{21}{25} and \frac{18}{20}\)

(iv) \(3\frac{3}{10} \; and \; 2\frac{7}{15}\)

Solution:

(i) We have, \(\frac{13}{24} \; and \;\frac{7}{16}\)

Taking out the LCM of 24 and 16, we get,

2 x 2 x 2 x 2 x 3 = 48

Now, we convert the given fractions to equivalent fractions by making the denominators 48

\(\frac{26}{48} – \frac{21}{48}\\ = \frac{26 – 21}{48}\\ = \frac{5}{48}\)

(ii) We have, 6 and \(\frac{23}{3}\)

The difference between 6 and \(\frac{23}{3}\)

= \(\frac{23}{3} – \frac{18}{3}\)

= \(\frac{23 – 18}{3}\\ = \frac{5}{3}\)

(iii) We have, \(\frac{21}{25} and \frac{18}{20}\)

Taking out the LCM of 25 and 20, we get,

5 x 5 x 4 = 100

Now, we convert the given fractions to equivalent fractions by making the denominators 100

\(\frac{21 \times 4}{25 \times 4}\; and \;\frac{18 \times 5}{20 \times 5 } \\ = \frac{84}{100}\; and \;\frac{90}{100} \\ The \; difference \; between \; both \; the \; fractions \; are \\ = \frac{90 – 84}{100}\\ = \frac{6}{100} \\ = \frac{3}{50}\)

(iv) We have, \(3\frac{3}{10} \; and \; 2\frac{7}{15}\)

= \(\frac{33}{10} \; and \; \frac{37}{15}\)

Taking out the LCM of 10 and 15, we get,

2 x 3 x 5 = 30

Now, we convert the given fractions to equivalent fractions by making the denominators 30

\(= \frac{33 \times 3}{10 \times 3} \; and \; \frac{37 \times 2}{15 \times 2} \\ = \frac{99}{30} \; and \; \frac{74}{30}\\ The \; difference \; between \; both \; the \; fractions \; are \\ = \frac{99 – 74}{30} \\ = \frac{25}{30} \\ = \frac{5}{6}\)

Q7. Find the difference:

(i) \(\frac{6}{7} – \frac{9}{11}\)

(ii) \(8 – \frac{5}{9}\)

(iii) 9 – \(5\frac{2}{3}\)

(iv) \(4\frac{3}{10} – 1\frac{2}{15}\)

Solution:

(i) We have, \(\frac{6}{7} – \frac{9}{11}\)

Taking out the LCM of 7 and 11, we get,

7 x 11 = 77

Now, we convert the given fractions to equivalent fractions by making the denominators 77

\(= \frac{6 \times 11}{7 \times 11} \; and \; \frac{9 \times 7}{11 \times 7} \\ = \frac{66}{77} \; and \; \frac{63}{77}\\ The \; difference \; between \; both \; the \; fractions \; are \\ = \frac{66 – 63}{77} \\ = \frac{3}{77} \\\)

(ii) We have, \(8 – \frac{5}{9}\)

= \(\frac{8 \times 9 – 5}{9} \\ = \frac{72 – 5}{9} \\ = \frac{67}{9}\)

(iii) We have, 9 – \(5\frac{2}{3}\)

= \(9 – \frac{17}{3} \\ = \frac{9 \times 3 – 17}{3} \\ = \frac{27 – 17}{3} \\ = \frac{10}{3}\)

(iv) We have, \(4\frac{3}{10} – 1\frac{2}{15}\)

= \(\frac{43}{10} – \frac{17}{15}\)

Taking out the LCM of 10 and 15, we get,

2 x 3 x 5 = 30

Now, we convert the given fractions to equivalent fractions by making the denominators 30

= \(\frac{43 \times 3}{10 \times 3} – \frac{17 \times 2}{15 \times 2} \\ = \frac{129}{30} – \frac{34}{30} \\ = \frac{129 – 34}{30} \\ = \frac{95}{30} \\ = \frac{19}{6}\)

Q8. Simplify:

(i) \(\frac{2}{3} + \frac{1}{6} – \frac{2}{9}\)

(ii) \(12 – 3\frac{1}{2}\)

(iii) \(7\frac{5}{6} – 4\frac{3}{8} + 2\frac{7}{12}\)

Solution:

(i) We have, \(\frac{2}{3} + \frac{1}{6} – \frac{2}{9}\)

Taking out the LCM of 3, 6 and 9, we get,

3 x 3 x 2 = 18

Now, we convert the given fractions to equivalent fractions by making the denominators 18, we get,

= \(\frac{12}{18} + \frac{3}{18} – \frac{4}{18} \\ = \frac{12 + 3 – 4}{18} \\ = \frac{11}{18}\)

(ii) We have, \(12 – 3\frac{1}{2}\)

= \(12 -\frac{7}{2} \\ = \frac{12 \times 2 -7}{2}\\ = \frac{24 – 7}{2} \\ = \frac{17}{2}\)

(iii) We have, \(7\frac{5}{6} – 4\frac{3}{8} + 2\frac{7}{12}\)

= \(\frac{47}{6} – \frac{35}{8} + \frac{31}{12} \\\)

Taking out the LCM of 6, 8 and 12, we get,

2 x 2 x 2 x 3 = 48

Now, we convert the given fractions to equivalent fractions by making the denominators 48, we get,

\(\frac{47 \times 8}{6 \times 8} – \frac{35 \times 6}{8 \times 6} + \frac{31 \times 4}{12 \times 4} \\ = \frac{376}{48} – \frac{210}{48} + \frac{124}{48} \\ = \frac{376 – 210 + 124}{48} \\ = \frac{290}{48} \\ = \frac{145}{24}\)

Q9. What should be added to \(5\frac{3}{7}\) to get 12?

Solution:

We have, \(5\frac{3}{7}\) = \(\frac{38}{7}\)

Let x be the number added to \(\frac{38}{7}\) to get 12

Therefore,

\(x + \frac{38}{7} = 12 \\ => x = 12 – \frac{38}{7} \\ => x = \frac{12 \times 7 – 38}{7} \\ => x = \frac{84 – 38}{7} \\ => x = \frac{46}{7}\)

Q10. What should be added to \(5\frac{4}{15}\) to get \(12\frac{3}{5}\)?

Solution:

\(5\frac{4}{15}\) = \(\frac{79}{15}\)

\(12\frac{3}{5}\) = \(\frac{63}{5}\)

Let x be the number added to \(\frac{79}{15}\) to get \(\frac{63}{5}\)

= \(\frac{79}{15} + x = \frac{63}{5} \\ => x = \frac{63}{5} – \frac{79}{15}\)

Taking out the LCM of 5 and 15, we get,

3 x 5 = 15

Now, we convert the given fractions to equivalent fractions by making the denominators 15, we get,

\(=> x = \frac{63 \times 3}{5 \times 3} – \frac{79}{15} \\ => x = \frac{189}{15} – \frac{79}{15} \\ => x = \frac{189 – 79}{15} \\ => x = \frac{110}{15} \\ => x = \frac{22}{3}\)

Q11. Suman studies for \(5\frac{2}{3}\) hours daily. She devotes \(2\frac{4}{5}\) hours of her time for science and mathematics . How much time does she devote for other subjects?

Solution:

Given,

Suman studies for \(5\frac{2}{3}\)  i .e, \(\frac{17}{3}\) hours daily.

She devotes \(2\frac{4}{5}\) i .e, \(\frac{14}{5}\) hours of her time for science and mathematics.

Let x be time she devotes for other subjects.

\(\frac{17}{3} = x + \frac{14}{5} \\ => x = \frac{17}{3} – \frac{14}{5}\)

Taking out the LCM of 3 and 5, we get,

3 x 5 = 15

Now, we convert the given fractions to equivalent fractions by making the denominators 48, we get,

\(=> x = \frac{17 \times 5}{3 \times 5} – \frac{14 \times 3}{5 \times 3} \\ => x = \frac{85}{15} – \frac{42}{15} \\ => x = \frac{85 – 42}{15} \; hours \\ => x = \frac{43}{15} \; hours \\ => x = 2\frac{13}{15}\; hours\)

Q12. A piece of wire of length \(12\frac{3}{4}\) m . If it is cut into two pieces in such a way that the length of one piece is \(5\frac{1}{4}\) m, what is the length of the other piece?

Solution:

Given,

A piece of wire of length \(12\frac{3}{4}\) m, one piece is \(5\frac{1}{4}\) m

\(12\frac{3}{4} = \frac{51}{4} \\ And \; 5\frac{1}{4} = \frac{21}{4}\)

Let the length of other piece be x m.

\(\frac{51}{4} = x + \frac{21}{4} \\ => x = \frac{51}{4} – \frac{21}{4} \\ => x = \frac{51 – 21}{4} \\ => x = \frac{30}{4} \\ => x = \frac{15}{2} \\ => x = 7\frac{1}{2}\)

Q13. A rectangular piece of paper is \(12\frac{1}{2} \; cm \; long \; and \; 10\frac{2}{3} \; cm\) wide. Find its perimeter?

Solution:

Given,

A rectangular piece of paper is \(12\frac{1}{2} \; cm \; long \; and \; 10\frac{2}{3} \; cm\) wide

\(12\frac{1}{2} = \frac{25}{2} \\ And \; 10\frac{2}{3} = \frac{32}{3}\)

Perimeter = 2 (length + width)

Perimeter = 2 (\(\frac{25}{2} \; cm \; + \frac{32}{3}\; cm\) )

Perimeter = 2 (\(\frac{75}{6}+ \frac{64}{6}\)) cm

Perimeter = 2 (\(\frac{139}{6}\)) cm

Perimeter = \(\frac{139}{3}\)

Perimeter = \(46\frac{1}{3}\)

Q14. In a “magic square”, the sum of numbers in each row, in each column and along the diagonal is same. Is this a “magic square”?

\(\frac{4}{11}\) \(\frac{9}{11}\) \(\frac{2}{11}\)
\(\frac{3}{11}\) \(\frac{5}{11}\) \(\frac{7}{11}\)
\(\frac{8}{11}\) \(\frac{1}{11}\) \(\frac{6}{11}\) 

Solution:

Given,

\(\frac{4}{11}\) \(\frac{9}{11}\) \(\frac{2}{11}\)
\(\frac{3}{11}\) \(\frac{5}{11}\) \(\frac{7}{11}\)
\(\frac{8}{11}\) \(\frac{1}{11}\) \(\frac{6}{11}\)

Along the 1st column = \(\frac{4}{11} + \frac{3}{11} + \frac{8}{11} = \frac{15}{11}\)

Along the 2nd column = \(\frac{9}{11} + \frac{5}{11} + \frac{1}{11} = \frac{15}{11}\)

Along the 3rd column = \(\frac{2}{11} + \frac{7}{11} + \frac{6}{11} = \frac{15}{11}\)

Along the 1st row = \(\frac{4}{11} + \frac{9}{11} + \frac{2}{11} = \frac{15}{11}\)

Along the 2nd row = \(\frac{3}{11} + \frac{5}{11} + \frac{7}{11} = \frac{15}{11}\)

Along the 3rd row = \(\frac{8}{11} + \frac{1}{11} + \frac{6}{11} = \frac{15}{11}\)

Diagonally = \(\frac{4}{11} + \frac{5}{11} + \frac{6}{11} = \frac{15}{11}\)

And, \(\frac{2}{11} + \frac{5}{11} + \frac{8}{11} = \frac{15}{11}\)

Therefore, the sum of numbers in each row, in each column and along the diagonal is same and the sum is \(\frac{15}{11}\)

Q15. The cost of Mathematics book is Rs \(25\frac{3}{4}\) and that of science book is \(20\frac{1}{2}\). Which costs more and by how much?

Solution:

Given,

The cost of mathematics book is Rs \(25\frac{3}{4}\) and that of science book is \(20\frac{1}{2}\).

We need to compare the cost of mathematics and science book,

\(25\frac{3}{4} = \frac{103}{4}, \\ And \; 20\frac{1}{2} = \frac{41}{2}\)

Taking out the LCM of 4 and 2, we get,

2 x 2 = 4

Now, we convert the given fractions to equivalent fractions by making the denominators 4, we get

\(\frac{103}{4} \\ And, \frac{41 \times 2}{2 \times 2} = \frac{82}{4}\)

As we know, 103 > 82

Therefore, \(\frac{103}{4} > \frac{82}{4}\)

Hence, the cost of mathematics book is more than that of the cost of the science book.

Q16. Provide the number in the box [] and also give its simplest form in each of the following:

(i) \(\frac{2}{3} \times [] = \frac{10}{30}\)

(ii) \(\frac{3}{5} \times [] = \frac{24}{75}\)

Solution:

(i) Given, \(\frac{2}{3} \times [] = \frac{10}{30}\)

= \(\frac{5}{10}\) is the answer

(ii) \(\frac{3}{5} \times [] = \frac{24}{75}\)

= \(\frac{8}{15}\)

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