RD Sharma Solutions Class 7 Operations On Rational Numbers Exercise 5.4

RD Sharma Class 7 Solutions Chapter 5 Ex 5.4 PDF Free Download

RD Sharma Solutions Class 7 Chapter 5 Exercise 5.4

Exercise 5.4

Q1. Divide:

(i) 1 by \(\frac{1}{2}\)

\(1\div\frac{1}{2}\\ =1\times2\\ =2\)

(ii) 5 by \(\frac{-5}{7}\)

\(5\div\frac{-5}{7}\\ =5\times\frac{-7}{5}\\ =-7\)

(iii) \(\frac{-3}{4}\) by \(\frac{9}{-16}\)

\(\frac{-3}{4}\div\frac{9}{-16}\\ =\frac{-3}{4}\div\frac{-9}{16}\\ =\frac{-3}{4}\times\frac{-16}{9}\\ =\frac{-4}{-3}\\ =\frac{4}{3}\\\)

(iv) \(\frac{-7}{8}\) by \(\frac{-21}{16}\)

\(\frac{-7}{8}\div\frac{-21}{16}\\ =\frac{-7}{8}\times\frac{-16}{21}\\ =\frac{2}{3}\\\)

(v) \(\frac{7}{-4}\) by \(\frac{63}{64}\)

\(\frac{7}{-4}\div\frac{63}{64}\\ =\frac{7}{-4}\times\frac{64}{63}\\ =\frac{-16}{9}\\\)

(vi) 0 by \(\frac{-7}{5}\)

\(0\div\frac{-7}{5}\\ =0\times\frac{-5}{7}\\ =0\)

(vii) \(\frac{-3}{4}\) by -6

\(\frac{-3}{4}\div -6\\ =\frac{-3}{4}\times\frac{-1}{6}\\ =\frac{1}{8}\)

(viii) \(\frac{2}{3}\) by \(\frac{-7}{12}\)

\(\frac{2}{3}\div\frac{-7}{12}\\ =\frac{2}{3}\times\frac{-12}{7}\\ =\frac{-8}{7}\)

Q2. Find the value and express as a rational number in standard form:

(i) \(\frac{2}{5}\div\frac{26}{15}\)

\(\frac{2}{5}\div\frac{26}{15}\\ =\frac{2}{5}\times\frac{15}{26}\\ =\frac{3}{13}\)

(ii) \(\frac{10}{3}\div\frac{-35}{12}\)

\(\frac{10}{3}\div\frac{-35}{12}\\ \frac{10}{3}\times\frac{-12}{35}\\ =\frac{-40}{35}\\ =\frac{-8}{7}\)

(iii) \(-6\div\frac{-8}{17}\)

\(-6\div\frac{-8}{17}\\ =-6\times\frac{-17}{8}\\ =\frac{102}{8}\\ =\frac{51}{4}\\\)

(iv) \(\frac{40}{98}\div -20\)

\(\frac{40}{98}\div -20\\ =\frac{40}{98}\times\frac{-1}{20}\\ =\frac{-2}{98}\\ =\frac{-1}{49}\\\)

Q3. The product of two rational numbers is 15. If one of the numbers is -10, find the other.

Let the number to be found be x

\(x\times-10=15\\ x=\frac{15}{-10}\\ x=\frac{3}{-2}\\ x=\frac{-3}{2}\\\)

Hence the number is \(x=\frac{-3}{2}\\\)

Q4. The product of two rational numbers is \(\frac{-8}{9}\). If one of the numbers is \(\frac{-4}{15}\), find the other.

Let the number to be found be x

\(x\times\frac{-4}{15}=\frac{-8}{9}\\ x=\frac{-8}{9}\div\frac{-4}{15}\\ x=\frac{-8}{9}\times\frac{15}{-4}\\ x=\frac{-8\times15}{9\times-4}\\ x=\frac{-120}{-36}\\ x=\frac{120}{36}\\ x=\frac{10}{3}\)

Hence the number is \(x=\frac{10}{3}\\\)

Q5. By what number should we multiply \(\frac{-1}{6}\) so that the product may be \(\frac{-23}{9}\)?

Let the number to be found be x

\(x\times\frac{-1}{6}=\frac{-23}{9}\\ -x=\frac{-23}{9}\times6\\ -x=\frac{-23\times6}{9}\\ -x=\frac{-138}{9}\\ x=\frac{138}{9}\\ x=\frac{46}{3}\\\)

Hence the number is \(x=\frac{46}{3}\)

Q6. By what number should we multiply \(\frac{-15}{28}\) so that the product may be \(\frac{-5}{7}\)?

Let the number to be found be x

\(x\times\frac{-15}{28}=\frac{-5}{7}\\ x=\frac{-5}{7}\div\frac{-15}{28}\\ x=\frac{-5}{7}\times\frac{-28}{15}\\ x=\frac{-8}{9}\times\frac{15}{-4}\\ x=\frac{4}{3}\\\)

Hence the number is \(x=\frac{4}{3}\)

Q7. By what number should we multiply \(\frac{-8}{13}\) so that the product may be 24?

Let the number to be found be x

\(x\times\frac{-8}{13}=24\\ x=24\div\frac{-8}{13}\\ x=24\times\frac{13}{-8}\\ x=-3\times13\\ x=-39\)

Hence the number is \(x=-39\)

Q8. By what number should \(\frac{-3}{4}\) be multiplied in order to produce \(\frac{-2}{3}\)?

Let the number to be found be x

\(x\times\frac{-8}{13}=24\\ x=24\div\frac{-8}{13}\\ x=24\times\frac{13}{-8}\\ x=-3\times13\\ x=-39\)

Hence the number is \(x=-39\)

 

Q9. Find (x + y)÷(x —y), if

(i) x= \(\frac{2}{3}\) y= \(\frac{3}{2}\)

\(\left(x+y \right )\div\left(x-y \right )\\ =\left(\frac{2}{3}+\frac{3}{2} \right )\div\left(\frac{2}{3}-\frac{3}{2} \right )\\ =\left(\frac{4+9}{6} \right )\div\left(\frac{4-9}{6} \right )\\ =\left(\frac{4+9}{6} \right )\times\left(\frac{6}{4-9} \right )\\ =\left(\frac{4+9}{4-9} \right )\\ =\left(\frac{13}{-5} \right )\\ =\left(\frac{-13}{5} \right )\\\)

(ii) x= \(\frac{2}{5}\) y= \(\frac{1}{2}\)

\(\left(x+y \right )\div\left(x-y \right )\\ =\left(\frac{2}{5}+\frac{1}{2} \right )\div\left(\frac{2}{5}-\frac{1}{2} \right )\\ =\left(\frac{4+5}{16} \right )\div\left(\frac{4-5}{16} \right )\\ =\left(\frac{4+5}{16} \right )\times\left(\frac{16}{4-5} \right )\\ =\left(\frac{4+5}{4-5} \right )\\ =\left(\frac{9}{-1} \right )\\ =\left(\frac{-9}{1} \right )\\ =9\)

(iii) x= \(\frac{5}{4}\) y= \(\frac{-1}{3}\)

\(\left(x+y \right )\div\left(x-y \right )\\ =\left(\frac{5}{4}+\frac{-1}{3} \right )\div\left(\frac{5}{4}-\frac{-1}{3} \right )\\ =\left(\frac{5\times3-1\times4}{12} \right )\div\left(\frac{5\times3+1\times4}{12} \right )\\ =\left(\frac{5\times3-1\times4}{12} \right )\times\left(\frac{12}{5\times3+1\times4} \right )\\ =\left(\frac{5\times3-1\times4}{5\times3+1\times4} \right )\\ =\left(\frac{11}{19} \right )\\\)

 

Q10. The cost of \(7\frac{2}{3}\) metres of rope is Rs. \(12\frac{3}{4}\). Find its cost per metre.

\(7\frac{2}{3}\) metres of rope cost= Rs. \(12\frac{3}{4}\)

=Rs.\(\frac{51}{4}\)

\(7\frac{2}{3}\)= \(\frac{23}{3}\)

Cost per metre= \(\frac{51}{4}\div\frac{23}{3}\\ =\frac{51}{4}\times\frac{3}{23}\\ =\frac{153}{92}\\ =Rs.1\frac{61}{92}\)

Q11. The cost of \(2\frac{1}{3}\) metres of cloth is Rs.\(75\frac{1}{4}\). Find the cost of cloth per metre.  

\(2\frac{1}{3}\) metres of rope cost= Rs. \(75\frac{1}{4}\)

=Rs.\(\frac{301}{4}\)

\(2\frac{1}{3}\)= \(\frac{7}{3}\)

Cost per metre= \(\frac{301}{4}\div\frac{7}{3}\\ =\frac{301}{4}\times\frac{3}{7}\\ =\frac{43\times3}{4}\\ =\frac{129}{4}\\ =Rs.32\frac{1}{4}\)

Q12. By what number should \(\frac{-33}{16}\) be divided to get  \(\frac{-11}{4}\)?

\(\frac{-33}{16}\div\;x=\frac{-11}{4}\\ x=\frac{-33}{16}\div\frac{-11}{4}\\ x=\frac{-33}{16}\times\frac{4}{-11}\\ x=\frac{3}{4}\\\)

The number is \(x=\frac{3}{4}\\\)

Q13. Divide the sum of \(\frac{-13}{5}\) and \(\frac{12}{7}\) by the product of \(\frac{-31}{7}\) and \(\frac{-1}{2}\)

\(\left(\frac{-13}{5}+\frac{12}{7} \right )\div\left(\frac{-31}{7}\times\frac{-1}{2} \right )\\ =\left(\frac{-13\times7}{5\times7}+\frac{12\times5}{7\times5} \right )\div\left(\frac{-31}{7}\times\frac{-1}{2} \right )\\ =\left(\frac{-91}{35}+\frac{60}{35} \right )\div\left(\frac{31}{14}\right )\\ =\left(\frac{-91+60}{35} \right )\div\left(\frac{31}{14}\right )\\ =\left(\frac{-31}{35} \right )\div\left(\frac{31}{14}\right )\\ =\left(\frac{-31}{35} \right )\times\left(\frac{14}{31}\right )\\ =\frac{-14}{35}\\ =\frac{-2}{5}\)589

 

Q14. Divide the sum of \(\frac{65}{12}\) and \(\frac{8}{3}\) by their difference.

\(\left(\frac{65}{12}+\frac{8}{3} \right )\div\left(\frac{65}{12}-\frac{8}{3} \right ) \\ =\left(\frac{65}{12}+\frac{8\times4}{3\times4} \right )\div\left(\frac{65}{12}-\frac{8\times4}{3\times4} \right ) \\ =\left(\frac{65}{12}+\frac{32}{12} \right )\div\left(\frac{65}{12}-\frac{32}{12} \right ) \\ =\left(\frac{65+32}{12}\right )\div\left(\frac{65-32}{12} \right ) \\ =\left(\frac{65+32}{12}\right )\times\left(\frac{12}{65-32} \right ) \\ =\frac{65+32}{65-32}\\ =\frac{97}{33}\\\)

 

Q15. If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?

Length of cloth required for each trouser=\(\frac{Total\;length\;of\;cloth}{number\; of\; trousers}\)

=\(\frac{54}{24}\)

=\(\frac{9}{4}\)metres

\(\frac{9}{4}\)mmetres of cloth is required to make each trouser

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