Write the following rational numbers in ascending order:
$\dfrac{{ - 3}}{7},\dfrac{{ - 3}}{2},\dfrac{{ - 3}}{4}$
Answer
541.5k+ views
Hint: Here we will arrange the rational numbers in ascending order by using the concept of LCM and by making the denominator the same for each rational number.
Complete step-by-step answer:
The L.C.M of 7, 2, and 4 is 28.So, convert the denominator to 28, in all rational numbers
$
\Rightarrow \dfrac{{ - 3}}{7} = \dfrac{{ - 3 \times 4}}{{7 \times 4}} = \dfrac{{ - 12}}{{28}} \\
\Rightarrow \dfrac{{ - 3}}{2} = \dfrac{{ - 3 \times 14}}{{2 \times 14}} = \dfrac{{ - 42}}{{28}} \\
\Rightarrow \dfrac{{ - 3}}{4} = \dfrac{{ - 3 \times 7}}{{4 \times 7}} = \dfrac{{ - 21}}{{28}} \\
$
Now it is clear that
$\dfrac{{ - 42}}{{28}} < \dfrac{{ - 21}}{{28}} < \dfrac{{ - 12}}{{28}} \Rightarrow \dfrac{{ - 3}}{2} < \dfrac{{ - 3}}{4} < \dfrac{{ - 3}}{7}$
So this is your required ascending order.
Note: In this type of questions always take the L.C.M of the denominator, then make all the rational numbers denominator same to the L.C.M value, by appropriate multiplication in numerator and denominator.
Complete step-by-step answer:
The L.C.M of 7, 2, and 4 is 28.So, convert the denominator to 28, in all rational numbers
$
\Rightarrow \dfrac{{ - 3}}{7} = \dfrac{{ - 3 \times 4}}{{7 \times 4}} = \dfrac{{ - 12}}{{28}} \\
\Rightarrow \dfrac{{ - 3}}{2} = \dfrac{{ - 3 \times 14}}{{2 \times 14}} = \dfrac{{ - 42}}{{28}} \\
\Rightarrow \dfrac{{ - 3}}{4} = \dfrac{{ - 3 \times 7}}{{4 \times 7}} = \dfrac{{ - 21}}{{28}} \\
$
Now it is clear that
$\dfrac{{ - 42}}{{28}} < \dfrac{{ - 21}}{{28}} < \dfrac{{ - 12}}{{28}} \Rightarrow \dfrac{{ - 3}}{2} < \dfrac{{ - 3}}{4} < \dfrac{{ - 3}}{7}$
So this is your required ascending order.
Note: In this type of questions always take the L.C.M of the denominator, then make all the rational numbers denominator same to the L.C.M value, by appropriate multiplication in numerator and denominator.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 7 Social Science: Engaging Questions & Answers for Success

Master Class 7 Science: Engaging Questions & Answers for Success

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE

List of coprime numbers from 1 to 100 class 7 maths CBSE

Convert 200 Million dollars in rupees class 7 maths CBSE

How many thousands make a crore class 7 maths CBSE

What is a subcontinent class 7 social science CBSE

Differentiate between map and globe class 7 social science CBSE


