# NCERT Solutions Class 8 Mathematics Solutions for Exercise 1.1 in Chapter 1 - Rational Numbers

Question 36 Exercise 1.1

Using appropriate properties find.

\text { (i) }-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}

\text { (ii) } \frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}

(i)

\begin{aligned} -\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6} & \\ =-\frac{2}{3} \times \frac{3}{5}-\frac{3}{5} \times \frac{1}{6}+\frac{5}{2} & \text { (by commutativity) } \end{aligned}

\begin{array}{l} =\frac{3}{5}\left(\frac{-2}{3}-\frac{1}{6}\right)+\frac{5}{2} \\ =\frac{3}{5}\left(\frac{-4-1}{6}\right)+\frac{5}{2} \\ =\frac{3}{5}\left(\frac{-5}{6}\right)+\frac{5}{2} \\ =\frac{-15}{30}+\frac{5}{2} \\ =\frac{-1}{2}+\frac{5}{2} \\ =\frac{4}{2} \\ =2 \end{array}

(ii)

\begin{array}{c} \frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5} \\ =\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5} \\ =\frac{2}{5} \times\left(-\frac{3}{7}\right)+\frac{1}{14} \times \frac{2}{5}-\left(\frac{1}{6} \times \frac{3}{2}\right) \text { (by commutativity) } \end{array}

\begin{array}{l} =\frac{2}{5} \times\left(-\frac{3}{7}+\frac{1}{14}\right)-\frac{3}{12} \\ =\frac{2}{5} \times\left(\frac{-6+1}{14}\right)-\frac{1}{4} \\ =\frac{2}{5} \times\left(\frac{-6+1}{14}\right)-\frac{1}{4} \\ =\frac{2}{5} \times\left(\frac{-5}{14}\right)-\frac{1}{4} \\ =\frac{2}{5} \times\left(\frac{-5}{14}\right)-\frac{1}{4} \\ =\left(\frac{-10}{70}\right)-\frac{1}{4} \\ =\frac{-1}{7}-\frac{1}{4} \\ =\frac{-4-7}{28} \end{array}

Video transcript
"hello students i am rita your math leader tutor and today's question is using appropriate properties find in the first part minus 2 by 3 into 3 by 5 plus 5 by 2 minus 3 by 5 into 1 by 6 first of all from the first and the last term 3 by 5 is common so we can take out 3 by 5 common from the first and the last term and within bracket we have minus 2 by 3 and minus 1 by 6 and plus 5 by 2 so in this step we apply the commutative property by communicative by commutative property we can take out common now we solve this by taking the lcm of three and six is six three to the sixth two twos of four six from the sixth one was the one plus five by two now three by five minus minus become plus so minus 5 pi 6 plus 5 by 2 now 5 and 5 cancel out each other three ones are three and three to the sixth now we get minus half plus five by 2 minus plus become minus so we got 4 by 2 it means the answer of this question is 2. so this is the answer of the first part now we go to the second part in second part we have 2 5 5 into minus 3 by 7 minus 1 by 6 into 3 by 2 plus 1 by 14 into 2 by 5. same in the first and the last term 2 by 5 is common so we can take out 2 by 5 common from the first and the last term and within bracket we have minus 3 by 7 plus 1 by 40 and we can cancel out 3 on the 3 and 3 to the 6 now we have minus 1 by 4 now we solve this and we apply here by commutative property we can take out common by commutative property we take out common from them now we take out lcm that is the lcm of seven and fourteen is fourteen seven twos are three to the six fourteen on the fourteen one ones are one minus one by four now two by five minus plus become minus minus five five fourteen minus five by fourteen minus one by four now five and five cancel out each other two ones are two two sevens are fourteen now we got minus one by seven and minus one by four now we take out the lcm of seven and four is twenty eight seven fours are twenty eight and four sevens are twenty eight now we got minus minus plus 7 plus 4 11 now the answer of the second part is minus 11 by 28 so this is our answer i hope you like this video so please subscribe youtube for more updates and do comment your questions thank you "
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