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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}

Answer:

(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}

"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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