Question

find the value of 2.45 bar +0.36 bar
1) 67/33
2) 24/11
3) 31/11
4) 167/110​

Answer 1

Answer:

The correct answer is 3) 31/11

Step-by-step explanation:

2.(45) bar ( Here . (45) bars represents that the bar is put over the two digits i.e 4 and 5 )

Let's consider 'x' = 2.(45) bar

= 2.45454545..............(i)

Now, 100x = 245.454545..............(ii)

Now, (ii)-(i)

100x-x= 245.4545....-2.4545........

=> 99 x = 243

=> x = 27/11

Similarly in 0.(36) bar, we get = 4/11

(NOTE :- THE SAME PROCESS SHOULD BE UTILISED SO AS TO GET THE VALUE FOR 0.(36) BAR AS WE DID IN FINDING THE VALUE FOR 2.(45) BAR)

Now, 2.(45) bar + 0.(36) bar

= 27/11 + 4/11

= 31/11

Therefore , the correct answer is 3) 31/11

Answer 2

Given: $\[2.\overline {45} + 0.\overline {36} \]$.

To find: the value of the given expression.

Solution:

Know that, the bar given above the number after the decimals of each number in the given expression indicates that each number has infinitely recurring decimals.

Assume that, $\[x = 2.\overline {45} \]$

$\[\begin{array}{l} \Rightarrow x = (2.454545 - - - - - - )------------(1)\\ \Rightarrow 100x = 2.454545 - - - - - - \times 100\\ \Rightarrow 100x =( 245.454545 - - - - - - )----------(2)\end{array}\]$

Subtract equation (1) and (2).

$\[\begin{array}{l}100x - x = \left( {245.454545 ------ } \right) - \left( {2.454545 ------ } \right)\\ \Rightarrow 99x = 243\end{array}\]$

$\[\begin{array}{l} \Rightarrow x = \frac{{243}}{{99}}\\ \Rightarrow x = \frac{{27}}{{11}}\end{array}\]$

Solve for the other number similarly.

Assume that, $\[y = 0.\overline {36} \]$

$\[\begin{array}{l} \Rightarrow y = (0.363636------) ------(3)\\100y = \left( {0.363636 - - - - } \right) \times 100\\ \Rightarrow 100y =(036.363636------) ------(4)\end{array}\]$

$\[\begin{array}{l} \Rightarrow 100y - y = \left( {036.363636 - - - - - } \right) - \left( {0.363636 - - - - } \right)\\ \Rightarrow 99y = 36\end{array}\]$

$\[\begin{array}{l} \Rightarrow y = \frac{{36}}{{99}}\\ \Rightarrow y = \frac{4}{{11}}\end{array}\]$

Find the value of $\[ 2.\overline {45} + 0.\overline {36} \]$.

$\[2.\overline {45} + 0.\overline {36} = \frac{{27}}{{11}} + \frac{4}{{11}}\]$

$\[ \Rightarrow 2.\overline {45} + 0.\overline {36} = \frac{{27 + 4}}{{11}}\]$

$\[ \Rightarrow 2.\overline {45} + 0.\overline {36} = \frac{{31}}{{11}}\]$

Therefore, the value of $\[ \Rightarrow 2.\overline {45} + 0.\overline {36} \]$ is $\frac{{31}}{{11}}$.