Question

what should be added to 1 upon 2 + 1 upon 3 + 1 upon 4 to get minus one​

Answer 1

Question:

What should be added to $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}$ to get -1?

Solution:

Let the number that should be added = x

So,

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + x = -1$

$\implies \dfrac{6}{12} + \dfrac{4}{12} + \dfrac{3}{12} + x = -1$

$\implies \dfrac{6+4+3}{12} = -1$

$\implies \dfrac{13}{12} + x = -1$

$\implies x = -1 - \dfrac{13}{12}$

$\implies \dfrac{-12-13}{12}$

$\implies x = \dfrac{-25}{12}$

Therefore, the number that should be added is $\dfrac{-25}{12}$

Answer 2

Question

$What\:sholud\:be\:added\:to\:\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\\\\to\:get\:-1\:?$

Solution

Let 'x' should be added to $\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}$

to get -1

According to the question

$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+x=-1\\\\Now\:we\:have\:to\:take\:the\:LCM\:of\\\\2\:,3\:,and\:4\\\\\begin{array}{r | 1} 2 & 2,3,4 \\ \cline{2-2} 2 & 1,3,2 \\ \cline{2-2} 3 & 1,3,1 \\ \cline{2-2} & 1,1,1 \end{array}\\\\\\LCM=2\times2\times3\\\\=12\\\\\dfrac{(1\times6)+(1\times4)+(1\times3)}{12}\\\\\\\implies\dfrac{6+4+3}{12}+x=-1\\\\\\\implies\dfrac{13}{12}+x=-1\\\\\\\implies x=-1-\dfrac{13}{12}\\\\LCM\:of\:1\:and\:12\:is\:12\\\\\\=\dfrac{-12-13}{12}\\\\\\\implies x=\dfrac{-25}{12}$