A person went to market to buy 1.5 kg of dried peas having 20% water content. He went home and soaked them for some time and the water ... A person went to market to buy 1.5 kg of dried peas having 20% water content. He went home and soaked them for some time and the water content in the peas become 60%. Find the final weight of the soaked peas . 

My approach : 

Since 1.5 kg contain 20% water , it means 1.5 x 20/100 = 300gm of water content. 

Now I am not getting idea how to use this or may be i m wrong here.. please guide will be of great help. Thanks.
1.5 kg leads to 300 grams of water being 20%. So, we have 1.2 kg of dried material.
By soaking we get this 1.2 kg of dried material to equate to 40% of our total mass ( 100% - 60% water weight). $$1.2 \text{kg} /40\% =3 \text{kg}$$ So, that means we have 3 kg of mass total. 

This of course assumes the percent water content is by mass and not by volume.
Direct Proportionality implies that increasing mass $m$ is linked to increasing water content $w$, $w \sim m$ or $w = c ~ \cdot ~m$, where $c$ is a constant (direct proportionality factor).
We are given $$(w_1;m_1)=(0.2;1.5)$$
From what's given, we can find $$c = w_1/m_1 = 2/15$$
If we like to find $m_2$ in $(0.6;m_2)$, then we use $c$ and the equation $ w = c ~ \cdot ~m$ to get
$$m_2 = w_2/c=0.6/(2/15)=4.5 ~\text{kg}.$$
EDIT:
Apparaently it's not that easy, because $4.5 ~\text{kg}$ is too much weight? So, we forget about direct proportionality, it's of no help in this case...
Let's check the chemistry and mechanics of the problem. The water content has many different definitions and it can be defined as the gravimetric water content (, which also has many types of sub-definitions depending on the application and on the point of time of investigation of the water content) $$u'=m_{\text{water}}/m_{\text{wet peas}}.$$
If $m_{\text{wet peas}}=1.5 ~\text{kg}$ of dried peas have $u'=20~\%$ water content, then we use above formula: $$m_{\text{water}}=u' \cdot m_{\text{wet peas}} = 20~\% \cdot 1.5~\text{kg}= 0.3~\text{kg}.$$
Now the water mass is just the difference of the masses of the wet peas and the dried peas:
$$m_{\text{water}}=m_{\text{wet peas}}-m_{\text{dried peas}}.$$
Solve above equation for $m_{\text{dried peas}}$ and plug in the numbers for $m_{\text{wet peas}}$ and $m_{\text{water}}$: $$m_{\text{dried peas}} = m_{\text{wet peas}} - m_{\text{water}}= 1.5~\text{kg}-0.3~\text{kg}=1.2~\text{kg}.$$
If now the water content becomes $u'=60\%$, then we still have $m_{\text{dried peas}} =1.2~\text{kg}$, but:
$$ m_{\text{wet peas}} = m_{\text{dried peas}}/(1-u') =\color{red}{3~\text{kg}} .$$
This result is achieved using the definition of the gravimetric water content before drying, which is: $$ u'=m_{\text{water}}/m_{\text{wet peas}}=(m_{\text{wet peas}}-m_{\text{dried peas}})/m_{\text{wet peas}}.$$
EDIT
The easiest and quickest method to solve this problem is by embracing the condition, that the dry mass of the peas before soaking ($=80\% \cdot 1.5~ \text{kg}$) and the dry mass of the peas after soaking ($=40\% \cdot m_{\text{wet peas}}~ \text{kg}$) still didn't change. This condition leads immediately to:
$$80\% \cdot 1.5=40\% \cdot m_{\text{wet peas}} \Rightarrow m_{\text{wet peas}} = (80\% \cdot 1.5)/40\%  = 3~\text{kg}.$$
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