Question 1207039

let the number of all fruits be {{{x}}}

we have:  

{{{pears + apples+ bananas+ kiwis+ oranges=x}}}

given that:

{{{1/3}}} of the fruits are pears=>{{{(1/3)x=pears}}}

There are half as many apples as pears=>{{{((1/3)/2)x=(1/6)x=apples}}}

{{{1/6}}} of the fruits are bananas=>{{{(1/6)x=bananas}}}

There are half as many kiwis as bananas=> {{{((1/6)/2)x=(1/12)x=kiwis}}}

There are {{{3}}} times as many oranges as kiwis.=>{{{3(1/12)x=(1/4)x=oranges}}}


now we have:

{{{x/3+x/6+x/6+x/12+x/4=x}}}


{{{(4x+2x+2x+x+3x)/12=x}}}

{{{12x/12=x}}}

{{{x=x}}}=> true


so, since common denominator is {{{12}}}, the LEAST number of fruits that can be in the bowl is {{{12}}}


{{{(1/3)12=4}}} pears
{{{(1/6)12=2}}} apples
{{{(1/6)12=2}}} bananas
{{{(1/12)12=1}}} kiwi
{{{(1/4)12=3}}} oranges


now create a bird's eye view drawing