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anastassius [24]
3 years ago
15

Jacob has some cookies. 3/4 of his cookies are chocolate chip. Out of the chocolate chip cookies 1/8 of them have nuts as well.

What fraction of Jacobs cookies are chocolate chip cookies with nuts?
Mathematics
2 answers:
ryzh [129]3 years ago
7 0
Well, we can find the answer by multiplying the two fractions. 3/4*1/8 equals 3/32.

Dahasolnce [82]3 years ago
4 0
The answer is:  " \frac{1}{6} " . 
_______________________________________________________
Explanation:
________________________________________________________
Given the fractions: 

"3/4" ;  and "1/8" ; 

Note the "denominators:  "4" and "8" .

We can easily convert "3/4" to its fraction value with a denominator of "8" ; 

→ "3/4" =  " (what value?) / 8"  ?

→  Look at the "denominators" :
 
 →  4 * (what value?) = "8" ? ;  → "8 ÷ 4 = "2" ; 

→ So, 4 * 2  = 8 ;  for the "denominator" ;  so we multiply by "2" in the 
"numerator" , as well:

→ "(3/4)" = (3*2)/(4*2) = "6/8" ;
__________________________________________________
→  "3/4"  =  "6/8" ; 
__________________________________________________
→  Note:  The entire batch of cookies is:  "8/8" ; or "1 whole" ; 
       {since:  "8/8" = "{8÷8 = 1 whole}" ; 
__________________________________________________

Given:  "3/4" of the {entire batch of} his cookies are chocolate chip." ; 

i.e.    "6/8 out of 8/8" are chocolate chips ;  
 
6/8 are chocolate chips;

1/8 out of 6/8 are {"chocolate chip with nuts"} ;
______________________________________________
  ==> What fraction of cookies are {"chocolate chip with nuts"} ?
__________________________________________________

To get the answer;  we simplify:  "1/8 out of 6/8";  

or, simplify, "1/8 out of 3/4" ; 

that is:

" \frac{1}{8}  ÷  \frac{3}{4} " = ? {our answer} ?? ; 

→  " \frac{1}{8}  ÷  \frac{3}{4} " ;

→   Note:  When we divide 2 (TWO)  fractions;  we find the equivalent by writing the expression with a multiplication sign; AND by "inverting" (or taking the "reciprocal" of) the "second" fraction ; 

   =      " \frac{1}{8}  *  \frac{4}{3} " ; 
_______________________________________________________

Note:  The "4" cancels out to a "1" ; and the "8" cancels out to a "2" ; 

since:  "{8 ÷ 4 = 2}" ;  and since:  "{4 ÷ 4 = 1 }" ,
________________________________________________________
And we can rewrite the expression as:
________________________________________________________
  
→     " \frac{1}{2}  *  \frac{1}{3} " ; 

And simplify further:

      =  " \frac{(1*1)}{(2*3)}  ; 

      =  " \frac{1}{6} " .
______________________________________________________

The answer is:  " \frac{1}{6} " .

______________________________________________________
  Variant:  At the point {above} which we have:
______________________________________________________

 →      " \frac{1}{8}  *  \frac{4}{3} " ; 
______________________________________________________
Simplify further; as following: 

       =  " \frac{(1*4)}{(8*3)} ; 

       =  " \frac{4}{24} " ;

       =  " (4 ÷ 4) / (24 ÷ 4) "  ;
 
       =  " \frac{1}{6} " .

_______________________________________________________

The answer is:  " \frac{1}{6} " . 
_______________________________________________________ 
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John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
2 years ago
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