How do you evaluate #4\frac { 1} { 3} \div 1\frac { 2} { 3} \times 2\frac { 3} { 4} \div \frac { 1} { 4}#?

3 Answers

Use the order of operations and change all of the terms to improper fractions so they can be divided and multiplied.

Explanation:

PEMDAS

Think if you get hurt in PE call an MD ASap

There are no Parenthesis or Exponents so the first step can be skipped.
Remember an MD a doctor is only one person so Multiplication and Division must be done at the same time working from left to right

Start with

#(4 1/3) /( 1 2/3)#

Change both to improper fractions and divide using the complex fraction theorem.

# 4 1/3 = 13/3" "and" " 1 2/3 = 5/3 # so

#(4 1/3) /( 1 2/3) = ( 13/3)/ ( 5/3)#

Now multiply both the top and the bottom by the inverse # 3/5#

# ( 13/3 xx 3/5)/(5/3 xx 3/5) = 13/3 xx 3/5 #

The bottom fraction is equal to 1: #rarr 5/3 xx 3/5 = 1 #

# 13/cancel3 xx cancel3/5 = 13/5#

Next multiply #13/5 xx 2 3/4" "# change #2 3/4 to 11/4 #

# 13/ 5 xx 11/ 4 = (13 xx 11)/ (5 xx 4 )#

Next divide by #1/4 #

# { (13 xx 11)/(5 xx 4)}/ ( 1/4) #

Multiply both the top and the bottom by the inverse : #4/1#

# { ( 13 xx 11) / ( 5 xx 4) xx 4/1}/ (1/4 xx 4/1)#

The bottom fraction and the factors of 4 divide out leaving

# ( 13 xx 11)/5# = #143/5# = #28 3/5#

There are no addition or subtraction operations ( ASap) so the problem is done

Jun 6, 2018

#4 1/3-:1 2/3xx2 3/4-:1/4=color(blue)(143/5)=color(teal)(28 3/5#

Explanation:

Given:

#4 1/3-:1 2/3xx2 3/4-:1/4#

According to the order of operations, multiply and divide in order from left to right.

Convert #4 1/3# and #1 2/3# to improper fractions by multiplying the denominator by the whole number and adding the numerator, placing the result over the same denominator.

#(3xx4+1)/3-:(3xx1+2)/3xx2 3/4-:1/4#

#13/3-:5/3xx2 3/4-:1/4#

Invert #5/3# and multiply.

#13/3xx3/5xx2 3/4-:1/4#

#39/15xx2 3/4-:1/4#

Convert #2 3/4# to an improper fraction.

#39/15xx(4xx2+3)/4-:1/4#

#39/15xx11/4-:1/4#

Simplify #39/15xx11/4# to #429/60#.

#429/60-:1/4#

Invert #1/4# and multiply.

#429/60xx4/1=#

#1716/60#

Simplify #1716/60# by dividing the numerator and denominator by #12#.

#(1716-:12)/(60-:12)=#

#143/5#

Convert to a mixed number by dividing #143# by #5# by long division. The whole number quotient is the whole number of the mixed number and the remainder is the new numerator, and the divisor #5# is the denominator.

#143-:5="28 remainder 3"#

#143/5=28 3/5#

Jun 6, 2018

#28 3/5#

Explanation:

Multiplication and division are equally important and can be done in any order as long as each factor keeps its own operation.
This is all one term with four factors and can be calculated in one step.

#4 1/3 div 1 2/3 xx 2 3/4 div 1/4" "larr#change to improper fractions

#=13/3 color(blue)(div 5/3) xx 11/4 color(red)(div 1/4)" "larr a b/c = (c xx a +b)/c#

To divide, multiply by the reciprocal

#=13/3color(blue)(xx3/5) xx 11/4 color(red)(xx4/1)#

#=13/cancel3 xxcancel3/5xx11/cancel4xxcancel4/1" "larr# cancel like factors

#= 143/5" "larr# multiply straight across

#= 28 3/5" "larr# change to mixed number as was given