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word problems with fractions


mike mcgarry
lesson by mike mcgarry
magoosh expert

summary
the essence of solving word problems with fractions in the gre exam context is understanding the translation of words to mathematical operations, specifically recognizing 'of' as multiplication and 'is' as equals.
  • the fundamental rule in translating word problems into mathematical equations involves interpreting 'is' as equals and 'of' as multiply.
  • a basic example illustrates this principle with a simple multiplication of fractions to find a part of a whole.
  • a more complex problem involves calculating monthly rent based on the fraction of the rent that represents a cable bill, demonstrating the application of reciprocal multiplication to isolate variables.
  • an advanced problem requires translating relationships between three variables into a fraction that represents one variable as a fraction of another, showcasing the elimination of irrelevant variables to focus on the relationship of interest.
  • understanding and applying these translation rules are crucial for solving fraction-based word problems effectively on the gre.
chapters
00:06
translating words to math
00:29
basic fraction multiplication
01:01
solving for variables in fraction word problems
02:02
advanced fraction comparison

q: in the last problem, i would have liked to see how the n was isolated. i did not follow the math on the reciprocal to isolate n.

a: not to worry! let's take a look: 

cathy's salary is 3/7 of nora's salary and (cathy's salary) is also 5/4 of teresa's salary.

c = (3/7)n

c is also equal to (5/4)t

so we can set:

(3/7)n = (5/4)t

now, i isolate n by multiplying both sides of the equation by (7/3), the reciprocal of 3/7.

(7/3)(3/7)n = (7/3)(5/4)t

(7/3)(3/7) = 1 so (7/3)(3/7)n just equals n.

n = (7/3)(5/4)t =

(35/12)t


q: the lesson doesn't really explain why we multiply by the reciprocal to get the answer in the second problem. i am very confused.

a: happy to help! this is just an extension of one of the rules of working with algebraic expressions: when trying to isolate a variable, you can perform a number of operations to both sides of an equation, but they have to be done to both sides.

so, for example, say i have this equation:

x/2 = 4

if i want to get x alone, i multiply it by 2. but i also have to do the same to the other side:

x/2 = 4
2x/2 = 2 * 4
x = 8

now, let's try that with this equation:

2x/7 = y

first, i multiply out that 7:

(7 * 2x)/7 = 7 * y
2x = 7y

then, i divide out the 2:

2x/2 = 7y/2
x = 7y/2

look again at my original and final equations:

2x/7 = y
x = 7y/2

in order to isolate x, i multiplied y by the reciprocal of the fraction on the left side. that's just the easier, faster way to do it. it's shorthand for the same process.

in the lesson video, there were two cases of expressions like that:

300 = 2r/7
7 * 300 = 2r
7 * 300 / 2= r

and here's the other:

c = 3n/7
7c = 3n
7c/3 = n