CHAPTER 7 / ESSENTIAL PRE-ALGEBRA SKILLS 281
Concept Review 3
- 19/35 (Use zip-zap-zup: it’s better than using your
calculator!) - 20/3 (Change to 5/2 ×8/3.)
- 8/3 (Divide numerator and denominator by their
common factor: 7.) - 8/45 (Change to 2/9 ×4/5.)
- (12x+14)/21x(Use zip-zap-zup.)
- 4x/3z(Change to 3x/2 ×8/9zand simplify.)
- (3m+1)/(2m+1) (Factor and cancel 4 from the
numerator and denominator.)
8.−1/(6x) (Change to −2/(9x^2 ) × 3 x/4 and simplify.) - 13/12 (Change to 6/12 +4/12 +3/12.)
- (3 − 2 x)/4 (Use zip-zap-zup and simplify.)
11.x+5 (As long as x≠5.) (Factor as
and cancel the common factor. For factoring re-
view, see Chapter 8, Lesson 5.)xx
x()− ()+
()−
55
5
- (2n+3)/4n(Divide numerator and denominator
by the common factor: 3. Don’t forget to “distrib-
ute” the division in the numerator!)
13.^1 ⁄ 4 14.^1 ⁄ 5 15.^1 ⁄ 10 16.^1 ⁄ 3
(Knowing how to “convert” numbers back and
forth from percents to decimals to fractions can
be very helpful in simplifying problems!) - Multiply by the reciprocal of the fraction.
- “Cross-multiply” to get the new numerators, and
multiply the denominators to get the new
denominator, then just add (or subtract) the
numerators. - Only common factors. (Factors = terms in
products.) - Just divide the numbers by hand or on a calculator.
- 4/9 (Not 4/5! Remember the fraction is a part of
the whole,which is 27 students, 12/27 =4/9.) - 27 (If 2/3 are girls, 1/3 are boys: t/3 =9, so t=27.)
- It must have a value between 0 and 1 (“bottom-
heavy”).
Answer Key 3: Fractions
SAT Practice 3
1.C 4/(4+ 5 +3) =4/12 =1/3
2.BMultiply by y: 1 =− 3 yDivide by −3:3.C
4.A Multiply by x:Multiply by :1
10
(^1) =x
5
1
2
= 5 x1
(^25)
x
=
1
2
3
3
3
2
3
1
3
0 333
1
2
3
1
3
2
15
2
3
2
3
2
3
−=−= =
÷=× =
÷=
..
.
..
××= =
÷= × =
3
2
6
6
1
2
3
1
2
3
1 0 666. ...
2
3
×= 1 0 666. ...
−=
1
3
y1
3
y=−
5.D Multiply by 6m: 3 m+ 6 = 5 m
Subtract 3m: 6 = 2 m
Divide by 2: 3 =m
6.A The quotient of opposites is always −1. (Try
x=2 and y=−2 or any other solution.)7.B To turn into m+ 2 n,we only need tomultiply by 4!8.E You can solve this by “plugging in” a number
for zor by simplifying algebraically. To plug in,
pick z=2 and notice that the expression equals^2 ⁄ 5
or 4. Substituting z=2 into the choices shows
that only (E) is .4.
Alternatively, you can simplify by just multiply-
ing numerator and denominator by z:9.C^5 ⁄ 8 are boys, so^3 ⁄ 8 must be girls. Of the girls,^2 ⁄ 3
do not have dark hair, so^1 ⁄ 3 do. Therefore,^1 ⁄ 3 of^3 ⁄ 8
of the class are girls with dark hair.^1 ⁄ 3 ×^3 ⁄ 8 =^1 ⁄ 8.1
1
1
z^121
zzzz
zz+ z=
×
+
⎛
⎝⎜
⎞
⎠⎟
=
×
mn
mn
+= +⎛
⎝⎜
⎞
⎠⎟
=
⎛
⎝⎜
⎞
⎠⎟
24 ==
42
4
7
8
28
8
7
2
mn
42+
1
2
15
6
+=
m