1001 Algebra Problems.PDF

45. b.Apply the order of operations:

1 + (^75 ^130 ) = 1 + (^75 ^53 ^2 )

= 1 + ^130 (–^170 )(^75  )

= 1 +  (– 1
7  0 )(^134 )

= 1 – ^73 ^134 

= 1 –^73  13

4 

= 1 –^12 

= ^12 

46. e.Since there are mmen in a class ofnstudents,
    there must be n – mwomen in the class. So, the
    ratio of men to women in the class is.


47. d.Compute the difference between ^12 and each
    of the four choices. Then, compare the absolute
    values of these differences; the choice that pro-
    duces the smallest difference is the one closest
    to ^12 . The differences are as follows:

^23 – ^12 = ^46 – ^36 = ^16 

 130 – ^12 =  130 –  150 = – 120 = –^15 

^56 – ^12 = ^56 – ^36 = ^26 = ^13 

^35 – ^12 =  160 –  150 =

The smallest absolute value of these four dif-

ferences is  110 . Of the four choices, the one

closest to ^12 is ^35 .

48. c.Applying the order of operations, first sim-
    plify the exponentiated term. Then, multiply
    left to right. Finally, compute the resulting dif-
    ference by first rewriting both fractions using
    the least common denominator, which is 12:

(^7) ^56 – 3^12 
2
= 7^56 – 3^14 = ^365 – ^34 =

^365  22 –  43 ^33 =^7102 –  192 = ^6112 

Set 4 (Page 7)

49. d.The exponent applies only to 5, not to the
    –1 multiplied in front. So, –5^3 = –(5 5 5)
    = –125.


50. a.By definition, (–11)^2 = (–11)(–11) = 121.


51. c.Using the fact that any nonzero base raised
    to the zero power is 1, we have 5(4^0 ) = 5(1) = 5.


52. a.Applying the exponent rules yields:
    (2^2 )–3= 2(2x– 3)= 2–6= =  614 


53. c.Applying the order of operations and the
    definition of an exponent yields:
    = = = –^4  8 = – 21 


54. b.Applying the order of operations and the
    definition of an exponent yields:
    –5(–1 – 5– 2) = –5(–1 –  215 ) = –5(–^2255 –  215 ) =
    –5(–^2265 ) = 5 (= ^256 


55. c.First, apply the definition of a negative
    exponent to simplify the first term within the
    brackets. Next, rewrite the resulting first term
    using the fact that “a product raised to a power
    is the product of the powers.” Then, simplify:
    
    
    • –^32 
    
    
    
    

–2

• ^23 

2

= ––

2

3 

2

• ^23 

2

=

• –1

2



2

3 

2

• ^23 

2

= –

2

3 

2

• ^23 

2

= 0

56. b.First, apply the definition of a negative
    exponent to simplify the two terms to which it
    applies. Then, apply the order of operations:

–(–^12 )–3 – = –(–2)^3 –

= –[(–2)(–2)(–2)] –

= –[–8] –

= 8 – ^19 ^811 

= 8 – 9

= – 1

^19 



 811 

–

(^13) 
–
(^13) 


(^19) 

(^19) 

(–^13 )^2

(^19 )^2
(–^13 )^2
 9 –2
^26
5 
5
(–2)(–2)
–8
(–2)^2
–8
(1 – 3)^2
–8
^1
26
^1
10
m
n – m
^10
3
^5 ^2
3
^130 

• ^170 

– 

(^53) 
( – 2)
–
(^170) 

ANSWERS & EXPLANATIONS–