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MATH1051: Exam 1
Vishal Saraswat
This exam covered some of the high school algebra that you must know how to do in order to be successful in
precalculus. If you had difficulty with this material you should talk to your instructor about switching to PsTL 0732,
which is a review of high school algebra.
√
32 · 16 + 9 + 3
.
1. (10 points) Simplify Completely: 11 −
12 − 8 ÷ 4 · 2 + 2 · 22
Follow the order of operations for simplifying arithmetic expressions:
Step 1. Simplify expressions inside grouping symbols, which include parentheses ( ), brack√
ets [ ], absolute value bars | |, the fraction bar , and radicals n .
Step 2. Simplify exponents and logarithms.
Step 3. Simplify multiplication and division, working left to right.
Step 4. Simplify addition and subtraction, working left to right.
Solution:
√
√
32 · 16 + 9 + 3
32 · 25 + 3
11 −
= 11 −
12 − 8 ÷ 4 · 2 + 2 · 22
12 − 8 ÷ 4 · 2 − 2 · 22
32 · 5 + 3
= 11 −
12 − 8 ÷ 4 · 2 − 2 · 22
9·5+3
= 11 −
12 − 8 ÷ 4 · 2 − 2 · 4
45 + 3
= 11 −
12 − 2 · 2 − 8
45 + 3
= 11 −
12 − 4 + 8
45 + 3
= 11 −
8+8
48
= 11 −
16
48
= 11 −
16
= 11 − 3
= 8.
Ans: 8
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 1
Vishal Saraswat
p
p
2. (10 points) Simplify Completely: 3 3 32x10 y 20 + 5x2 y 5 3 4x4 y 5 .
Solution:
Follow the order of operations for simplifying arithmetic expressions:
Step 1. Simplify expressions inside grouping symbols, which include parentheses ( ), brack√
ets [ ], absolute value bars | |, the fraction bar , and radicals n .
Step 2. Simplify exponents and logarithms.
Step 3. Simplify multiplication and division, working left to right.
Step 4. Simplify addition and subtraction, working left to right.
p
p
p
p
3 3 32x10 y 20 + 5x2 y 5 3 4x4 y 5 = 3 3 32x3×3+1 y 6×3+2 + 5x2 y 5 3 4x3+1 y 3+2
p
p
= 3 3 (8 × 4)x3×3 x1 y 6×3 y 2 + 5x2 y 5 3 4x3 x1 y 3 y 2
p
p
= 3 3 23 4(x3 )3 x(y 6 )3 y 2 + 5x2 y 5 3 4x3 xy 3 y 2
p
p
√
√
p
√
√
√ p
3
3
3
3
= 3 23 4 3 (x3 )3 3 x 3 (y 6 )3 3 y 2 + 5x2 y 5 4 x3 x 3 y 3 y 2
p
p
√
√
√
√
3
3
= 3(2) 4(x3 ) 3 x(y 6 ) 3 y 2 + 5x2 y 5 4(x) 3 x(y) 3 y 2
√
√
√ p
√ p
3
3
= 6x3 y 6 4 3 x 3 y 2 + 5x3 y 6 4 3 x 3 y 2
p
p
= 6x3 y 6 3 4xy 2 + 5x3 y 6 3 4xy 2
p
= 11x3 y 6 3 4xy 2 .
p
3 6 3
Ans: 11x y
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
4xy 2
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MATH1051: Exam 1
Vishal Saraswat
3. (10 points) Calculate the area of the following shape. It is an equilateral triangle on top of a square. The
perimeter of the entire shape is 10 feet. Round your answer to the nearest tenth.
Solution:
Let A be the area of the given shape and let us denote by As the area of the
square at the bottom and by At the area of the equilateral triangle at the top.
Then,
A = As + At .
Since the shape is an equilateral triangle on top of a square, all the five sides
are equal. Let the length of each side be x feet. Then the perimeter of the
shape is x + x + x + x + x = 10 so that
x=
1
× 10 = 2 .
5
Then the area of the square is:
As = x2 = 22 = 4
and the area of the equilateral triangle is:
√
√
3 2
3 2 √
At =
x =
2 = 3.
4
4
Thus,
A = As + At = 4 +
√
3 = 4 + 1.7 = 5.7 .
Thus the area of the given shape is 5.7 square feet.
Ans: Area = 5.7 sq ft
One can also use the formula:
area =
1
× base × height .
2
to find the area of the triangle. We know its base is x = 2. The height of the triangle is same the
length of the altitude from the top vertex to the base. We note that the altitude forms a right
triangle with the hypotenuse x = 2 (length of the side of the triange), and the base x/2 = 1
(half the base of the triangle) and we can use the Pythagorean Theorem to find the height h:
p
√
√
h = x2 − (x/2)2 = 22 − 12 = 3 .
So the area of the triangle is
At =
Vishal Saraswat
√
√
1
× 2 × 3 = 3.
2
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 1
Vishal Saraswat
4. (10 points) Find the quotient and remainder of 4x5 − 3x3 + 2x − 1 divided by 2x − 3.
Solution:
Use long division:
2x − 3
2x4 + 3x3 + 3x2 + 92 x + 31
4
4x5
− 3x3
+ 2x − 1
− 4x5 + 6x4
6x4 − 3x3
− 6x4 + 9x3
6x3
− 6x3 + 9x2
9x2 + 2x
− 9x2 + 27
x
2
31
x −1
2
31
− 2 x + 93
4
89
4
31
89
9
Ans: Quotient = 2x4 + 3x3 + 3x2 + x + ; Remainder =
2
4
4
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 4 of 10
MATH1051: Exam 1
Vishal Saraswat
5. (10 points) Factor completely: 42x3 − 7x2 − 105x .
Solution:
First, factor out the GCF, which is 7x:
42x3 − 7x2 − 105 = 7x(6x2 − x − 15) .
Since 6x2 − x − 15 has the form ax2 + bx + c, find two integers whose product is a · c and whose
sum is b, replace the bx term using these integers, and factor by grouping.
For this problem a · c = (6)(−15) = −90. Here are the pairs of positive integers whose product
is 90:
(1, 90), (2, 45), (3, 30), (5, 18), (6, 15), (9, 10) .
The pair whose sum or difference is 1 is 9 and 10. That is,
9 + (−10) = −1 and 9 × (−10) = (6)(−15) = −90 .
Now, replace the middle term using the two integers we just found:
7x(6x2 − x − 15) = 7x(6x2 − 10x + 9x − 15)
and then factor by grouping:
42x3 − 7x2 − 105 = 7x(6x2 − x − 15)
= 7x(6x2 − 10x + 9x − 15)
= 7x (2x(3x − 5) + 3(3x − 5))
= 7x ((2x + 3)(3x − 5))
= 7x(2x + 3)(3x − 5) .
Ans: 7x(2x + 3)(3x − 5)
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 1
Vishal Saraswat
6. (10 points) Perform the indicated operations and simplify:
3x − 2
2
−
.
2x2 − 5x − 3 x3 − 27
Solution: First, find the LCD. To do that, factor each denominator and then multiply the
largest group of each factor:
2x2 − 5x − 3 = (2x + 1)(x − 3) ;
and x3 − 27 = (x − 3)(x2 + 3x + 9) ;
so that LCD = (2x + 1)(x − 3)(x2 + 3x + 9) .
Now convert each fraction into an equivalent fraction with the LCD as the denominator and do
indicated operations:
2x2
2
3x − 2
2
3x − 2
− 3
=
−
− 5x − 3 x − 27
(2x + 1)(x − 3) (x − 3)(x2 + 3x + 9)
2(x2 + 3x + 9)
(3x − 2)(2x + 1)
=
−
(2x + 1)(x − 3)(x2 + 3x + 9) (x − 3)(x2 + 3x + 9)(2x + 1)
2(x2 + 3x + 9) − (3x − 2)(2x + 1)
=
(x − 3)(x2 + 3x + 9)(2x + 1)
2x2 + 6x + 18 − (6x2 − 4x + 3x − 2)
=
(x − 3)(x2 + 3x + 9)(2x + 1)
2x2 + 6x + 18 − (6x2 − x − 2)
=
(x − 3)(x2 + 3x + 9)(2x + 1)
2x2 + 6x + 18 − 6x2 + x + 2
=
(x − 3)(x2 + 3x + 9)(2x + 1)
−4x2 + 7x + 20
.
=
(x − 3)(x2 + 3x + 9)(2x + 1)
Now, reduce if possible. The numerator cannot be factored on integers because there are no two
integers whose product is (−4)(20) = −80 and whose sum is 7. So, this cannot be simplified.
−4x2 + 7x + 20
Ans:
(x − 3)(x2 + 3x + 9)(2x + 1)
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 1
7. (10 points) Solve:
Vishal Saraswat
1
1
x
+ 2 = −3x + (2x + 1) + .
2
3
5
First, clear the fractions by multiplying each term by the LCD of all the fractions
and then solve using the properties of equality.
Solution:
x
1
1
+ 2 = −3x + (2x + 1) +
2
3
5
or 15x + 60 = −90x + 10(2x + 1) + 6
or 15x + 60 = −90x + 20x + 10 + 6
or 15x + 90x − 20x = 10 + 6 − 60
or 85x = −44
−44
or x =
.
85
multiplying each term by 30
distributing 10 over 2x + 1
separating the variables and constants
We can check this answer by replacing x with −44/85 in the original equation and seeing if a
true statement results.
Ans: x = −
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
44
85
Page 7 of 10
MATH1051: Exam 1
Vishal Saraswat
8. (10 points) Solve: x +
1
= 2.
x
Solution: Note that x cannot be 0 so we can clear the fractions by multiplying each term by x
and then solve using the properties of equality.
1
=2
x
or x2 + 1 = 2x
x+
multiplying each term by x
2
or x − 2x + 1 = 0
or (x − 1)2 = 0
or x = 1 .
We can check this answer by replacing x with 1 in the original equation and seeing if a true
statement results.
Ans: x = 1
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 8 of 10
MATH1051: Exam 1
Vishal Saraswat
9. (10 points) A freight train and a passenger train must share the same track. The freight train leaves a station
at 9:00 am and travels with a speed of 30 mph. One hour later, the passenger train leaves the station and travels
with a speed of 40 mph in the same direction. To avoid a crash, at what time and where (distance down the
track) must the railroad dispatcher move the freight train to a side track?
Solution:
This problem involves rates, times and distances so we will need the formula:
d=r×t
where d is the distance, r is the rate, and t is the time.
The problem asks for a time of day which does not fit the formula so we will look at raw time.
Let d be the distance down the track where the dispatcher must move the freight train to a side
track to avoid a crash and let the passenger train reaches there after t hours.
Then,
d = 40t .
(1)
Since the trains leave from the same place and take the same route they travel the same distance
to reach the point of crash. Thus, since the freight train leaves an hour early at 30mph, we get
d = 30(t + 1) .
(2)
Thus, from equations 1 and 2,
40t = 30(t + 1)
or 40t = 30t + 30
or 40t − 30t = 30
or 10t = 30
or t = 3 .
So, the passenger train will catch the freight train in 3 hours, that is, at 10:00 am + 3 hours =
13:00 hours = 1:00 pm
They will meet at d = 40 × 3 = 120 miles from the station.
Ans: Time = 1:00 pm; Distance = 120 miles.
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 1
Vishal Saraswat
10. (10 points) Tickets to a show cost $2 for kids, $3 for seniors, and $5 for adults. If there were twice as many kids
as seniors, and if 91 tickets were sold and $311 was collected, how many of each type of ticket was sold?
Solution:
Let the no. of seniors’ tickets sold be x.
Since there were twice as many kids as seniors, the number of kids’ tickets sold is 2x and the
remaining # of tickets 91 − x − 2x = 91 − 3x were adults’ tickets.
Then the proceeds from these tickets is
2 × 2x + 3 × x + 5 × (91 − 3x) = 4x + 3x + 455 − 15x = 455 − 8x .
Thus,
311 = 455 − 8x
solving which we get
1
1
x = (455 − 311) = 144 = 18 .
8
8
Then 2x = 36 and 91 − 3x = 37.
# of kids’ tickets sold
= 36
Ans: # of seniors’ tickets sold = 18
# of adults’ tickets sold = 37
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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