Solusinya berbentuk bahasa Inggris. Semoga bisa dipahami.
Terimakasih.
Nomor 1. At certain hospital, 75% of the interns receive fewer than 6 hours of sleep and report feeling
tired during their shifts. At the same time, 70% of the interns who receive 6 or more hours of
sleep no feelings of tiredness. If 80% of the interns receive fewer than 6 hours of sleep, what
percent of the interns report no feelings of tiredness during their shifts ?
A). $ 6 \, $
B). $ 14 \, $
C). $ 19 \, $
D). $ 20 \, $
E). $ 81 $
Nomor 2.
A school’s annual budget for the purchase of student computers increased by 60% this year over
last year. If the price of student computers inceased by 20% this year, then the number of
computers it can purchase this year is what percent greater than the number of computers
it purchase last year ?
A). $ 33\frac{1}{3} \, $
B). $ 40 \, $
C). $ 42\frac{1}{3} \, $
D). $ 48 \, $
E). $ 60 $
$\spadesuit $ Jawaban : A
$\clubsuit $ Pembahasan :
Let’s denote the formula for the money spent on computers as $ pq = b $, where
$ p = \, $ price of computers
$ q = \, $ quantity of computers
$ b = \, $ budget
We can solve a percent question that doesn’t involve actual values by using smart numbers. Let’s assign a smart number of 1000 to last year’s computer
budget ($b$) and a smart number 100 to last year’s computer price ($p$). 1000 and 100 are easy numbers to take a percent of.
This year’s budget will equal $ 1000 \times 1,6 = 1600 $
This year’s computer price will equal $ 100 \times 1,2 = 120 $
Now we can calculate the number of computers purchased each year, $ \frac{b}{p} $
Number of computers purchased last year $ = \frac{1000}{100} = 10 $
Number of computers purchased this year $ = \frac{1600}{120} = 13\frac{1}{3} $ (while $ \frac{1}{3} $ of a computer doesn’t make sense it won’t affect
the calculation)
The question is asking for the percent increase in quantity from last year to this year
$ \begin{align}
& = \frac{\text{new – old}}{\text{old}} \times 100\% \\
& = \frac{13\frac{1}{3} – 10}{10} \times 100\% \\
& = 33\frac{1}{3} \%
\end{align} $
The correct answer is A.
$ \heartsuit $
Nomor 3.
Machine A and machine B can produce 1 widget in 3 hours working together at their respective
constant rates. If machine A’s speed were doubled, the two machines could produce 1 widget in 2
hours working together at their respective rates. How many hours does it currently take machine
A to produce 1 widget on its own ?
A). $ 1 \, $
B). $ 2 \, $
C). $ 3 \, $
D). $ 5 \, $
E). $ 6 $
$\spadesuit $ Jawaban : E
$\clubsuit $ Pembahasan :
Let $ a $ be the number of hours it takes Machine A to produce 1 widget on its own. Let $ b $ be the number of hours it takes Machine B to produce 1 widget
on its own.
The question tells us that Machines A and B together can produce 1 widget in 3 hours. Therefore, in 1 hour, the two machines can
produce $\frac{1}{3} $ of a widget. In 1 hour, Machine A can produce $ \frac{1}{a} $ widgets and Machine B can
produce $\frac{1}{b} $ widgets. Together in 1 hour, they produce $ \frac{1}{a} + \frac{1}{b} = \frac{1}{3} $ widgets.
If Machine A’s speed were doubled it would take the two machines 2 hours to produce 1 widget. When one doubles the speed, one cuts the amount of time it
takes in half. Therefore, the amount of time it would take Machine A to produce 1 widget would be $ \frac{a}{2} $. Under these new conditions, in 1 hour
Machine A and B could produce $ \frac{1}{\frac{a}{2}} + \frac{1}{b} = \frac{1}{2} $ widgets. We now have two unknowns and two different equations.
We can solve for $ a $.
The two equations:
$ \frac{1}{\frac{a}{2}} + \frac{1}{b} = \frac{1}{2} \rightarrow
\frac{2}{a} + \frac{1}{b} = \frac{1}{2} \text{ ….(i)} $
$ \frac{1}{a} + \frac{1}{b} = \frac{1}{3} \, $ ….(ii)
Subtract the bottom equation from the top:
$ \begin{align}
\frac{2}{a} – \frac{1}{a} & = \frac{1}{2} – \frac{1}{3} \\
\frac{1}{a} & = \frac{3}{6} – \frac{2}{6} \\
\frac{1}{a} & = \frac{1}{6} \\
a & = 6
\end{align} $
The correct answer is E.
$ \heartsuit $
Nomor 4.
Wati walks 5 miles from point A to point B in ane hour, then bicyles back to point A along the
same route at 15 miles per hour. Budi makes the same round trip, but does so at half of Wati’s
average speed. How many minutes does Budi spend on his round trip ?
A). $ 40 \, $
B). $ 80 \, $
C). $ 120 \, $
D). $ 160 \, $
E). $ 180 $
$\spadesuit $ Jawaban : D
$\clubsuit $ Pembahasan :
We begin by figuring out Wati’s average speed. On her way from A to B, she travels 5 miles in one hour, so her speed is 5 miles per hour. On her way back
from B to A, she travels the same 5 miles at 15 miles per hour. Her average speed for the round trip is NOT simply the average of these two speeds.
Rather, her average speed must be computed using the formula $ RT = D $, where $ R $ is rate, $T $ is time and $ D $ is distance. Her average speed for
the whole trip is the total distance of her trip divided by the total time of her trip.
We already know that she spends 1 hour going from A to B. When she returns from B to A, Wati travels 5 miles at a rate of 15 miles per hour, so our
formula tells us that $ 15T = 5$, or $T = \frac{1}{3} $ . In other words, it only takes Wati $ \frac{1}{3} $ of an hour, or 20 minutes, to return from
B to A. Her total distance traveled for the round trip is $ 5+5=10 $ miles and her total time is $ 1+\frac{1}{3} = \frac{4}{3} $ of an hour,
or 80 minutes.
We have to give our final answer in minutes, so it makes sense to find Wati’s average rate in miles per minute, rather than miles per hour.
$ \frac{10 \, \text{miles}}{80 \, \text{minutes}} = \frac{1}{8} \, $ miles per minute. This is Wati’s average rate.
We are told that Budi’s rate is half of Wati’s, so he must be traveling at $ \frac{1}{16} $ miles per minute. He also travels a total of 10 miles,
so $ \frac{1}{16}T = 10 $ , or $ T = 160 $. Budi’s round trip takes 160 minutes.
Alternatively, we could use a shortcut for the last part of this problem. We know that Budi’s rate is half of Wati’s average rate.
This means that, for the entire trip, Budi will take twice as long as Wati to travel the same distance. Once we determine that Wati will take 80
minutes to complete the round trip, we can double the figure to get Budi’s time. $ 80 \times 2 = 160$.
The correct answer is D.
$ \heartsuit $
Nomor 5.
At Denpasar Utara Elementary School, the number of teachers and students totals 510
(there are six grade levels). The ratio of students to teachers is 16 to 1. SD Tulang ampiang
students (SD Tulang ampiang is one of the school in Denpasar Utara) make up $ \frac{1}{5} $ of
the student population. Fifth and sixth graders account for $ \frac{1}{3} $ of the remainder.
Student in first and second grades account for $ \frac{1}{4} $ of all the students. If there
are an equal number of students in the third and fourth grades, then the number of students in
third grade is how many greater or fewer than the number of students in SD Tulang ampiang ?
A). 12 greater
B). 17 fewer
C). 28 fewer
D). 36 fewer
E). 44 fewer
$\spadesuit $ Jawaban : C
$\clubsuit $ Pembahasan :
We know that the student to teacher ratio at the school is 16 to 1, and the total number of people is 510. Therefore:
Number of students $ = \frac{16}{17} \times 510 = 480 $
Number of teachers $ = \frac{1}{17} \times 510 = 30 $
SD Tulang ampiang students make up $ \frac{1}{5} $ of the student population, so:
Number of SD Tulang ampiang students $ = \frac{1}{5} \times 480 = 96 $
Fifth and sixth graders account for $ \frac{1}{3} $ of the remainder (after SD Tulang ampiang students are subtracted from the total), therefore:
Number of 5th and 6th grade students $ = \frac{1}{3} \times (480 – 96) = \frac{1}{3} \times (384) = 128 $
Students in first and second grades account for $ \frac{1}{4} $ of all the students, so:
Number of 1st and 2nd grade students $ = \frac{1}{4} \times (480) = 120 $
So far, we have accounted for every grade but the 3rd and 4th grades, so they must consist of the students left over:
Number of 3rd and 4th grade students = Total students – students in other grades
Number of 3rd and 4th grade students $ = 480 – 96 – 128 – 120 = 136 $
If there are an equal number of students in the third and fourth grades, then:
Number of 3rd grade students $ = \frac{136}{2} = 68 $
The number of students in third grade is 68, which is fewer than 96, the number of students in SD Tulang ampiang. The number of students in 3rd grade
is thus $ 96 – 68 = 28 $ fewer than the number of SD Tulang ampiang students.
The correct answer is C.
$ \heartsuit $
Nomor 6.
Set S consist of integers 7, 8, 10, 12, and 13. If integer $ n $ is included in the set, the
average (arithmetic mean) of the set S will increase by 20%. What is the value of
integer $ n $ ?
A). 16
B). 20
C). 22
D). 23
E). 24
Nomor 7.
If $ x $ and $ n $ are integers, is the sum of $ x $ and $ n $ less than zero ?
Decide whether the following statements (1) and (2) are sufficient to answer the question.
(1). $ x + 3 < n – 1 $
(2). $ – 2x > 2n $
A). Statement (1) alone is sufficient, but statement (2) alone is not sufficient
B). Statement (2) alone is sufficient, but statement (1) alone is not sufficient
C). Both statements together are sufficient, but neither statement alone is sufficient
D). Each statement alone is sufficient
E). Statement (1) and (2) together are not sufficient
$\spadesuit $ Jawaban : B
$\clubsuit $ Pembahasan :
The question asks: is $ x + n
(1) INSUFFICIENT: This statement can be rewritten as $ x + n < 2n – 4$. This rephrased statement is consistent with $ x + n $ being either negative or
non-negative. (For example if $2n – 4 = 1.000 $ , then $ x + n $ could be any integer, negative or not, that is less than 1.000).
Statement (1) is insufficient because it answers our question by saying “maybe yes, maybe no”.
(2) SUFFICIENT: We can divide both sides of this equation by $ -2 $ to get $ x < -n $ (remember that the inequality sign flips when we multiply or divide
by a negative number). After adding n to both sides of resulting inequality, we are left with $ x + n < 0 $ .
The correct answer is B.
$ \heartsuit $
Nomor 8.
if $ n $ is a multiple of 5 and $ n = p^2q$, where $ p $ and $ q $ are prime numbers,
which of the following must be a multiple of 25 ?
A). $ p^2 \, $
B). $ q^2 \, $
C). $ pq \, $
D). $ p^2q^2 \, $
E). $ p^3q $
Nomor 9.
if $ a, b, \, $ and $ c $ are integers and $ \frac{ab^2}{c} $ is a positive even integer,
which of the following must be true ?
I. $ ab $ is even
II. $ ab > 0 $
III. $ c $ is even
A). I only
B). II only
C). I and II
D). I and III
E). I, II, and III
$\spadesuit $ Jawaban : A
$\clubsuit $ Pembahasan :
If $ab^2 $ were odd, the quotient would never be divisible by 2, regardless of what $ c $ is. To prove this try to divide an odd number by any integer to
come up with an even number; you can’t. If $ab^2 $ is even, either $ a $ is even or $ b $ is even.
(I) TRUE: Since $a$ or $b$ is even, the product $ab$ must be even
(II) NOT NECESSARILY: For the quotient to be positive, $a$ and $c$ must have the same sign since $b^2$ is definitely positive. We know nothing about
the sign of $b$. The product of $ab$ could be negative or positive.
(III) NOT NECESSARILY: For the quotient to be even, $ab^2$ must be even but $c$ could be even or odd. An even number divided by an odd number could
be even (ex: $\frac{18}{3} $), as could an even number divided by an even number (ex: $\frac{16}{4}$).
The correct answer is A.
$ \heartsuit $
Nomor 10.
if $(x\# y) $ represents the remainder that results when the positive integer $ x $ is divided
by the positive integer $y$, what is the sum of all the possible values of $ b $ such that
$ (16 \# b) = 1 $ ?
A). $ 23 \, $
B). $ 22 \, $
C). $ 20 \, $
D). $ 16 \, $
E). $ 9 $
$\spadesuit $ Jawaban : A
$\clubsuit $ Pembahasan :
The definition given tells us that when $x$ is divided by $y$ a remainder of $(x \# y)$ results. Consequently, when 16 is divided by $ b $ a remainder of
$(16 \# b)$ results. Since $(16 \# b) = 1$, we can conclude that when 16 is divided by $ b $ a remainder of 1 results.
Therefore, in determining the possible values of $ b $, we must find all the integers that will divide into 16 and leave a remainder of 1.
These integers are 3 , 5, and 15. The sum of these integers is 23.
The correct answer is A.
$ \heartsuit $
Nomor 11.
In a certain game, a large bag is filled with blue, green, purple and red chips
worth 1, 5, $ x $ and 11 points each, respectively. The purple chips are worth more than the
green chips, but less than the red chips. A certain number of chips are then selected from the
bag If the product of the point values of the selected chips is 704.000 , how many purple chips
were selectied ?
A). $ 2 \, $
B). $ 3 \, $
C). $ 4 \, $
D). $ 5 \, $
E). $ 6 $
$\spadesuit $ Jawaban : B
$\clubsuit $ Pembahasan :
704.000 is the product of an unknown number of 1’s, 5’s, 11’s and $x$’s. To figure out how many $x$’s are multiplied to achieve this product, we have to
figure out what the value of $x$ is. Remember that a number’s prime box shows all of the prime factors that when multiplied together produce that number:
704.000’s prime box contains one 11, three 5’s, and nine 2’s, since $ 704.000 = 11 \times 5^3 \times 2^9 $.
The 11 in the prime box must come from a red chip, since we are told that $ 5 < x < 11 $ and therefore $ x $ could not have 11 as a factor. In other words,
the factor of 11 definitely did not come from the selection of a purple chip, so we can ignore that factor for the rest of our solution.
So, turning to the remaining prime factors of 704.000: the three 5’s and nine 2’s. The 2’s must come from the purple chips, since the other colored chips
have odd values and thus no factor of two. Thus, we now know something new about $x$: it must be even. We already knew that $5 < x < 11$, so now we know
that $x$ is 6, 8, or 10.
However, $x$ cannot be 6: $6 = 2 \times 3$, and our prime box has no 3’s.
$x$ seemingly might be 10, because $10 = 2 \times 5$, and our prime box does have 2’s and 5’s. However, our prime box for 704.000 only has three 5’s, so a
maximum of three chips worth 10 points are possible. But that leaves three of the six factors of 2 unaccounted for, and we know those factors of two must
have come from the purple chips.
So $x$ must be 8, because $8 = 2^3$ and we have nine 2’s, or three full sets of three 2’s, in the prime box. Since $x$ is 8, the chips selected must have
been 1 red (one factor of 11), 3 green (three factors of 5), 3 purple (two factors of 8, equivalent to nine factors of 2), and an indeterminate number of
blue chips.
The correct answer is B.
$ \heartsuit $
Nomor 12.
In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2
units respectively. Given that $ \angle ADC = 60^o $ and $ \angle ABD = 45^o $ , what is the
measure of angle $ x $ in degrees ? (Note : Figure is not drawn to scale.)
A). $ 60 \, $
B). $ 70 \, $
C). $ 75 \, $
D). $ 80 \, $
E). $ 85 $
$\spadesuit $ Jawaban : C
$\clubsuit $ Pembahasan :
As first, it appears that there is not enough information to compute the rest of the angles in the diagram. When faced with situations such as this,
look for ways to draw in new lines to exploit any special properties of the given diagram.
For example, note than the figure contains a $60^o $ angle, and two lines with lengths in the ratio of 2 to 1. Recall that a $ 30-60-90 $ triangle also
has a ratio of 2 to 1 for the ratio of its hypotenuse to its short leg. This suggests that drawing in a line from C to line AD and forming a right triangle
may add to what we know about the figure. Let’s draw in a line from C to point E to form a right triangle, and then connect points E and B as follows:
Triangle CED is a $30-60-90 $ triangle. Using the side ratios of this special triangle, we know that the hypotenuse is two times the smallest leg.
Therefore, segment ED is equal to 1.
From this we see that triangle EDB is an isosceles triangle, since it has two equal sides (of length 1). We know that $ \angle EDB = 120^o$;
therefore angles DEB and DBE are both $30^o$.
Now notice two other isosceles triangles:
(1) Triangle CEB is an isosceles triangle, since it has two equal angles (each 30 degrees). Therefore segment CE = segment EB.
(2) Triangle AEB is an isosceles triangle, since CEA is 90 degrees, angles ACE and EAC must be equal to 45 degrees each. Therefore angle
$ x = 45 + 30 = 75 $ degrees.
The correct answer is C.
$ \heartsuit $
Nomor 13.
In the picture, quadrilateral ABCD is a parallelogram and quadrilateral DEFG is a rectangle.
What is the area of parallelogram ABCD ? (figure not drawn to scale)
Decide whether the following statements (1) and (2) are sufficient to answer the question.
(1). the area of rectangle DEFG is $ 8\sqrt{5} $
(2). Line AH, the altitude of parallelogram ABCD, is 5
A). Statement (1) alone is sufficient, but statement (2) alone is not sufficient
B). Statement (2) alone is sufficient, but statement (1) alone is not sufficient
C). Both statements together are sufficient, but neither statement alone is sufficient
D). Each statement alone is sufficient
E). Statement (1) and (2) together are not sufficient
$\spadesuit $ Jawaban : A
$\clubsuit $ Pembahasan :
At first, it looks as if there is not enough information to solve this problem. Whenever you have a geometry problem that does not look solvable,
one strategy is to look for a construction line that will add more information.
Let’s draw a line from point E to point C as shown in the picture below:
Now look at triangle DEC. Note that triangle DEC and parallelogram ABCD share the same base (line DC). They also necessarily share the same height
(the line perpendicular to base DC that passes through the point E). Thus, the area of triangle DEC is exactly one-half that of parallelogram ABCD.
We can also look at triangle DEC another way, by thinking of line ED as its base. Notice that ED is also a side of rectangle DEFG. This means that triangle
DEC is exactly one-half the area of rectangle DEFG.
We can conclude that parallelogram ABCD and DEFG have the same area!
Thus, since statement (1) gives us the area of the rectangle, it is clearly sufficient, on its own, to determine the area of the parallelogram.
Statement (2) gives us the length of line AH, the height of parallelogram ABCD. However, since we do not know the length of either of the bases, AB or DC,
we cannot determine the area of ABCD. Note also that if the length of AH is all we know, we can rescale the above figure horizontally, which would change
the area of ABCD while keeping AH constant. (Think about stretching the right side of parallelogram ABCD.) Hence, statement (2) is not sufficient on
its own.
The correct answer is A: Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
$ \heartsuit $
Nomor 14.
Four spheres and three cubes are arranged in a line according to increasing volume, with no two
solids of the same type adjacent to each ather. The ratio of the volume of one solid to that of
the next largest is constant. If the radius of the smallest sphere is $ \frac{1}{4} $ that of
the largest sphere, what is the radius of the smallest sphere ?
Decide whether the following statements (1) and (2) are sufficient to answer the question.
(1). The volume of the smallest cube is $ 72\pi $
(2). The volume of the second largest sphere is $ 576\pi $
A). Statement (1) alone is sufficient, but statement (2) alone is not sufficient
B). Statement (2) alone is sufficient, but statement (1) alone is not sufficient
C). Both statements together are sufficient, but neither statement alone is sufficient
D). Each statement alone is sufficient
E). Statement (1) and (2) together are not sufficient
$\spadesuit $ Jawaban : D
$\clubsuit $ Pembahasan :
The question stem tells us that the four spheres and three cubes are arranged in order of increasing volume, with no two solids of the same type adjacent
in the lineup. This allows only one possible arrangement: sphere, cube, sphere, cube, sphere, cube, sphere.
Then we are told that the ratio of one solid to the next in line is constant. This means that to find the volume of any solid after the first, one must
multiply the volume of the previous solid by a constant value. If, for example, the volume of the smallest sphere were 2, the volume of the first cube
(the next solid in line) would be $2x$, where $x$ is the constant. The volume of the second sphere (the third solid in line) would
be $ 2(x) \times (x) = 2x^2 $ and the volume of the second cube (the fourth solid in line) would then be $2(x)\times(x)\times(x) = 2x^3 $, and so on.
By the time we got to the largest sphere, the volume would be $ 2x^6 $ .
We are not given the value of the constant, but are told that the radius of the smallest sphere is $ \frac{1}{4} $ that of the largest. We can use this
information to determine the value of the constant. First, if the radius of the smallest sphere is $r$, then the radius of the largest sphere must be $4r$.
So if the volume of the smallest sphere is $ \frac{4}{3} \pi r^3 $ , then the volume of the largest sphere must
be $ \frac{4}{3} \pi (4r)^3 $ or $ \frac{4}{3} \pi 64 r^3 $ . So the volume of the largest sphere is 64 times larger than that of the smallest.
Using the information about the constant from the question stem, we can set up and simplify the following equation:
$ \begin{align}
\frac{4}{3} \pi r^3 x^6 & = \frac{4}{3} \pi 64 r^3 \\
x^6 & = 64 \\
x & = 2
\end{align} $
Therefore, the value of the constant is 2. This means that the volume of each successive solid is twice that of the preceding solid. We are ready
to look at the statements.
Statement (1) tells us that the volume of the smallest cube is $ 72\pi $ . This means that the volume of the smallest sphere (the immediately preceding
solid) must be half, or $ 36\pi $ . If we have the volume of the smallest sphere, we can find the radius of the smallest sphere:
$ \begin{align}
\frac{4}{3} \pi r^3 & = 36\pi \\
r^3 & = \frac{3}{4} \times 36 \\
r^3 & = 27 \\ r & = 3
\end{align} $
Therefore, the radius of the smallest sphere is 3. Statement (1) is sufficient.
Statement (2) tells us that the volume of the second largest sphere is $ 576\pi $ . Applying the same logic that we used to evaluate statement (1),
we can find the radius of the smallest sphere by dividing $ 576\pi $ by 2 four times (because there are four solids smaller than the second largest sphere)
until we get to the volume of the smallest sphere: $ 36\pi $ . At this point we can solve for the radius of the smallest sphere as we did in our analysis
of statement (1) above. Statement (2) is sufficient.
The correct answer is D: either statement alone is sufficient to answer the question.
$ \heartsuit $
Nomor 15.
A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin,
and the square is to have an area of 100. If all coordinates of the vertices must be integers,
how many different ways can this square be drawn ?
A). 4
B). 6
C). 8
D). 10
E). 12
$\spadesuit $ Jawaban : E
$\clubsuit $ Pembahasan :
Each side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn,
because each side, in order to have integer coordinates, would have to be drawn on the $x-$ or $y-$axis. What makes a length of 10 different is that it
could be the hyptoneuse of a pythagorean triple, meaning the vertices could have integer coordinates without lying on the $x-$ or $y-$axis.
For example, a square could be drawn with the coordinates (0,0), (6,8), ($-2, 14$) and ($-8, 6$). (It is tedious and unnecessary to figure out all four
coordinates for each square).
If we lable the square $abcd$, with $a$ at the origin and the letters representing points in a clockwise direction, we can get the number of possible
squares by figuring out the number of unique ways $ab$ can be drawn.
$a$ has coordinates (0,0) and $b$ could have coordinates:
$ (-10,0) $
$(-8,6) $
$(-6,8) $
(0,10)
(6,8)
(8,6)
(10,0)
$(8, -6) $
$(6, -8) $
(0, 10)
$(-6, -8) $
$(-8, -6) $
There are 12 different ways to draw ab, and so there are 12 ways to draw $abcd$.
The correct answer is E.
$ \heartsuit $
Nomor 16.
How many different 5-person teams can be formed from a group of $ x $ individuals ?
Decide whether the following statements (1) and (2) are sufficient to answer the question.
(1). If there had been $ x+2 $ individuals in the group, exactly 126 different 5-person teams
could have been formed.
(2). If there had been $ x+1 $ individuals in the group, exactly 56 different 3-person teams
could have been formed.
A). Statement (1) alone is sufficient, but statement (2) alone is not sufficient
B). Statement (2) alone is sufficient, but statement (1) alone is not sufficient
C). Both statements together are sufficient, but neither statement alone is sufficient
D). Each statement alone is sufficient
E). Statement (1) and (2) together are not sufficient
$\spadesuit $ Jawaban : D
$\clubsuit $ Pembahasan :
In order to answer this question, we need to be able to determine the value of $x$. Thus, this question can be rephrased: What is $x$?
(1) SUFFICIENT: In analyzing statement (1), consider how many individuals would have to be available to create 126 different 5 person teams.
We don’t actually have to figure this out as long as we know that we could figure this out. Certainly by testing some values, we could figure this out.
It turns out that if there are 9 available individuals, then we could create exactly 126 different 5-person teams
$ \left( \text{since } C^9_5 = \frac{9!}{4! \times 5!} = 126 \right) $. This value (9) represents $x + 2$. Thus $x$ would equal 7.
(2) SUFFICIENT: The same logic applies to statement (2). Consider how many individuals would have to be available to create 56 different 3-person teams.
Again, we don’t actually have to figure this out as long as we know that we could figure this out. It turns out that if there are 8 available individuals,
then we could create exactly 56 different 3-person teams $ \left( \text{since } C^8_3 = \frac{8!}{5! \times 3!} = 56 \right) $. This value (8)
represents $x + 1$. Thus $x$ would equal 7.
Statement (2) alone IS sufficient.
The correct answer is D.
$ \heartsuit $
Nomor 17.
Ms. Yuli has four children. You are told correctly that she has at least two girls but you are
not told which two of her four children are those girls. What is the probability that she
also has two boys? (Assume that the probability of having a boy is the same as the probability
of having a girl)
A). $ \frac{1}{4} $
B). $ \frac{3}{8} $
C). $ \frac{5}{11} $
D). $ \frac{1}{2} $
E). $ \frac{6}{11} $
$\spadesuit $ Jawaban : E
$\clubsuit $ Pembahasan :
Since each of the 4 children can be either a boy or a girl, there are $ 2\times 2 \times 2 \times 2 = 2^4 = 16 $ possible ways that the children might
be born, as listed below:
BBBB (all boys)
BBBG, BBGB, BGBB, GBBB, (3 boys, 1 girl)
BBGG, BGGB, BGBG, GGBB, GBBG, GBGB (2 boys, 2 girls)
GGGB, GGBG, GBGG, BGGG (3 girls, 1 boy)
GGGG (all girls)
Since we are told that there are at least 2 girls, we can eliminate 5 possibilities : the one possibility in which all of the children are boys
(the first row) and the four possibilities in which only one of the children is a girl (the second row).
That leaves 11 possibilities (the third, fourth, and fifth row) of which only 6 are comprised of two boys and two girls (the third row). Thus,
the probability that Ms. Barton also has 2 boys is $ \frac{6}{11} $ and the correct answer is E.
$ \heartsuit $
Nomor 18.
A political candidate collected $ \$ 1.749 $ from a fund raising dinner. If each supporter contributed
at least $ \$ 50 $ , what is the greatest possible number of contributors at the dinner ?
A). 33
B). 34
C). 35
D). 36
E). 37
$\spadesuit $ Jawaban : B
$\clubsuit $ Pembahasan :
To determine the greatest possible number of contributors we must assume that each of these individuals contributed the minimum amount, or $\$ 50$.
We can then set up an inequality in which $n$ equals the number of contributors:
$50n$ is less than or equal to $\$ $1,749
Divide both sides of the equation by 50 to isolate $n$, and get
$n$ is less than or equal to 34,98
Since $n$ represents individual people, it must be the greatest whole number less than 34,98 . Thus, the greatest possible value of $n$ is 34.
Alternately, we could have assumed that the fundraiser collected $\$ $1,750 rather than $\$ $1,749. If it had, and we assumed each individual contributed
the minimum amount, there would have been exactly 35 contributors ($ \$ 50 \times 35 = \$ 1,750$). Since the fundraiser actually raised one dollar less
than $\$ $1,750, there must have been one fewer contributor, or 34.
The correct answer is B.
$ \heartsuit $
Nomor 19.
The three-digit positive integer $ x $ has the hundreds, tens, and units digits of $a$, $b$,
and $c$, respectively. The three-digit positive integer $ y $ has the hundreds, tens, and
units digit of $k$, $l$, and $m$, respectively. If
$(2^a )(3^b )(5^c )= 12(2^k )(3^l )(5^m )$ , then which is the correct relationship between
the following quantities P and Q based on the information provided?
$\begin{array}{|c|c|}
\hline
\text{P} & \text{Q} \\
\hline
\text{The value of } x – y & 200 \\
\hline
\end{array} $
A). Quantity P is greater
B). Quantity Q is greater
C). The two quantities are equal
D). The relationship cannot be determined from the information given
$\spadesuit $ Jawaban : A
$\clubsuit $ Pembahasan :
First, let us simplify the exponential equation:
$\begin{align}
(2^a )(3^b )(5^c ) & = 12(2^k )(3^l )(5^m ) \\
(2^a )(3^b )(5^c ) & = (3)(4)(2^k )(3^l )(5^m ) \\
(2^a )(3^b )(5^c ) & = (2^{k+2} )(3^{l+1} )(5^m )
\end{align} $
When the bases on both sides of an equation are equal and the bases are prime numbers, the exponents of the respective bases must also be
equal: $a = k + 2$; $b = l + 1$; and $c = m$. Now recall that $a$, $b$, and $c$ represent the hundreds, tens, and units digits of the three-digit
integer $x$; similarly, $k$, $l$, and $m$ represent the hundreds, tens, and units digits of the three-digit integer $y$.
Therefore, the hundreds digit of $x$ is 2 greater than the hundreds digit of $y$; the tens digit of $x$ is 1 greater than the tens digit of $y$;
finally, the units digit of $x$ is equal to the units digit of $y$. Using this information, we can set up our subtraction problem and find the
value of $(x – y)$:
$ \begin{array}{cc}
abc & \\
klm & – \\
\hline
210 &
\end{array} $
So, the value of $ x – y = 210 $.
The correct answer is A.
$ \heartsuit $
Nomor 20.
Which of the following graphs is symmetric with respect to the $y$-axis ?
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