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SAT Practice Test 2015. A botanist is cultivating a rare species of plant in a controlled environment and currently has 3000 of these plants. The population of this species that the botanist expects to grow next year, $N_{next.year}$, can be estimated from the number of plants this year, $N_{this.year}$, by the equation below.
$N_{next.year}=N_{this.year}+0.2(N_{this.year})(1-\frac{N_{this.year}}{K})$
The constant K in this formula is the number of plants the environment is able to support.
According to the formula, what will be the number of plants two years from now if K = 4000? (Round your answer to the nearest whole number.)
N in year 1 will be:
$N_1=3000+0.2(3000)(1-\frac{3000}{4000})$
$N_1=3000+0.2(3000)(1-0,75)$
$N_1=3000+0.2(3000)(0,25)$
$N_1=3000+0.2(750)$
$N_1=3000+150=3150$
And N in year 2 will be:
$N_2=3150+0.2(3150)(1-\frac{3150}{4000})$
$N_2=3150+0.2(3150)(1-0.7875)$
$N_2=3150+134$
$N_2=3284$
Answer: 3284
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SAT Practice Test 2015. This question refers to the data in the previous question.
The botanist would like to increase the number of plants that the environment can support so that the population of the species will increase more rapidly. If the botanist’s goal is that the number of plants will increase from 3000 this year to 3360 next year, how many plants must the modified environment support?
Answer:
It is given that $N_{next.year}=3360$ and that $N_{this.year}=3000$. What is the value of K?
$N_{next.year}=N_{this.year}+0.2(N_{this.year})(1-\frac{N_{this.year}}{K})$
$3360=3000+0.2(3000)(1-\frac{3000}{K})$
$360=600(1-\frac{3000}{K})$
$360=600-600\frac{3000}{K}$
$600\frac{3000}{K}=240$
$600(3000)=240K$
$K=7500$
Answer: 7500
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SAT Practice Test 2015. A radioactive substance decays at an annual rate of 13 percent. If the initial amount of the substance is 325 grams, which of the following functions f models the remaining amount of the substance, in grams, t years later?
A) $f(t)= 325(0.87)^t$
B) $f(t)= 325(0.13)^t$
C) $f(t)= 0.87(325)^t$
D) $f(t)= 0.13(325)^t$
After one year, the amount of the radioactive substance is reduced by 13 percent, that is, 87 percent of 325 grams remains:
$R_1=325(0.87)$.
After two years, the amount of the radioactive substance is again reduced by 13 percent, that is, 87 percent of the previous year’s amount remains:
$R_2=R_1(0.87)=325(0.87)(0.87)=325(0.87)^2$.
So, after t years: $R_t=325(0.87)^t$
Answer: A
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SAT Test 2015. If $\frac{x^{a^2}}{x^{b^2}}=x^{16}$, x>1, and a+b=2, what is the value of a-b?
A) 8
B) 14
C) 16
D) 18
$\frac{x^{a^2}}{x^{b^2}}=x^{16}$
$x^{a^2-b^2}=x^{16}$
$a^2-b^2=16$
$(a+b)(a-b)=16$
$(2)(a-b)=16$
$a-b=8$
Answer: A
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SAT Test 2015. If $a=5\sqrt{2}$ and $2a=\sqrt{2x}$, what is the value of x?
$a=5\sqrt{2}$
$2a=10\sqrt{2}$
$\sqrt{2x}=\sqrt{100}\sqrt{2}$
$\sqrt{2x}=\sqrt{(100)(2)}$
$2x=(100)(2)=200$
$x=100$
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SAT Test 2015. If $3x-y=12$, what is the value of $\frac{8^x}{2^y}$?
A) $2^{12}$
B) $4^4$
C) $8^2$
D) The value cannot be determined from the information given.
$\frac{8^x}{2^y}=$
$\frac{(2^3)^x}{2^y}=$
$2^{3x}.2^{-y}=2^{3x-y}=2^{12}$
Answer: A
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SAT Practice Test.
The figure, titled “Bacteria Growth,” presents a graph of two curved lines. The horizontal axis is labeled “Time” in hours and the vertical axis is labeled “Area covered” in square centimeters. Both axes are labeled from 0 to 10 in increments of one with grid lines extending from each labeled increment.
The curved line labeled “Dish 1” begins on the vertical axis at 1 and curves steeply up and to the right passing through the point with coordinates (2, 4), and the point with coordinates (3, 8). The curved line labeled “Dish 2” begins on the vertical axis at 2 and moves to the right before curving gradually up and to the right passing through the point with coordinates (3, 3), and the point with coordinates (5, 6). The two curved lines intersect at a point with approximate coordinates (1.2, 2.1).
A researcher places two colonies of bacteria into two petri dishes that each have area 10 square centimeters. After the initial placement of the bacteria (t=0), the researcher measures and records the area covered by the bacteria in each dish every ten minutes. The data for each dish were fit by a smooth curve, as shown in the figure above, where each curve represents the area of a dish covered by bacteria as a function of time, in hours. Which of the following is a correct statement about the preceding data?
A. At time t=0, both dishes are 100% covered by bacteria.
B. At time t=0, bacteria covers 10% of Dish 1 and 20% of Dish 2.
C. At time t=0, Dish 2 is covered with 50% more bacteria than Dish 1.
D. For the first hour, the area covered in Dish 2 is increasing at a higher average rate than the area covered in Dish 1.
Answer:The curved line labeled “Dish 1” begins on the vertical axis at 1 and curves steeply up and to the right passing through the point with coordinates (2, 4), and the point with coordinates (3, 8). The curved line labeled “Dish 2” begins on the vertical axis at 2 and moves to the right before curving gradually up and to the right passing through the point with coordinates (3, 3), and the point with coordinates (5, 6). The two curved lines intersect at a point with approximate coordinates (1.2, 2.1).
A researcher places two colonies of bacteria into two petri dishes that each have area 10 square centimeters. After the initial placement of the bacteria (t=0), the researcher measures and records the area covered by the bacteria in each dish every ten minutes. The data for each dish were fit by a smooth curve, as shown in the figure above, where each curve represents the area of a dish covered by bacteria as a function of time, in hours. Which of the following is a correct statement about the preceding data?
A. At time t=0, both dishes are 100% covered by bacteria.
B. At time t=0, bacteria covers 10% of Dish 1 and 20% of Dish 2.
C. At time t=0, Dish 2 is covered with 50% more bacteria than Dish 1.
D. For the first hour, the area covered in Dish 2 is increasing at a higher average rate than the area covered in Dish 1.
Dish 1 has 1 sq cm (10%) covered by bacteria at time t=0, and Dish 2 has 2 sq cm (20%) covered by bacteria at time t=0. These percentages make alternative B correct.
Alternative D is wrong, because for the first hour the slope of the curve for Dish 2 is lower than that for Dish 1.
Answer: B
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SAT Practice Test. A biology class at Central High School predicted that a local population of animals will double in size every 12 years. The population at the beginning of 2014 was estimated to be 50 animals. If P represents the population n years after 2014, then which of the following equations represents the class’s model of the population over time?
A. $P=12+50n$
B. $P=50+12n$
C. $P=50(2)^{(12n)}$
D. $P=50(2)^{(n/12)}$
If the population doubles every 12 years, every year it multiplies by $2^{(1/12)}$.
One year after the first measure, $P_1=50.2^{(1/12)}$
Two years after the first measure, $P_2=50.2^{(1/12)}.2^{(1/12)}=50.2^{(2/12)}$
......
Twelve years after the first measure, $P_12=50.2^{(12/12)}=50.2^1=50.2$
Therefore in year n: $P_n=50.2^{(n/12)}$
Answer: D
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SAT Practice Test. If $a^{-\frac{1}{2}}=x$ (a, to the power of negative one half, equals x), where $a>0$ (a is greater than zero) and $x>0$ (x is greater than zero), which of the following equations gives a in terms of x?
A. $a=\frac{1}{\sqrt{x}}$ (a, equals the fraction 1 over the square root of x)
B. $a=\frac{1}{x^2}$ (a, equals the fraction 1 over x squared)
C. $a=\sqrt{x}$ (a, equals the square root of x)
D. $a=-x^2$ (a, equals negative x squared)
Answer:
$a^{-\frac{1}{2}}=x$
$a^{\frac{1}{2}}=x^{-1}$
$a=x^{-2}$
$a=\frac{1}{x^2}$
Answer: B
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