Problem Solving

1)A wildlife preserve is being planned for 3,000 rhinoceroses. The preserve is to contain a total of 10,000 acres of watering area, plus 100 acres of grazing area for each rhinoceros. If the number of rhinoceroses is expected to increase by 10 percent, how many thousand acres should the preserve have in order to provide for the increased population?
(A) 340
(B) 330
(C) 320
(D) 310
(E) 300

2)In a certain animal population, for each of the first 3 months of life, the probability that an animal will die during that month is . For a group of 200 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?
(A) 140
(B) 146
(C) 152
(D) 162
(E) 170


3) The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is parts per million, where 0 < x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?
(A) 2.4 ft
(B) 2.5 ft
(C) 2.8 ft
(D) 3.0 ft
(E) 3.2 ft

Answers
1 - A
2 - B
3 - E

gmatrant wrote:1)A wildlife preserve is being planned for 3,000 rhinoceroses. The preserve is to contain a total of 10,000 acres of watering area, plus 100 acres of grazing area for each rhinoceros. If the number of rhinoceroses is expected to increase by 10 percent, how many thousand acres should the preserve have in order to provide for the increased population?
(A) 340
(B) 330
(C) 320
(D) 310
(E) 300

2)In a certain animal population, for each of the first 3 months of life, the probability that an animal will die during that month is . For a group of 200 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?
(A) 140
(B) 146
(C) 152
(D) 162
(E) 170


3) The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is parts per million, where 0 < x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?
(A) 2.4 ft
(B) 2.5 ft
(C) 2.8 ft
(D) 3.0 ft
(E) 3.2 ft

Answers
1 - A
2 - B
3 - E

hey gurus...pls let me know how to solve the above problems

S-42 Q-9 ( Sum - 1)

Number of rhinoceroses is expected to increase by 10 percent and wildlife preserve is being planned for 3,000 rhinoceroses:

So the New number of Rhinocerouses 10/100 * 3000 = 300
Total number of Rhinocerouses = 3000 + 300 = 3300

Now, 100 acres of grazing area for each rhinoceros:
3300 * 100 = 330000

and 10,000 acres of watering area

So,
330000 + 10000 = 340000 (340 Thousand)

Ans A.

S-42 Q-12 ( Sum - 2)
* I hope you meant question 12 and not 11 of Section 42.

Given: For each of the first 3 months of life, the probability that an animal will die during that month is 1/10
and 200 newborn members of the population

First Month: 1/10 * 200 = 20
Members left = 200 - 20 = 180

Second Month: 1/10 * 180 = 18
Members left = 180 - 18 = 162

Third Month: 1/10 * 162 = 16.2
Members left = 162 - 16 = 146 ( truncated .2 as newborn member can not be .2 )

Ans B.

S-42 Q-16 ( Sum - 3)

Given: At a depth that is x feet below the top of the tank, the concentration is 3 + 4/root (5 - x); where 0 < x < 4

We can not form any equation so we should begin with the options given.
Start with the easy option i.e option 4:

4 - 3.0 ( Easy to calculate)
3 + 4/root(5 - 3)
= 3 + 4/1.42
= 3 + 2.8
= 5.8...No..Not close to 6 with the range of 0.1

Here u can go and select option E as all the other options will give lesser values so E is the answer

or
5 - 3.2 and
3 + 4/root(5 - 3.2)
= 3 + 4/root(1.8 )
= 3 + 4/1.34
= 3 + 2.98
= 5.98

Ans E.