MATHEMATICS

5 points:

Find all real solutions of the equation .x −17x −111

= x −13x −17

Solution: Answer: or x = 0 −x = 1 By multiplying both sides of equation by we obtainx )(x ) ( 7 − 1 3 − 1

.x x14 − x11 − x3 + 1 = x14 − 2 7 + 1 We simplify

− x − x11 − x3 = 2 7 x x11 − 2 7 + x3 = 0

(x x ) x3 8 − 2 4 + 1 = 0 (x ) x3 4 − 1 2 = 0

or x3 = 0 x4 − 1 = 0 or or x = 0 − x = 1 x = 1

However, 1 would lead to division by zero in the initial equation. Thus, the only valid solutions are 0 and ­1. 10 points: Suppose that Find all real solutions of the equation .(x) 0x 0. F = x2 + 1 + 2 (x) F 2014 = 0 Here we denoted the function applied to exactly times. For example,(x) F k F x k

.(x) (F (F (x))) F 3 = F

Solution 1: Answer: −x = 5 ± √22014 5 By using the quadratic formula, we find the inverse of is(x) x 0x 0 F = 2 + 1 + 2 x = − 5 ±√5 F (x) + Applying the inverse function to F(x) = 0 gives the roots . − 5 ± √5 Applying the inverse a second time only gives real roots for . These roots are − 5 + √5

. − 5 ± √4 5 The nth step of this process gives . The final solution is . − 5 ± √2n 5 − 5 ± √22014 5

Solution 2: By completing the square, we see .(x) (x 5) 5 F = + 2 − Then we have (F (x)) (x 5) 5 F = + 4 − and (F (F (x))) (x 5) 5 F = + 8 − and proceeding similarly

(x) (x 5) 5 . F 2014 = + 22014 − = 0 We find x 5) 5 ( + 22014 = taking square root of this we get x 5) ± , ( + 22013 = √ 5 and taking one more square root (using only positive value from above) we obtain x 5) ± ( + 22012 = √4 5 and proceeding similarly x 5) ( + = ± √22014 5 Therefore, the solutions are . − 5 ± √22014 5

PHYSICS 5 points: Inordertofindthedistancetotheepicenterofanearthquake,seismologistsusethefact thatdifferentwavestravelatdifferentspeed.Forinstance,so­calledP­waves(Pstands forprimary)travelatspeedabout5000m/s.TheSecondary,socalledS­waves,move slower,about3000m/s.FindthedistanceD(inkm)fromtheepicenterofanearthquake totheseismologicalstationwhereS­wavesweredetected20secondslaterthan P­waves. Solution: LetDbethe(unknown)distancetotheepicenterfrom thestation, and km/s V P = 5

be speeds of P and S waves,respectively.km/sV S = 3 P­wave reaches the station in time

t1 = DV P

S­wave reaches it in

t2 = DV S

The delay between the two signals, T=20s is the difference of the two times:

T = t2 − t1 =DV S− DV P

We can now solve this equation to find D:

0s 2 = D( 13km/s −

15km/s) = 2D

15km/s Therefore, D=150km. 10 points: Theboatmovesrelativetothewateratthespeedwhichis2timessmallerthanthe speedoftheriver.Atwhatangletothedirectionoftheflowshouldtheboataimtocross the river so that it is taken down the river the least? Solution: Fromthepointofviewofapersonstandingontheriverbank,theboatmoveswitha velocity whichisavectorsumofthevelocityoftheriver andthatoftheboatwith V V R respect to water, (see picture).V B

Accordingtothevectoradditionrule,theendofthisvector (green)mustbeapointC V onacirclewhosecenterB istheendofthevector (blue).Theradiusofthecircle V R equals to the length of vector (speed of the boat in water).V B Inorderfortheboattobemoveddownstreamaslittleaspossible,thevector mustbe V beasclosetoperpendiculartotheflowaspossible.Thishappenswhenittouchesthe circle.Therefore,thethreevectorsformaright­angletriangle(angleCis900).Since speedofboatinwater(vectorBC)is2timessmallerthanthespeedoftheriver(AB), angleCABmustbe30o,andanglebetweenvectors and is180o­60o=120o.This V R V B is the direction along which the boat must be aimed.

CHEMISTRY

5 points: Introduction I. An areometer is an instrument used to measure the specific gravity (relative density) of liquids. An areometer (a.k.a. hydrometer) is a glass cylindrical stem and a bulb weighted with lead shot to make it float upright (Fig.1). To measure the density of a liquid, you pour it into a tall container, and gently lower the areometer into the liquid until it floats freely (Fig.2). The point at which the

surface of the liquid touches the stem of the hydrometer corresponds to the liquid’s density. Usually, areometers contain a scale inside the stem, so you simply read the number off the scale, and this number equals to the density of your liquid (Fig. 2). Introduction II. As we know, the density of water at 25ºC is 0.9971 grams per cubic centimeter (it is 1 g/cm3 at 0ºC). The density of acetic acid is slightly higher, 1.044 g/cm3. Interestingly, 69% aqueous acetic acid and 85% acetic acid have the same density, 1.0635 g/cm3 (69% means that 100 grams of the acetic acid – water mixture contain 69 grams of acetic acid per 100 grams of the mixture. Chemists usually say the concentration of this solution is 69%). You have a solution of acetic acid and the areometer. You measured the density of the solution and found it equals 1.0635 g/cm3. Can you tell whether the concentration of this solution is 69% or 85%, if the only chemical you have is water, and the only equipment you have is the areometer and a tall glass container? Solution From the data provided in the problem #1 we can conclude that the acetic acid density gradually increases when the concentration increases from 0 to 69%, but for the 85% acid it is again the same as for the 69% acid. That means the density passes through the maximum in between 69 and 85%, and the concentration­density dependence is as shown on this figure:

This plot gives a solution: if we dilute dilute the 85% solution with water (add a small amount of it), the density will increase, but if we do the same with the 69% solution, the density will decrease. Therefore, the answer is “Yes, we can.”

However, there is one more consideration we need to take into account. Although it is correct to say that the density of the acetic acid passes through a maximum, we cannot rule out a possibility that the plot is as shown on the second figure:

Although this situation is rare, theoretically it is possible. Therefore, to make sure the concentration of our solution is 85%, we need to convert in into the 69% solution by adding a calculated amount of water: if the density remains the same, our solution is 85% acetic acid, otherwise, it is 69% acid. However, the answer is still the same: “Yes, we can tell whether the concentration is 69 of 85 percent” 10 points: You have 10 grams of the mixture of zinc and tin shavings, and 1 L of 5% solution of cupric sulfate (a.k.a. copper (II) sulfate). Can you determine a composition of the zinc­tin mixture, if the only materials and equipment available to you are filter paper, flasks, beakers, filtration funnel, and scales? Solution Both zinc and tin react with soluble copper salts (for example, copper sulfate) to produce copper metal and soluble salts of zinc or tin. Zn + CuSO4 → Cu + ZnSO4

Sn + CuSO4 → Cu + SnSO4

In both cases, one atom of the metal produces one atom of copper, however, whereas one zinc atom weighs 65 Da (atomic mass units), one atom of tin weighs 119 Da. That means 10 grams of zinc shavings will produce 10 * 64/65 = 9.8 grams of copper, and 10

grams of tin shavings produce just 10 * 64/119 = 5.4 grams of copper (the mass of one copper atom is 64 Da). Obviously, 10 grams of the mixture of tin and zinc produce some intermediate amount of copper, so if the mass of the copper obtained is M, the percentage of tin in the mixture equals to: %of tin = 100*(M ­ 5.4)/(9.8­5.4) I think I don’t need to explain this formula, because you all studied elementary algebra in school, and it is not a problem for you to derive it by yourself. However, it is important that this problem can be solved without using the term “mole”, because knowing atomic weights and the equations of these two reactions (plus logical reasoning) is pretty sufficient. Importantly, all these considerations are valid provided, but only provided, that all shavings react completely with the copper sulfate solution. That means we have to make sure the amount of copper sulfate is sufficient for that. To check if that is the case, lets, calculate how much of copper sulfate is needed, and how much of it we have. Assuming the density of 5% solution of copper sulfate is approximately 1 g/mL (which is a very reasonable assumption), the amount of copper sulfate is 10*5=50 grams. To calculate the amount of copper sulfate needed for our reaction, let’s calculate the amount needed for the reaction with 10 g of zinc (because the greater is the percentage of tin in the sample, the smaller is the amount of copper sulfate needed for the reaction). From the above equation we can conclude that: 65 Da of zinc react with 64+32+16*4=160 Da of copper sulfate, so 10 grams of zinc react with X grams of copper sulfate. By solving this proportion we find X=10 * 160/65 = 24.6 grams, which is less than 50 grams we have. That means we have an excess of copper sulfate, so we can determine a composition of the zinc­tin mixture. Please, note: before you started solving the problem of that kind, it is EXTREMELY important to make sure the chemical you are using is in an excess. Imagine, the amount of the copper sulfate solution were just 0.3 L: in that situation the correct answer on the question: “Can you determine a composition of the zinc­tin mixture?” would be “No”, because the amount of the copper sulfate solution would be insufficient for making any definite conclusion.

BIOLOGY 5 points: Some plants have seeds in form of burs/prickles. Such seeds can stick to animals' fur, and then the animals spread seeds to new areas. But how do these seeds fall off and get into the soil, if they are stuck in the fur? Suggest as many explanations as possible. What are other ways that animals can spread seeds? Solution: ­ animals scratch (groom) to remove prickles; ­ shedding fur together with seeds; ­ prickles become dry in 2­3 days, and drying changes their properties: thorns are broken, bumpy surfaces become smooth, sticky secretions on the surface dry and become non­sticky. The seeds cannot hold to the fur any more, and fall off. ­ in the rain many of those burs/prickles will be washed away. ­seeds, etc can fall off of dead animals As for the other ways to spread seeds: ­ Animals can eat seeds, and carry them large distances, and provide fertilizer when seeds finally come out of digestive tract. ­ Carry seeds and stash them for future use (like squirrels and mice). Unused seeds can germinate. 10 points: In contrast to their negative reputation as disease causing agents, many viruses play very important and often overlooked biological roles (for example, in evolution and ecosystem maintenance). Overall, viruses helped shaping the world we live in today.

1. Please give at least 4 examples of viruses' beneficial roles in particular biological systems, and explain how that serves the system overall.

2. Please give at least 2 examples of viral infections in a) plants, b) animals, c) bacteria, and explain if those viruses are beneficial or harmful.

3. In what conditions, and how, can viruses be used by humans elsewhere? Please explain which virus features will be most important in your proposed application(s). Solution:

Part I. A virus is a microscopic organism that can replicate only inside the cells of a host

organism. Viruses infect all types of organisms, including animals and plants, as well as bacteria and archaea. Approximately 5000 different viruses have been described in detail at the current time, although it is known that there are millions of distinct types. Viruses are found in virtually every ecosystem on Earth, and these minute life forms are thought to be the most abundant type of biological entity.

Viruses are best known for their role as pathogens. In reality, many kinds of viruses are beneficial to certain species and provide services to the ecosystems they inhabit.

1. Viruses help expand our understanding of biology. The first genome sequences that were revealed were those of viruses

(RNA bacteriophage MS2 and DNA bacteriophage X174). Methods for genetic manipulation were developed from studying bacterial

viruses, and viral elements are still widely used in such work. The reverse flow of genetic information (RNA to DNA) was discovered

from work with the Retroviridae. 2. Viruses produce direct benefits for human health. The most obvious of

these are vaccines against viruses themselves (usually just a weakened version of the pathogenic virus), and vaccine vectors, which use a virus as a Trojan horse to introduce bits of microbial DNA into animals. As a result, the infected cells themselves produce small amounts of microbial proteins that can stimulate the immune system into acquiring immunity. Viral vectors are not used in humans yet, and are used primarily in veterinary medicine.

3. Viruses are routinely used to modify model organisms in research. In regards to human health, an emerging new technology is gene therapy, or the introduction of functioning genes into cells where such genes are missing or defective. Gene therapy is meant for application in patients suffering as a result of acquired or inborn mutations, i.e. cancer and genetic disorders, respectively.

4. In agriculture, viruses that infect insects are used for control of pest populations.

5. Bacteria­specific types of viruses, also known as Bacteriophages, can effectively target, infect, and destroy pathogenic bacteria. Before the discovery of antibiotics, bacteriophages were considered as a potential way to control disease spread by bacteria. With the numbers of antibiotic­resistant bacterial strains steadily rising since the 1990s and

currently approaching critical, viruses are once more viewed as a promising means to fight bacterial infection.

6. Viruses are key to the carbon cycle. The deeper layers of the ocean are full of bacterial species and the viruses that infect and kill these species. When bacteriophages destroy bacteria, they rip their hosts wide open, releasing their content. This process releases bioavailable, highly­reactive organic carbon compounds from the bacterial cells into the ocean, where due to a completely different chemical environment, it is converted to unreactive forms of dissolved organic carbon (also known as refractory DOC). The shallower ocean waters act as a source of bioavailable carbon compounds to replenish and maintain the bacterial population, while some of the refractory DOC itself is slowly deposited on the ocean floor or evaporated as CO2 into the atmosphere.

7. Horizontal gene transfer – evolution. Viruses that can incorporate into the genetic material of their host organisms are known as retroviruses (Retroviridae discussed in 1.), and are found in nearly all eukaryotes. Approxmately 8% of our own human genome is of retroviral origin.

Part II

Plants: Tobacco mosaic virus (TMV) was the first virus to be described. Since then, a

large number of diverse viruses have been found in plants, animals, fungi, and bacteria. The current estimate of recognized viruses is approaching 4,000, of which about 1,000 are plant viruses.

When most people think of viral infection in plants, they think of withered, dying crops.Recent studies suggest that most viruses may not be causing obvious disease, and some are clearly beneficial. For plants, several known acute virusal infections induce drought tolerance; an important trait in a changing environment and Cucumber mosaic virus (CMV) confers cold tolerance in red beets. Although the mechanisms for drought and cold tolerance are not clear, a number of potential osmoprotectants are increased in virus­infected plants.

Plants are capable of adaptation to extreme environments, such as the geothermal soils in Yellowstone National Park. Plants in this environment harbor fungal endophytes that are infected with a virus, and the presence of all three partners is required for thermal tolerance. Viruses, with their extreme diversity, are likely to be capable of contributing new genetic material to a host system, and may be common factors in rapid adaptation. With our changing climate, making use of this potential role for viruses could become important for agriculture.

When Nicotiana benthamiana plants (a relative of tobacco) are infected with TMV, (CMV), brome mosaic virus (BMV) or tobacco rattle virus (TRV), they survive longer after water withdrawal than do uninfected plants. The same is true for rice infected with BMV, for tobacco infected with TMV, and for beet, cucumber, pepper, water­ melon, squash, tomato, Chenopodium amaranticolor and Solanum habrochaites (wild relative of tomato) infected with CMV. In addition, beets infected with CMV survive cold treatments that kill uninfected plants.

Animals and insects: Drosophila spp. can be infected by several viruses. Drosophila C virus (DCV)

can be either a pathogen or a mutualist, depending on the age of the infected fly. In young flies, DCV is a pathogen and reduces the survival of prepubescent flies after infections, which occur when the flies ingest infected food. However, infected adult flies get a boost in their reproductive activity.

Almost every animal uses a layer of mucus produced by specialized cells known as epithelia, to make a barrier that protects tissues that are exposed to the environment, such as the gut or lungs. Animal mucus — whether from humans, fish or corals — is loaded with bacteria­killing viruses called phages. These protect their hosts from infection by destroying incoming bacteria. In return, the phages are exposed to a steady torrent of microbes in which to reproduce. The idea that a phage can be viewed as part of the innate immune system is original, new and exciting (Nature, May 2013)

Another example of a benefit conferred by viral genes comes from humans. A specific DNA sequence installed into the human genome by a retrovirus regulates the amylase gene cluster, allowing us to produce amylase in our saliva. This sequence, which we share with a few other primates, enables us to eat starchy foods we otherwise couldn't.

Certain viruses are found to work against cancer. It has been found that “Seneca Valley Virus­001″ has 10,000 times more cancer killing activity than chemotherapeutics and has been found to have no toxicity.

Part 3.

The use of materials derived from natural sources in material science has allowed us to harness complex structures that resulted from eons of evolutionary fine­tuning. A better understanding of the structure and function of viruses has revealed a collection of natural molecular assemblies and containers with a variety of shapes, sizes, stabilities, dynamic properties, and chemical reactivities. Viruses are increasingly being used in material science, engineering, and nanotechnology as tools and building blocks for electronics, chemistry, and biomedical science.

Artificial viruses can be used in biomedical and biotechnological applications, such as targeted delivery of nucleic acids for gene therapy and as scaffolds in material science.

Virus­like particles represent one of the most exciting new technologies to produce vaccines.

Useful Links: http://web.gps.caltech.edu/classes/ge246/roossinck_natrev2011_goodvi.pdf http://www.garlandscience.com/res/pdf/9780815341505_ch02.pdf http://www.news­medical.net/health/Virus­Uses.aspx http://www.immpressmagazine.com/viruses­and­us­evolving­by­infection/ http://www.ncbi.nlm.nih.gov/books/NBK21523/ http://www.nature.com/nrmicro/journal/v8/n1/full/nrmicro2240.html http://www.sciencemag.org/site/products/scor_aaas.pdf

COMPUTER SCIENCE

Solutions must be typed and submitted in one of following formats: .txt .c .cpp .java .py Solutions written in Java, C, C++, Python and pseudo­code are accepted. Pseudo­code

guidelines are at ttp://users.csc.calpoly.edu/~jdalbey/SWE/pdl_std.html 5 points: Write a program that would survey the popularity of 2 apps ­ SnapChat and Instagram ­ among 20 students: 1. Input the results (1: Instagram or 2: SnapChat). 2. Count the number of results of each type. 3. Display a summary of the results : number of students who prefer Instagram and number of students who like SnapChat more. 4. If more than 10 students prefer a particular app, print a message about a winning app. Solution: Set Instagram count to 0. Set SnapChat count to 0. Set student counter to 1. While student counter is less than 20:

Input the next result. If it is an Instagram Add 1 to Instagram count else Add 1 to SnapChat count Add 1 to student counter. Print number of Instagram votes. Print number of SnapChat votes. If more than 10 students prefer a particular app: Print a message about a winning app. 10 points: You have a 10­step staircase that is 10 ft from railing­to­railing with tall walls on either side of it, parallel to the railings. Each step is 1 ft wide. It is a special staircase, in that the steps don’t just go up ­­ any step can either be one foot higher or one foot lower than the previous step. You are given the heights of all steps in the staircase as an array of non­negative integers, where the lowest step is at height 0. For instance, normal, ascending stairs would be represented by 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, descending stairs by 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, and stairs that go down and then back up would be represented by 4, 3, 2, 1, 0, 1, 2, 3, 4, 5. Question: One day, it rains really hard. Given the array on the input, determine how much rain is trapped in the staircase. You may assume the input is valid (all steps are separated by one foot in height, all elements in the array are integers, array has size 10, etc.). For example, on the input 0, 1, 2, 3, 4, 3, 4, 3, 2, 1, the output should be 10. This is because the only step that traps water is the step between in the section 4, 3, 4 (see diagram), and the volume of that is 1 ft high by 1 ft wide by 10 ft long = 10 cubic feet. Solution : You can check for each square in the 10 by 10 coordinate space (x = step number, h = height), whether or not water would get trapped in that square (if the square is not part of the staircase itself). That is, water will NOT get trapped if, as you scan right OR left, the height of all steps is less high than that of the square. This is not a very efficient algorithm, but it still gets the job done.

//ALGORITHM IN PSEUDOCODE BELOW height[x] = input array that contains the value of the height of the staircase at position x. total water = 0; For all integer x from 0 to 9: For all integer h from 0 to 9: water_trapped_left = false; water_trapped_right = false; if height[x] < h: For all integer x’ from 0 to x: if height[x’] = h: water_trapped_left = true; For all integer x’ from x to 9: if height[x’] = h: water_trapped_right = true; If both water_trapped_right and water_trapped_left are true: add 10 to total_water return total_water;