Free image from Pixabay The Verbal Reasoning section features Sentence Equivalence questions on the GRE. In each sentence, one word will be missing, and you must identify two correct words to complete the sentence. The correct answer choices, when used in the sentence, will result in the same meaning for both sentences. This question type tests your ability to figure out how a sentence should be completed by using the meaning of the entire sentence. In the questions below, select the two answer choices that, when inserted into the sentence, fit the meaning of the sentence as a whole and yield complete sentences that are similar in meaning. Questions1. Her lastminute vacation was _______________________ compared to her usual trips, which are planned down to the last detail. A. expensive B. spontaneous C. predictable D. satisfying E. impulsive F. atrocious 2. After staying up all night, she felt extremely _____________________; however, she still an three miles with her friends. A. apprehensive B. lethargic C. controversial D. sluggish E. vigorous F. energetic 3. Although the lab assistant openly apologized for allowing the samples to spoil, her _________________ did not appease the research head, and she was let go. A. insincerity B. frankness C. falsehoods D. candor E. inexperience F. hesitation 4. He was unable to move his arm after the stroke; in addition, the stroke ____________________ his ability to speak. A. appeased B. satisfied C. impeded D. helped E. hindered F. assisted 5. The firefighter, desperate to save the children on the second floor of the fiery house, rushed into their bedroom; his colleagues, more wary of the ____________________ structure, remained outside.
A. stalwart B. precarious C. stout D. irrefragable E. tottering F. fecund
0 Comments
Free image from Pixabay
\[\frac{1}{(a + b)^\frac{1}{2}} = (a + b)^{\frac{1}{2}}\]
If the above is true, which of the following must be true?
\[(A) \quad a = 0\]
\[(B) \quad \sqrt{a + b} =  1\]
\[(C) \quad \sqrt{a + b} = 0\]
\[(D) \quad a + b = 1\]
\[(E) \quad (a + b)^2 = 0\]
One of the most amazing properties of the natural numbers  1, 2, 3, 4, 5, 6, 7 ...  is that they go on forever. However, you can't write down the largest number. If you think a number is the largest, you can always add 1 to it and make it larger. Hence, the sequence of natural numbers is infinite. So how to we deal with large numbers? For example, the earth has a mass of about 5.98 sextillion metric tons. Wait a minute! What is sextillion?? It is 1,000,000,000,000,000,000,000. And the Great Lakes contain roughly 52.92 duodecillion water molecules !! What is duodecillion? It is 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000. How do you represent such large numbers easily? And what if we have to add or multiply them? Archimedes, the great scientist and mathematician of antiquity, came up with a solution. Can you guess it? Exponents!! Yes, you can use exponents to reduce the number of zeroes to deal with. Archimedes wrote a letter to a local monarch, in which he proposed some tools and methods to handle arbitrarily large numbers. He informed the monarch that he could use this method to count the number of grains of sand in the universe. Archimedes was fed up with people saying you couldn’t calculate the number of grains of sand on a beach. He believed that one could, and he calculated not just how many grains of sand there were on the beach, but how many there were in the universe. The trouble Archimedes faced was the Greek number system. It was a primitive system in which letters became numbers: A = 1, B = 2, C = 3, etc. So Archimedes invented a new classification of numbers: the exponents. Archimedes thought that to count the number of grains of sand in the universe he needed numbers up to the eighth order, i.e. (10^8)^8 = 10^64, which is equal to: 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000. We call this great innovation that Archimedes developed the Exponential Notation. He uses exponents of powers of 10. And the idea is to express large numbers using powers of 10, by keeping track of the number of digits in the number. So using exponential notation, we would write 100 as 10 squared. And 1,000 as 10 cubed. 10,000  10 to the 4th. 100,000  10 to the 5th power. So a number like 4,270,000 would be written as 4.27 times 10 to the 6th, or 4.27 x 10^6. The advantage of this exponential notation that Archimedes discovered is, that it makes it very easy to multiply arbitrarily large numbers. And that's for the following reason. If we wanted to multiply a million (10^6) times 100 million (10^8), all we have to do is add the exponents! So 6 + 8 = 14 and you get 10^14 as the result. In other words, 10^6 x 10^8 = 10^(6+8) = 10^14. How would we apply the exponential notation to estimating the time since the Big Bang? Astronomers have estimated that the Big Bang is about is about 13.8 billion years old. That was a number that was obtained from the recent Planck mission. We can write it in exponential notation, as 13.8 x 10^10, as 1 Billion is 10^9. We can rewrite it in the standard form as 1.38 x 10^9 years. That's the number of years in the history of the universe. What about the number of seconds in the history of the universe? Well, we have to take that large number 1.38 times 10 to the 10th power, or 1.38 x 10^10  and multiply it by the number of seconds in every year. Well, how many seconds are there in a year? There are 60 seconds in a minute, 60 minutes in an hour, 24 hours every day, and 365 days in a normal year. So the total number of seconds in a year is 60 times 60 times 24 times 365, which works out exactly to 31,536,000 seconds in a year. In exponential notation, we have to convert that number. And we get 3.15 times 10 to the 7th power, or 3.15 x 10^7. So to calculate the number of seconds in the history of the universe, we take the number of years in the history of the universe 1.38 x 10^10  and multiply it by the number of seconds in a year  3.15 x 10^7. We multiply the constants, 1.38 times 3.15, which is about 4.4. And then we add the exponents, 10 and 7 to get 17. So, approximately, the number of seconds in the history of the Big Bang as we know it is 4.4 x 10^17 seconds. Isn't that impressive? References:
1. https://sites.google.com/site/largenumbers/home/21/Larger_Numbers_in_Science 2. https://www.famousscientists.org/howarchimedesinventedthebeastnumber/ Image from Pixabay The Fibonacci sequence was first observed by the Italian mathematician Leonardo Fibonacci in 1202. He was investigating how fast rabbits could breed under ideal circumstances. He made the following assumptions:
Fibonacci asked how many pairs of rabbits would be produced in one year. Can you create the numbers yourself? Remember to count the 'pairs' of rabbits and not the individual ones. Try it. Were you able to come up with the Fibonacci numbers? If not, here is how you would do it.
The pattern comes out to be 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233. Fibonacci numbers are of interest to biologists and physicists because they are frequently observed in various natural objects and phenomena. For example, the branching patterns in trees and leaves are based on Fibonacci numbers. On many plants, the number of petals is a Fibonacci number: buttercups have 5 petals; lilies and iris have 3 petals; some delphiniums have 8; corn marigolds have 13 petals; some asters have 21 whereas daisies can be found with 34, 55 or even 89 petals. How can we create a rule (algorithm) for the fibonacci series (sequence)? First, the terms are numbered from 0 onwards like this: n = 0 1 2 3 4 5 6 7 8 9 10 ... xn =0 1 1 2 3 5 8 13 21 34 55 ... What rule can we create here? Well, if you look, x3 = x2 + x 1 (2 = 1 + 1) and x4 = x2 + x3 (3 = 1 + 2), etc. So we can write the rule (algorithm) as: xn = x(n1) + x(n2). where:
Example: term 7 is calculated as: x7= x(71) + x(72) = x6 + x5 = 13 + 8 = 21 Let's write programs in Python to calculate the Fibonacci numbers. 1. With looping: def fib(n): a,b = 1,1 for i in range(n1): a,b = b,a+b return a print(fib(1)) print(fib(2)) print(fib(3)) print(fib(4)) print(fib(5)) print(fib(6)) print(fib(7)) print(fib(8)) 1. With recursion: def fibR(n): if n==1 or n==2: return 1 return fibR(n1)+fibR(n2) print(fibR(1)) print(fibR(2)) print(fibR(3)) print(fibR(4)) print(fibR(5)) print(fibR(6)) print(fibR(7)) print(fibR(8)) N.B: No not copy and paste the python code as identation is important.
The greatest common divisor (GCD) or the highest common factor (HCF) of two numbers is the largest positive integer that perfectly divides the two given numbers. Solving this problem for a specific set of numbers is easy. For example, find the GCD of 12 and 18. The The divisors of 12 are 1, 2, 3, 4, 6, 12 and for 18 are 1, 2, 3, 6, 9, 18. The common factors are 1, 2, 3, and 6. So the greatest common factor is 6. How would you find the GCD for any number? Here the problem is more challenging. Here is one solution. Let's take two integers a and b passed to a function which returns the GCD. In the function, we first determine the smaller of the two number since the GCD (HCF) can only be less than or equal to the smallest number. For example, the GCD of 12 and 14 can only be less than 12 and not greater. We then use a for loop to go from 1 to that number. In each iteration, we check if our number perfectly divides both the input numbers. If so, we store the number as the GCD. At the completion of the loop we end up with the largest number that perfectly divides both the numbers. Below is the algorithm in python. def computeGCD(a, b): if a < b: smaller = a else: smaller = b for i in range(1, smaller+1): if (a % i == 0) & (b % i == 0): gcd = i return gcd print(computeGCD(24, 16)) print(computeGCD(48, 256)) N.B: Do not cut and paste the above code. Make sure the indentation is correct. The above method is easy to understand and implement but not efficient. A much more efficient method to find the GCD (HCF) is the Euclidean algorithm. The Euclidean algorithm is based on the principle that the greatest common divisor of two numbers does not change if the larger number is replaced by its difference with the smaller number. That is a mouthful! Let's make it simple by taking an example. 21 is the GCD of 252 and 105 (as 252 = 21 × 12 and 105 = 21 × 5), and the same number 21 is also the GCD of 105 and 147 (252 − 105). Since this replacement reduces the larger of the two numbers, repeating this process gives successively smaller pairs of numbers until the two numbers become equal. When that occurs, they are the GCD of the original two numbers. A more efficient version of the algorithm shortcuts these steps, instead we divide the greater by smaller and take the remainder. Now, divide the smaller by this remainder. Repeat until the remainder is 0. For example, if we want to find the H.C.F. of 54 and 24, we divide 54 by 24. The remainder is 6. Now, we divide 24 by 6 and the remainder is 0. Hence, 6 is the required GCD. Python code for Euclidean Algorithm def euclidAlgo(a, b): while (b): a, b = b, a % b return a print(euclidAlgo(24, 16)) print(euclidAlgo(48, 256)) Python code for Euclidean Algorithm using recursion: def euclidAlgo(a, b): if (b == 0): return a else: return euclidAlgo(b, a % b) print(euclidAlgo(24, 16)) print(euclidAlgo(48, 256)) Sources: Wikipedia; https://www.programiz.com/pythonprogramming/examples/hcf
Free Image from Pixabay Finding Prime NumbersPrime numbers are very important, yet many students do not see the value of learning them. Primes have several applications, most importantly in information technology, such as publickey cryptography, which relies on the difficulty of factoring large numbers into their prime factors. One key challenge is to find prime numbers. Interestingly, Prime numbers and their properties were first studied extensively by the ancient Greek mathematicians. Euclid, for example, proved that there are infinitely many prime numbers. Just to refresh our memory, a number greater than 1 is called a prime number, if it has only two factors, namely 1 and the number itself. Proof by Contradiction One of the first known proofs is the method of contradiction. It is used to calculate prime factors of large numbers. Calculating prime factors of small numbers is easy. For example, the factors of 17 is 1 and 17, so it is a prime number. What about large numbers? Let's look at the proof by contradiction method. If a number n is not a prime, it can be factored into two factors a and b, such that n = a*b. For example, let's say a * b = 100, for various pairs of a and b. If a = b, then they are equal, we have a*a = 100, or a^2 = 100, or a = 10, the square root of 100. If one of the numbers is less than 10, then the other has to be greater to make it to 100. For example, take 4 x 25 = 100. 4 is less than 10, the other number has to be greater than 10. In other words, if a * b, if one of them goes down, the other number has to get bigger to compensate so the product stays at 100. Put mathematically, the numbers revolve around the square root of their product. Let's test if 101 is prime number. You could start dividing 101 by 2, 3, 5, 7, etc, but that is very tedious. A better way is to take the square root of 101, which is roughly equal to 10.049875621. So you only need to try the integers up through 10, including 10. 8, 9, and 10 are not themselves prime, so you only have to test up through 7, which is prime. Because if there's a pair of factors with one of the numbers bigger than 10, the other of the pair has to be less than 10. If the smaller one doesn't exist, there is no matching larger factor of 101. Let's now build an algorithm using this method to test any number for primality. Algorithm in Pythonimport math def isPrime(num): if (num < 2): return False else: for i in range(2, int(math.sqrt(num)) + 1): if num % i == 0: return False return True print(isPrime(33)) print(isPrime(0)) print(isPrime(47)) print(isPrime(1047)) print(isPrime(11)) print(isPrime(59392847)) N.B: Do not just copy the code because you have to be careful with indentation in python. Try the above algorithm and let us know if you found it useful or have alternative solutions.
Free Image from Pixabay
Problem
N.B: This problem is for SAT Subject Math Level 2
\(If \quad x_0 \quad and \quad x_{x+1} = \sqrt {4 + x_n}, \quad then \quad x_3 = \)
(A) 2.65
(B) 2.58 (C) 2.56 (D) 2.55 (E) 2.54 Solution
Answer is C.
This is a simple but tricky problem. It is simple to apply but you have to think recursively. For n = 0, we have:
\(x_{0+1} = \sqrt {4 + x_0} \)
\(x_1 \quad = \quad \sqrt {4 + 3} \quad = \quad \sqrt{7} \quad = \quad 2.65 \)
\(x_2 \quad = \quad \sqrt {4 + x_1} \quad = \quad \sqrt {4 + 2.65} \quad = \quad 2.58 \)
\(x_3 \quad = \quad \sqrt {4 + x_2} \quad = \quad \sqrt{4 + 2.58} \quad = \quad 2.56\)
Problem:In the equation r = 4/(2 + k), k represents a positive integer. As k gets larger without bound, the value of r: F. gets closer and closer to 4. G. gets closer and closer to 2. H. gets closer and closer to 0. J. remains constant. K. gets larger and larger Answer:Answer is H.
As k gets larger and larger without bound, the expression 4/(2+k) becomes 4 divided by an increasingly large number. For example, think about the trend between the following fractions: 4/100, 4/10,000, 4/1,000,000, ... Looking at it this way, you can see that the expression for r gets closer and closer to zero. Seeing me, she roused herself: she made a sort of effort to smile, and framed a few words of congratulations; but the smile expired, and the sentence was abandoned unfinished. She put up her spectacles and pushed her chair back from the table. “I feel so astonished,” she began, “I hardly know what to say to you, Miss Eyre. I have surely not been dreaming, have I? Sometimes I half fall asleep when I am sitting alone and fancy things that have never happened. It has seemed to me more than once when I have been in a doze, that my dear husband, who died fifteen years since, has come in and sat down beside me; and that I have even heard him call me by my name, Alice, as he used to do. Now, can you tell me whether it is actually true that Mr. Rochester has asked you to marry him? Don’t laugh at me. But I really thought he came in here five minutes ago, and said that in a month you would be his wife.” [10] “He has said the same thing to me,” I replied. “He has! Do you believe him? Have you accepted him?” “Yes.” She looked at me bewildered. “I could never have thought it. He is a proud man; all the Rochesters were proud: and his father at least, liked money. He, too, has always been called careful. He means to marry you?” “He tells me so.” She surveyed my whole person: in her eyes I read 30 that they had there found no charm powerful enough to solve the enigma. “It passes me!” she continued; “but no doubt it is true since you say so. How it will answer I cannot tell: I really don’t know. Equality of position and fortune is often advisable in such cases; and there are twenty years of difference in your ages. He might almost be your father.” [22] “No, indeed, Mrs. Fairfax!” I exclaimed, nettled; “he is nothing like my father! No one, who saw us 40 together, would suppose it for an instant. Mr. Rochester looks as young, and is as young, as some men at five and twenty.” “Is it really for love he is going to marry you?” she asked. I was so hurt by her coldness and skepticism, that the tears rose to my eyes. “I am sorry to grieve you,” pursued the widow; “but you are so young, and so little acquainted with men, I wished to put you on your guard. It is an old saying that ‘all is not gold that glitters’; and in this case I do fear there will be something found to be different to what either you or I expect.” [30] “Why?—am I a monster?” I said: “Is it impossible that Mr. Rochester should have a sincere affection for me?” “No: you are very well; and much improved of late; and Mr. Rochester, I dare say, is fond of you. I have always noticed that you were a sort of pet of his. There are times when, for your sake, I have been a little uneasy at his marked preference, and have wished to put you on your guard; but I did not like to suggest even the possibility of wrong. I knew such an idea would shock, perhaps offend you; and you were so discreet, and so thoroughly modest and sensible, I hoped you might be trusted to protect yourself. Last night I cannot tell you what I suffered when I sought all over the house, and could find you nowhere, nor the master either; and then, at twelve o’clock, saw you come in with him. “Well never mind that now,” I interrupted impatiently; “it is enough that all was right.” [40] “I hope all will be right in the end,” she said: “but, believe me, you cannot be too careful. Try and keep Mr. Rochester at a distance: distrust yourself as well as him. Gentlemen in his station are not accustomed to marry their governesses.” Questions:1. When Mrs. Fairfax says, “Gentlemen in his station are not accustomed to marry their governesses,” she is expressing her belief that: A. Mr. Rochester is incapable of loving Miss Eyre. B. Mr. Rochester will treat Miss Eyre like a governess when they are married. C. Mr. Rochester may not be sincere about his feeling towards Miss Eyre D. Mr. Rochester may not really have asked Miss Eyre to marry him. 2. It can be reasonably inferred from the conversation that Mrs. Fairfax believes Miss Eyre will: F. recognize that Mr. Rochester actually wants to marry Mrs. Fairfax. G. marry Mr. Rochester much sooner than originally planned. H. no longer desire to marry Mr. Rochester. J. potentially regret her decision to agree to marry Mr. Rochester. 3. Mrs. Fairfax’s opinion about Miss Eyre and Mr. Rochester’s relationship can best be exemplified by which of the following quotations from the passage? A. “Mr. Rochester looks as young, and is as young, as some men at five and twenty.” B. “How it will answer I cannot tell: I really don’t know.” C. “He is a proud man; all the Rochesters were proud.” D. “But I really thought he came in here five minutes ago, and said that in a month you would be his wife.” 4. The phrase “you were so discreet, and so thor oughly modest and sensible” (lines 36–37) is used by Mrs. Fairfax to: F. explain why Miss Eyre should not marry Mr. Rochester. G. explain why it is likely that Mr. Rochester really does not plan on marrying Miss Eyre. H. explain why Mrs. Fairfax had not discussed Mr. Rochester’s feelings toward Miss Eyre before. J. insult Miss Eyre and let her know that Mrs. Fairfax was disappointed in her. 5. The passage makes it clear that Miss Eyre and Mr. Rochester: A. get married. B. do not really know each other well enough to become engaged. C. will not live happily because they will be shunned by society. D. have a relationship that is not typical in their society. Answers:
Ref: McGraw Hill
QuestionWhich of the following is a factored form of 3x^3y^3 + 3xy? A. 3xy(x^2y^2 + 1) B. 3(3x^2y^2) C. (3x + 3y)(3x + 3y) D. 3x^2y^2(xy) E. 3x(x^2y^2 + 3) AnswerThe correct answer is A. This problem requires you to find the Greatest Common Factor. The Greatest Common Factor is 3xy, because each term has at least 1 factor of 3, 1 factor of x, and 1 factor of y. When you factor 3xy out of 3x^3y^3 you are left with x^2 y^2 , and when you factor 3xy out of 3xy, you are left with 1. Therefore, when factored, 3x^3 y^3 + 3xy = 3xy(x^2 y^2 + 1). Astronomers have found over 400 planets orbiting stars. The discovered planets have a variety of compositions, masses, and orbits. Despite the variety, the universal rules of physics and chemistry allow scientists to broadly categorize these planets into just a few types: Gas Giant, Carbon Orb, Water World, and Rocky Earth. Table 1 shows the composition of the various planet types and typical mass ranges relative to Earth. Table 2 shows a sampling of planets orbiting various stars described in Table 1. These planets are merely numbered 17. Table 2 details the masses and orbital radii of the planets. Questions:1. The data in Table 1 and Table 2 support which of the following statements? A. Gas Giant planets have the largest orbital radii. B. Orbital radius is directly related to mass. C. Orbital radius is inversely related to mass. D. The data does not support a correlation between mass and orbital radius. 2. According to Table 1 and Table 2, which of the following stars has the most massive Gas Giant planet orbiting it? F. Gliese 777 G. OGLE TR 132 H. PSR 1257 J. Gleise 581 3. If a new planet were discovered, with a mass of 325, an orbital radius of 1.5, and a composition of mostly hydrogen, what would be its most likely classification? A. Carbon Orb B. Water C. Rocky Earth D. Gas Giant Answers1. D
2. F 3. D QuestionA waffle ice cream cone is pictured above. If the volume of the cone is 14.4pi cubic inches, what is the diameter of the cone (in inches)? AnswerAnswer is 6.
Solution: The volume of a cone is found by using the formula V= 1/3 x pi x r^2 x h, where r is the radius of the circular base and h is the height. Input the known data and solve for r. 14.4pi = 1/3 pi r^2 (4.*) 14.4 = 1.6 r^2 9 = r^2 3 = r or, d, the diameter is 6. Question:A gymnast has a routine in which he sways back and forth on a high bar, making an arc that measures 135 deg. As he swings, the bottom of his shoes create an arc that measures 9 feet. At the conclusion of his routine, he swings completely around for one full circle around the bar. What is the circumference of that circle (answer in feet)? Answer:Answer is 24.
Solution: A circle measures 360 deg. Find what portion a 135 deg arc is of a circle and then use that information to create a proportion. We get: 135/260 which can be reduced to 3/8. So, 3/8 = 9 feet/x feet, or 3/8 = 9/x, or, 3x = 72 or, x = 24 Image from Pixabay Questions:1. An element consists of three isotopes in the relative abundance given below. What is the atomic mass of this element? 
AuthorWrite something about yourself. No need to be fancy, just an overview. Archives
June 2018
Categories
All
