My fifth GMAT practice, and I’m still on target. A total of 106 questions, 61 verbal (with one ‘not scored’) and 52 quantitative. 5 wrong in sentence correction (oops), 1 wrong in reading comprehension, and 1 wrong in critical reasoning. 11 wrong in maths, of which 4 were data sufficiency and 7 problem solving. Which means proportionately my data sufficiency has improved, suggesting the practice with these questions has paid off. Corrected raw scores of 51V and 38Q lead to my highest simulated score yet, 690.
Problems: I still keep running out of time, having to rush the last few q’s in all sections. But I feel a definite improvement in ‘test taking abilities’; I’m now approaching each question on the basis of the list of answers, rather than ‘working it out’ and seeing if the answer I get is in the list. I’m also resisting my natural urge to complete each problem and get a final result, rather than stopping the moment I have enough info to answer the question. So some good progress yesterday.
If x+y / xy = 1, then y =
(A) x / x-1
(B) x / x+1
(C) x-1 / x
(D) x+1 / x
I chose D. I missed a trick: x+y must equal xy, since anything divided by the same thing is 1. So y must equal x+y / x. But that’s not among the answers. The only answer that suggests something ‘taken away’ from x, leaving behind y, is A.
I keep getting algebraic rearrangement questions wrong – need to study these this afternoon, since they should be simple. There must be patterns to these sorts of questions; I’ve tripped up so many times on not knowing what to do with this precise situation of evaluating xy and x+y.
How many integers n are there such that 1 < 5n + 5 < 25 ?
I chose C, falling into a trap. Getting rid of the 5, 5n must be less than 20, so n must be less than 4. And 5n must also be greater than -4 (1-5). So the range of integers must be at most 6: -3, -2, -1, 0, 1, 2, 3. Nothing less than 1 works, since negative integers for n produce results less than 1 when 5 is added. So n can only be 1, 2, 3. Or – my mistake – zero. B is the answer. Lesson learned: 0 is an integer.
In 1985 a company sold a brand of shoes to retailers for a fixed price per pair. In 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent. If the company’s revenue from the sale of the shoes in 1986 was $3.0 million, what was the approximate revenue from the sale of the shoes in 1985?
(A) $2.4 million
(B) $2.9 million
(C) $3.0 million
(D) $3.1 million
(E) $3.6 million
I chose B, with seconds to spare before the time limit. Let’s see: a figure that if decreased by 20% units, each selling at 20% more, equals $3m. That’s $3m = 0.8 x 1.2, which is 0.96. So the 1986 total is 96% of the 1985 total… we’re looking for a figure a bit higher than $3m. The answer is D.
With these questions, it’s necessary to grok precisely what the question’s asking and focus on the sums that’ll get that answer. It’s too easy to tie your brain in knots with these. Need practice; on a GMAT exam today I wouldn’t have got it in time.
What is the value of the two-digit integer x ?
(1) The sum of the two digits is 3.
(2) x is divisible by 3.
I chose C. Statement 1 alone could mean x is 21 or 12, so not enough. Statement 2 alone could mean x is 3, 6, 9, 12…. 21… so even both statements aren’t enough. The answer is E. Should have got this.
Is y = 6 ?
(1) = y^2 = 36
(2) y^2 – 7y + 6 = 0
I chose D. But statement 1 alone isn’t enough; y could be 6 or -6. Statement 2 suggests y is less than 7, since 7y is more than y times y (if y is positive – which it must be, since y^2 must be positive and if y were negative it’d make -7y positive. Three positive terms added together equalling 0? Not likely). So Statement 2 confirms y is positive, making y = 6. C is the answer. Again, should have got this one.
The figure above represents the floor plan of an art gallery that has a lobby and 18 rooms. If Lisa goes from the lobby into room A at the same time that Paul goes from lobby into room R, and each goes through all of the rooms in succession, entering by one door and exiting by the other, which room will they be in at the same time?
(1) Lisa spends 2x minutes in each room and Paul spends 3x minutes in each room.
(2) Lisa spends 10 minutes less time in each room than Paul.
I chose D. You get the same result regardless of whether Lisa and Paul linger for 2 and 3 minutes, or 20 and 30, or days on end, since both quantities are increasing at a linear rate. But in statement 2, we have no idea of how long either is lingering, save that Lisa’s moving faster – a non-linear relationship, so they’ll end up in different rooms depending on how long Paul mooches around. A is the answer.
Remember, there’s no need to work it out; I just needed to establish that the rate at which they move rooms is constant for all values of x. Simple – but I wouldn’t have got this, nor would I have recognised how to solve it in two minutes. Another one for the books this afternoon.
If xy = – 6, what is the value of xy ( x + y )?
(1) x – y = 5
(2) xy^2 = 18
I chose A. Another rearranging algebra question wrong. (Note to self: xy^2 is x times y^2, not the square of xy.) Knowing x-y is 5 and xy = -6 doesn’t tell us what x and y are; just that one of them is negative (because only a minus times a plus is negative). We don’t even know if they’re both integers, so can’t assume they’re 2 and 3. Statement 1 is out.
Is statement 2 alone sufficient? It does tell us y is the negative one, since 18 is positive, which means x and y^2 are positive and y is the only one that could be negative. Knowing 18/x equals y^2 and the original xy = -6 gives us enough. The answer is B. Slow at these, but they’re solvable. Simultaneous equations are my weak point.
If the average (arithmetic mean) of the four numbers K, 2K + 3, 3K – 5 and 5K + 1 is 63, what is the value of K ?
(B) 15 3/4
(E) 25 3/10
I chose B, but should have got this right. K + 2K + 3 + 3K – 5 + 5K + 1 / 5 is 11K-1 = 63 x 4 = 252. Shifting the -1 across, 11K = 253…. quick bit of division… K = 23. The answer is D.
If x + y = a and x – y = b, then 2xy =
(A) (a^2 – b^2) / 2
(B) (b^2 – a^2) / 2
(C) a-b / 2
(D) ab / 2
(E) (a^2 + b^2) / 2
I guessed E. Surprise, another wrong algebra one. It’s a case of expressing 2xy in terms of a and b. First, what does xy equal? That’ll give us the top line of the answer.
Well, we know (x+y)(x-y) must equal ab. Which is the same as x^2 -xy +yx -y^2 = ab. So x^2 -y^2 = ab. That looks just like a^2 – b^2 = xy, so a guess of A would be correct. But I need to learn these properly.
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?
(A) w^2 + pw + k = 0
(B) w^2 – pw + 2k = 0
(C) 2w^2 + pw + 2k = 0
(D) 2w^2 – pw – 2k = 0
(E) 2w^2 – pw + 2k = 0
More algebra! I chose D, since the width squared minus the perimeter times the width seemed to cancel out. But it was a guess; barely knew how to start on this one, and made the mistake of not substituting numbers. Let’s fix that: say the board is 4 x 6 inches, which makes w = 4, p = 20, and k = 24. Does (A) 16 + 80 + 24 = 0 work? No – and nor can any answer that involves all adding, so C’s out too. (B) gives 16 – 80 + 48, which doesn’t equal zero either; nor does D, 32 – 80 – 48. It must be E, which gives 32 – 80 + 48.
On a certain road 10 percent of the motorists exceed the posted speed limit and receive speeding tickets, but 20 percent of the motorists who exceed the posted speed limit do not receive speeding tickets. What percent of the motorists on the road exceed the posted speed limit?
(A) 10 1/2 %
(B) 12 1/2 %
(C) 15 %
(D) 22 %
(E) 30 %
I chose E, outatime. This question’s badly phrased, so read carefully. The 20% isn’t 20% of all motorists, or even 20% of the 10%; the question means how many drivers broke the speed limit, if 10% of the total got tickets and that 10% is 80% of those who broke it? The answer is 12.5%, which is B. Should have got this one.
In 1527 King Henry VIII sought to have his marriage to Queen Catherine annulled so as to marry Anne Boleyn.
(A) so as to marry
(B) and so could be married to
(C) to be married to
(D) so that he could marry
(E) in order that he would marry
I chose A, but this one’s borderline: strict grammar is the answer, not everyday usage. The sentence raises a conditional, which is only answered by making the second half dependent on the first half, which means D.
Two new studies indicate that many people become obese more due to the fact that their bodies burn calories too slowly than overeating.
(A) due to the fact that their bodies burn calories too slowly than overeating
(B) due to their bodies burning calories too slowly than to eating too much
(C) because their bodies burn calories too slowly than that they are overeaters
(D) because their bodies burn calories too slowly than because they eat too much
(E) because of their bodies burning calories too slowly than because of their eating too much
I chose B. But that implies it’s the body doing the eating, which is slightly wrong: it’s the person. E is too wordy. D satisfies: ‘their’ and ‘they’ make it correct by referring to the same thing, the fat individual.
Because of the enormous research and development expenditures required to survive in the electronics industry, an industry marked by rapid innovation and volatile demand, such firms tend to be very large.
(A) to survive
(B) of firms to survive
(C) for surviving
(D) for survival
(E) for firms’ survival
(E) in order to keep from
I chose A, but it misses an important point: introducing the firm as the subject. B is the answer.
Consumers may not think of household cleaning products to be hazardous substances, but many of them can be harmful to health, especially if they are used improperly.
(A) Consumers may not think of household cleaning products to be
(B) Consumers may not think of household cleaning products being
(C) A consumer may not think of their household cleaning products being
(D) A consumer may not think of household cleaning products as
(E) Household cleaning products may not be thought of, by consumers, as
I got E. ‘Think of … to be’ isn’t a grammatically correct construction, so eliminate A. B creates ambiguity over whether it’s the substances or consumers that are hazardous. C is wrong too, since it lacks agreement between the singular ‘consumer’ and the plural possessive ‘their’. D is left over, and must be correct.
Manifestations of Islamic political militancy in the first period of religious reformism were the rise of the Wahhabis in Arabia, the Sanusi in Cyrenaica, the Fulani in Nigeria, the Mahdi in the Sudan, and the victory of the Usuli “mujtahids” in Shiite Iran and Iraq.
(A) Manifestations of Islamic political militancy in the first period of religious reformism were the rise of the Wahhabis in Arabia, the Sanusi in Cyrenaica, the Fulani in Nigeria, the Mahdi in the Sudan, and
(B) Manifestations of Islamic political militancy in the first period of religious reformism were shown in the rise of the Wahhabis in Arabia, the Sanusi in Cyrenaica, the Fulani in Nigeria, the Mahdi in the Sudan, and also
(C) In the first period of religious reformism, manifestations of Islamic political militancy were the rise of the Wahhabis in Arabia, of the Sanusi in Cyrenaica, the Fulani in Nigeria, the Mahdi in the Sudan, and
(D) In the first period of religious reformism, manifestations of Islamic political militancy were shown in the rise of the Wahhabis in Arabia, the Sanusi in Cyrenaica, the Fulani in Nigeria, the Mahdi in the Sudan, and
(E) In the first period of religious reformism, Islamic political militancy was manifested in the rise of the Wahhabis in Arabia, the Sanusi in Cyrenaica, the Fulani in Nigeria, and the Mahdi in the Sudan, and in
Bloody hell, what a question! I chose A. All the choices sound wrong to me; it’s a case of picking the least worst. E is clumsy, but it’s the only one that makes clear distinctions between the terms separated by commas, by using ‘and in’ at the end. The (hard) answer is E.
It was once assumed that all living things could be divided into two fundamental and exhaustive categories. Multicellular plants and animals, as well as many unicellular organisms, are eukaryotic – their large, complex cells have a well-formed nucleus and many organelles. On the other hand, the true bacteria are prokaryotic cells, which are simple and lack a nucleus. The distinction between eukaryotes and bacteria, initially defined in terms of subcellular structures visible with a microscope, was ultimately carried to the molecular level. Here prokaryotic and eukaryotic cells have many features in common. For instance, they translate genetic information into proteins according to the same type of genetic coding. But even where the molecular processes are the same, the details in the two forms are different and characteristic of the respective forms. For example, the amino acid sequences of various enzymes tend to be typically prokaryotic or eukaryotic. The differences between the groups and the similarities within each group made it seem certain to most biologists that the tree of life had only two stems. Moreover, arguments pointing out the extent of both structural and functional differences between eukaryotes and true bacteria convinced many biologists that the precursors of the eukaryotes must have diverged from the common ancestor before the bacteria arose.
Although much of this picture has been sustained by more recent research, it seems fundamentally wrong in one respect. Among the bacteria, there are arganisms that are significantly different both form the cells of eukaryotes and from the true bacteria, and it now appears that there are three stems in the tree of life. New techniques for determining the molecular sequence of the RNA of organisms have produced evolutionary information about the degree to which organisms are related, the time since they diverged from a common ancestor, and the reconstruction of ancestral versions of genes. These techniques have strongly suggested that altough the true bacteria indeed from a large coherent group, certain other bacteria, the archaebacteria, which are also prokaryotes and which resemble true bacteria, represent a distinct evolutionary branch that far antedates the common ancestor of all true bacteria.
The author’s attitude toward the view that living things are divided into three categories is best described as one of
(A) tentative acceptance
(B) mild skepticism
(C) limited denial
(D) studious criticism
(E) wholehearted endorsement
I chose E. It can’t be B, C, or D, since he’s clearly supporting the point of three categories; A or E are the only options. He’s not roaring with support – this is a scientific passage, not a lawyer’s advocacy – so it’s A.
Roland: The alarming fact is that 90 percent of the people in this country now report that they know someone who is unemployed.
Sharon: But a normal, moderate level of unemployment is 5 percent, with 1 out of 20 workers unemployed. So at any given time if a person knows approximately 50 workers, 1 or more will very likely be unemployed.
Sharon’s argument relies on the assumption that
(A) normal levels of unemployment are rarely exceeded
(B) unemployment is not normally concentrated in geographically isolated segments of the
(C) the number of people who each know someone who is unemployed is always higher than 90% of the population
(D) Roland is not consciously distorting the statistics he presents
(E) Knowledge that a personal acquaintance is unemployed generates more fear of losing one’s job than does knowledge of unemployment statistics
I chose D. But in fact Sharon’s assertion doesn’t rely at all on whether Roland’s got his facts right: her stated principle exists independently. A and E are irrelevant, and C tries to tie her down to a 90% number, which she’s not asserting. B is left over, and does have a bearing on Sharon’s argument: if unemployment were only high in a few geographical areas, there’s a high chance that many people wouldn’t know anyone without a job.
So: learnt some new things, on this test. If I hadn’t made silly mistakes and wasn’t so poor at algebraic stuff, there are at least nine questions here I could have scored, which would have put me up to 730. Got to hit GMAT for Dummies – hard.