How to ace the GMAT in 28 days: Day 8 (analysing Day 7)
So, that 680. I'm on target - actually slightly ahead of target. No questions wrong in reading comprehension and critical reasoning; not bad on verbal, just 2 wrong in 56. But I keep running out of time on the maths sections: 16 wrong of 52, and at least 8 of those due to cramming the last few on each section.
Correcting the raw scores (take the number you got wrong, divide by 4, subtract that from the number you got right, then add 0.5 and ignore anything on the right of the decimal point) gives 54V/32Q.(The practice papers I'm doing allow slightly less time than in the real test, and there are more of them, but I hope they're not easier; I need to practice the difficult ones.)
Quant
Is 7 < √n < 8 ?
(1) n > 50
(2) n < 60
Once again, the standard set for data sufficiency is:
A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D EACH Statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
I chose E. On a basic misunderstanding: in a sense, this is a critical reasoning question! I thought it was asking the value of n, and it's not: it's asking whether 7 is less than the square root of n. We can answer that if we know (from both statements) that n is between 50 and 60. The answer is C.
If a total of 84 students are enrolled in two sections of a calculus course, how many of the 84 students are female?
(1) 2/3 of the students in Section 1 are female.
(2) 1/2 of the students in Section 2 are male.
I chose C here. Again, I missed the sense of the question. Knowing something about each section doesn't compensate for not knowing how many students were enrolled in each section. It's E.
The figure above shows the shape of a flower bed. If arc QR is a semicircle and PQRS is a rectangle with QR > RS, What is the perimeter of the flower bed?
(1) The perimeter of rectangle PQRS is 28 feet.
(2) Each diagonal of rectangle PQRS is 10 feet long.
I chose D. But Statement 1 isn't sufficient, because the perimeter doesn't give the length and breadth of the rectangle; it's not square. Nor, despite appearances, does knowing the diagonal. But both together give the answer: the rectangle is a pair of 3:4:5 triangles. C is the answer.
What is the value of y ?
(1) y^2 - 7y +12 = 0
(2) y > 0
I chose A. because statement 1 looks like a solvable equation. But there's no x term or other equation to make a pair of simultaneous equations, so it's unsolvable. Which knowing y is positive doesn't solve. The answer is E, but I'd have got this wrong; my tendency in such cases is to guess things are solvable.
Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5?
I. Multiplying the 5 numbers and then finding the 5 th root of the product.
II. Adding the 5 numbers, doubling the sum, and then moving the decimal point one place to the left.
III. Ordering the 5 numbers numerically and then selecting the middle number.
(A) None
(B) I only
(C) II only
(D) III only
(E) I and III
These variants of data sufficiency are HARD. I chose A. Looking again, III can't be true, since we don't know if the numbers are sequential or evenly spaced (or even integers.) As for I though - that sounds promising... on second thoughts, no it isn't; the 5th root of anything is really, really small. That leaves II. Let's test: 2,3,4,5,6... 20... 40... 4. II is deceptively complex-looking; it's actually finding a fifth of the total, just like adding them all and dividing by 5. (Doubling anything then moving the decimal point one place left is the same as dividing by 5.) The answer is C.
If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?
(A) 2/225
(B) 1/111
(C) 1/110
(D 1/100
(E) 1/50
Aha, probability. I guessed B. Let's choose a random number: 136. The first digit has a 1 in 10 chance of being higher than the middle digit; so does the last. It can't be that simple, surely? Isn't there some matrix problem here, where the probability of the middle digit being high decreases the probability of the first and last digits beating it? No there isn't; the clue is that would take too much calculation for a two-minute question. It's D.
Mr. Kramer, the losing candidate in a two-candidate election, received 942,568 votes, which was exactly 40 percent of all the votes cast. Approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast?
(A) 10%
(B) 12%
(C) 15%
(D) 17%
(E) 20%
I chose E, thinking if he needs another 10% of 100%, he needs 20% of 50%, tying myself in needless (and wrong) knots. But of course it's D. Of the 60% remaining votes, he needs to gain 10 percentage points or 10/60, which is 1/6 or approximately 17%. The clues: 'approximately' and the awkward 942,568, which suggests there's no calculation involved here.
– 2 < x < 4 ?
(A) x - 2 < 4
(B) x - 1 < 3
(C) x + 1 < 3
(D) x + 2 < 4
(E) None of the above
I chose D. But simple substitution works. If -2 is less than x, then x+2 is more than 0 and less than 6. Which rules out A and D. x - 1 must be one less than 4. It's B. Again, simple stuff - but I slipped up.
If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be
(A) 6x
(B) 4x
(C) 5x/3
(D) 3x/2
(E) 3x/5
I chose E. Clue: what the temp 'could be', not 'what it is'. Again, an easy question if you spot the twist. The sum of 3 average temps will be 3x, but that's not an answer. It does, however, rule out C, D, and E for being too small. A or B? Well it can't be 6x, since we only had 5 temps to add. It's B.
In a marketing survey for products A, B, and C, 1,000 people were asked which of the products, if any, they use. The three circular regions in the diagram above represent the numbers of people who use products A, B, and C, according to the survey results. Of the people surveyed, a total of 400 use A, a total of 400 use B, and a total of 450 use C.
What percent of the people surveyed use product A or product B or both, but not product C ?
(A) 12.5%
(B) 17.5%
(C) 30%
(D) 40%
(E) 60%
I chose E. Because with 400 people in set A and another 400 in set B, with 200 common (125+75) leaves 600 of the 1000. But 125 + 75 + 100 of those are in set C (with the 75 previously included), so that's wrong.
75 people (400 - 125+125+75) use A exclusively. 100 people (400 - 125+75+100) use B exclusively. A further 125 use A and B. That's all we need; everyone else uses C. 300 people. The answer is C.
The equation M+6 / 36 = P-7 / 21 relates two temperature scales. Where M is the number of degrees on one scale and P is the number of degrees on the other scale. Which of the following equations can be used to convert temperatures from the P scale to the M scale?
(A) M = 7/12 P + 13
(B) M = 7/12 P + 21
(C) M = 12/7 P -12
(D) M = 12/7 P - 13
(E) M = 12/7 P -18
OK, a hard one. I chose A. So what do I do to isolate M? Get rid of the 36 first... 36(p-7)/21... equals (36p-252) / 21... which cancels down to 36/21 p - 252/21, ending up with 12/7p - 12. Now add in the -6 to the -12, giving -18. The answer is E.
The incomplete table above shows a distribution of scores for a class of 20 students. If the average (arithmetic mean) score for the class is 78, what score is missing from the table?
(A) 73
(B) 75
(C) 77
(D) 79
(E) 81
I chose E. Haha. Great 'trap' question. They want you to forget about the weighting effect of the number of students scoring each number. E would be correct if only one student scored each number. This question tests speed of mental arithmetic. Let's see if I can do it in under two minutes:
First, work out the total aggregate scores: 415, 420, 276, x, 64. Add them together: 1175. Now work out the total aggregrate: 78 x 5, which is 1560. Subtract: leaving 385 (whoops, easy to make the mistake it's 375.) Which of the answers divides neatly into 385? It's 77, which is C. This calculation took me under a minute - so if I'd just paced the test, I'd have been fine.
Here's where I started to run out of time; the next 4 questions, all of which I got wrong, are largely guesses.
Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took 1/2 hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
I guessed B. Let's do this by substituting numbers, taking the middle value first. 320km at 80km/h will take 4 hours. How far does 4.5 hours at 70 take Carl? 315km. Not far enough. Try D. He'll travel 280km in 3.5 hours at 80, and in 4 hours at 70. Bingo. The answer is D.
If 5x = 6y and xy ≠ 0, what is the ratio of 1/5x to 1/6y?
(A) 25/6
(B) 36/25
(C) 6/5
(D) 5/6
(E) 25/36
Another brilliant question to tie your brain in knots. I chose C after a vague guess the ratio would be a higher to a lower, ruling out D and E. Put in some numbers instead: x= 6, making y = 5. So x is 6/5y. A fifth of x (1 1/5) to a sixth of y (5/6). Get the same denominator: 6/5 and 5/6 = 36/30 to 25/30, which is 36 to 25. Answer B.
The figure above shows a cord around two circular disks. If the radii of the two disks are 80 centimeters and 60 centimeters, respectively, what is the total length, in centimeters, of the cord?
(A) 210 ∏
(B) 210 ∏ + 280
(C) 280 ∏
(D) 280 ∏ + 80
(E) 280 ∏ + 280
More fun. 2 x Pi'ing both radii to get the perimeters of the circles then adding 2 lengths of 80cm and 2 lengths of 60 cm to account for the connecting bits, I hurriedly chose E. Which if I'd had time to think, assumes there's a length of cord around the entire circumference of the discs, which is wrong: the cord clearly lifts away in the centre. And because it lifts away at a tangent, the amount of perimeter uncovered on each disc is precisely a quarter, making it not 280∏ + 280, but 210∏ + 280. B is correct.
If x, y, and z are positive integers and 3x = 4y = 7z, then the least possible value of x + y + z is
(A) 33
(B) 40
(C) 49
(D) 61
(E) 84
Straightforward, but out of time I chose A. Instantly, you need the lowest common denominator of 3, 4, and 7, which is 84. 3 x 28 = 4 x 21 = 7 x 12. Lowest possible value of these three: 61. It's D.
Verbal
Two sentence correction errors:
Like the color-discriminating apparatus of the human eye, insects' eyes depend on recording and comparing light intensities in three regions of the electromagnetic spectrum.
(A) insects' eyes depend on
(B) an insect eye depends on
(C) that of insects depend on the
(D) that of an insect's eye depends on
(E) that of an insect's is dependent on the
I chose B. The pedantic reason is that you're not comparing eyes with eyes; you're comparing eye apparatus to eye apparatus, so we need to make clear that insects' eyes aren't being compared with the machinery within human eyes: it's the machinery within both being compared. C, D, and E would work - the 'that' is the clue - but C doesn't provide the article being talked about (eye) and E makes it passive voice, always a bad idea on the GMAT. The answer is D.
The computer software being designed for a project studying Native American access to higher education will not only meet the needs of that study, but also has the versatility and power of facilitating similar research endeavors.
(A) but also has the versatility and power of facilitating
(B) but also have the versatility and power to facilitate
(C) but it also has the versatility and power to facilitate
(D) and also have the versatility and power of facilitating
(E) and it also has such versatility and power that it can facilitate
I chose A. Don't know why: B jumps out at you. 'Versatility of facilitating' isn't good English, so it isn't A or D; C and E adds an unneeded article 'it' and E is too wordy anyway. Initially I rejected B for the 'have' - but it needs to be 'have' not 'has', since it follows on from 'will' earlier. B is the answer.
2 sentence correction... but no reading comp or critical reasoning errors! Got to concentrate on maths. One big lesson learned today: I've got to start substituting numbers more instead of working things out on the maths section. After all, the answer is there in A-E; multiple choice is about eliminating wrong answers as much as finding right ones.
So, 20 days and 10 practice tests to go. I'm on the way, but from here it gets harder.
Correcting the raw scores (take the number you got wrong, divide by 4, subtract that from the number you got right, then add 0.5 and ignore anything on the right of the decimal point) gives 54V/32Q.(The practice papers I'm doing allow slightly less time than in the real test, and there are more of them, but I hope they're not easier; I need to practice the difficult ones.)
Quant
Is 7 < √n < 8 ?
(1) n > 50
(2) n < 60
Once again, the standard set for data sufficiency is:
A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D EACH Statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
I chose E. On a basic misunderstanding: in a sense, this is a critical reasoning question! I thought it was asking the value of n, and it's not: it's asking whether 7 is less than the square root of n. We can answer that if we know (from both statements) that n is between 50 and 60. The answer is C.
If a total of 84 students are enrolled in two sections of a calculus course, how many of the 84 students are female?
(1) 2/3 of the students in Section 1 are female.
(2) 1/2 of the students in Section 2 are male.
I chose C here. Again, I missed the sense of the question. Knowing something about each section doesn't compensate for not knowing how many students were enrolled in each section. It's E.
The figure above shows the shape of a flower bed. If arc QR is a semicircle and PQRS is a rectangle with QR > RS, What is the perimeter of the flower bed?
(1) The perimeter of rectangle PQRS is 28 feet.
(2) Each diagonal of rectangle PQRS is 10 feet long.
I chose D. But Statement 1 isn't sufficient, because the perimeter doesn't give the length and breadth of the rectangle; it's not square. Nor, despite appearances, does knowing the diagonal. But both together give the answer: the rectangle is a pair of 3:4:5 triangles. C is the answer.
What is the value of y ?
(1) y^2 - 7y +12 = 0
(2) y > 0
I chose A. because statement 1 looks like a solvable equation. But there's no x term or other equation to make a pair of simultaneous equations, so it's unsolvable. Which knowing y is positive doesn't solve. The answer is E, but I'd have got this wrong; my tendency in such cases is to guess things are solvable.
Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5?
I. Multiplying the 5 numbers and then finding the 5 th root of the product.
II. Adding the 5 numbers, doubling the sum, and then moving the decimal point one place to the left.
III. Ordering the 5 numbers numerically and then selecting the middle number.
(A) None
(B) I only
(C) II only
(D) III only
(E) I and III
These variants of data sufficiency are HARD. I chose A. Looking again, III can't be true, since we don't know if the numbers are sequential or evenly spaced (or even integers.) As for I though - that sounds promising... on second thoughts, no it isn't; the 5th root of anything is really, really small. That leaves II. Let's test: 2,3,4,5,6... 20... 40... 4. II is deceptively complex-looking; it's actually finding a fifth of the total, just like adding them all and dividing by 5. (Doubling anything then moving the decimal point one place left is the same as dividing by 5.) The answer is C.
If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?
(A) 2/225
(B) 1/111
(C) 1/110
(D 1/100
(E) 1/50
Aha, probability. I guessed B. Let's choose a random number: 136. The first digit has a 1 in 10 chance of being higher than the middle digit; so does the last. It can't be that simple, surely? Isn't there some matrix problem here, where the probability of the middle digit being high decreases the probability of the first and last digits beating it? No there isn't; the clue is that would take too much calculation for a two-minute question. It's D.
Mr. Kramer, the losing candidate in a two-candidate election, received 942,568 votes, which was exactly 40 percent of all the votes cast. Approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast?
(A) 10%
(B) 12%
(C) 15%
(D) 17%
(E) 20%
I chose E, thinking if he needs another 10% of 100%, he needs 20% of 50%, tying myself in needless (and wrong) knots. But of course it's D. Of the 60% remaining votes, he needs to gain 10 percentage points or 10/60, which is 1/6 or approximately 17%. The clues: 'approximately' and the awkward 942,568, which suggests there's no calculation involved here.
– 2 < x < 4 ?
(A) x - 2 < 4
(B) x - 1 < 3
(C) x + 1 < 3
(D) x + 2 < 4
(E) None of the above
I chose D. But simple substitution works. If -2 is less than x, then x+2 is more than 0 and less than 6. Which rules out A and D. x - 1 must be one less than 4. It's B. Again, simple stuff - but I slipped up.
If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be
(A) 6x
(B) 4x
(C) 5x/3
(D) 3x/2
(E) 3x/5
I chose E. Clue: what the temp 'could be', not 'what it is'. Again, an easy question if you spot the twist. The sum of 3 average temps will be 3x, but that's not an answer. It does, however, rule out C, D, and E for being too small. A or B? Well it can't be 6x, since we only had 5 temps to add. It's B.
In a marketing survey for products A, B, and C, 1,000 people were asked which of the products, if any, they use. The three circular regions in the diagram above represent the numbers of people who use products A, B, and C, according to the survey results. Of the people surveyed, a total of 400 use A, a total of 400 use B, and a total of 450 use C.
What percent of the people surveyed use product A or product B or both, but not product C ?
(A) 12.5%
(B) 17.5%
(C) 30%
(D) 40%
(E) 60%
I chose E. Because with 400 people in set A and another 400 in set B, with 200 common (125+75) leaves 600 of the 1000. But 125 + 75 + 100 of those are in set C (with the 75 previously included), so that's wrong.
75 people (400 - 125+125+75) use A exclusively. 100 people (400 - 125+75+100) use B exclusively. A further 125 use A and B. That's all we need; everyone else uses C. 300 people. The answer is C.
The equation M+6 / 36 = P-7 / 21 relates two temperature scales. Where M is the number of degrees on one scale and P is the number of degrees on the other scale. Which of the following equations can be used to convert temperatures from the P scale to the M scale?
(A) M = 7/12 P + 13
(B) M = 7/12 P + 21
(C) M = 12/7 P -12
(D) M = 12/7 P - 13
(E) M = 12/7 P -18
OK, a hard one. I chose A. So what do I do to isolate M? Get rid of the 36 first... 36(p-7)/21... equals (36p-252) / 21... which cancels down to 36/21 p - 252/21, ending up with 12/7p - 12. Now add in the -6 to the -12, giving -18. The answer is E.
The incomplete table above shows a distribution of scores for a class of 20 students. If the average (arithmetic mean) score for the class is 78, what score is missing from the table?
(A) 73
(B) 75
(C) 77
(D) 79
(E) 81
I chose E. Haha. Great 'trap' question. They want you to forget about the weighting effect of the number of students scoring each number. E would be correct if only one student scored each number. This question tests speed of mental arithmetic. Let's see if I can do it in under two minutes:
First, work out the total aggregate scores: 415, 420, 276, x, 64. Add them together: 1175. Now work out the total aggregrate: 78 x 5, which is 1560. Subtract: leaving 385 (whoops, easy to make the mistake it's 375.) Which of the answers divides neatly into 385? It's 77, which is C. This calculation took me under a minute - so if I'd just paced the test, I'd have been fine.
Here's where I started to run out of time; the next 4 questions, all of which I got wrong, are largely guesses.
Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took 1/2 hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
I guessed B. Let's do this by substituting numbers, taking the middle value first. 320km at 80km/h will take 4 hours. How far does 4.5 hours at 70 take Carl? 315km. Not far enough. Try D. He'll travel 280km in 3.5 hours at 80, and in 4 hours at 70. Bingo. The answer is D.
If 5x = 6y and xy ≠ 0, what is the ratio of 1/5x to 1/6y?
(A) 25/6
(B) 36/25
(C) 6/5
(D) 5/6
(E) 25/36
Another brilliant question to tie your brain in knots. I chose C after a vague guess the ratio would be a higher to a lower, ruling out D and E. Put in some numbers instead: x= 6, making y = 5. So x is 6/5y. A fifth of x (1 1/5) to a sixth of y (5/6). Get the same denominator: 6/5 and 5/6 = 36/30 to 25/30, which is 36 to 25. Answer B.
The figure above shows a cord around two circular disks. If the radii of the two disks are 80 centimeters and 60 centimeters, respectively, what is the total length, in centimeters, of the cord?
(A) 210 ∏
(B) 210 ∏ + 280
(C) 280 ∏
(D) 280 ∏ + 80
(E) 280 ∏ + 280
More fun. 2 x Pi'ing both radii to get the perimeters of the circles then adding 2 lengths of 80cm and 2 lengths of 60 cm to account for the connecting bits, I hurriedly chose E. Which if I'd had time to think, assumes there's a length of cord around the entire circumference of the discs, which is wrong: the cord clearly lifts away in the centre. And because it lifts away at a tangent, the amount of perimeter uncovered on each disc is precisely a quarter, making it not 280∏ + 280, but 210∏ + 280. B is correct.
If x, y, and z are positive integers and 3x = 4y = 7z, then the least possible value of x + y + z is
(A) 33
(B) 40
(C) 49
(D) 61
(E) 84
Straightforward, but out of time I chose A. Instantly, you need the lowest common denominator of 3, 4, and 7, which is 84. 3 x 28 = 4 x 21 = 7 x 12. Lowest possible value of these three: 61. It's D.
Verbal
Two sentence correction errors:
Like the color-discriminating apparatus of the human eye, insects' eyes depend on recording and comparing light intensities in three regions of the electromagnetic spectrum.
(A) insects' eyes depend on
(B) an insect eye depends on
(C) that of insects depend on the
(D) that of an insect's eye depends on
(E) that of an insect's is dependent on the
I chose B. The pedantic reason is that you're not comparing eyes with eyes; you're comparing eye apparatus to eye apparatus, so we need to make clear that insects' eyes aren't being compared with the machinery within human eyes: it's the machinery within both being compared. C, D, and E would work - the 'that' is the clue - but C doesn't provide the article being talked about (eye) and E makes it passive voice, always a bad idea on the GMAT. The answer is D.
The computer software being designed for a project studying Native American access to higher education will not only meet the needs of that study, but also has the versatility and power of facilitating similar research endeavors.
(A) but also has the versatility and power of facilitating
(B) but also have the versatility and power to facilitate
(C) but it also has the versatility and power to facilitate
(D) and also have the versatility and power of facilitating
(E) and it also has such versatility and power that it can facilitate
I chose A. Don't know why: B jumps out at you. 'Versatility of facilitating' isn't good English, so it isn't A or D; C and E adds an unneeded article 'it' and E is too wordy anyway. Initially I rejected B for the 'have' - but it needs to be 'have' not 'has', since it follows on from 'will' earlier. B is the answer.
2 sentence correction... but no reading comp or critical reasoning errors! Got to concentrate on maths. One big lesson learned today: I've got to start substituting numbers more instead of working things out on the maths section. After all, the answer is there in A-E; multiple choice is about eliminating wrong answers as much as finding right ones.
So, 20 days and 10 practice tests to go. I'm on the way, but from here it gets harder.
Labels: gmat
posted by chrisworth at 9:23 AM
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