Yesterday’s test was under test conditions, so it was good to score 650; shows actual conditions aren’t a big factor for me. 13 wrong quant questions and 3 wrong verbals – pleased with that verbal; more practice means I’m capable of a perfect score there.

But while I hit the 85th percentile on verbal, I’m only beating 55% of test takers on maths: not good. With 14 of the 37 quant questions on data sufficiency (less than might be expected; they’re often about half) and me getting 50% of them wrong, I need to concentrate hard on data sufficiency. It’s symptomatic of the broader personality fault that’s prompting me to try the GMAT in the first place: a tendency to ‘wing it’, trying to get by on native intelligence rather than applied methodology.

Quant

Since this was a Kaplan test, I got a nice little report card showing how many questions there were of each type. 23 problem solving, 14 data sufficiency; within this 37 were 2 arithmetic (both of which I got right), 14 algebra (8 right, 6 wrong), 9 on number properties (3 wrong), 5 on proportions (2 wrong), 3 on sets (1 wrong), and 4 on geometry (1 wrong.) So my weak area within data sufficiency is algebra. Here are the data sufficiency options:

(A) if statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked;

(B) if statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked;

(C) if BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient;

(D) if EACH statement ALONE is sufficient to answer the question asked;

(E) if statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

And my catalog of errors.

If a > 0, b > 0, and b+y ≠ 0, is a + y / b + y < a / b ?

(1) a < b

(2) y < 0

I chose B. Substituting 1 for a and b gives you the real question, which is can we be sure y is less than 1? (1) doesn’t tell us much – substituting a = 1 and b = 2 gives us whether 1/2y is less than 1/2, the same question. So take (2) and substitute y = -1. Is 0/-1 less than 1/2? It is, and I went no further.

But if you do – substituting a=1, b=2, and y=-3, you get whether -2/-1 is less than 1/2 … which it isn’t, despite the negative signs: it’s 2. (A negative divided by a negative is a positive.) So both statements together aren’t enough. The answer is E.

The operation ** is defined for all a and b by a**b = (a^2 + b)^2. If 2**c = 81 and c>0, then c= :

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5

It asks what you can do to 2 to make it 81. I chose C, since if ** meant ‘raise the next term by the power of the previous digit plus 2′, previous digit to this power’ were the exponent sign, 81 is 3 to the power of 4. Guesswork.

Substituting some numbers, a=2 and b=3, you get a**b = 49. What can you do to 2 and 3 that equals 49? Wild idea: you can square each and put the results side by side, 4 and 9. 2 something 5 is 81. Foxed so far! Need some help on solving this one.

What is the value of st / u ?

(1) s = 3t/4 and u = 2t

(2) s = u-10 and u = s+t+2

I chose E, which is totally wrong. Substituting a few numbers:

Switching around: 4s = 3t. That means 4s also equals 1.5u, since u=2t. So substituting in the question, it asks the value of 3t/4 * t / 2t, which is 3/4 divided by 2. (1) is sufficient.

Looking at (2), the question can be rephrased as (u-10) * (u-s-2) divided by u, which multiplies out to u^2-10u * u^2 * u^2-10u -2u… or something. It’s data sufficiency, you don’t need to work it out – all you need to do is notice that the expression’s reduced to one with a single term in it which cancels itself out. Either statement alone is sufficient. D is correct. I need to practice these.

A local restaurant recently renovated its dining space, purchasing new tables and chairs to use in addition to the originals. The new tables seat 6 customers, while the original tables seat 4 customers. Altogether, the restaurant now has 40 tables and is capable of seating 220 customers. How many more new tables than original tables does the restaurant have?

(a) 10

(b) 20

(c) 30

(d) 34

(e) 36

My choice – C. Wrong-o. But I should have got it: careless, careless. Substituting from the middle value: 30 new tables (all seating 6) would seat 180 people. which would leave 10 old tables, which seat 40. That makes 220. How’s that wrong?

Kaplan gives the answer as B. That’d mean 20 new tables (seating 120 people) and 20 old tables (seating 80.) That’s 200 people. But wait: the question asks for the difference between the number of old and new tables, which is 20! So B is correct. Brilliant question – an instant trap.

If x and y are positive integers, is 2x/y an integer?

(1) All factors of x are also factors of y.

(2) All prime factors of x are also prime factors of y.

My choice was D. Quick test: x is 12 and y is 24, with factors 3, 4, 6, and 12. The result, 1, is an integer.

But prime factors? Let’s try x = 21 (prime factors 3 and 7) and y = 231 (prime factors 3, 7, and 11.) I don’t know what 42/231 is, but it doesn’t sound like an integer, even when reduced to 2/11. Statement 2 can’t hold. A guess of A would have been correct.

The positive integer x has how many different positive factors?

(1) x is a multiple of the same number of positive integers that 7^5 is.

(2) x = a^2 b where a and b are different prime numbers.

I answered E, and it was a guess. Let’s see what I could have done better. I’m not sure what statement 1 is saying; is it saying x is equal to 7^5? It doesn’t seem so. I assume there’s a rule I’m unaware of: where two numbers are multiples of the same number of positive integers, they have the same number of positive factors. Doesn’t sound right, but I’ll check it.

Statement 2 looks no easier. Those two primes could be big; the only rule is that any prime squared is od, so x must be odd. The correct answer is D: either statement is sufficient. But I don’t know how to get there. Another odd one.

An exam is given in a certain class. The average (arithmetic mean) of the highest score and the lowest score is equal to x. If the average score for the entire class is equal to y and there are z students in the class, where z > 5, then in terms of x, y, and z, what is the average score for the class excluding the highest and lowest scorers?

(a) zy – 2x / z

(b) zy – 2 / z

(c) zx – y / z – 2

(d) zy – 2x / z -2

(e) zy – x / z + 2

I chose C. But this question should be straightforward. y = x / z (arithmetic mean) which means zy = x. And we want y(z-2) to equal… 2x less than y – 2. Well, there’s z-2 in several answers, so I’ll narrow it down to C and D. My guess of C was wrong because 2x equals the total of both the lowest and highest scorers, so D must be correct. Damn.

A salesperson received N dollars commission for selling a radio. What is the value of N?

(1) The salesperson’s commission was equal to 20 percent of the radio’s price.

(2) The radio’s price was $40 more than the salesperson’s commission.

I chose B, thinking statement 2 alone was sufficient. It isn’t since we don’t know the commission amount, and nor is 1: we have no idea of the radio’s price. Let’s see if both together work, with p as the price and N. p = N + 40 and N = 0.2p. Definitely enough info there (p = 5N and p = N+40, which means 5N = N + 40 therefore 4N = 40, making the commission $10, which you don’t need to work out) so it’s answer C.

In the figure above, the length of line segment CD is twice the length of line segment BC. The ratio of the area of triangle ABC to the area of triangle ABD is

(a) 1:4

(b) 1:3

(c) 1:2

(d) 1:√3

(e) 1: √2

I chose C, making an obvious error: the ‘base’ (horizontal length) of ABD is three times as long as ABC, not two times (call BC x and BD 3x, with common height AB y.) A triangle’s area is 1/2base x height, so ABC is area 0.5xy and ABD is 1.5xy. It’s a 1:3 ratio, which is B.

A train travelled 960 miles from Town A to Town B. What was the train’s average speed during that journey?

(1) If the train had travelled from A to B at an average speed that was 12 miles per hour slower, it would have taken 20 hours.

(2) If the train had travelled from A to B at an average speed that was 60 percent faster, it would have taken 6 fewer hours.

I chose A. Statement 1 is certainly sufficient: 960 / (s -12) = 20, therefore 20(s-12) 20s-240 = 960 and s is 60mph. (I’m taking the time to work it out; I shouldn’t, I need to get out of the habit.) Statement 2 seems to be missing something.

If it were 60% faster, it’d travel 960/1.6 miles in time t. 600 miles. So if it can travel 600 miles in time t-6 (i.e. at the slower speed) and 960 miles in time t-6 at the higher speed, it travels 360 miles in those 6 hours, i.e. 60mph, and at 60% faster (96mph) it travels 960 miles in 10 hours, which would take 16 hours (960/60) at the slower speed. Bingo. Either statement is sufficient; the answer is D.

I’m seeing a pattern to this type of question, where it seems one statement doesn’t have enough info. with practice, I should get better at them.

If x is a number in digital representation, and y is the tenths digit of x, such that x = 1.y5, what is the value of x, rounded to the nearest integer?

(1) 2x > 3

(2) x < y

Hardly know where to start on this, so I guessed E. Let’s look again: if 2x > 3, then x is greater than 1.5, which means y must be equal to or greater than 5. Any value of y that satisfies this gives a rounded-up integer of 2, so (1) alone is sufficient.

Looking at (2), for the smallest value of y, x would be 1.05, and 1.05 isn’t less than 0. 1.15 isn’t less than 1, but 1.25 is less than 2, so x can be less than y. But a rounded-up x can’t be greater than 2, and if y is 9, x rounds up to 2, so rounded x could be 1 or 2 depending on y. (2) isn’t enough to answer, so the answer is A.

If x = (3z)^2 / y, by what number will x be multiplied if z is multipled by 2 and y is divided by 3?

(a) 2

(b) 4

(c) 6

(d) 8

(e) 12

I chose B. Multiplying z by 2 and dividing y by 3, we get 36z^2 / 1/3y. Slap in some numbers: 1 for z and y. 36 over 1/3 = x= 108. In the original equation, x = 9. x must be multiplied by 12, which is E, the correct answer.

A local museum is setting up a new exhibit. 5 oil paintings and 6 watercolours are available for the exhibit. The museum director must choose 3 oil paintings and 4 watercolours. How many different combinations of paintings for the exhibit are possible?

(a) 12

(b) 30

(c) 90

(d) 120

(e) 150

I chose A. The question is asking how many different subsets of 3 are possible in a set of 5 oil paintings and how many sets of 4 are in a set of 6 watercolours.

This is a permutations and combinations question, so I need the factorial n!. 5 oil paintings can be arranged in 5 x 4 x 3 x 2 x 1 ways – but we’re looking for arrangements of 3, not all 5. The number of permutations of n objects taken r at a time is n!/(n-r). Which here is 5 x 4 x 3 x 2 x 1/5-3. Which is 120/2, or 60. For the watercolours, it’s 6 x 5 x 4 x 3 x 2 x 1 / 6-4, which is 240/2, or 120.

But it’s a second-order effect; this doesn’t give us the total combinations of both types of painting together. We don’t need to work it out though; it’s got to be more than 120, so E is the answer.

Verbal

Three questions wrong here: a good result. And none of them were on reading comprehension, my weakness to date – even better. The wrong ‘uns were:

According to the National Institutes of Health, much of the sleep deprived people would benefit from a short nap during the day.

(a) much of the sleep-deprived people

(b) much sleep-deprived people

(c) many people who are sleep-deprived

(d) many sleep-deprived people

(e) most of the sleep-deprived people

A sentence correction error. My mistake (choosing C) – trying to keep to the beat of the original paragraph, when there’s a shorter option that omits no information. D is correct.

A worthwhile reason to become a skilled user of the Internet is because Internet skills are increasingly valued in today’s workplace.

(a) A worthwhile reason to become a skilled user of the Internet is because Internet skills

(b) A worthwhile reason to acquire Internet skills is that they

(c) Internet skills are worthwhile for acquiring because they

(d) It is worthwhile to become a skilled user of the Internet because Internet skills

(e) Internet skills are worthwhile for the reason that said skills

I chose D. Wrong, because it sounds slightly off base: it doesn’t link the activity of becoming a skilled user with the benefit of those skills in the workplace as strong as B. Discount A, C, and E immediately for being verbose. B is the answer.

So: my tips tonight will be on the permutations and combinations equations, n!/(n-r) and n!/r!(n-r), for discovering the number of ways n things taken r at a time can be put together.

” The operation ** is defined for all a and b by a**b = (a^2 + b)^2. If 2**c = 81 and c>0, then c= :

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5………………. “

rewriting the question

———————————-

if a**b = (a^2 + b)^2 then what is the value of c when 2**c = 81 (if c>0 ) ?

———————————-

approach is ..

replace a by 2 and b by c

so

2**c = (2^2 + c )^2 =81

—> (4 + c)^2=81

—> 4+c = 9 taking sqrt

—> c =5 ( coz c is +ve )

i hope its right

– vikrammanagoli@gmail.com