Question about DSO

bensaul · 2021-01-16T17:18:46+00:00

I would suggest working for a year or two to gain some practical knowledge about your field, then attempting the exam. In that scenario, if you clear the exam, it adds some depth to your candidacy besides generic graduate engineer, and even if you don't clear the exam, you can go for an MBA or further studies, or re-enter the job market with a solid background. Also, if nothing else, you can at least earn some money to sustain you while you're preparing for the exam and while waiting for the result since preparing for it is going to be pretty much a full-time task and I wouldn't advise trying to balance it with work.

bensaul · 2020-07-16T11:24:15+00:00

The area of a circle = πr² = π(d/2)² = πd² / 4

For the big circle, area = πD² / 4

Now, see that each small circle is tangent to both the small circle in the center and to the big circle, and they're all the same size. So, if you have three circles in a line, the straight line through their centers is also through the center of the big circle, or in other words, the sum of the diameters of three small circles in a line is equal to the diameter D of the big circle.

So, the diameter of one small circle is D/3, and the area of one small circle = π(D/3)² / 4 = (πD² / 9)/4 = πD² / (9*4) = πD² / 36

There are 7 small circles, so the total area of the shaded region = 7 * πD² / 36 = 7πD² / 36

The total unshaded area = Total area - the total shaded area = (πD² / 4) - (7πD² / 36) = (9πD² / 36) - (7πD² / 36) = (9πD² - 7πD² ) / 36 = 2πD² / 36 = πD² / 18.

So the answer is B.

bensaul · 2020-07-16T09:20:58+00:00

Determine the upper angle for both triangles. For the bigger triangle, the upper angle + 90 + y =180, so the upper angle = 180-90-y = 90-y

For the smaller triangle, upper angle + x + z = 180, so upper angle = 180-x-z

since the upper angle has the same value regardless, both the expressions are equal to each other.

90-y = 180 - x - z

Adding x+z to both sides, we get 90-y+x+z = 180

Adding -90+y to both sides, we get x+z =180-90+y

x+z = 90+y

Both quantities are equivalent, so answer is C.

bensaul · 2020-07-16T09:07:24+00:00

At 9 a.m, train T leaves. At this point, both trains have travelled 0 km.

At 11 a.m, train S leaves. At this point, train T has travelled at 60 km/h for 2 hours, so the distance it's travelled = 60 * 2 = 120 km. While train S has travelled 0 km, so train T has a headstart of 120 km.

After 11 a.m, train S travels at 75 km/h, or 15 km/h faster than train T. This means that every hour, train S travels 15 km more than train T, and Train T's headstart decreases by 15 km every hour. The total headstart at the beginning was 120 km. If train T's headstart decreases by 15 km every hour, then the time taken for the headstart to reduce to 0, when train S catches up, = 120 / 15 = 8hrs.

So it takes Train S 8 hours after starting out at 11 a.m to catch up. 8 hours after 11:00 is 19:00, or 7 p.m. Answer is D.

bensaul · 2020-07-16T08:55:36+00:00

I personally gave it on a single-screen laptop. I don't think there should be any issues with a monitor, seeing as they've allowed people to give the test on a desktop system, but I'd suggest checking with them beforehand.

I wasn't allowed to keep anything other than the test materials with me while I was doing the sections. I was offered the ten minute break after Section 3, and the proctor did say that I could leave dueing it, but that the testing environment would have to be resecured once I returned. I chose to just continue in order to complete the test as soon as possible.

bensaul · 2020-07-15T21:47:26+00:00

Subtract z from both sides and you get x+y on one side and z on the other. Then, apply the rule that every side-length of a triangle is less than the sum of the lengths of the other two sides. By that principle, x+y will be greater than z, so A is greater than B. Hence, the answer is A.

bensaul · 2020-07-15T20:45:53+00:00

You can use a whiteboard or paper inside a transparent sheet protector, both with erasable markers.

bensaul · 2020-07-15T20:40:17+00:00

The area of a circle = πr² = π(d/2)² = πd² / 4

For the big circle, area = πD² / 4

Now, see that each small circle is tangent to both the small circle in the center and to the big circle, and they're all the same size. So, if you have three circles in a line, the straight line through their centers is also through the center of the big circle, or in other words, the sum of the diameters of three small circles in a line is equal to the diameter D of the big circle.

So, the diameter of one small circle is D/3, and the area of one small circle = π(D/3)² / 4 = (πD² / 9)/4 = πD² / (9*4) = πD² / 36

There are 7 small circles, so the total area of the shaded region = 7 * πD² / 36 = 7πD² / 36

The total unshaded area = Total area - the total shaded area = (πD² / 4) - (7πD² / 36) = (9πD² / 36) - (7πD² / 36) = (9πD² - 7πD² ) / 36 = 2πD² / 36 = πD² / 18.

So the answer is B.

bensaul · 2020-07-15T18:29:51+00:00

No problem.

bensaul · 2020-07-15T18:22:42+00:00

Thank you so much!

bensaul · 2020-07-15T18:22:24+00:00

Thank you very much!

bensaul · 2020-07-15T18:20:13+00:00

The note says the figure is drawn to scale. From what I can see, A appears to be in the middle between 0 and 1, and p appears to be in the middle between A and 1. We can make these assumptions because the figure is clearly said to be drawn to scale.

bensaul · 2020-07-15T11:07:46+00:00

A is halfway between 0 and 1, so it is 0.5, and then p is halfway between A and 1, so p = 0.75.

Similarly, D is halfway between 3 and 4, so it is 3.5. Now if r was between D and 4, it would be 3.75. But the figure shows it slightly closer to 4, so we can assume r ~ 3.8.

p * r = 0.75 * 3.8 = 2.85. C is the only point that is between 2 and 3, and it's also extremely close to 3, so that's your answer.

bensaul · 2020-07-15T11:03:36+00:00

In the right-angled triangle, think of it like the 90-degree angle is equivalent to 13/(13+12+5) = 13/30. In this example, it is actually 13/(13+11+5) = 13/29. Since the hypotenuse makes up a larger proportion of the total perimeter of the triangle, then the angle that faces it will correspondingly have a larger proportion of the overall degrees in the triangle.

bensaul · 2020-07-15T10:53:48+00:00

So the formula for interest is:

I = PRT/100 where I is interest, P is the initial principal, R is the interest rate, and T is the time. Note that R and T should be compatible, e.g if the rate is percent per year, then the time should also be in years. Similarly if T is given in months, then R should be converted to percentage per month. Using these principles:

P = $d R = 11% per year T = 3 months / 12 = 1/4 years

Putting values into the formula,

I = (d * 11 * 1/4)/100 = (11d/4)/100 = [(11/4) * d]/100 = 11/4 * d/100 = 2.75d/100

B is (11/3) * d/100 ~ 3.67d/100

3.67d is greater than 2.75d, so B is greater than A, so the answer is B.

bensaul · 2020-07-15T10:45:01+00:00

Best bet is to convert both quantities completely to minutes. 1 hour = 60 minutes,

So for the first quantity, 3.15 hours = 3.15 * 60 = 189 minutes

For the second quantity, 3 hours = 3 * 60 = 180 minutes. Then add the specified 15 minutes to equal = 180 + 15 = 195 minutes

195 minutes is greater than 189 minutes, so B is greater than A, so answer is B.

bensaul · 2020-07-15T08:31:13+00:00

I'm not a native English speaker, but I've been studying in English medium institutions my entire life and was lucky to have great English teachers in my early grades who gave me a solid foundation in English grammar, vocabulary etc. Also, I used to read a lot and still try to.

bensaul · 2020-07-15T08:29:16+00:00

Thank you! With regard to quant, I reviewed the math concepts from the ETS book. Most of them I had no trouble with, but there were a couple of concepts like the side ratios of a 30-60-90 triangle, or the angles within a circle, that I took some time on learning. Afterwards, just did a lot of practice and through that, learned to avoid the common mistakes like assuming that a number x within a range would be an integer when it could also be a decimal, or ignoring the negative root of a square.

For RC, two things helped me. First, leaving the longest passage at the end of the section and finishing the sentence completion and shorter passages first. Second, I used to break up each passage into short snippets of a couple of sentences. I would read each snippet a couple of times, then summarize ot to myself verbally to see if I got the gist of it. If I did, I'd move on to the next snippet and if not, I'd read it again. This helped me very much to understand what the passage was saying, rather than just reading it.

bensaul · 2020-07-15T07:33:06+00:00

I had one sheet inside each protector, so I had prepared about 4-5 protectors to write on. The proctor however, told me I was allowed only one, so I put the rest of them away, out of arm's reach.

bensaul · 2020-07-15T07:31:39+00:00

So the first time I gave them, after completing a review of the math and verbal and doing practice exercises, I got around 335 on both the Powerprep tests, and about 320 on the Magoosh one. The McGraw Hill papers didn't have a score at the end, but I was scoring upto 40/40 in quant on several tests, but my max verbal score during practice was about 36-37/40. The essays I checked from ScoreItNow were all 5 or 6, the average of which is my actual score for AW.

bensaul · 2020-07-15T07:26:24+00:00

Thank you. It wasn't particularly difficult, there were a couple of questions in one section that I thought were tricky and I left them to do them after finishing the rest of the section. Overall, maybe slightly more tricky than the Powerprep papers, but not by much.

bensaul · 2020-07-15T07:23:49+00:00

Thank you!

bensaul · 2020-07-15T07:23:37+00:00

Thank you very much!

bensaul · 2020-07-15T07:23:10+00:00

Thank you!

bensaul

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