How To Not Get A Job At Google

The release of the film ‘The Internship’ has arrived on our screens recently, following the job interview process required to become a Google employee (I think? Haven’t actually seen it…)
I thought it would be a good time to attempt some of those so-called ‘brain teaser’ questions Google supposedly has all applicants go through during a job interview.

 I think I did pretty well…

1. How many golf balls can fit in a school bus?

As Australian schoolkids rely on public transportation buses, we will assume a ‘school bus’ is one of those big yellow ones seen in the USA. These school buses are usually 36 feet in length (national standards do not permit a school bus over 40 feet in length), 8 feet wide and under 11 feet in height (usually 10 feet) including tyres and rooftop escape hatches.

Stock image courtesy of US national Road Safety Organisation. 

Taking into account engine space and hull thickness, we will have an interior bus space that is about 34 long x 8 high x 7 feet wide (1,904 cubic feet). We must then deduct the amount of space taken up by seating to find the available area remaining; so 23 shared seats for passengers at 3 x 2 x 3 feet (414 cubic feet) plus the backseat at 7 x 2 x 3 (42 cubic feet) and a driver’s seat of 1.5 x 2 x 3 (9 cubic feet) leaves us with 1,439 cubic feet of available bus space. However, it wouldn’t be a school bus without being filled with students now, would it?

Due to rising levels of childhood obesity, with rates of overweight or obese children surpassing 1 in 4 White kids by 2010, the average size of an American schoolkid these days is about 115 pounds. Overweightness and obesity in Black children is even higher, at around 40% averaged between boys and girls, so we will presume an average weight of 120 pounds.
 (Hispanic children are also at similar levels of obesity to Black kids, but since when have you ever seen a Hispanic kid on a schoolbus?)

You know, apart from a residential address.

Water weighs 62 pounds per cubic foot, with a human body at roughly 63 pounds per cubic foot (sinking in water with empty lungs, before a build-up of bodily gases causes it to float to the surface) hence the average White schoolkid would take up approximately 1.8 cubic feet of space. Assuming 23 White American schoolkids in the seating area, one sitting up front annoying the bus driver, this would be around 41.5 cubic feet of pasty pudge. Adding an additional 4 Black children sitting at the back of the bus would consume a further 7.5 cubic feet of space. A 300lb Chris Farley bus driver a la Billy Madison, another 4.75 cubic feet would be taken up at the front. The amount of free space available inside your typical American school bus interior at full passenger capacity is thus around 1385.25 cubic feet.

A golf ball with standard diameter of 1.68 inches takes up almost exactly 2.5 cubic inches.
One cubic inch equals ~0.00058 of a cubic foot, thus a single golf ball takes up 0.00145 cubic feet. With 1385.25 cubic feet to spare on our school bus, one could therefore fit 955,344 golf balls in a school bus, or 955,372 if all the passengers had a golf ball stuffed in their mouths.

 

I surpassed my racist joke quota *minutes* ago.



2. How much should you charge to wash all the windows in Seattle?

Window washing is considered an unskilled profession, with no formal training mandatory, though rope training and working at heights training is part of job orientation. Thus, while regular building cleaners earn around $11.81 per hour, a unionised exterior window washer can earn over double that at $23.65 per hour due to added risk. Assuming that every window in Seattle requires an internal and external clean, every hour should be logically charged at a rate of $17.73, alternating between an exterior and interior window cleaning job.

Seattle has a population of 635,000. Assuming there are 4 people per household/apartment at an average of 15 windows, plus around one window per person in all the skyscrapers and places of business/education/recreation/dining in the city, Seattle assumedly has around 3 million windows to be cleaned. Professional window cleaners use half cold/half warm water with detergent to lather over a window to remove dirt and marks, followed by squeegeeing away suds, then wiping away excess drips with a cloth around the edges. This process takes about 30 seconds for a small window, with obviously more time required for a full glass pane wall in an office building; not to mention the time spent setting up any ladders, safety rigging or platforms and changing into your Spiderman outfit when washing the windows of children’s hospitals. 

With great power comes great glass visibility.

 

With a full interior and exterior clean of a window thus averaged out to around 3 minutes each, or 20 windows per hour, washing every window in Seattle would take 9 million minutes of labour, or 150,000 man hours, at a cost of $2,659,500. This is a far cheaper method of billing that the alternative payment plan of professional pricing per double-sided clean, at $3.75 for a small pane and $6 for large sliding doors. Assuming larger office windows would cost even more, the city would clock up a total bill of around $20 million going from this pricing method.

About 0.0006% of the Australian workforce were listed as window cleaners in the 2011 Australian census. Applying this ratio to the population of Seattle, we can assume about 380 window cleaners will be faced with the task. This amounts to 394 hours of employment for each cleaner for a complete city clean, or 49 working days each, assumedly carried out over 10 working weeks for an individual wage of $6,985. Assuming windows would be washed five times a year, with two weeks off for holidays, the task would provide a respectable $34,925 annual wage to the average Seattle window cleaner. Though, they would probably wash their own windows, so deduct 380 payments from that equation.

  On a related note, I hear that some homeless guy standing at

the CBD traffic lights of Seattle with a dirty old t-shirt in one hand

and a bucket of stale water in the other has got a multi-million-dollar

windscreen washing monopoly going on at the moment.

 3. Why are manhole covers round?

Our streets are bound by the rules of gravity. The best shape for a protective cover over a hole is a round one, as you exert the same amount of even force when lifting it upwards from whichever angle you approach from, and it can also then be placed down from any angle without having to twist your wrist or elbow to align it, thus risking an injury.

It's like trying to fit some kind of peg into some kind of hole!

Furthermore, a round cover cannot fall through the gap below even when tilted sideways, which could potentially occur with a square or otherly-shaped cover. A circular manhole also provides the same diameter of space for a human to enter and exit from at any angle, without any corners to bump or scrape against whilst climbing up and down.

Most importantly of all, however, is the fact that a circular pizza can be easily carried down a circular manhole to the sewers by the Teenage Mutant Ninja Turtles.

Square pizza boxes are a whole different question entirely.

4. How many piano tuners are there in the entire world?

Forte.

5. How many times a day do a clock’s hands overlap?

Sigh. There is a reason I dropped maths in high school, and it’s because of questions like this.
Brace yourself…

An irritating paradox occurs in the movement of clock hands, for as the second hand ‘ticks’ around a clock 60 times every minute in visually noticeable 6 degree intervals, it must therefore be presumed that a minute hand similarly ‘ticks’ 60 times across the 6 degree interval that it covers every minute, though the jerky ticking movement may not be as perceptible to human sight from notch to tiny notch. If 60 ‘arcminutes’ comprise 1 degree, a minute hand thus ‘ticks’ a distance of 10 arcminutes per second. (I will use the terms ‘arcminute’ and ‘arcsecond’ here so as to not confuse these units of angle measurement with the ‘second’ and ‘minute’ units of temporal measurement.)

An hour hand must thus ‘tick’ at an interval distance that is 60 times smaller than a minute hand; 3,600 times that of a second hand’s movements. After one hour (or 3,600 ‘ticks’) a second hand will have completed 21,600 degrees of rotation around a clock, a minute hand will have moved 360 degrees around, whilst an hour hand will have moved just 6 degrees to the 1 symbol (360 arcminutes, or 21,600 arcseconds). Each ‘tick’ of an hour hand is thus a tiny 6 arcsecond movement.

Why this pedantic analysis of arcminutes and arcseconds? Because keeping these miniscule movements in mind, a second hand will not land perfectly on top of a minute or hour hand if they have both shifted forward by the time the second hand has circled around the clock one rotation. An ‘overlap’ thus only really occurs ‘mid-tick’ as the second hand moves between one visual notch to the next; the minute and hour hands are often somewhere in between.

It's 720 minutes past two. Get it? Eh? GET IT?!

 

So, starting at midnight, we have the second hand, minute hand and hour hand perfectly aligned straight upwards. Let us allow 1 minute to pass. Many people assume that 60 seconds later, the second hand has completed one ‘overlap’ of the other hands. However, during those 60 ticks, the minute hand has moved forward 6 degrees to the first notch on the clock, and the hour hand has moved forward 6 tiny arcminutes from due north.
 

Google Images: providing you with 850,000 pictures of clocks
that are completely irrelevant to the time you wish to depict.

 

Hence, at 00:01:00am:

The second hand is at the 0 degree, 0 arcminute, 0 arcsecond mark.
The minute hand is at the 6 degree, 0 arcminute, 0 arcsecond mark.
The hour hand is at the 0 degree, 6 arcminute, 0 arcsecond mark.
The second hand has yet to overlap any of the other hands. Another second ticks past.
At 00:01:01am:
The second hand is now at the 6 degree, 0 arcminute, 0 arcsecond mark.
The minute hand is now at the 6 degree, 10 arcminutes, 0 arcsecond mark.
The hour hand is at the 0 degree, 6 arcminute, 6 arcsecond mark.

The second hand has landed PAST the hour hand, and the minute hand has shifted slightly forward of the 1 so that the second hand is still behind it by a distance of 10 arcminutes.
A second hand will take one entire rotation of a clock PLUS another 1.0167 seconds before it has actually ‘overlapped’ the minute hand ‘mid-tick’ between the 1 and 2 second mark.

Tick, tick, tick... coming up next on 60 Minutes...

The minute hand on a clock overlaps an hour hand 22 times each day between midnight and 11:59:59pm, before the 23rd overlap marks a new day beginning at midnight again. However, as these clock hands do not smoothly rotate around a clock like a swimming pool timer or a stopwatch, they TICK by jumping across notched intervals, I believe that a traditional clock *does not measure any interval of time shorter than one second*. With this in mind, moments of time on a clock are limited to single second increments. For example, it is said that the first ‘overlap’ moment for an hour and a minute hand occurs 1 hour, 5 minutes and 27.272727 seconds after midnight. The second hand would be ‘midtick’ at this moment, therefore I do not acknowledge this as a moment of time measurable on a standard clock. By comparison, at midnight or midday there is a distinct 0 degree, 0 arcminute, 0 arcsecond gap to be found between the hour, minute and second hand. For second hand and minute hand overlapping, this only happens every hour on the hour.

Let us see what happens 6 hours after midnight, as the second hand marks 6.00am:
The second hand is at the 0 degree, 0 arcminute, 0 arcsecond mark.
The minute hand is at the 0 degree, 0 arcminute, 0 arcsecond mark. PERFECT OVERLAP!
The hour hand is at the 180 degree, 0 arcminute, 0 arcsecond mark.
 30 seconds later from this point, at 6:00:30am:
The second hand is at the 180 degree, 0 arcminute, 0 arcsecond mark.
The minute hand is at the 3 degree, 0 arcminute, 0 arcsecond mark.
The hour hand is at the 180 degree, 3 arcminutes, 0 arcsecond mark. OVERLAP MISSED.

From this analysis, I suggest that from an inclusive period of midnight until 11:59:59pm that same day, the second hand lands perfectly overlapped on a minute hand at 24 hourly occasions, along with an hour hand overlap only twice.
You can shove daylight savings up your arse.


6. In a country in which people only want boys, every family continues to have children until they have a boy. If they have a girl, they have another child. If they have a boy, they stop. What is the proportion of boys to girls in the country?

To begin with, the ratio will be 1:1.

For firstborns, the chance of a boy is 50% (assuming hermaphrodites and sexually ambiguous genitalia are assigned a gender by systems of oppressive dichotomous social norms) so that the number of firstborns who are boys will be the same as the number of firstborn girls.
 For secondborns (not to be confused with temporal ‘seconds’ or angular ‘arcseconds’) the chance of a boy is, again, still 50%. However, there must have already been a girl firstborn in order to necessitate a secondborn child, so once more the boy/girl ratio is perpetuated at 50%. There is no chance of two boys ever being born to a family if they always stop after the first boy, so there will only ever be a 1:1 ratio of girls to boys. For every firstborn boy to no girls born, assumedly there will be neighbours with two girls born before a boy, in perfect 1:1 ratio across the country according to basic probability.

Just consider yourself lucky you aint in Game Of Thrones, kiddo.

For third-borns, the chance of a boy is 50% again. And again, there must have been two girls born already to necessitate a third born, and so on and so on…
This problem would have been figured out by China shortly after their one-child policy was introduced, which most probably implies that ‘having another child’ involves the removal of previous girls from the family via drowning, abandonment in the wild, or transnational exchange for livestock and grain.

A country that wants only boys (really WANTS only boys) will eventually end up with only boys once their maternal mothers have all died out. The country at this point will then be made up entirely of boys, as all the girls would have died out, creating a 0% procreation rate for the country, all those boys will eventually die out and the ratio of boys to girls within the country will go from 1:1 to 0:0.




7. Design an evacuation plan for San Francisco.

This depends on what state of emergency is being called here. An earthquake, for example, would jeopardise the Golden Gate Bridge as an escape alternative, if thousands of cars were to suddenly plunge into the sea as the bridge collapsed. However, a nerve gas released in the epicentre of the city would not rule out this as part of the escape routes.


Has a swarm of Africanised Bees landed in Beuna Vista park? 
Is there a terrorist bombing of one of the many international consulates in San Fran?
Was someone smoking in the mayor’s toilet and accidentally set off the town fire alarm? Some specifics here would be nice.

If a tidal wave is sweeping in over the area from the west, obviously the evacuation plan would involve different population movements than if nuclear material was spilled along the Dwight D. Eisenhower Highway.

Design an evacuation plan for San Francisco? How about, if X is the location that will be affected by threat Y, evacuate citizens in an orderly radius away from this area according to what the limitations of roads out of the city will allow, to get people away from said threat to safety. Everyone meet up at Sacramento for a head count and roll call. How’s that for your damned evacuation plan.

Stay calm
and
GIT TO DA CHOPPA!!!

 8. A man pushed his car to a hotel and lost his fortune. What happened?

What the fuck do you think happened, he just pushed his car down the driveway into a freaking hotel!

Navigating the pool whilst maintaining speed
was *quite* the sight, let me tell you...

As he is a man with a fortune, one can assume that the hotel he was attending was a really posh place with expensive furnishings. A lot of structural damage was inevitably caused as his car rolled through the entrance, perhaps even taking out a few casualties. There would have been security cameras everywhere capturing the moment when some pompous rich guy pushed his Rolls Royce into a hotel, causing tens of thousands of dollars’ worth of damage and bodily harm to innocent bystanders. His actions would have been seen as negligent and/or malicious destruction of property, hence his insurance agency would not pay out for anything and he would have been arrested.

Such a bizarre occurrence would have garnered lots of media attention and his implied high standing in society would have been jeopardised in the public eye, causing his stock investors  to lose faith in his character and business. Company shares would plummet. Legal fees would then drain the rest of his money as he fought the case, causing several other discrepancies to surface into the public eye as people appeared out of the woodwork to sell the media their stories and cash in on the hype. Eventually the man would be bankrupt and have to sell everything he owned, leaving him with nothing.

You know, that or NERRRRRRR PLAYING MONOPOLY this riddle was old in primary school.

9. Add any standard arithmetic signs to this equation to make it true: 3 1 3 6 = 8

3! + 1/3(6) = 8

… what do you mean a ! isn’t a standard arithmetic sign? Fine.

3 – 1 + (√36) = 8

... oh, so now √ is getting too fancy? Well, how about a little alphabetical arithmetic, then?

 T H R E E     O N E     T H R E E     S I X     =     E I G H T
 20 8 18 5 5    15 14 5   20 8 18 5 5   19 9 24   =    5 9 7 8 20

20 8 18 5 5 15 14 5 20 8 18 5 5 19 9 24 = 5 9 7 8 20

20 {[(8 + 18 – 5 – 5 – 15) + (14 x 5)] x [20 x (8 + 18)] + 5 – (5 x 19) - (9 x 24º)]} = 597,820

= 20 {[1 + 70] x [20(26)] + 5 - (95) – (9)]}

= 20 (71) x [(520) + 5 – 95 – 9]
= 20 (71) x (421)

= 597,820

Or a plain old boring (3 + 1) ÷ (3/6) = 8 will do the trick.
NEXT!


 10. You have eight identical balls all of the same size. One ball weighs heavier than the others. How can you find the heavy ball using a balance and only 2 weighings?

Psht, really? I remember doing this in primary school, too.

Place 3 balls on one side of the balance, place 3 balls on the other. If the balance is even, you then weigh the 2 leftover balls on the scale to see which one is the heavy ball.

I am strangely hungry all of a sudden.

 

If the balance tilts down to one side to begin with, you know one of those three balls is the heavy one. Weigh two of those balls from the heavy side against each other for your second weighing. The balance will either tilt again, showing you the heavy ball, or it will remain balanced, ergo the last remaining ball from that group must be the heavy one.

Either that, or go to stack 4 balls on either side of the balance, one by one. At one point in this process, you will notice an imbalance occur after an even stacking, and it will technically count as only one weighing having taken place to identify the heavy ball.



11. You are shrunk to the height of a nickel and your mass is proportionally reduced so as to maintain your original density. You are then dropped into an empty glass blender. The blades will be switched on in 60 seconds. What do you do to survive?

 

I doubt I would survive long enough to escape. If my mass is proportionally reduced, I assume there will be all kinds of sudden issues regarding sensory ability and brain function, blood pressure, organ behaviour, rapid warmth issues from the sudden dramatic shift in surface area, basic respiration complications, and other problems that I won’t even begin to delve into now.

Rick Moranis: renowned for his inept shrinkage.

 

Being shrunk down to half of one’s usual height reduces one’s bodily mass by significantly more than half, much like how a 6 foot cube is 216 cubic feet in volume with 216 square feet of surface area, but reducing this to a 3 foot cube results in only 27 cubic feet of volume and 54 square feet of surface area. A nickel being scaled up to the weight of a 78kg human would only be 33cm in diameter, as metal is more dense than a human is. If we simplify the human body into a cylindrical shape for calculation’s sake, a human who stands 178.16cm tall has the equivalent height of 84 nickels, thus an 84:1 scaling ratio. A nickel coin expanded to our height would be 16.38cm thick, with the nickel/copper alloy weighing about 3.65 tonnes.

One cent is a nickel, right?
God damn these imperial measurements...

 As aforementioned in the school bus question, the human body is roughly the same density as water, with a weight of 1g per cubic cm. If our person weighs 75kg, this gives them 75,000 cubic cm of volume. The volume of a cylinder is calculated as height x πr², thus our 178.16cm tall human must have a round diameter of 23.16cm.

Shrinking this human cylinder down to 1/84th of original size involves a little bit of magic from Galileo, thanks to his observations on biological sizes and weight ratios, so we can calculate that this process would reduce a human to 127mg. By comparison, a grasshopper weighs about 300mg. It would not be an unrealistic scenario that one would be able to scale the side of the empty blender, perhaps by running in a circular upward motion like a motorcycle in a thunderdome, with the accumulated centrifugal force allowing one to scale upwards and out over the lip without too much difficulty within 60 seconds.

Wait, do we get a mini motorcycle too?

 12. You're the captain of a pirate ship. You and your crew get to vote on how the gold is divided up. If fewer than half of the pirates agree with you, you die. How do you recommend apportioning the gold in such a way that you get a good share of the booty, but still survive?

 

Of *COURSE* I could source a pirate at Occupy Wall Street.

 

I would initially keep 50% of the gold for myself, then divide the remaining gold into exponentially decaying increments around the remaining crew whilst claiming a ‘trickle down’ effect. By circulating investment exchanges only with the wealthiest of select individuals, this would thus perpetuate a cycle of income among the fiscal elite on deck.
 Any individuals supporting an equal dispersal of the gold would be labelled a communist and made to walk the plank as a public example to other dissenters.

Although the majority of the crew will be impoverished, I would offer them extra gold upon request if they paid it back in regular instalments with interest. After an initial boost in assets, these dubloon-pinchers will soon be struggling to pay back with gold that does not exist within their combined financial strata, resulting in an inevitable debt cycle to the empowered rich. Eventually, the majority will be unable to cover basic scurvy costs and everyday buccaneer essentials, let alone fund additional resources required to successfully revolt against this system and be dependent on gold investments from the elite for survival.

I would put off allowing them to vote for as long as possible, and even after that time arrived, this way of apportioning gold will be so ingrained into the crew hierarchy that it wouldn’t make any difference what alternate system was proposed.

(Source: United States of America.)


13. You have two identical eggs and a 100-storey building. The eggs may break from a 1 storey height or a 100-storey height or any number in between, assuming they survive every fall that they can safely over and over again. You must find the safest height at which an egg can be dropped and not break, using the least amount of attempts. You may break both eggs to find your answer.

2 eggs.
Was this image necessary?

 

If we start by dropping an egg from the 50th floor and it breaks, we know the height must be below 50 storeys, and we would then need to go all the way down to the first storey with our second egg and go up the levels one-by-one until we found the answer. At best, the egg would break at the first storey and we would know after two attempts that the answer is 0 storeys. However, at worst, the egg would survive being dropped from storey 1 all the way up to storey 49. This would be 50 attempts before we found our answer.

If we dropped the first egg at level 50 and it survived, then at level 100 and it broke, then again we would need to drop our second egg from level 51 all the way up to 99 until we found our answer, potentially taking up to 50 attempts again to find the answer.
 It thus makes sense that we find the highest increment for our first drop such that the number of attempts with the second egg after this would be the least amount of ‘catching up’ attempts required to be certain of the level in between our ‘jumping up’ increments.

For example, if our first egg survives level 10, then level 20, then level 30, then level 40, then level 50, then level 60, then level 70, then cracks at level 80, we would only have to test levels 71-79 with our second egg for, at most, 17 attempts to be certain of our answer. (A vast improvement compared with the 50 attempts required using our initial method.) The worst case scenario should be accounted for in every method chosen, to find the smallest maximum number of attempts required for each method.

Every 10th storey
Egg 1: 10, 20, 30, 40, 50, 60, 70, 80, 90, *100 breaks*
Egg 2: 91, 92, 93, 94, 95, 96, 97, 98, *99 breaks*
ATTEMPTS: 19

Every 9th storey
Egg 1: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, *99 breaks*
Egg 2: 91, 92, 93, 94, 95, 96, 97, *98 breaks*
ATTEMPTS: 19

An interesting observation is that even though we have just moved the ‘jumping’ increment down from 10 to 9, we still need 19 attempts to find the answer in worst case scenario. Because we are dealing with a 100 storey building, it is probably best to work with factors of 100.

Every 5th storey
Egg 1: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, *100 breaks*
Egg 2: 96, 97, 98, *99 breaks*
ATTEMPTS: 23

Every 20th storey
Egg 1: 20, 40, 60, 80, *100 breaks*
Egg 2: 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, *99 breaks*
ATTEMPTS: 24

Hmm, no, that didn’t improve things either, we either lost too many attempts with a small first egg jumping increment or catching up the large gap with our second egg after testing a large first egg jumping increment. What would happen if we were to vary the number of storeys we ascend each time, so that we still land on storey 100 at the end for maximum numeric efficiency? Logically, a transcending series of numbers that adds up to 100 would be the most effective method of minimising the amount of attempts in a worst case scenario. Our best count so far has been 19, so let’s start at 18 and deduct an increment each time we go up. This way, we maximise 18 as the most amount of attempts that will be used up by our second egg in a worst case scenario.

18 + (n-1) increments:
Egg 1: 18 (+17) 35 (+16) 51 (+15) 66 (+14) 80 (+13) 93, 94, 95, 96, 97, 98, 99 *100 breaks*
ATTEMPTS: 13

In this scenario, we required only 13 attempts and one egg to find our answer.
What would happen in worst case scenario that our first egg broke first go?

18 + (n-1) increments:
Egg 1: *18 breaks*
Egg 2: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, *17 breaks*
ATTEMPTS: 18

At worst case scenario, we have still improved our attempt count to 18. However, our first egg is left with a large number of individual catch-up levels to manage in the final stretch. By starting with a lower number, can we manage to fill that gap more effectively?
Let’s start with 1 and tally upwards until our increments hit 100 storeys, that sounds sensible:
1 + 2 (3) + 3 (6) + 4 (10) + 5 (15) + 6 (21) + 7 (28) + 8 (36) + 9 (45) + 10 (55) + 11 (66) + 12 (78) + 13 (91) 92, 93, 94, 95, 96, 97, 98, 99, *100 breaks*.
ATTEMPTS: 22, if we start with 13 storeys. Let’s try 14 then:

14 (+13) 27 (+12) 39 (+11) 50 (+10) 60 (+9) 69 (+8) 77 (+7) 84 (+6) 90 (+5) 95 (+4) 99, *100 breaks*.
ATTEMPTS: 12

Huzzah! By starting with our first egg from 14 storeys, the maximum number of attempts we would need to make in a worst case scenario of breakage would be 14 attempts:

14 + (n-1) increments:
Egg 1: *14 breaks*
Egg 2: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, *13 breaks*
ATTEMPTS: 14.

If our first egg breaks anywhere from 14-100, it will only ever take 14 attempts to catch up from the previous storey, as we have methodically reduced the increment each time by one in order to counter the previous number of attempts lost. We are left with only one extra attempt needed at the end to reach storey 100 if needed rather than our previous string of turns.

14 + (n-1) increments:
Egg 1: 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, *99 breaks*
Egg 2: 96, 97, *98 breaks*
ATTEMPTS: 14.

The answer is therefore 14 storeys to begin with, then one storey less than this each time up until the 99th storey, when there will be either one more storey to attempt with the first egg or three more storeys to attempt with the second egg to be certain of our answer. 

For science!

14. Imagine you have a closet full of shirts. It’s very hard to find a shirt. What can you do to organise your shirts for easy retrieval?

If my closet is full of shirts and it is very hard to find a shirt, I would first suggest switching on my bedroom light. Hey presto! Shirts found!

After that, I would pull them all out and scatter them all over my bedroom floor so they are now both visible and accessible when needed.

 (Source: My bedroom.)


15. Using only a four-minute hourglass and a seven-minute hourglass, measure exactly nine minutes without the process taking longer than nine minutes.

Turn both hourglasses over.

 

Next up: paint drying on growing grass!

After 4 minutes, the small one will be empty and the big one will have 3 minutes of grain remaining at the top. Flip them both over again.

After 3 more minutes (7 minute mark), the small hourglass will have 1 minute remaining and the big hourglass will be empty. Flip them both over again.

After 1 more minute (8 minute mark), the small hourglass will be empty and the big hourglass will have 1 minute of sand at the bottom and 6 minutes remaining up top. Flip the big hourglass.

After 1 more minute (9 minute mark), the big hourglass will be empty again.

Of course, we could always assume that the 4 minute hourglass is 36.36cm in height and the 7 minute hourglass is 63.63cm in height; so by stacking one atop the other with the sand at the bottom and dangling them from a hook off the ceiling, we would create a 1m pendulum with a 2 second period rate, with the shifting of the sand along with gravity as each swing reached the extent of its arc causing a perpetual movement backwards and forwards until 270 swings had been counted.


16. You're in a car with a helium balloon on a string that is tied to the floor. The windows are closed. When you step on the gas pedal, what happens to the balloon - does it move forward, move backward, or stay put?

That depends. If the car is in Drive, the car will move forward and the backward movement of the dense air will push the lighter helium balloon forward.

If the car is in Reverse when you step on the gas pedal, the car will move backwards and the movement of dense air forwards will push the helium balloon backwards also.

If you are not in gear, or step on the gas pedal really hard and do a sick burnout, the balloon will stay put.

I believe this experiment will be tested out in Fast & Furious 8.

Thank you for your time, I anticipate my job offer in the mail by the end of the hour.
I am available to start working on Monday.

Yours drolly and trolly,

Nib Oswald