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30/.5 = 60 and then add by 12 then its 72

Test one battery with 4 others, and if it doesn't work, that battery is discharged. Repeat this process, but with one less test each time. That's 10 attempts. Now you're done while the pedantic people are just figuring out how to do it in 6 attempts.

First give two samples at same time, if it dies after 10 hours then one of the two must be poisonous which you can find, else the two must be good, and give one of other two samples

5,4,3,2,2

why can't we try two bottle at a time, if it works then gud and try 3rd bottle after 10hrs if it works then 4th one is poisonous

How do you prove that this solution is minimal, though?

0:30 Well 2 bottles in the first test, then if the rat is dead bingo, if not give a random bottle in the second 2:49 okay it's a way better solution but the 2bottle in the first is also viable just take more time

nahhh saying 2 planes then immediately showing NYC 💀💀💀💀💀💀

I originally got 2, 2, 3, 4, 4, 3, 2

Very nice

8.15 into degree

where is the weight of the coin in you formula?

a-87.5 b-160

13 balls by pigeonhole principle

I solved it just with 2 moves!!

Whichever company asked these questions in their job interview must be bankrupt by now. 😂

Thank you 🙏🙏

Fun fact: if n is the ratio bike speed/ walking speed. The time deduction of the optimize method will be equal 8/(3n+9) Meaning the faster the bike (at constant walking speed, the less useful is the optimisation (because they need to walk less since the bike covers more of the distance)

A=87.5° B=160°

why cant there be multiple pick up and drop off to get shorter time

Amazing!!!

Amazing solution!!!

I completely understood the solution to the question, but the problem is that a bird can go and come between two trains an infinite number of times. If we consider this, the two trains will not collide. How can this problem be solved? Of course, in reality, the bird must first accelerate from zero to three times, but is there anything wrong with this simplification?

That 10 To 10 shit is stupid.

I don't think a human being can even RUN at 15km per hour. A fast paced walker can only walk 8km per hour. So, both problem AND solution are ... problematic at best. A person would need ridiculously long legs to walk 15km/hour. Now THAT'S a logic test! How long would a person's legs need to be to walk that freaking fast? And let's include an assumption: that a person's stride (each foot taking one step) is the same as the length of one of their legs. Go ahead math whiz!

You're solution SUCKS. It has scientific margins for error. Without going into them my solution is far better. Pay attention. Feed the rat samples from two bottles immediately and after ten hours you will KNOW for sure that two bottles are good leaving you fourteen hours for a second test from a third bottle or take your chances with basically a coin flip. Just don't drink the one that smells sweet, sour or like almonds, BOOM! Solved with four hours to spare. Assuming the margins for error from a scientific POV didn't kick your ass first. Not only better, but faster. Drink that bottle of MILK!

Super class

I will open the switch board and check with multimeter as it was not mentioned so we can use multimeter and check current😊

45 is the correct answer lmao

The coin dropped from top of Burz Khalifa, will never reach the ground. It will strike on some diff floor due to pyramidal structure.

I wish to start a Clock-Radio Company with the following concepts 1- 60℅ problem with 50℅ Solution 2- 60℅ problem with 40℅ Solution 3- 60℅ problem with 30℅ Solution 4- 60℅ problem with 20℅ Solution 5- 60℅ problem with 10℅ Solution 6- 60℅ Problem with 0℅ Solution 7- 40℅ Job using 30℅ Protractor 8- 40℅ Job using 20℅ Protractor 9- 40℅ Job using 10℅ Protractor 10- 40℅ Job using 0℅ Protractor So kindly suggest a suitable plan for me to start upon

A. 87.5° B. 160°

Thank you ❤

2000 loss, He gave 200 bag + 800 change. 1000 for next shop Total of 2000

excellent proof!!

Mind blowing 🤯

Immediately thought, “this isn’t hard. Just stagger the feedings.” I could work for google apparently.

Answer is 3

nonsense question if the rat drinks all the milk then how we drink at last? or at last instead of you die you have to mention rat die

How can you prove that 7 is minimal amount of races?

This is not true. The phrase "Divide 30 by half" is at best linguistically imprecise. But if you want to be pedantic 1/2 is "a half" not "half". Similarly, dividing by 2 is defined by the verb "halve". The question did not say "divide 30 by A HALF" which is 30 / 0.5 = 60. It also did not say "HALVE 30" which is 30/2 = 15. It said, "divide by half" provoking the question "Half of what?". What is the whole you are dividing by half? The idea you can simply assume you are dividing by half of 1 is wrong. This means divide 30 by 30's half. which is 30 / 15 = 2 You then add 12 to get to a final answer of 14.

11hours

WRONG ANSWER I gave this problem to my hs classes 20 years ago. The correct answer is 534. Rather than leave that 1 banana stranded in the desert, you go from that point, walk forward 1/3 mile, dump some bananas, go back, get the rest, then go forward to the pile. You just traveled 1 mi (1/3 + 1/3 + 1/3). You eat 1 banana, the one that would have been stranded. Then you travel the remaining 466 + 2/3 miles to the market with 1000 bananas, eating 466 along the way, bringing 534 to market.

A=87.5° B=160°

Actually, the shortest time can be found, without calculating any distances, explicitly. Lets say: T0: time, when the first guy is dropped off the bike, T1: time, which this guy needs to get to the destination, T2: time, for the bike to get back to the third guy, who is already walking toward the destination, T3: time, for the two guys on bike to get to the destination. As you said, they all arrive at the same time. We have the following equations: a) 60 T0 +15 T1 = 300 (the dropped-off guy) b) 15 T0 +15 T2 + 60 T3 = 300 (the picked-up guy) c) 60 T0 - 60 T2 + 60 T3 = 300 (the bike-guy), d) T1 = T2 + T3 (simultaneous arrival) 4 equations, can be solved for T0..3, to obtain the total shortest time: T = T0 + T1 ~ 9,28 h

My solution, check holes 2, 2, 2, 3, 4 On the last day - if the fox isn't in hole 4, then it's in hole 5.

For bonus riddle 1: I would write a 6 on the paper. For bonus riddle 2: Put a diagonal line through the equal sign, changing it to an inequality sign.

This fails at step 1 The thief would run off with $700 as that’s more he’d get if he stayed with them $700 > (1000+700+300)/3 Fails at step 2 The thief would run off with $1000 Fails at step 3 for the same reason as step 2 Fails at step 4 for the same reason as step 1 Fails at step 5 as the two would run off with $1,700 Fails horribly at step 6 as one thief would definitely run off with $1000

the two approaches are just piles of 4 2 2 in different testing order

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