Total Distance of the Race!

 

What was the PUZZLE?

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Suppose B’s speed is 1 m/s, then A’s speed will be 8/3 m/s.

A allowed B to go 80 meters ahead, meaning the distance between them is 80 meters.

Now, after A starts running, every second B increases this distance (80 m) by 1 meter, but A (because his speed is higher) reduces this distance by 8/3 meters.
So, each second the distance between them decreases by:

8/3 − 1 = 5/3 meters per second.

This means that to close the 80-meter gap, the time required is:

80 ÷ (5/3) = 48 seconds.

Both finish the race at the same time, meaning the gap becomes zero at the finish line, and from A’s perspective, it took him 48 seconds to complete the race at a speed of 8/3 m/s.

Therefore, the total race distance must be:

(8/3) × 48 = 128 meters.

Or

B run for 48 seconds at speed of 1m/s means covered distance of 48m. Since he started 80m ahead of A the total distance of race is 80 + 48 = 128m.

                                                    

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IPL Matches For The Charity

For Charity, five IPL teams were competing in cricket where they face each other exactly once. After the tournament, the following is the point table.

DareDevils 6
SuperKings 5
Royals 4
Sunrisers 2
Riders ?

How many points did Riders end up with? 

Click to know points of Riders! 

Source 

Points Scored By Riders in Charity Matches

Look at the point table first!

There are 5 teams playing in the competition for the charity. Hence, there are 10 possible matches between the teams like below.

DareDevils vs SuperKings
DareDevils vs Royals
DareDevils vs Sunrisers
DareDevils vs Riders
SuperKings vs Royals
SuperKings vs Sunrisers
SuperKings vs Riders
Royals vs Sunrisers
Royals vs Riders
Sunrisers vs Riders

For winning team, 2 points must have been given & for playing a draw 1 point must have been given to each team.

So there are 10 matches & total points distributed must be 10 x 2 = 20.

Adding up points of all teams 6 + 5 + 4 + 2 + Riders  = 20. That gives Riders = 3.

Hence point score of Riders is 3! 

 

Puzzle : Draw The Maximum Sum

Assume you are blindfolded and placed in front of a large bowl containing currency in $50, $20, $10, and $5 denominations. You are allowed to reach in and remove bills, one bill at a time. The drawing stops as soon as you have selected four-of-a-kind— four bills of the same denomination. 

What is the maximum sum of money you could accumulate before the drawing ends?

THIS should be the maximum sum! 

Source

The Maximum That You Can Get!

What was the question?

The quick response to the question by any body would be 3 x 50 + 1 x 20 = $170 would be the maximum as next pick of highest currency of $50 will stop drawing of currencies. But that 5th attempt may not be of $50 & could be of $20,$10 or $5. So this case doesn't count those possibilities. This is worst case scenario where 3 out of 4 picks resulted into accumulation of 3 currencies of same denomination of $50.

So ideally, after drawing three $50 s, three $20 s, three $10 s, three $5 s and finally drawing one anything will stop drawing attempts as that will accumulate 4 bills of same denomination (of $50/$20/$10/$5). The best that can be picked is $50 to get the maximum.

That is total maximum of 50 x 3 + 20 x 3 + 10 x 3 + 5 x 3 + 50 = $305 can be accumulated in 12 attempts & stopping after 13th attempt. And this will be the best case.

Spot The Liar Among

There is a party of 100 politicians. All of them are either honest or liars. You walk in knowing two things:

1. At least one of them is honest.

2. If you take any two politicians, at least one of them is a liar.

From this information, can you know how many are liars and how many are honest?

Know the total number of liars here! 

Source

Knowing The Number of Liars

What is the question?

The information that we have - 

1. At least one of them is honest.

2. If you take any two politicians, at least one of them is a liar. 

Only one of them must be honest and other 99 must be liars. 

Those are the only numbers that can justify the second piece of information. In the case, if any 2 are selected at least 1 of them liar while other may or may not liar (i.e. may be honest).