The Spotting Contest

During a recent plane and train spotting contest, five eager entrants were lined up ready to be tested on their spotting ability. 

They had each spotted a number of planes (26, 86, 123, 174, 250) and a number of trains (5, 42, 45, 98, 105). From the clues below, can you determine what colour shirt each was wearing, their position, their age (21, 23, 31, 36, 40) and the number of trains and planes spotted?

1. Bob spotted 44 less trains than planes. 

2. Tom was 36 years old. 

3. The person on the far right was 8 years younger than Bob, and spotted 174 planes. 

4. Josh was wearing a yellow shirt and spotted 37 trains fewer than Bob. 

5. The person who was wearing a green shirt, was 19 years younger than the person to his left. 

6. Steven spotted 105 trains and 250 planes.

7. The person in the center was 31 years old, was wearing a blue shirt and spotted 42 trains.

8. Doug, who was on the far left, spotted 26 planes, and spotted 72 trains more than planes. 

9. The person who was wearing a red shirt was 4 years older than Tom and was not next to the person wearing a blue shirt. 

10.The person who was next to the 31 year old, but not next to the person who spotted 26 planes, was wearing a orange shirt, and spotted 45 trains. 

Here are final STATS of the contest! 

Final Stats of The Spotting Contest

What was the contest?

Five participants having ages (21, 23, 31, 36, 40) had each spotted a number of planes (26, 86, 123, 174, 250) and a number of trains (5, 42, 45, 98, 105). 

Below are the clues given - 


------------------------------------------

1. Bob spotted 44 less trains than planes. 

2. Tom was 36 years old. 

3. The person on the far right was 8 years younger than Bob, and spotted 174 planes. 

4. Josh was wearing a yellow shirt and spotted 37 trains fewer than Bob. 

5. The person who was wearing a green shirt, was 19 years younger than the person to his left. 

6. Steven spotted 105 trains and 250 planes. 

7. The person in the center was 31 years old, was wearing a blue shirt and spotted 42 trains. 

8. Doug, who was on the far left, spotted 26 planes, and spotted 72 trains more than planes. 

9. The person who was wearing a red shirt was 4 years older than Tom and was not next to the person wearing a blue shirt. 

10. The person who was next to the 31 year old, but not next to the person who spotted 26 planes, was wearing a orange shirt, and spotted 45 trains. 

------------------------------------------

STEPS : 

Let's make at table like below and fill the data one by one.

 
STEP 1 : 

As per (3), the person on the far right that is at position 5 had spotted 174 planes & he must be 23 years old so that Bob aged 31 years (only pair of ages having 8 years difference.

 
STEP 2 : 

As per (7), the person at position 3 who is wearing blue shirt and is 31 year old. He has spotted 42 trains.

STEP 3 : 

For (5) to be true, the ages of those participants occupying consecutive positions must be 40 and 21 respectively. Only, positions lefts for (5) to be true are 1 & 2.

STEP 4 : 

And as per (8), the person at position 1 must be Dough who spotted 26 planes and spotted 72  trains more than planes i.e. 26 + 72 = 98 trains.

STEP 5 : 

As per (10), the person at position 4 must be wearing orange shirt and must have spotted 45 trains. His age must be 36 years (only number in ages list left).

STEP 6 :

As per (9) suggests, Tom must be 36 years old positioned at 4 and Dough must be the person wearing red shirt. With that, Josh in (4) must be wearing yellow shirt and positioned at no.5. 

STEP 7 : 

The hint (4), also suggests that Josh must had spotted 5 trains out of 42-5 possible spotted trains pair shared with Bob. Hence, the person at 2nd position must had spotted 105 trains (only number left in list of number of trains spotted).
 

STEP 8 : 

As per (3), the person having age 31 must be Bob. Hence, at no.2, Steven must be there. 

STEP 9 : 

As per (6), Steven have spotted 250 planes. As per (1), Bob must have spotted 42 + 44 = 86 planes. And the only number left as number of planes spotted for Tom is 123.

STEP 10 :

Final Stats look like as - 

The Allotment Challenge?

Five bankers are sharing 12 golden ingots. They decide to proceed that way : 

The elder one will suggest an ingots allotment. The rest will vote for or against it. If the majority accepts, the sharing is ratified. If not, the elder will be dismissed. So, the sharing would be done between the remaining banker with the same rules. 

Knowing that they are set from left to right in a diminishing order of their ages, how would be the allotment ?

THIS should be the UNDENIABLE Allotment!

Source 

The Undenial Allotment Proposition!

What were rules of allotment process?

The eldest should allot ingot like 9, 0, 1, 0, 2 among 5 bankers.

Let's name bankers as Banker 5, Banker 4 ...... Banker 1 according to decreasing order of their ages.

------------------------------------------------------------------------------

CASE 1 : 

Suppose there are only 2 bankers left then the youngest will always deny whatever elder offers so that he can take away all 12 ingots on his turn.

------------------------------------------------------------------------------

CASE 2 : 

If case is reduced to 3 bankers then the eldest knows that the youngest is not going to agree with him in any case. With that, the eldest will be dismissed and case reduced to CASE 1 where youngest can take away all. 

So, the eldest here proposes allotment as 11, 1, 0. The Banker 2 has no option than to accept this proposal otherwise he won't get anything if case is reduced to 2 bankers as in CASE 1 above.

------------------------------------------------------------------------------

CASE 3 : 

With 4 bankers, the eldest would propose allotment 9, 0, 2, 1.

In the case, Banker 3 will never accept any proposal as after banker 4 is dismissed he would be getting 11 ingots as in CASE 2 above.

The Banker 2 will happily agree with the eldest as he would be getting 1 more ingot than the CASE 2. 

And Banker 1 knows he will be getting nothing if the case is reduced to 3 bankers as in CASE 2. So, he too will agree with the eldest in the case.

------------------------------------------------------------------------------

CASE 4 : 

With 5 bankers, the eldest i.e. Banker 5 should propose allotment 9,0,1,0,2.

Obviously, any how Banker 4 is going to deny any proposal as he wants the distribution among 4 bankers where he will be getting 9 ingots as in CASE 3 above. 

And if Banker 5 is eliminated and the case is reduced to CASE 3 where 4 bankers are left then Banker 3 knows he won't be getting anything. So, better he should be happily agree this proposition where he is getting at least 1 ingot.

Finally, offering Banker 5 one more extra ingot than the case where 4 bankers will be left, makes him in favor of this proposition.

Notice that the Banker 5 has to give 3 ingots at least to banker 2 to get his vote as he will be getting 2 ingots in case of 4 bankers as in CASE 3. Whereas, in the same case Banker 3 is not getting anything & would be happily agree if getting 1 ingot at least in this case. 


------------------------------------------------------------------------------

 

The Wedding Anniversary Puzzle

Recently I attended the twelfth wedding anniversary celebrations of my good friends Mohini and Jayant. Beaming with pride Jayant looked at his wife and commented, ‘At the time we were married Mohini was 3/4 of my age, but now she is only 5/6 th.
We began to wonder how old the couple must have been each at the time of their marriage!

Can you figure it out?

Know their ages! 

The Wedding Anniversary Puzzle's Solution

What was the puzzle?

Let 'x' be the age of the Jayant & 'y' be the age of Mohini 12 years ago.

So 12 years ago,

y = (3/4) x          .........(1)

And now 12 years later, the proportion is - 

y + 12 = (5/6) (x+12)  

Putting (1) in above,

(3/4) x + 12 = (5/6) x + 10

(1/12) x = 2

x = 24

Again putting this value in (1),

y = (3/4) 24 = 18

So, the age of Jayant was 24 and that of Mohini was 18 at the time of marriage. And now after 12 years, they are 36 and 30 years old respectively.  

The Exact Age?

Tommy: "How old are you, Mamma?"

Mamma: "Let me think, Tommy. Well, our three ages add up to exactly seventy years."

Tommy: "That's a lot, isn't it? And how old are you, Papa?"

Papa: "Just six times as old as you, my son."

Tommy: "Shall I ever be half as old as you, Papa?"

Papa: "Yes, Tommy; and when that happens our three ages will add up to exactly twice as much as to-day."

Tommy: "And supposing I was born before you, Papa; and supposing Mamma had forgot all about it, and hadn't been at home when I came; and supposing..."

Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'll have a headache."

Now, if Tommy had been some years older he might have calculated the exact ages of his parents from the information they had given him. Can you find out the exact age of Mamma?

Here is CALCULATION of exact age!

Calculation Of The Exact Age

What was the problem? 

Let us suppose T be the age of Tommy, M be of the Mamma and P be that of Papa.

Sum of their ages is 70.

T + M + P = 70  ......(1)

and Papa is 6 times as old as Tommy,

P = 6T  .....(2)

In unknown number of years X, Papa will be twice old as Tommy,

P + X = 2 (T + X)  ....(3)

and the sum of ages at that time is 70 x 2 = 140,

(T + X) + (P + X) + (M + X) = 140.

T + P + M + 3X = 140

From (1), above equation becomes,

70 + 3X = 140

X = 70/3     .....(4)

Putting (4) and (2) in (3),

P + X = 2 (T + X) 

6T + 70/3 = 2(T + 70/3)

T = 70/12    .....(6)

Using (6) in (2),

P = 6T

P = 6(70/12)

P = 70/2    ......(7)

Putting (6) and (7) in (1),

M = 70 - 70/2 - 70/12

M = 29.1666 = 29 years 2 months.

P = 70/2 = 35 years.

T = 70/12 = 5.8333 = 5 years 10 months.

 
To summarize, the Tommy is 5 years 10 months old, Mama is 29 years 2 months old and Papa is 35 years old.