Puzzles Contest - 9 With Answers

[HELP]

1. I have five sections of a necklace-each section consisting of four links. I took the sections to a goldsmith and asked him to give me an estimate to join the 5 sections - into a one piece necklace. The goldsmith wanted-Re.1 to cut open a link and Re. 1 to solder it
together again.
What is the cheapest method and how much should it cost me to get the five pieces joined together into one full necklace?

2. I have three square boards, the surface of the first containing five square feet more than the second, and the second containing five square feet more than the third." " Can you find the exact measurement for the sides of the boards?

3. ThiS is an ancient problem dating back to about 310 A.D. Demochares had lived one-fourth of his life as a boy, one-fifth as a youth. one-third as a man, and has spent thirteen years in his dotage- How old is Demo-chares?

4. The combined ages of Reena and Seena are 44 years and Reena is twice as old as Seena was when Reena was half as old as Seena will be when Seena is three times as old as Reena was when Reena was three times as old as Seena. How old is Reena?


5. How can you make a total of 1,000 by using sixteen 4's


6.here are two whole numbers, difference of their squares is a cube and the difference of their cubes is a square. These are the smallest possible numbers. Can you find the numbers?


7. I have two 10 paise coins. If 4/5 of what I have is 8/9 of what you have. how much do you have?
8. Four married couples 'played a tennis tournament of 'mixed doubles'. A man and a woman always played against a man and a woman. However, no person zver played with or against any other person more than once. They all played together in two courts on three successive days.
Can you show how they could have done it?


9. Which would you say is heavier, a pound of cotton or a pound of gold?


10. Last time I visited a friend's farm near Bangalore he gave me a bag containing 1000 peanuts. From this I took out 230 peanuts for myself and gave away the bag with the remainder of peanuts to three little brothers who live in my neighbourhood and told them to distribute the nuts among, themselves in proportion to their ages-which together amounted to 171/2 years.
Tinku, Rinku and Jojo, the three brothers, divided the nuts in the following manner: As often as Tinku took four Rinku took three and as often as Tinku took six Jojo took seven. With this data can you find out what were the respective ages of the bQYs and how many nuts each got?


Answers

1. There is only one cheapest method and that is to open the 4 links of one section and then use these links to join the other 4 sections together, which should cost altogether Rs. 8.

2. The sides of the three boards measure 31 inches, 41 inches and 49 inches.

3.Demochares must be sixty years of age.

4.The ratio of Reena' s age to Seena's must be as 5 to 3. Since the sum of their age is 44, Reena must be 271/2 and Seena 161h




5. 444 + 444 + 44 + 44 + 4 + 4 + 4 + 4 + 4 +4

6. 102 - 62 = 100 - 36 = 64 = 43
103 - 63 = ]000 - 216 = 784 = 282


7. 18 Paise.

8.Let's call the men ABC D and their wives E F G H They must play in such a way that no person ever play~twice with or against another person.
First Court
Second Court
1st day ADagainst BF CE against DF
2nd day AH against CF DE against BH
3rd day AF against DG BE against CH
In this way no man ever plays with or against his
Ownlwife




9.A pound of cotton is heavier than a pound of gold because cotton is weighed by the voirdupois pound, which consists of 16 ounces, whereas gold, being a precious metal is weighed by the troy pound 1 \Jhich contains 12 ounces (5760 grams).

10.When Tinku takes 12, Rinku and Jojo will take 9 and 14, respectively-and then they would have taken altogether thirty-five nuts. Thirty-five is contained in 770 twenty-two times which means all one has to do now is merely multiply 12, 9 and 14 by 22 to find that Tinku's share was 264, Rinku's 198 and Jojo's 30B. Now as the total of their ages is 171/2 years or half
the sum of 12, 9 and 14, their respective ages must be 6,4.1/2 and 7 years.
Edited by: HELP

[Hemanth_2u]

2. I have three square boards, the surface of the first containing five square feet more than the second, and the second containing five square feet more than the third." " Can you find the exact measurement for the sides of the boards?


The sides of the three boards measure 31 in., 41 in., and 49 in. The
common difference of area is exactly five square feet. Three numbers
whose squares are in A.P., with a common difference of 7, are 113/120,
337/120, 463/120; and with a common difference of 13 are 80929/19380,
106921/19380, and 127729/19380. In the case of whole square numbers the
common difference will always be divisible by 24, so it is obvious that
our squares must be fractional. Readers should now try to solve the case
where the common difference is 23. It is rather a hard nut.

[Hemanth_2u]

3. ThiS is an ancient problem dating back to about 310 A.D. Demochares had lived one-fourth of his life as a boy, one-fifth as a youth. one-third as a man, and has spent thirteen years in his dotage- How old is Demo-chares?
Ans: Let Demochares lived in x years
As a boy had lived = x/4
As a youth had lived = x/5
As a man had lived = x/3
spent thirteen years in his dotage
x/4+x/5+x/3+13 = x
x = 130 Years

[Hemanth_2u]

5. How can you make a total of 1,000 by using sixteen 4's


444+444+44+44+4+4+4+4+4+4

[Hemanth_2u]

6.here are two whole numbers, difference of their squares is a cube and the difference of their cubes is a square. These are the smallest possible numbers. Can you find the numbers?




Ans:


Let the two positive numbers be a and b, with a > b. Now,
a^2 - b^2 = m^3 ==> (a+b)*(a-b) = m^3

It implies that (a-b) = m and (a-b)^2 = m^2. a = (m^2 + m)/2 and b =
(m^2 - m)/2. Now, there would be an infinite series satisfying this
condition that follows:

m a b
------------------
1 1 0
2 3 1
3 6 3
4 10 6
5 15 10
6 21 15
7 28 21
8 36 28
9 45 36
10 55 45
and, so on.

But, a^3 - b^3 = n^2 => n^2 = (a-b)*(a^2 + ab + b^2)
= (m^2/4)*(3*m^3 + m)

So, {3*(m^3) + m} should be a square as (m^2/4) is of that form. I am
completely stuck at this point. Please, help me out.

I could find out the following by just trial and error. The lowest
possible solution is (a,b) = (10,6)

10^2 - 6^2 = 64 = 4^3
10^3 - 6^3 = 784 = 28^2

But, I am nowhere near to finding a generic solution.

Eagerly waiting for your reply.

Thanks and regards

[Hemanth_2u]

8. Four married couples 'played a tennis tournament of 'mixed doubles'. A man and a woman always played against a man and a woman. However, no person zver played with or against any other person more than once. They all played together in two courts on three successive days.


Can you show how they could have done it?
Ans:
Call the men A, B, D, E and their wives a, b, d, e. Then they may play
as follows without any person ever playing twice with or against any
other person:--
First Court. Second Court.
1st Day | A d against B e | D a against E b
2nd Day | A e against D b | E a against B d
3rd Day | A b against E d | B a against D e
It will be seen that no man ever plays with or against his own wife--an
ideal arrangement. If the reader wants a hard puzzle, let him try to
arrange eight married couples (in four courts on seven days) under
exactly similar conditions.


Edited by: Hemanth_2u

[Hemanth_2u]

9. Which would you say is heavier, a pound of cotton or a pound of gold?



Ans : Those are same in heavier. a pound of cotton have greate volume campare with a pound of gold

[keerthi]

7. I have two 10 paise coins. If 4/5 of what I have is 8/9 of what you have. how much do you have?
Ans) 18 paise
Sol: let the amount i have = 'x'
4/5(20) = 8/9(x)
x =18

[meera]

9. Which would you say is heavier, a pound of cotton or a pound of gold?


soln:


a pound of cotton weighs equal to the pound of gold

[meera]

7. I have two 10 paise coins. If 4/5 of what I have is 8/9 of what you have. how much do you have?
soln
i have 20 paise
let ur's amount be x
4/5(20)=8/9(x)
16=8x/9
144=8x
x=144/8
x=18

[keerthi]

3. ThiS is an ancient problem dating back to about 310 A.D. Demochares had lived one-fourth of his life as a boy, one-fifth as a youth. one-third as a man, and has spent thirteen years in his dotage- How old is Demo-chares?
Let the age of the person be 'x'
x/4+x/5+x/3+13 = x
solving we get x = 60

[keerthi]

9. Which would you say is heavier, a pound of cotton or a pound of gold?



Ans) the pound of cotton is heavier.


The pound used for measuring cotton has 16 ounces. where as the pound used for measuring gold 12 ounces.

[keerthi]

hi hemanth,


for 6th one the least possible answer is 10 and 6.well i also dont know generalised procedure for that.

[keerthi]

1. I have five sections of a necklace-each section consisting of four links. I took the sections to a goldsmith and asked him to give me an estimate to join the 5 sections - into a one piece necklace. The goldsmith wanted-Re.1 to cut open a link and Re. 1 to solder it
together again.
What is the cheapest method and how much should it cost me to get the five pieces joined together into one full necklace?
Ans)8 Rs
the cheapest way is to cut open 4 links and join them.
so 4 cuts cost =4Rs and
4 joinings cost 4 Rs
total = 8Rs

[keerthi]

10. Last time I visited a friend's farm near Bangalore he gave me a bag containing 1000 peanuts. From this I took out 230 peanuts for myself and gave away the bag with the remainder of peanuts to three little brothers who live in my neighbourhood and told them to distribute the nuts among, themselves in proportion to their ages-which together amounted to 171/2 years.
Tinku, Rinku and Jojo, the three brothers, divided the nuts in the following manner: As often as Tinku took four Rinku took three and as often as Tinku took six Jojo took seven. With this data can you find out what were the respective ages of the bQYs and how many nuts each got?
Ans) ages were tinku =6yrs , rinku =4 1/2 yrs ,jojo = 7yrs
T:R = 4:3 = 12:9
T:J = 6:7 = 12:14
T:R:J = 12:9:14 = 6:4 1/2:7
Nuts to tinku = 264
rinku = 198
jojo = 308