Turbo C++ download

Click here to download Turbo C++.zip

(3.2 MB only)

Linux Ebooks for free

Click to DOWNLOAD Ω

Linux Configuration & Installation (Slackware).zip

Redhat Linux Rhce Cramsession.pdf

Project Management Quick Tips – Video Tutorial

Challenging Mathematical Puzzles

1. Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided.

Answer 18

Assume that initial there were 3*X bullets.

So they got X bullets each after division.

All of them shot 4 bullets. So now they have (X – 4) bullets each.

But it is given that,after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X

Therefore, the equation is
3 * (X – 4) = X
3 * X – 12 = X
2 * X = 12
X = 6

Therefore the total bullets before division is = 3 * X = 18

Find sum of digits of D.
Let
A= 19991999
B = sum of digits of A
C = sum of digits of B
D = sum of digits of C
(HINT : A = B = C = D (mod 9))

Answer
The sum of the digits od D is 1.

Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)

Consider, A = 19991999
< 20002000
= 22000 * 10002000
= 1024200 * 106000
< 10800 * 106000
= 106800
i.e. A < 106800
i.e. B < 6800 * 9 = 61200
i.e. C < 5 * 9 = 45
i.e. D < 2 * 9 = 18
i.e. E 1).

On the first day 1 medal and 1/7 of the remaining m – 1 medals were awarded.
On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?

Answer

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6

10. A number of 9 digits has the following properties:

The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
Each digit in the number is different i.e. no digits are repeated.
The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.
Find the number.

Answer
The answer is 381654729

One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.

11. 1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day.
What part of the contents of the container is left at the end of the second day?

Answer

Assume that contents of the container is X

On the first day 1/3rd is evaporated.
(1 – 1/3) of X is remaining i.e. (2/3)X

On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X

Hence 1/6th of the contents of the container is remaining

12. Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.

For how long did Vipul study in candle light?

Answer

Vipul studied for 3 hours in candle light.

Assume that the initial lenght of both the candle was L and Vipul studied for X hours.

In X hours, total thick candle burnt = XL/6
In X hours, total thin candle burnt = XL/4

After X hours, total thick candle remaining = L – XL/6
After X hours, total thin candle remaining = L – XL/4

Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.
(L – XL/6) = 2(L – XL/4)
(6 – X)/6 = (4 – X)/2
(6 – X) = 3*(4 – X)
6 – X = 12 – 3X
2X = 6
X = 3

Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.

13. If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, & so on.
How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?
Answer

Rs.333,062,500

To begin with, you want to know the total number of days: 365 x 50 = 18250.

By experimentation, the following formula can be discovered, & used to determine the amount earned for any particular day: 1 + 2(x-1), with x being the number of the day. Take half of the 18250 days, & pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, & so on, & you will see that if you add these pairs together, they always equal Rs.36500.

Multiply this number by the total number of pairs (9125), & you have the amount you would have earned in 50 years.

Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11…..upto 18250 terms)