Make 24 with 8,8,3,3

Make 24 with 8,8,3,3

24 by the use of 8,8,3,3:-

[(3! * 8) - (8 * 3)]

= (6 * 8) - 24

= 48 - 24

= 24.......

Simple coded math made complex

Supply the digits from 0-9 to decode the given problem.

seven+seven+seven+seven+seven+seven+seve

n+seven+seven+seven+
three+three+three+three+nine+nine=hundred

10 times seven,

4 times three
2 times nine


...SEVEN
...SEVEN
...SEVEN
...SEVEN
...SEVEN
...SEVEN
...SEVEN
...SEVEN
...SEVEN
...SEVEN
..THREE
..THREE
..THREE
..THREE
...NINE
...NINE

............

HUNDRED

..87375
..87375
..87375
..87375
..87375
..87375
..87375
..87375
..87375
..87375
..41977
..41977
..41977
..41977
...5657
...5657
----------
1052972

Here
H=1,
D=2
V=3
T=4
N=5
I=6
E=7
S=8
R=9
U=0


Pls check if correct.
I=6

Cricket Puzzle

Some boys are playing cricket. Wasim is at the striker' s end of the pitch and
Rachit is at the non-striker' s end. Wasim hits the ball far into the outfield and the
batsmen start running. Each of them runs at a constant speed, though their speeds are
not necessarily equal. They first cross each other at a distance of 9 ft from the striker' s
end while running the first run. They immediately turn back for the second run and cross each other at a distance of 3 ft from the non-striker' s end while running for the
second run. After this, they turn back for the third run and so on. What is the total
distance (In feet) Rachit has run when they cross each other for the third time?

Let Wasim and Rachit are running at a speed of 'S1','S2' repectively.
Assume that the length of the pitch is 'x'

When they crossed each other for 1st time,
Wasim covered a distance of 9 feet and
Rachit covered (x-9) feet.
Since both are running at their own individual constant speeds, the distance covered by them is proportional to their speed

S2/S1 = (x-9)/9 --- (1)

When they crossed each other for 2nd time,
Wasim covered a distance of (x+3) feet and
Rachit covered (2x-3) feet.

S2/S1 = (2x-3)/(x+3) --- (2)

Solving both the equation we get x = 24 feet.

From the explanation given in the above post I am taking 'y' as the distance from the non striker end where Wasim and Rachit crossed fo the 3 rd time.
Since wasim is still trying to complete his 2nd run he covered a distance of (x+y) feet
Rachit already completed his 2nd run and going to complete his 3rd run he covered a distace of (2x+y) feet

S2/S1 = (2x+y)/(x+y) ..... (3)

Solving the equations and substituting the value of 'x' we get 60 feet as the answer.

Insurance company

Insurance company

A man would like to take a new health insurance. An officer taking care of these matters says to the man: "Please tell me how many children you have." The man answers: "I have three of them." The officer: "What are the ages of your children?". The man answers: "The product of the ages is equal to 36." The officer replies: "This is not enough information Sir!". "Sorry that I was a little bit unclear, but the sum of the ages is equal to the number of shops in front of your office," says the man. The officer: "This still isn't enough information Sir!". The man replies: "My oldest child loves chocolate." The officer: "Thanks for your cooperation, I now know the ages."
Are you as smart as the officer? Then give the ages of the children.




The product of the ages is 36. Using this one can make the following combination of ages:

1,36, 1 sum = 38
1,18, 2 sum = 22
1,12, 3 sum = 16
1, 9, 4 sum = 14
1, 6, 6 sum = 13
2, 9, 2 sum = 13
2, 6, 3 sum = 11
3, 3, 4 sum = 10

After the man had said that the product of the ages is equal to 36, the officer didn't have enough information. Then he was told that the sum is equal to number of shops in front of the office. He replied by saying that this still isn't enough information. So the sum of the ages should be 13, because otherwise he would have known the ages immediately. The last statement is that that the oldest child loves chocolate. So there is an oldest child. Hence the officer concludes that the ages of the children are 2, 2 and 9 years.

Fake coins

Fake coins

Suppose you have five bags filled with the same type of coins and a weigthing machine. One of this bags is completely filled with fake coins, the other bags contain only real ones. A real coin weights 10 gram and a fake one 11 gram.
What is the minimal number of times you need to use the weighting machine in order to find the bag with the fake coins?




Take 1 coin from the first bag, 2 from the second, 3 out of the third, 4 from the fourth and 5 coins from the fifth bag. Now weigh all these coins togeheter at once. If the fake coins are in the first bag, the result of the weighting will be 11 + 20 + 30 + 40 + 50 = 151 gram. Likewise, if the fake coins are in respectively the second, the third, the fourth and the fifth bag, the result of the weighting will be 152, 153, 154 and 155 gram. Hence in this way it is possible to find the bag with the fake coins with weighting only once.

Whats the next number?

Whats the next number?

17, 29, 101, 1541, _____ ?



(17*2)-5*1=29
(29*4)-5*3=101
(101*16)-15*5=1541


(1541*256)-75*5=393971

instead of 75*5 it will b 75*7...final ans right

the logic is 5*1 gets carry forward in the next line and now 5*3=15...nw 15*5=75....den 75*7=525...den 525*9=4725


The 6th term will be ====25819278731

Children.

A number of children are standing in a circle. They are evenly spaced and the 4th child is directly opposite the 18th child. How many children are there altogether?


28