Brain Teaser No : 00022
There were N stations on a railroad. After adding X stations 46 additional tickets have to be printed.
Find N and X.
Answer
Let before adding X stations, total number of tickets
t = N(N-1)
After adding X stations total number of tickets are
t + 46 = (N+X)(N+X-1)
Subtracting 1st from 2nd
46 = (N+X)(N+X-1) - N(N-1)
46 = N2 + NX - N + NX + X2 - X - N2 + N
46 = 2NX + X2 - X
46 = (2N - 1)X + X2
X2 + (2N - 1)X - 46 = 0
Now there are only two possible factors of 46. They are (46,1) and (23,2)
Case I: (46,1)
2N - 1 = 45
2N = 46
N = 23
And X = 1
Case II: (23,2)
2N - 1 = 21
2N = 22
N = 11
And X = 2
Hence, there are 2 possible answers.
Brain Teaser No : 00023
There is a grid of 20 squares by 10 squares. How many different rectangles are possible?
Note that square is a rectangle.
Answer
11550
The Generic solution to this is:
Total number of rectangles = (Summation of row numbers) * (Summation of column numbers)
Here there are 20 rows and 10 columns or vice versa. Hence, total possible rectangles
= ( 20 + 19 + 18 + 17 + 16 + .... + 3 + 2 + 1 ) * ( 10 + 9 +8 + 7 + .... + 3 + 2 + 1)
= ( 210 ) * (55)
= 11550
Hence, total 11,550 different rectangles are possible.
If you don't believe it, try formula on some smaller grids like 4x2, 3x2, 3x3 etc...
Brain Teaser No : 00024
A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X
paise. Neither the person nor the cashier noticed that.
After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.
Find X and Y. (1 Rupee = 100 Paise)
Answer
As given, the person wanted to withdraw 100X + Y paise.
But he got 100Y + X paise.
After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is
2 * (100X + Y) = 100Y + X - 20
200X + 2Y = 100Y +X - 20
199X - 98Y = -20
98Y - 199X = 20
Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we
interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1
Case I : Y=2X
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 0
We get X = - 20/3 & Y = - 40/2
Case II : Y=2X+1
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 1
We get X = 26 & Y = 53
Now, its obvious that he wanted to withdraw Rs. 26.53
Brain Teaser No : 00025
What is the remainder left after dividing 1! + 2! + 3! + … + 100! By 7?
Think carefully !!!
Answer
A tricky one.
7! onwards all terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms i.e. remainder left
after dividing 7! + 8! + 9! + ... + 100! is 0.
The only part to be consider is
= 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873
The remainder left after dividing 873 by 7 is 5
Hence, the remainder is 5.
Brain Teaser No : 00026
Find the last digit of summation of the series:
199 + 299 + 399 + 499 + ……… + 9899 + 9999
Answer
The last digit of the series is 0.
We group the sum as follow:
(199 + 1199 + ... + 9199) + (299 + 2299 + ... 9299) + ...... + (999 + 1999 + ... + 9999) + (1099 + 2099 + 3099 + ... 9099)
All the terms in a single group have the same last digit (i.e. last digits of 199 + 1199 + ... + 9199 are same, is 1, & similarly for the
other groups).
Also, there are 10 terms in each group except for the last one. Therefore the last digit of the sum of terms in first 9 groups is 0.
(as whatever be the last digit, we have to multiply it by 10) And the last digit of the sum of the terms in the group 10 is obviously
0.
Hence, the last digit of the series is 0.
Brain Teaser No : 00027
Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger
than the original number.
Answer
The answer is 285714.
If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714.
The simplest way is to write a small program. And the other way is trial and error !!!
Brain Teaser No : 00028
There are 10 boxes containing 10 balls each. 9 boxes contain 10 balls of 10 kg each and one box contains 10 balls of 9 kg each.
Tool is available for proper weighing. How can you find out the box containing 9 kg balls?
You are allowed to weigh only once. You can remove balls from the boxes. All balls are of same size and color.
Answer
1. Mark the boxes with numbers 1, 2, 3, 4, ... upto 10
2. Take 1 ball from box 1, take 2 balls from box 2, take 3 balls from box 3, take 4 balls from box 4 and so on
3. Put all of them on the scale at once and take the measurement.
4. Now, subtract the measurement from 550 ( 1*10 + 2*10 + 3*10 + 4*10 + 5*10 + 6*10 + 7*10 + 8*10 + 9*10 + 10*10)
5. The result will give you the box number which has a ball of 9 Kg
Brain Teaser No : 00029
A fly is flying between two trains, each travelling towards each other on the same track at 60 km/h. The fly reaches one engine,
reverses itself immediately, and flies back to the other engine, repeating the process each time.
The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?
Answer
Initially, the trains were 240 km apart.
The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started.
It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of
60 km/h. So, each train traveled 120 km before they met.
Hence, the trains were 240 km apart initially.
Brain Teaser No : 00030
A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he
need?
Answer
The sign-maker will need 192 zeroes.
Divide 1000 building numbers into groups of 100 each as follow:
(1..100), (101..200), (201..300), ....... (901..1000)
For the first group, sign-maker will need 11 zeroes.
For group numbers 2 to 9, he will require 20 zeroes each.
And for group number 10, he will require 21 zeroes.
The total numbers of zeroes required are
= 11 + 8*20 + 21
= 11 + 160 + 21
= 192