1. A man is going to an Antique Car auction. All purchases must be paid for in cash. He goes to the bank and draws out $25,000.

Since the man does not want to be seen carrying that much money, he places it in 15 evelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (i.e. no two tens in place of a twenty).

At the auction he makes a successful bid of $8322 for a car. He hands the auctioneer envelopes number(s) 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount.

How many ones did the auctioneer find in the envelopes?

Answer :

Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of bills possible of any available US currency" is only to misguide you. This is always possible for any amount !!!

One more thing to notice here is that the man must have placed money in envelopes in such a way that if he bids for any amount less than $25000, he should be able to pick them in terms of envelopes.

First envelope contains, 2^0 = $1
Second envelope contains, 2^1 = $2
Third envelope contains, 2^2 = $4
Fourth envelope contains, 2^3 = $8 and so on...

Hence the amount in envelopes are $1, $2, $4, $8, $16, $32, $64, $128, $256, $512, $1024, $2048, $4096, $8192, $8617

Last envelope (No. 15) contains only $8617 as total amount is only $25000.

Now as he bids for $8322 and gives envelope number 2, 8 and 14 which contains $2, $128 and $8192 respectively.

Envelope No 2 conrains one $2 bill
Envelope No 8 conrains one $100 bill, one $20 bill, one $5 bill, one $2 bill and one $1 bill
Envelope No 14 conrains eighty-one $100 bill, one $50 bill, four $10 bill and one $2 bill

2. Decipher this sentence.

B R W Q H L F K W H J K Q I B W K

Q I C E D W Z B G W K K M I K E

Z B G Q H S K Z B G J K Z K W

B U U Z B G J D B H F W

Answer : 

Start with ZBG and ZBGJ. It should be either "the/then" or "you/your" combination as they appear more.

B R W Q H L F K W H J K Q I B W K
o b s t a c l e s a r e t h o s e

Q I C E D W Z B G W K K M I K E
t h i n g s y o u s e e w h e n

Z B G Q H S K Z B G J K Z K W
y o u t a k e y o u r e y e s

B U U Z B G J D B H F W
o f f y o u r g o a l s

3. Because cigars cannot be entirely smoked, a Bobo who collects cigar butts can make a cigar to smoke out of every 3 butts that he finds. Today, he has collected 27 cigar butts. How many cigars will he be able to smoke?

Answer :     13 not 12

He makes 9 originals from the 27 butts he found, and after he smokes them he has 9 butts left for another 3 cigars. And then he has 3 butts for another cigar.

So 9+3+1=13

4. In the General meeting of "Friends Club", Sameer said, "The repairs to the Club will come to a total of Rs 3120 and I propose that this amount should be met by the members, each paying an equal amount."

The proposal was immediately agreed. However, four members of the Club chose to resign, leaving the remaining members to pay an extra Rs 26 each. How many members did the Club originally have?

Answer :

The Club originally had 24 members.

Assume that there were initially N members.

As 4 members resigned and remaining members paid Rs 26 each, it means that total amount of 4 members is equal to Rs 26 each from remaining (N-4) members. Thus,

4 * (3120 / N) = 26 * (N - 4)
12480 = 26N2 - 104N
26N2 - 104N - 12480 = 0

Solving the quadratic equation we get N=24.

Hence, the Club originally had 24 members.

5. 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

6. There is a number that is 5 times the sum of its digits. What is this number? Answer is not 0.

Answer :

The number is 45, simply because
45 = 5 * (4 + 5)
How does one find this number?

Let T be the digit in the tens place and U be the digit in the units place. Then, the number is 10*T + U, and the sum of its digits is T + U.

The following equation can be readily written:
10*T + U = 5*(T + U) or
10*T + U = 5*T + 5*U or
5*T = 4*U

Thus, T / U = 4 / 5

Since T and U are digits, T must be 4 and U must be 5.

7. There is a family party consisting of two fathers, two mothers, two sons, one father-in-law, one mother-in-law, one daughter-in-law, one grandfather, one grandmother and one grandson. What is the minimum number of persons required so that this is possible?

Answer :

There are total 2 couples and a son. Grandfather and Grand mother, their son and his wife and again their son. So total 5 people.

Grandfather, Grandmother
|
|
Son, wife
|
|
Son
 

8. He paid five rupees per day to each man, four ruppes to each woman and one rupee to each child. He gave out 200 rupees every evening. How many men, women and children were working with the constructor?

Answer :

16 men, 12 women and 72 children were working with the constructor.

Let's assume that there were X men, Y women and Z children working with the constructor. Hence,

X + Y + Z = 100
5X + 4Y + Z = 200

Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z

As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get

if Y = X,
300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428

if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727

But Z must be an integer, hence Z=72. Also, X=16 and Y=12

There were 16 men, 12 women and 72 children working with the constructor.

9. In training for a competition, you find that swimming downstream (with the current) in a river, you can swim 2 miles in 40 minutes, & upstream (against the current), you can swim 2 miles in 60 minutes. How long would it take you to swim a mile in still water?

Answer :

You are able to swim downstream at 3 miles an hour, & upstream at 2 miles an hour. There is a difference of 1 mile an hour, which is the river helping you in 1 direction, & slowing you in the other direction.

Average the 2 rates, & you have the rate that you can swim in still water, which is 2.5 miles an hour.

You can thus swim a mile in still water in 24 minutes.

10. There is a shortage of tube lights , bulbs and fans in a village - Kharghar. It is found that

All houses do not have either tubelight or bulb or fan.

exactly 19% of houses do not have just one of these.

atleast 67% of houses do not have tubelights.

atleast 83% of houses do not have bulbs.

atleast 73% of houses do not have fans.

What percentage of houses do not have tubelight, bulb and fan?

Answer :

42% houses do not have tubelight, bulb and fan.

Let's assume that there are 100 houses. Hence, there should be total 300 items i.e. 100 tubelights, 100 bulbs and 100 fans.

From the given data, we know that there is shortage of atleast (67+83+73) 223 items in every 100 houses.

Also, exactly 19 houses do not have just one item. It means that remaining 81 houses should account for the shortage of remaining (223-19) 204 items. If those remaining 81 houses do not have 2 items each, there would be a shortage of 162 items. But total of 204 items are short. Hence, atleast (204-162) 42 houses do not have all 3 items - tubelight, bulb and fan.

Thus, 42% houses do not have tubelight, bulb and fan.