1. In a certain year, the number of girls who graduated from City High School was twice the number of boys. If 3/4 of the girls and 5/6 of the boys went to college immediately after graduation, what fraction of the graduates that year went to college immediately after graduation?

Answer :   7/9

Assume that number of boys graduated from City High School = B
Therefore, number of girls graduated from City High School = 2*B

It is given that 3/4 of the girls and 5/6 of the boys went to college immediately after graduation.
Hence, total students went to college
= (3/4)(2*B) + (5/6)(B)
= B * (3/2 + 5/6)
= (7/3)B

Fraction of the graduates that year went to college immediately after graduation
= [(7/3)B] / [3*B]
= 7/9

Therefore, the answer is 7/9

2. A 3 digit number is such that it's unit digit is equal to the product of the other two digits which are prime. Also, the difference between it's reverse and itself is 396. What is the sum of the three digits?

Answer :

The required number is 236 and the sum is 11.

It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers - 224, 236, 326 and 339.

Now, it is also given that - the difference between it's reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.

3. One of the four people - Mr. Clinton, his wife Monika, their son Mandy and their daughter Cindy - is a singer and another is a dancer. Mr. Clinton is older than his wife and Mady is older than his sister.

If the singer and the dancer are the same sex, then the dancer is older than the singer.

If neither the singer nor the dancer is the parent of the other, then the singer is older than the dancer.

If the singer is a man, then the singer and the dancer are the same age.

If the singer and the dancer are of opposite sex then the man is older than the woman.

If the dancer is a woman, then the dancer is older than the singer.

Whose occupation do you know? And what is his/her occupation?

Answer :

Cindy is the Singer. Mr. Clinton or Monika is the Dancer.

From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman.

CASE I : Singer is a woman and Dancer is also a woman
Then, the dancer is Monika and the singer is Cindy.

CASE II : Singer is a woman and Dancer is also a man
Then, the dancer is Mr. Clinton and the singer is Cindy.

In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.

4. 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

5. There is a 4-character code, with 2 of them being letters and the other 2 being numbers. How many maximum attempts would be necessary to find the correct code? Note that the code is case-sensitive.

Answer :    The maximum number of attempts required are 16,22,400

There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 4C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).

Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@

Hence, the required answer is
= 52*52*10*10*6
= 16,22,400 attempts
= 1.6 million approx.

6. Two people enter a race in whick you run to a point and back. Person A runs 20 mph to and from the point. Person B runs to the point going 10 mph and 30 mph going back. Who came in first?

Answer :   Person A came in first.

Let's assume that the distance between start and the point is D miles.

Total time taken by Person A to finish
= (D/20) + (D/20)
= D/10
= 0.1D

Total time taken by Person B to finish
= (D/10) + (D/30)
= 2D/15
= 0.1333D

Thus, Person A is the Winner.

7. A, B, C and D are related to each other.

One of the four is the opposite sex from each of the other three.

D is A's brother or only daughter.

A or B is C's only son.

B or C is D's sister.

Answer :

A, B & D are males; C is female. B is C's only son. A & D are C's brothers.

A(male) --- C(female) --- D(male)

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B(male)

Work out which relation can hold and discard the contradictory options.

From (2) and (4), D can not be a only daughter and have a sister (B or C). Hence, D is A's brother i.e. D is a Male.

From (4), let's say that B is D's sister i.e. B is Female.
From (3), A is C's only son i.e. A is Male.
But D is A's brother which means that A is not C's only son. Hence, our assumption was wrong.

Thus, C is D's sister i.e. C is Female. And B must be C's only son.

Now it is clear that D & B are Males and C is Female. A must be a Male as only one of them is of opposite sex from each of the other three. And he is C & D's brother.How are they related to each other?

8. Two identical pack of cards A and B are shuffled throughly. One card is picked from A and shuffled with B. The top card from pack A is turned up. If this is the Queen of Hearts, what are the chances that the top card in B will be the King of Hearts?

Answer :    52 / 2703

There are two cases to be considered.

CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B

Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 2/53

So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)

CASE 2 : King of Hearts is not drawn from Pack A

Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 1/53

So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)

Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378

9. In a sports contest there were m medals awarded on n successive days (n > 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. Scientist decided to do a study on the population growth of rabbits. Inside a controlled environment, 1000 rabbits were placed.

Six months later, there were 1000Z rabbits. At the beginning of the 3rd year, there were roughly 2828Z rabbits, which was 4 times what the scientists placed in there at the beginning of the 1st year.

If Z is a positive variable, how many rabbits would be there at the beginning of the 11th year?

Answer :

At the beginning of the 11th year, there would be 1,024,000 rabbits.

At the beginning, there were 1000 rabbits. Also, there were 4000 rabbits at the beginning of third year which is equal to 2828Z. Thus, Z = 4000/2828 i.e. 1.414 (the square root of 2)

Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further simplified as 1000*Z*Z*Z*Z

Also, it is given that at the end of 6 months, there were 1000Z rabbits.

It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(Z^(2N)) i.e. 1000*(2^N)

Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 rabbits.