Puzzle

1.In a lake, there are 10 steps labelled using alphabets from A to J. Starting from step A, every minute a frog jumps to the 4th step from where it started - that is from the step A it would go to the step E and from E it would go to the step I and from I it would go to C etc. Where would the frog be at the 60th minute if it starts at the step A ?

**Answer:**Option** 1**

Solution:

The steps are labelled using alphabets from A to J.TO MAKE IT EASY, WE HAVE ASSIGNED NUMBERS TO THE STEPS AS FOLLOWS:

Labelled with alphabets : A B C D E F G H I J

Labelled with numbers : 1 2 3 4 5 6 7 8 9 10

The frog takes total 60 minutes and takes 4 step length jumps every time.

Thus,

1st minute : 1 + 4 = 5th step (E)

2nd minute : 5 + 4 = 9th step (I)

3rd minute : 9 + 4 = 3rd step (C)

4th minute : 3 + 4 = 7th step (G)

5th minute : 7 + 4 = 11 = 10+1 = 1st step (A)

The same process is repeated 12 x 5 times.

Then, the jumping positions are

1 5 9 3 7

1 5 9 3 7

1 5 9 3 7 and so on.

After 15 cycles, frog will be in the 1st position i.e., at 5th minute, 10th minute,

15minutes....60th minute frog will be in the 1st position.

i.e., at 60t

2.In a family there are several brothers and sisters. Every 2 boys have brothers as many as sisters and each girl has 2 brothers less than twice as many brothers as sisters. Now find the number of boys and girls.

**Answer:**Option** 1**

Solution:

Let B be the number of brothers and S be the number of sisters in the family.Consider any two boys. They would be having (B - 2) brothers (excluding the two). But this number is equal to the number of sisters they have.

Therefore,

B - 2 = S

or , B - S = 2 ............(1) Each girl will have (S - 1) sisters. Twice the number of sisters = 2(S - 1).

Since, each girl has twice as many brothers as sisters, we have, 2(S-1)-2 = B

2S - 4 = B ........... (2)

Substituting, eqn (2) in Eqn (1), we get

2S - 4 - S = 2

S = 6

On substituting S = 6 in eqn (1) , we get

B - 6 = 2

B = 8.

3.In a row of trees, a tree is 7^{th} from left end and 14^{th} from right end. How many tree are there in the row ?

**Answer:**Option** 1**

Solution:

Total number of trees,= 7+14-1

= 20

4.Ramesh ranks 13^{th} in the class of 33 students. There are 5 students below Suresh rankwise. How many student are there between Ramesh and Suresh ?

**Answer:**Option** 1**

Solution:

1 | 2 | 3 | 4 | 5 |

6 | 7 | 8 | 9 | 10 |

11 | 12 | Ramesh | 14 | ... |

... | ... | ... | ... | ... |

... | ... | ... | ... | Suresh |

5 | 4 | 3 | 2 | 1 |

Number of student in between Ramesh and Suresh is,

= 33 - (13+6)

=

5.Shan is 55 years old, Sthian is 5 years junior to Shan and 6 years senior to Balan. The youngest brother of Balan is Devan and he is & years junior to him. So what is the age difference between Devan and Shan ?

**Answer:**Option** 1**

Solution:

Age of Shan = 55 years Age of Sathian = 55-5 = 50 years

Age of Balan = 50-6 = 44 years

Age of Deven = 44-7 = 37 years

Thus, difference between Devan and Shan,

= 55 -37

= 18.

6.Rohit was walking on the street, one boy requested him to donate for cancer patients welfare fund. He gave him a rupee more than half the money he had. He walked a few more steps. Then came a girl who requested him to donate for poor people's fund for which he gave two rupees more than half the money he had then. After that, again a boy approached him for an orphanage fund. He gave three rupees more than half of what he had. At last he had just one rupee remaining in his hand.
How much amount did Ram have in his pocket when he started?

**Answer:**Option** 1**

Solution:

Let X be the rupees he initially had.He gave for the cancer fund one rupee more than half of what he had.

i.e.,[1 +(X/2)].

Remaining money = X-(1+X/2) = [(X/2) - 1.

he gave for poor people's, rupee 2 more than half what he remain with,

= [2+{1/2*(X/2-1)}]

= [2+{(X-2)/4}]

= (6+X)/4

Now, remaining money = ((X/2)-1) - ((6+X)/4)

= (X-10)/4.

Again he gave 3 rupees more than half of what he had for orphanage,

[3+(1/2*((X-10)/4))]

= 3+[(X-10)/8]

= (14+X)/8

now left money,[{(X-10)/4]-[(14+X)/8]}

= [(2X-X-20-14)/8]

= (X-34)/8

As given, finally he had one rupee remaining so (X-34)/8 = 1

So,

X-34 = 8

X = 8+34 = 42

Hence, Rohit had Rs. 42 initially in his pocket.

7.If Football is called Cricket, Cricket is called Basketball, Basketball is called Badminton, Badminton is called Volleyball, Volleyball is called Hockey and Hockey is called Golf, then which of the following games is not played using a ball?

**Answer:**Option** 1**

Solution:

Badminton is played without any ball and in the question Badminton is called Volleyball.
8.Each vowel of the GLADIOLUS word is substituted with the next letter of the English alphabetical series and each consonant is substituted with the letters preceding it. How many vowels will be present in the new arrangement?

**Answer:**Option** 1**

Solution:

G → F L → K

A → B

D → C

I → J

O → P

L → K

U → V

S → R

Thus,

GLADIOLUS = FKBCJPKVR.

9.How many such pairs of letters are there in the word SENDING, each of which has as many letters between its two letters as there are between them in the English alphabets?

**Answer:**Option** 1**

Solution:

Only Three such pairs,I

D

S

10.A lad was asked his age by his friend . The lad said, "The number you get when you subtract 25 times my age from twice the square of my age will be thrice your age". If the friend's age is 14, then the age of the lad is :

**Answer:**Option** 1**

Solution:

Considering option 3,

2 *(14)

Let the age of the lad be x then,

2x

2x

Either solve the quadratic equation or put the options to get the value. Option 3 is correct.