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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 ?

Discuss
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 positioni.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.

Discuss
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 7th from left end and 14th from right end. How many tree are there in the row ?

Discuss
Solution:
Total number of trees,
= 7+14-1
= 20
4. B is twice as old as A but twice younger than F. C is half the age of A but is twice older than D. Who is the second oldest ?

Discuss
Solution:
Let, A = x
Then, B = 2x
and, F = 4x
C = x/2
and D = x/4
Thus, The second oldest is B.
5. Ramesh ranks 13th in the class of 33 students. There are 5 students below Suresh rankwise. How many student are there between Ramesh and Suresh ?

Discuss
Solution:
1 2 3 4 5
6 7 8 9 10
11 12 Ramesh 14 ...
... ... ... ... ...
... ... ... ... Suresh
5 4 3 2 1
Total Number of Student = 33
Number of student in between Ramesh and Suresh is,
= 33 - (13+6)
= 14.
6. 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 7 years junior to him. So what is the age difference between Devan and Shan ?

Discuss
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.
7. 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?

Discuss
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.
8. 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?

Discuss
Solution:
Badminton is played without any ball and in the question Badminton is called Volleyball.
9. If each of the digits in the number 92581473 is arranged in ascending order, what will be the difference between the digits which are fourth from the right and third from left in the new arrangement? The ascending order of the number is 123456789.

Discuss
Solution:

The ascending order of the number is 123456789.
Required difference,
6-3 = 3.

10. 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?

Discuss
Solution:
G → F
L → K
A → B
D → C
I → J
O → P
L → K
U → V
S → R
Thus,
GLADIOLUS = FKBCJPKVR.

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