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.
(A) 8 , 6
(B) 6 , 4
(C) 6 , 8
(D) 12 , 10
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.
2.
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?
(A) Rs.58
(B) Rs.35
(C) Rs.42
(D) Rs.72
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.
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 ?
(A) 18
(B) 19
(C) 20
(D) 21
Solution:
Total number of trees, = 7+14-1 = 20
4.
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?
(A) One
(B) Two
(C) Three
(D) No Vowel
Solution:
G → F L → K A → B D → C I → J O → P L → K U → V S → R Thus, GLADIOLUS = FKBCJPKVR.
5.
Mohan is older than Prabir. Suresh is younger than Prabir. Mihir is older than Suresh, but younger than Prabir. Who among the four is youngest?
(A) Prabir
(B) Mihir
(C) Mohan
(D) Suresh
Solution:
Mohan is older than Prabir. Prabir < Mohan Suresh is younger than Prabir means Prabir is older than Suresh. Suresh < Prabir < Mohan Mihir is older than Suresh but younger than Prabir. Suresh < Mihir < Prabir < Mohan Suresh is the youngest.
6.
A man is 24 years, older than his son. In two years, his age will be twice the age of his son. The present age of his son is
(A) 14 years
(B) 22 years
(C) 20 years
(D) 18 years
Solution:
Let the present age of his son be X years. So, present age of his father = X+24 (as he is 24 years older than his son) After 2 years, father's age become (X+24+2) years and Son's age will be (X+2). Now, according to question, X+24+2 = 2*(X +2) Or, X = 22 years. Son's present age = 22 years.
7.
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.
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
The ascending order of the number is 123456789. Required difference, 6-3 = 3.
8.
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 ?
(A) B
(B) F
(C) C
(D) D
Solution:
Let, A = x Then, B = 2x and, F = 4x C = x/2 and D = x/4 Thus, The second oldest is B.
9.
Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass. How many cows are needed to eat the grass in 96 days?
(A) 22 Cows
(B) 20 Cows
(C) 18 Cows
(D) 16 Cows
Solution:
Let initially X grass was present there,and it is increasing by Y grass per day, then for the first condition We get,X+24*y = 24*70 ----(1)For the 2nd condition, we have,X+60*Y = 60*30----(2) Now, On solving equation (1) and (2), we get X = 1600 and Y = 10 /3 Third Condition,X+96*Y = 96 *N -----(3) [N = Number of Cows required] Putting the values of X and Y in equation (3), We get N = 20.
10.
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?