Answer: A) 60 Explanation: Let, no. of rice bowl = xno. of dal bowl = yno. of meat bowl = z Then, x + y + z = 65 ....(1) Also given that, 2x = 3y = 4z substituting value of y & z in terms of x i.e. y = 2x/3 and z = x/2 in eq (1),We get x = 30. As 1 rice bowl is shared between 2 guests, Therefore, there are 60 guests.

Answer: B) 14 Explanation: 1 2 3 4 5 6(Shankar) 7(Nitu) 8(Althaf) 9 10 11 12 13 14Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between themSo minimum no. of boys standing in the queue = 14

Answer: B) Rs. 830 Explanation: Let the total amount in the partnership be 'x'.Then A's share = x/3B's share = x/4C's share = x/5D's share = x - (x/3 + x/4 +x/5) = 13x/60 A : B : C : D = x/3 : x/4 : x/5 : 13x/60 = 20 : 15 : 12 : 13 A's share in the profit of Rs. 2490 = 20 (2490/60) = Rs. 830.

Answer: A) 9!/(2!)^{2}x3! Explanation: n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.The letter pattern 'MESMERISE' consists of 10 letters of which there are 2M's, 3E's, 2S's and 1I and 1R.Number of arrangements = 9!(2!)2×3!

Answer: A) Rs. 620 Explanation: Let the C.P of one item is Rs. P and that of other is Rs. (7500 - P) According to the data given C.P = S.P => Px(116/100) + (7500-P)x(86/100) = 7500 => 30P = 105000 => P = 3500 Required difference between selling prices = Rs. [(3500/100) x 116] - [(4000/100) x 86] = 4060-3440 = Rs. 620

Answer: A) 2893 Explanation: For solving this problem first we would break the whole range in 5 sections 1) From 1 to 9Total number of zero in this range = 0 2) From 10 to 99Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer) 3) From 100 to 999 - three type of numbers are there in this range a) x00 b) x0x c) xx0 (here x represents a non zero number)Total possibilities for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18for x0x = 9*1*9 = 81, hence total zeros = 81similarly for xx0 = 81 total zeros in three digit numbers = 18 + 81 +81 = 180 4) From 1000 to 9999 - seven type of numbers are there in this rangea)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xxTotal possibilities for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27 for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162 for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162 for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)= 2893

Answer: B) 40 yrs Explanation: Let the present ages of Ajay and Vijay be 'A' and 'V' years respectively. => A - 8 = 4/3 (V - 8) and A + 8 = 6/5 (V + 8) => 3/4(A - 8) = V - 8 and 5/6(A + 8) = V + 8 V = 3/4 (A - 8) + 8 = 5/6 (A + 8) - 8 => 3/4 A - 6 + 8 = 5/6 A + 20/3 - 8 => 10 - 20/3 = 10/12 A - 9/12 A => 10/3 = A/12 => A = 40.