Answer: B) 3/4 hrs Explanation: Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time   so 1/3rd of the usual time = 15minor usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.

Answer: A) 3:2 Explanation: Concentration of glucose are in the ratio = 12:35:45 Quantity of glucose taken from A = 1 liter out of 2 Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit Quantity of glucose taken from C = 0.8 lit So, total quantity of glucose taken from A,B and C = 3.6 lit So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit Ratio of glucose to alcohol = 3.6/2.4 = 3:2

Answer: B) 31 Explanation: Given  65×359×113 ⇒2×35×5×79×113  ⇒25×35×59×79×113 Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.

Answer: C) Rs. 115 Explanation: Amount to be paid = Rs.100+200×5×1100+100×5×1100= Rs. 115.

Answer: A) 10 days Explanation: After 10 days, the remaining food would be sufficient for the 1000 students for 20 more days -->If 1000 more students are added, it shall be sufficient for only 10 days (as the no. of students is doubled, the days are halved).

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: D) Rs. 616 Explanation: Let the average expenditure per head be Rs. pNow, the expenditure of the mess for old students is Rs. 44pAfter joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33Therefore, 59(p-3) = 44p + 3359p - 177 = 44p + 3315p = 210 => p = 14Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.