Answer: A) 20% p.a Explanation: Rs.1440 - 1200 = Rs.240 is the interest on Rs.1200 for one year. Rate of interest = (100 x 240) / (1200) = 20% p.a

Answer: D) Data is not sufficient Explanation: Let the sum be Rs. p, rate be R% p.a. and time be T years. Then,  P×T×(R+2)100-P×T×R100=108  ⇒2PT= 10800  ...... (1)  And,   P×R×(T+2)100-P×R×T100=180   ⇒2PR = 18000 ......(2) Clearly, from (1) and (2), we cannot find the value of pSo, the data is not sufficient.

Answer: C) 81 Explanation: 729 = 9 x 9 x 9 => Cube root of 729 = 9   Now, required square of 9 = 9 x 9 = 81.

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: 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: C) 2029 Explanation: How to find the years which have the same Calendars :   Leap year calendar repeats every 28 years. Here 28 is distributed as 6 + 11 + 11.   Rules: a) If given year is at 1st position after Leap year then next repeated calendar year is Givenyear+6. b) If given year is at 2nd position after Leap year then next repeated calendar year is Givenyear+11. c) If given year is at 3rd position after Leap year then next repeated calendar year is Givenyear+11.   Now, the given year is 2018 We know that 2016 is a Leap year. 2016    2017    2018    2019  2020 Lp Y      1st     2nd      3rd     Lp Y   Here 2018 is at 2 nd position after the Leap year. According to rule b) the calendar of 2018 is repeated for the year is 2018 + 11 = 2029.

Answer: A) 22/7 x 24/35 Explanation: Since level of water increased in cylinder by height 4.  This is because of the rectangular block .   Therefore , area of rectangular block immersed in water is  = 227×6×6×4   Thats why portion of block immersed in water is  = (22/7 * 36 * 4) / total vol. of rectangle  = (22/7 * 36 * 4)/ (7*5*6)  = (22/7 * 24)/ 35.