Answer: B) 0 Explanation: = log tan10+log tan890 + log tan20+ log tan880+⋯⋯+log tan450     = log [tan10 × tan890] + log [tan20 × tan880 ] +⋯⋯+log1     ∵ tan(90-θ)=cotθ and tan 450=1     = log 1 + log 1 +.....+log 1    = 0.

Answer: C) p > q Explanation: kl>k+1l+1 for (k,l) > 0 and  k > l        Let     k = x+1    and   l = x       Therefore, x+1x>(x+1)+1(x)+1        (x + 1) > x       Therefore, log(x+1)log(x)>log(x+2)log(x+1)       ⇒logxx+1 >logx+1x+2

Answer: D) 45.61 sec Explanation: Given loom weaves 1.14 mts of cloth in one second then 52 mts of cloth can be weaved by loom in, 1.14 ----- 1 52.0 ------?  ⇒521.14= 45.61 sec

Answer: B) 1.2040 Explanation: Given that, log 64 = 1.8061  i.e log43=1.8061  --> 3 log 4 = 1.8061 --> log 4 = 0.6020 --> 2 log 4 = 1.2040 ⇒log42=1.2040 Therefore, log 16 = 1.2040

Answer: B) 2 Explanation: Given a2 + b2 = c2   Now  1logc+ab  + 1logc-ab    = logbc+a + logbc-a   = logbc2-a2 = 2logbb = 2

Answer: B) (1+2m)/2 Explanation: log4928 = 12log77×4   = 12+122log72= 12+log72= 12 + m= 1+2m2.

Answer: C) Rs. 340 Explanation: Assume first child (the youngest) get = Rs. x According to the question ;each son having Rs. 30 more than the younger one Second child will get = Rs. x + 30 Third child will get = Rs. x + 30 + 30 = x + 60 Forth child will get = Rs. x + 30 + 30 + 30 = x + 90 Fifth child will get = Rs. x + 30 + 30 + 30 + 30 = x + 120 Total amount they got = Rs. 2000 x + (x+30) + (x+60) + (x+90) + (x+120) = 20005x + 300 = 20005x = 1700x = Rs. 340So the youngest child will get Rs. 340.

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