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: B) 3/2 Explanation:  = log 33 + log 23+ log 103log10×3×22        =log33 12+log 23+log 10312log(10×3×22)                 =12log 33+3 log 2+12 log103log10+log3+log22                 =32log 3 + 2 log 2 + log 10log 3 + 2 log 2 + log 10 = 32

Answer: C) 108 m Explanation: We know the rule that,   At particular time for all object , ratio of height and shadow are same.   Let the height of the pole be 'H'   Then  188=H48   => H = 108 m.

Answer: A) -1/2 Explanation: Rewrite equation as 122x+1 = 120   Leads to 2x + 1 = 0    Solve for x : x = -1/2

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) 13 Explanation:  Let   Let x=333 = 333    Then, logx = 33 log3     = 27 x 0.47712 = 12.88224    Since the characteristic in the resultant value of log x is 12   ∴The number of digits in x is (12 + 1) = 13    Hence the required number of digits in 333is 13.