Answer: B) 4 Explanation: 4 eyes. Here in the question, it is asked how many Eyes We have so that means here the person who has asked the question is also including the person who is suppose to give the answer. In a clear understanding, the Conversation is happening between 2 people 1st who asked the question and 2nd to whom it has been asked, which means there are 4 eyes.

Answer: D) 70 Explanation: Here the given series is 1, 3, 6, 11, 20, 37, ??, and the logic behind the series is 1 x 2 + 1 = 3 3 x 2 + 0 = 6 6 x 2 + (-1) = 11 11 x 2 + (-2) = 20 20 x 2 + (-3) = 37 37 x 2 + (-4) = 70   Hence, the next number in the given series is 70.

Answer: D) April, July Explanation: If the period between the two months is divisible by 7, then that two months will have the same calender . Now,(a). October + November = 31 + 30 = 61 (not divisible by 7) (b). Apr. + May + June + July + Aug. + Sep. + Oct. = 30 + 31 + 30 + 31 + 31 +30 + 31 = 213 (not divisible by 7) (c). June + July + Aug. + Sep. = 30 + 31 + 31 + 30 = 122 (not divisible by 7) (d). Apr. + May + June = 30 + 31 + 30 = 91 (divisible by 7)   Hence, April and July months will have the same calendar.

Answer: B) 217.50 Explanation: Given expression = (45 x 750/100) - (25 x 480/100) = (337.50 - 120) = 217.50

Answer: D) Data is not sufficient Explanation: From the given data,how many goals scored = 5397....(1)many more matches = 982.....(2)he scored five = 163.....(3)from (1)&(2), many = 9 ....(4)from (1)&(3), scored = 3 ....(5)Using (4)&(5)in(1), we get goals = 5 or 7.

Answer: C) 1 time Explanation: How many times can you subtract 10 from 100, for this if we go through logically, it is only 1 time. For the first time if we subtract 10 from 100, then there will no 100 to subtract from.   Hence, it is only 1 time we can subtract 10 from 100.

Answer: D) Can't be determined Explanation: We cannot find out the answer because the number of days of the current month is not given.

Answer: D) 5/21 Explanation: Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ x ⁶C₁) As remaining person can be any one among three couples left. Required probability = (⁵C₂ x ⁶C₁)/¹⁰C₅= (10 x 6)/252 = 5/21