Answer: A) 50 days Explanation: 50 men can build a tank in 40 days  Assume 1 man does 1 unit of work in 1 day Then the total work is 50×40 = 2000 units   50 men work in the first 10 days and completes 50×10 = 500 units of work 45 men work in the next 10 days and completes 45×10 = 450 units of work 40 men work in the next 10 days and completes 40×10 = 400 units of work 35 men work in the next 10 days and completes 35×10 = 350 units of work So far 500 + 450 + 400 + 350 = 1700 units of work is completed and   Remaining work is 2000 - 1700 = 300 units   30 men work in the next 10 days. In each day, they does 30 units of work.   Therefore, additional days required = 300/30 =10   Thus, total 10+10+10+10+10 = 50 days required.

Answer: C) 13 days Explanation: Ratio of times taken by A and B = 100 : 130 = 10 : 13.Suppose B takes x days to do the work. Then, 10 : 13 :: 23 : x     =>     x = ( 23 x 13/10 )     =>     x = 299 /10. A's 1 day's work = 1/23 ;B's 1 day's work = 10/299 . (A + B)'s 1 day's work = ( 1/23 + 10/299 ) = 23/299 = 113 . Therefore, A and B together can complete the work in 13 days.

Answer: C) 12.5 cm Explanation: Let the required height of the building be x meter More shadow length, More height(direct proportion) Hence we can write as (shadow length) 40.25 : 28.75 :: 17.5 : x⇒ 40.25 × x = 28.75 × 17.5⇒ x = 28.75 × 17.5/40.25= 2875 × 175/40250= 2875 × 7/1610= 2875/230= 575/46= 12.5 cm

Answer: C) 166252 Explanation: Given 92613 x 62742241 = ? + 89   Using unit digit method,   On LHS,   1 x 1 x 1 = 1   sqrt of last digit 1 => it must end with either 9 or 1   so now   1 x 1 = 1   1 x 9 = 9   On RHS,   ? + 89   => ? = ...1 - 89 = last digit is 11- 9 = 2   or   => ? = ...9 - 89 = last digit is 9 - 9 = 0   Here in the options last digit 2 is option C.

Answer: B) Rs. 480 Explanation: Let the sum be Rs. P. Then,   P1+252x1002-P= 510   P[982- 1] = 510.   Sum = Rs. 1920   So, S.I. = (1920 x 25 x 2) / (2 x 100) = Rs. 480

Answer: D) 84.8 Explanation: Let 'M' be the maximum marks in the examination. Therefore, Madhu got 32% of M = 32M/100 = 0.32MAnd Kumar got 42% of M = 42M/100 = 0.42M.In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1) The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2) Now, Equating (1) and (2), we get0.1M = 24 => M = 24/0.1 = 240'M' is the maximum mark and is equal to 240 marks.We know that Madhu got 32% of the maximum marks.Therefore, Madhu got 32 x 240/100 = 76.8 marks.We also know that Madhu got 8 marks less than the cut-off marks. Therefore, the cut-off marks will be 8 marks more than what Madhu got = 76.8 + 8 = 84.8.

Answer: B) Rs. 3 Explanation: Let wholesaler dealers marked price = 100%, Retailer's C.P = 80% And the retailer sells at 5% less than the marked price => S.P = 95%If S.P of 95% of the retailer costs Rs.19 to customer,so its C.P of 80% will cost 80 x 19/95 = 16 Profit made by the retailer = 19-16 =  Rs.3

Answer: C) 606 Explanation: 56% of 870 = 56x870/100 = 487.20 82% of 180 = 82x180/100 = 147.60 32% of 90 = 32x90/100 487.20 + 147.6 - 28.8  =  ? ? = 634.8  - 28.8 ? = 606.