Quantitative Aptitude - Arithmetic Ability Questions

30 + (.0825 x 30) + (15% x 30)

Now

=> 0.0825 x 30 = 2.475 -----(1)

=> 15% of 30 = 450/100 = 4.5 -----(2) 

Hence, from (1) & (2)

30 + (.0825 x 30) + (15% x 30)

= 30 + 2.475 + 4.5

30 + (.0825 x 30) + (15% x 30) = 36.975.

Case I :  MW  MW  MW  MW

Case II:  WM  WM  WM  WM

Let us arrange 4 men in 4! ways, then we arrange 4 women in 4P4 ways at 4 places either left of the men or right of the men. Hence required number of arrangements

   4! × 4P4 + 4! × 4P4 = 2 × 576 = 1152

You can find the integers which when divided by 5 have a remainder 2 by adding 2 to all multiples of 5. So we have n = 7 , 12, 17, 22 etc.

From this series we can see that n does not have to be odd.

Also n + 1 can be a prime because, for example, 12 + 1 = 13

And (n + 2) / 7 has a remainder 2 in some cases but not all.

Remember the question asks us for what MUST be true, and we see that none of the statements are true in all cases. However, adding 3 to any of the values of n will always give a multiple of 5.

Speed in still water = Average of Speed in Upstream and speed in Downstream

= 1/2 (12 + 6) kmph = 9 kmph.

Average = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99) / 9

             =( (11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55) / 9

             = (4 * 110 + 55)/9 =  495 / 9 = 55. 

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days

167 days = (23 weeks + 6 days) =6 odd days.

Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.

Given day is Wednesday.