Quantitative Aptitude - Arithmetic Ability Questions

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) = 6 odd days

(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 dayss = (21 weeks + 1 day) = 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 = 0 odd days.

Given day is Sunday

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,
(2x+4000) / (3x+4000) = 40 / 57
⇒ 57 × (2x + 4000) = 40 × (3x+4000)
⇒ 6x = 68,000
⇒ 3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x. 

S.P. of x articles = Rs. 20. 

Profit = Rs. (20 - x). 

20-xx*100=25 

=> 2000 - 100x = 25x 

=> 125x=2000 

=> x=16

Average = total runs / no.of innings = 32 

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs 

Total runs = new avg x new no. of innings = 36 x 11 = 396 

Runs made in the 11th inning = 396 - 320 = 76

Least Cost Price = Rs. (200 * 8) = Rs. 1600.

Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.

Required profit = Rs. (3400 - 1600) = Rs. 1800.

Let the cost price of 1-liter pure milk be Re.1, then 

 6 liters (milk) → C.P = Rs. 62 liters (water) → C.P = Rs. 0→CP = Rs.6 only

8 litre mixture  =>

SP =>  8 x 2 = Rs. 16

Profit % = 16 - 66x100 = 10006 = 166.66%

 Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

 Required sale = Rs.[(6500 x 6) - 34009]

= Rs. (39000 - 34009) 

= Rs.  4991.