Quantitative Aptitude - Arithmetic Ability Questions

Average of 20 numbers = 0. 

Sum of 20 numbers = (0 * 20) = 0. 

It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

In this type of problems the formuae is
(5*x+ or - t)*12/11
Here x is replaced by the first interval of given time. Here x is 5.
t is spaces apart

  Case 1 : (5*x + t) * 12/11

 (5*5 + 3) * 12/11

 28 * 12/11 = 336/11= 31511 min

therefore the hands will be 3 min apart at 31 5/11 min past 5.

Case 2 : (5*x - t) * 12/11

(5*5 -3 ) * 12/11

 22 *12/11 = 24 min 

 therefore the hands will be 3 min apart at 24 min past 5

The given number series follows a pattern that

5 x 5 – 1 = 24

24 x 4 – 2 = 94

94 x 3 – 3 = 279

279 x 2 – 4 = 554

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. 

Jan   Feb   Mar   Apr   May   Jun   Jul  

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) 

Total number of odd days = (0 + 2) = 2.  

Required day was 'Tuesday'.

Step (i): Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100 -------------- (i)

Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.60x + 2.40y = 312 -------------- (ii)

Step (iii):
Solving (i) and (ii)
3.60x + 3.60y = 360 --------- Multiply (i) by 3.60
3.60x + 2.40y = 312 --------- (ii)
1.20y = 48
y = 48 / 1.20
= 40

 Number of girls = 40

Alternatively 23 is added and 17 is subtracted from the terms. So, 634 is wrong.

By the rule of alligation:

Profit of first part                         Profit of second part

So, ratio of 1st and 2nd parts = 4 : 6 = 2 : 3. 

Therefore, Quantity of 2nd kind = (3/5 x 1000)kg = 600 kg.