Quantitative Aptitude - Arithmetic Ability Questions

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

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).

132 = 4 x 3 x 11, So if the number is divisible by all three numbers 4,3 and 11,then the number is divisible by 132 also.
264   => 4,3,11(/)
396   => 4,3,11(/)
462   => 11,3
792   => 4,3,11(/)
968   => 11,4
2178 => 11,3
5184 => 3,4
6336 => 4,3,11(/)
Required number of numbers=4.

Here the given number series is 2 4 27 256 3125 46656

It follows a pattern that,

$$\begin{gathered} 11 = 1 \\ 22 = 4 \\ 33 = 27 \\ 44 = 256 \\ 55 = 3125 \\ 66 = 46656 \end{gathered}$$

Hence the wrong number in the given number series is 2 instead of 1.

Let there be x pupils in the class.

Total increase in marks = (x * 1/2) = x/2.

x/2 = (83 - 63) => x/2 = 20 => x = 40.

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