Quantitative Aptitude - Arithmetic Ability Questions

Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 2005, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 2005 + 6) = 2011

Therfore, The calendar for the year 2005 will be same for the year 2011

Loss%= (common gain or loss % / 10)² = (10/10)² % = 1%.

Let the original rate be R%. Then, new rate = (2R)%.

Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e.1/3 year(s).

 725*R*1100+362.50*2R*1100*3=33.50

=> (2175 + 725) R = 33.50 x 100 x 3

=>  (2175 + 725) R = 10050

=>  (2900)R = 10050

 => R=100502900=3.46

Original rate = 3.46%

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7

  PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 

                          = 12×47+12×26=1942