Quantitative Aptitude - Arithmetic Ability Questions

Given 15% of x = 20% of y

=> 15x = 20y

=> x/y = 20/15

=> x : y = 4 : 3

When B runs 25 m, A runs 45/2 m

When B runs 1000 m, A runs (45/2*1/25*1000)m = 900 m.

B beats A by 100 m.

Compound Interest on P at 10% for 2 years when interest is compounded half-yearly

=P1+R21002T-P=P1+1204-P=P21204-P

Simple Interest on P at 10% for 2 years = PRT100=P×10×2100=P5

Given that difference between compound interest and simple interest = 124.05

 P*21204-P-P5=124.05

=>P21204-1-15=124.05

 P=8000

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 1897, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903

Therfore, The calendar of the year 1897 can be used again in the year 1903

                       A   :   C    

Efficiency      5    :   3    

No of days   3x   :  5x     

Given that, 5x-6 =3x  => x = 3  

Number of days taken by A = 9  

Number of days taken by C = 15     

           B  :  C    

Days   2  :  3  

Therefore, Number of days taken by B = 10   

Work done by B and C in initial 2 days = 2110+115= 1/3  

Thus,  Rest work =2/3  

Number of days required by A to finish 2/3 work = (2/3) x 9 = 6 days