Quantitative Aptitude - Arithmetic Ability Questions

Let original Cost price is x
Its Selling price = (105/100) * x = 21x/20
New Cost price  = (95/100) * x  = 19x/20
New Selling price = (110/100 )* (19x/20 )= 209x/200
[(21x/20) - (209x/200)] = 1
=> x = 200

Let the length of the train be x metres and its speed be y m/sec.
Then, (x/y)= 15   y =(x/15)

(x+100)/25 = x/15

=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

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 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

Let x men can do the in 12 days and the required number of days be z

More men, Less days     [Indirect Proportion]

Less work, Less days     [Direct Proportion  ]

 men2x:xwork1:12 :: 12 : z

∴2x×1×z=x×12×12

 ⇒z=3

Let the original speed  be x km/h, then

600x-600x+20=1 

⇒600x+20-xxx+20=1  

⇒x2+120x-12000=0

⇒x-120x-100=0

Therefore,  original speed = 100km/h

Total quantity of mixture = 75 litre

Milk  :  Water  =  4  :  1   

                   3       :       1