Quantitative Aptitude - Arithmetic Ability Questions

C.P of article = 6500 * (100/80) = Rs. 8125 S.P for a gain of 20% = (8125*120)/100 = Rs. 9750

Let the distance between A and B = D

Given the speed of maruthi = 60 kmph

Maruthi reaches from A to B in 6 hrs 20 min = 6 + 20/60 = 6 1/3 hrs

Now, the distance between A & B = D = speed x time = 60 x 19/3 = 380 kms

As the distance is same for toyato also,

given time for toyato = 4 hrs 45 min = 4 + 45/60 = 4 3/4 = 19/4 hrs

speed of toyato = distance/time = 380/19/4 = 380 x 4/19 = 80 kmph.

We know that,  Speed = Distance/time

Let the speed of the train be = x kmph

           NORMAL SPEED                           SPEED 5 KMPH MORE

          Distance = 360 Km                         Distance =360
 
          speed    =  X Kmph                        Speed = (x+5)
 
            Time =360/X                              Time =360/(x+15)
          

For the same journey if the speed increased 10 kmph more it will take 1 hour less

Time with original speed - time with increased speed = 1

 => (360/x) - (360/x+10) =  1                               

 =>  (360(x+5)-360x)/x(x+5) =1

 =>   360X + 1800 -360X  = X(X+5)

  =>   X^2 + 5x - 1800 = 0

  X = 40 or -45

 X cannot be negetive value
                  

 X = 40 kmph.

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 4 x 5 = 20), so there will be 6 significant digits to the right of the decimal point.

(3x + 2) (2x - 5) = ax2+kx+n ...............(1)

But (3x + 2)(2x - 5) = 6x2-11x-10........(2)

so by comparing (1) & (2),

we get a= 6, k= -11 , n= -10

(a - n + k) = 6 + 10 - 11 = 5.

i=j/m