Quantitative Aptitude - Arithmetic Ability Questions

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.

The Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 upto 40. Mutiples of 4 means which can be divided by 4 leaving remainder '0'.

Common Multiples of 4 & 6 are 12, 24, 36, 48, 60 upto 60.

Let the required two numbers be a, b

From the given data,

2a + 3b = 36 .....(1)

3a + 2b = 39 .....(2)

Now, by solving 1 and 2 eqnts, we get

5a = 45

a = 9 ....(3)

Put (3) in (1), we get

2(9) + 3b = 36

=> 18 + 3b = 36

3b = 36 - 18

=> b = 18/3 = 6

Required, difference between the numbers = a + b = 9 + 6 = 15.

Principle amount = Rs. 29000

Interest = Rs. 10440

Let rate of interest = r%

=> So, time = r years

According to the question,

10440 = 29000 x r x r/100

290 x r x r = 10440

r x r = 1044/29 = 36

r = 6

Hence, the rate of interest = 6% and time = 6 yrs.

Let the marked price of the router be Rs. P

From the given data,

$$\begin{gathered} P x 85100 - P x 80100 = 51 \\ P x 5100 = 51 \\ ⇒ P = 51005 = 1020 \end{gathered}$$

Hence, the original marked price of the router is Rs. 1020.